tag:blogger.com,1999:blog-49876091144152055932023-03-30T15:01:32.983+01:00M-PhiA blog dedicated to mathematical philosophy.Jeffrey Ketlandhttp://www.blogger.com/profile/01753975411670884721noreply@blogger.comBlogger594125tag:blogger.com,1999:blog-4987609114415205593.post-86813677937424472302023-03-17T17:00:00.002+00:002023-03-17T17:06:35.215+00:00The Robustness of the Diversity Prediction Theorem II: problems with asymmetry<p>There is a PDF of this post <a href="https://drive.google.com/file/d/1-Kn1cyPAgYIuB3sdxRIHHCkLPpbswiBv/view?usp=sharing" target="_blank">here</a>.</p><p>Take a quantity whose value you wish to estimate---the basic reproduction number for a virus; the number of jelly beans in a jar at the school fête; the global temperature rise caused by doubling the concentration of CO2 in the atmosphere; the number of years before humanity goes extinct. Ask a group of people to provide their estimate of that value, and take the mean of their answers. The Diversity Prediction Theorem says that, if you measure distance as squared difference, so that, for example, the distance from $2$ to $5$ is $(2-5)^2$, the distance that the mean of the answers lies from the true value will be equal to the average distance from the answers to the true value less a quantity that measures the diversity of the answers, namely, the average distance from the answers to the mean answer. </p><p>In a previous blogpost, I asked: to what extent is this result a quirk of squared difference? Extending a result due to David Pfau, I showed that it is true of exactly the Bregman divergences. But there's a problem. In the Diversity Prediction Theorem, we measure the diversity of estimates as the average distance <i>from</i> the individual answers <i>to</i> the mean answer. But why this, and not the average distance <i>to</i> the individual answers <i>from </i>the mean answer? Of course, if we use squared difference, then these values are the same, because squared difference is a symmetric measure of distance: the squared difference from one value to another is the same as the squared difference from the second value to the first. And so one set of answers will be more diverse than another according to one definition if it is more diverse according to the other definition. But squared difference is in fact the only symmetric Bregman divergence. So, for all other definitions of Bregman divergence, the two definitions of diversity come apart.</p><p>This has an odd effect on the Diversity Prediction Theorem. One of the standard lessons the theorem is supposed to teach is that the mean of a more diverse group is more accurate than the mean of a less diverse group. In fact, even the original version of the theorem doesn't tell us that. It tells us that the mean of a more diverse group is more accurate than the mean of a less diverse group <i>when the average distance from the truth is the same for both groups</i>. But, if we use a non-symmetric distance measure, i.e., one of the Bregman divergences that isn't squared error, and we use the alternative measure of diversity mentioned in the previous paragraph---that is, the average distance <i>to</i> the individual answers <i>from </i>the mean answer---then we can get a case in which the mean of a less diverse group is more accurate than the mean of a more diverse group, even though the average distance from the answers to the truth is the same for both groups. So it seems that we have three choices: (i) justify using squared difference only; (ii) justify the first of the two putative definitions of diversity in terms of average distance between mean answer and the answers; (iii) give up the apparent lesson of the Diversity Prediction Theorem that diversity leads to more accurate average answers. For my money, I think (iii) is the most plausible.</p><span><a name='more'></a></span><p>Let me finish off by providing an example. First, define two measures of distance:</p><p><i>Squared difference</i>: $q(x, y) = (x-y)^2$</p><p><i>Generalized Kullback-Leibler divergence</i>: $l(x, y) = x\log(x/y) - x + y$</p><p>Then the Diversity Prediction Theorem says that, for any $a_1, \ldots, a_n, t$, if $a^\star = \frac{1}{n}\sum^n_{i=1} a_i$,$$q(a^\star, t) = \frac{1}{n}\sum^n_{i=1} q(a_i, t) - \frac{1}{n}\sum^n_{i=1} q(a_i, a^\star)$$And the generalization I discussed in the previous blogpost entails that$$l(a^\star, t) = \frac{1}{n}\sum^n_{i=1} l(a_i, t) - \frac{1}{n}\sum^n_{i=1} l(a_i, a^\star)$$But drawing from this the conclusion that groups with the same average distance to the truth are more accurate if they're more diverse relies on defining the diversity of the estimates $a_1, \ldots, a_n$ to be $\frac{1}{n}\sum^n_{i=1} l(a_i, a^\star)$, rather than $\frac{1}{n}\sum^n_{i=1} l(a^\star, a_i)$. Suppose that we define it in the second way instead. And take two groups each containing two individuals. Here are their estimates of a quantity (perhaps the R number of a virus):$$\begin{array}{c|c|c|c|c|c} a_1 & a_2 & a^\star & b_1 & b_2 & b^\star \\ \hline 0.5 & 0.1 & 0.3 & 0.3 & 0.9 & 0.6 \end{array}$$Then $a_1, a_2$ is less diverse then $b_1, b_2$ according to the original definition of diversity, but more diverse according to the second definition. That is,$$\frac{l(0.5, 0.3) + l(0.1, 0.3)}{2} < \frac{l(0.3, 0.6) + l(0.9, 0.6)}{2}$$and$$\frac{l(0.3, 0.5) + l(0.3, 0.1)}{2} > \frac{l(0.6, 0.3) + l(0.6, 0.9)}{2}$$What's more, if $t = 0.44994$, then the average distance to the truth is the same for both groups. That is,$$\frac{l(0.5, 0.44994) + l(0.1, 0.44994)}{2} = \frac{l(0.3, 0.44994) + l(0.9, 0.44994)}{2}$$But then it follows that the mean of $a_1, a_2$ is less accurate than the mean of $b_1, b_2$, even though, according to one seemingly legitimate definition of diversity, $a_1, a_2$ is more diverse than $b_1, b_2$.$$l(0.3, 0.44994) > l(0.6, 0.44994)$$</p><p><br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com0tag:blogger.com,1999:blog-4987609114415205593.post-44497137950938722942023-03-14T08:40:00.006+00:002023-03-17T17:00:26.242+00:00The Robustness of the Diversity Prediction Theorem I: generalizing the result<p>There is a <a href="https://drive.google.com/file/d/1C8cIxAjMuGaSbiGZQH_1C-_V7cgw08CQ/view?usp=sharing" target="_blank">PDF</a> version of this post here.</p><p>Pick a question to which the answer is a number. What is the population of Weston-super-Mare? How many ants are there in the world? Then go out into the street and ask the first ten people you meet for their answers. Look how far those answers lie from the truth. Now take the average of the answers---their mean. And look how far it lies from the truth. If you measure the distance between an answer and the true value in the way many statisticians do---that is, you take the difference and square it---then you'll notice that the distance of the average from the truth is less than the average distance from the truth. I can predict this with confidence because it is no coincidence, nor even just a common but contingent outcome---it's a mathematical fact. And it remains a mathematical fact for many alternative ways of measuring the distance from an answer to the truth---essentially, all that's required is that the measure is strictly convex in its first argument (<a href="http://students.aiu.edu/submissions/profiles/resources/onlineBook/d3N8T8_Social_Judgment_and_Decision_Making.pdf#page=246" target="_blank">Larrick, et al. 2012</a>). That's a neat result! In expectation, you're better off going with the average answer than picking an answer at random and going with that.</p><p>But, if we keep using squared difference, we get an even more interesting result. The distance the average answer lies from the truth is equal to the average distance the answers lie from the truth less a quantity that measures the diversity of the answers. Scott Page calls this the Diversity Prediction Theorem and uses it as part of his general argument that diversity is valuable in group reasoning and decision-making. The quantity that measures the diversity of the answer is the average distance the answers lie from the mean answer: the answers in a homogeneous set will, on average, lie close to their mean, while those in a heterogeneous set will, on average, lie far from their mean. This has an intriguing corollary: given two collections of answers with the same average distance from the truth, the mean of the more diverse one will be more accurate than the mean of the less diverse one. That's another neat result! It isn't quite as neat as people sometimes make out. Sometimes, they say that it shows that more diverse sets of answers are more accurate; but that's only true when you're comparing sets with the same average distance from the truth; increasing diversity can often increase the average distance from the truth. But it's still a neat result!</p><span></span><span><a name='more'></a></span><p><br /></p><p>A natural question: is it a quirk of mean squared error? In fact, it's not. There are many many ways of measuring the distance between answers, mean answers, and the truth for which the Diversity Prediction Theorem holds. As I'll show here, they are exactly the so-called Bregman divergences, which I'll introduce below.</p><p>That every Bregman divergence gives the Diversity Prediction Theorem was shown by <a href="http://davidpfau.com/assets/generalized_bvd_proof.pdf" target="_blank">David Pfau (2013)</a> in an unpublished note. I haven't been able to find a proof that only those functions give it, but it follows reasonably straightforwardly from a characterization of Bregman divergences due to <a href="https://ecommons.cornell.edu/bitstream/handle/1813/9264/TR001387.pdf;sequence=1" target="_blank">Banerjee, et al. (2005)</a>.</p><p>Having talked informally through the results I want to present, let me now present them precisely.</p><p>First, the measure of distance between two numbers that is used in the standard version of the Diversity Prediction Theorem:</p><p><b>Definition 1</b> (Squared difference)$$q(x, y) := (x-y)^2$$</p><p>So our first result is this:</p><p><b>Proposition 1</b> Suppose $a_1, \ldots, a_n, t$ are real numbers with $a_i \neq a_j$ for some $i, j$. And let $a^\star = \frac{1}{n} \sum^n_{i=1} a_i$. Then$$q(a^\star, t) < \frac{1}{n} \sum^n_{i=1} q(a_i, t)$$</p><p>As I mentioned, this generalises to any measure of distance that is strictly convex in its first argument.</p><p><b>Definition 2 </b>(Strictly convex function) Suppose $X$ is a convex subset of the real numbers and $d : X \times X \rightarrow \mathbb{R}$. Then $d$ is strictly convex in its first argument if, for any $x_1, x_2, y$ in $X$ and any $0 < \lambda < 1$,$$d(\lambda x_1 + (1-\lambda) x_2, y) < \lambda d(x_1, y) + (1-\lambda) d(x_2, y)$$</p><p><b>Proposition 2 </b>Suppose $X$ is a convex subset of the real numbers and $d : X \times X \rightarrow [0, \infty]$ is strictly convex in its first argument. Suppose $a_1, \ldots, a_n, t$ are real numbers in $X$ with $a_i \neq a_j$ for some $i, j$. And let $a^\star = \frac{1}{n} \sum^n_{i=1} a_i$. Then$$d(a^\star, t) < \frac{1}{n} \sum^n_{i=1} d(a_i, t)$$</p><p><i>Proof.</i> This follows immediately from Jensen's inequality, which entails that, for any strictly convex function $f$, and any $a_1, \ldots, a_n$ with $a_i \neq a_j$ for some $i, j$,$$f\left (\frac{1}{n}\sum^n_{i=1} a_i \right ) < \frac{1}{n} \sum^n_{i=1} f(a_i).$$$\Box$</p><p>Next, we meet the Diversity Prediction Theorem:</p><p><b>Proposition 3</b> Suppose $a_1, \ldots, a_n, t$ are real numbers with $a_i \neq a_j$ for some $i, j$. And let $a^\star = \frac{1}{n} \sum^n_{i=1} a_i$. Then$$q(a^\star, t) = \frac{1}{n} \sum^n_{i=1} q(a_i, t) - \frac{1}{n} \sum^n_{i=1} q(a_i, a^\star)$$</p><p>I won't offer a proof, since it follows from the more general version below. To state that, we need the notion of a <a href="https://en.wikipedia.org/wiki/Bregman_divergence" target="_blank">Bregman divergence</a>:</p><p><b>Definition 3</b> (Bregman divergence) Suppose $X$ is a convex subset of the real numbers and $\varphi : X \rightarrow \mathbb{R}$ is a continuously differentiable, strictly convex function. Then, for $x, y$ in $X$, $$d_\varphi(x, y) = \varphi(x) - \varphi(y) - \varphi'(y)(x-y)$$We say that $d_\varphi$ is the Bregman divergence generated by $\varphi$.</p><p>To calculate the Bregman divergence from $x$ to $y$ generated by $\varphi$, you take the tangent to $\varphi$ at $y$ and calculate the difference between $\varphi$ at $x$ and this tangent at $x$. This is illustrated in Figure 1. Note that, for any $x, y$ in $X$, $d_\varphi(x, y) \geq 0$, with equality iff $x = y$.</p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiSLZ7zLqa-WmpUe8CaQyUGBT0pslpONBfoz5EivsYNPn7qT7-VfaIyCj8-ew5847FzzB1Ff3VLmRHfonz2uKIpWRJxidTGHXg7um8SJBzpBXXnwdeqyowH92cP0P-X-mV88WmEjWywrt_L2iXgAE2T2hPRaeDcuJf8Qe9kuvJV063o2goNGE7dXOtvcw/s720/bregman.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="428" data-original-width="720" height="190" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiSLZ7zLqa-WmpUe8CaQyUGBT0pslpONBfoz5EivsYNPn7qT7-VfaIyCj8-ew5847FzzB1Ff3VLmRHfonz2uKIpWRJxidTGHXg7um8SJBzpBXXnwdeqyowH92cP0P-X-mV88WmEjWywrt_L2iXgAE2T2hPRaeDcuJf8Qe9kuvJV063o2goNGE7dXOtvcw/s320/bregman.png" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 1: $\varphi(x) = x\log(x)$ is plotted in blue; the tangent to $\varphi$ at $y$ is plotted in yellow; $d_\varphi(x, y)$ is the length of the dashed line.</td></tr></tbody></table><p>Two examples of Bregman divergences, with the strictly convex functions that generate them:</p><p><i>Squared Euclidean distance</i> Suppose $\varphi(x) = x^2$. Then$$d_\varphi(x, y) = (x-y)^2$$</p><p><i>Generalized Kullback-Leibler divergence</i> Suppose $\varphi(x) = x\log(x)$. Then$$d_\varphi(x, y) = x\log \left ( \frac{x}{y} \right ) - x + y$$</p><p><b>Proposition 4</b> (<a href="http://davidpfau.com/assets/generalized_bvd_proof.pdf" target="_blank">Pfau 2013</a>) Suppose $X$ is a convex subset of the real numbers and $\varphi : X \rightarrow \mathbb{R}$ is a continuously differentiable, strictly convex function. Suppose $a_1, \ldots, a_n, t$ are real numbers in $X$ with $a_i \neq a_j$ for some $i, j$. And let $a^\star = \frac{1}{n} \sum^n_{i=1} a_i$. Then$$d_\varphi(a^\star, t) = \frac{1}{n} \sum^n_{i=1} d_\varphi(a_i, t) - \frac{1}{n} \sum^n_{i=1} d_\varphi(a_i, a^\star)$$</p><p><i>Proof.</i> The result follows easily from the following three equations:</p><p>\begin{eqnarray*} d_\varphi(a^\star, t) & = & \varphi(a^\star) - \varphi(t) - \varphi'(t)(a^\star - t) \\ & & \\ \frac{1}{n}\sum^n_{i=1} d_\varphi(a_i, t) & = & \frac{1}{n} \sum^n_{i=1} \varphi(a_i) - \varphi(t) - \varphi'(t)\left (\frac{1}{n} \sum^n_{i=1} a_i - t \right ) \\ & = & \frac{1}{n} \sum^n_{i=1} \varphi(a_i) - \varphi(t) - \varphi'(t)\left (a^\star - t \right ) \\ & & \\ \frac{1}{n} \sum^n_{i=1} d_\varphi(a_i, a^\star) & = & \frac{1}{n}\sum^n_{i=1}\varphi(a_i) - \varphi(a^\star) - \varphi'(a^\star)\left (\frac{1}{n}\sum^n_{i=1} a_i - \varphi(a^\star) \right ) \\ & = & \frac{1}{n}\sum^n_{i=1}\varphi(a_i) - \varphi(a^\star)\end{eqnarray*}</p><p>$\Box$</p><p>Next, we prove that the Bregman divergences are the only functions that give the Diversity Prediction Theorem. The proof relies on the following characterization of Bregman divergences.</p><p><b>Lemma 5 </b>(<a href="https://ecommons.cornell.edu/bitstream/handle/1813/9264/TR001387.pdf;sequence=1" target="_blank">Banerjee et al. 2005</a>) Suppose $X$ is a convex subset of the real numbers and $d : X \times X \rightarrow [0, \infty]$. And suppose that, for any $a_1, \ldots, a_n$ in $X$, with $a^\star = \frac{1}{n}\sum^n_{i=1} a_i$, the function $t \mapsto \frac{1}{n} \sum^n_{i=1} d(a_i, t)$ is minimized at $t = a^\star$. Then there is a continuously differentiable and strictly convex function on $X$ such that $d = d_\varphi$.</p><p><b>Proposition 6</b> Suppose $X$ is a convex subset of the real numbers and $d : X \times X \rightarrow [0, \infty]$. And suppose that, for any $a_1, \ldots, a_n, t$ in $X$ with $a_i \neq a_j$ for some $i, j$, with $a^\star = \frac{1}{n} \sum^n_{i=1} a_i$, $$d(a^\star, t) = \frac{1}{n} \sum^n_{i=1} d(a_i, t) - \frac{1}{n} \sum^n_{i=1} d(a_i, a^\star)$$Then there is a continuously differentiable and strictly convex function on $X$ such that $d = d_\varphi$.</p><p><i>Proof.</i> Suppose that, for any $a_1, \ldots, a_n, t$ are real numbers in $X$ with $a_i \neq a_j$ for some $i, j$, with $a^\star = \frac{1}{n} \sum^n_{i=1} a_i$, $$d(a^\star, t) = \frac{1}{n} \sum^n_{i=1} d(a_i, t) - \frac{1}{n} \sum^n_{i=1} d(a_i, a^\star)$$Then$$\frac{1}{n} \sum^n_{i=1} d(a_i, t) = d(a^\star, t) + \frac{1}{n} \sum^n_{i=1} d(a_i, a^\star)$$And so the function $t \mapsto \frac{1}{n} \sum^n_{i=1} d(a_i, t)$ is minimal when $t \mapsto d(a^\star, t)$ is minimal; and since, for all $x, y$ in $X$, $d(x, y) \geq 0$ with equality iff $x = y$, $t \mapsto d(a^\star, t)$ is minimal at $t = a^\star$. So, by Lemma 5, there is a continuously differentiable and strictly convex function on $X$ such that $d = d_\varphi$. $\Box$</p><h4 style="text-align: left;">References</h4><a href="https://ecommons.cornell.edu/bitstream/handle/1813/9264/TR001387.pdf;sequence=1" target="_blank">Banerjee, A., Guo, X., & Wang, H. (2005). On the Optimality of Conditional Expectation as a Bregman Predictor. <i>IEEE Transactions of Information Theory,</i> 51, 2664–69. </a><br /><br /><a href="http://students.aiu.edu/submissions/profiles/resources/onlineBook/d3N8T8_Social_Judgment_and_Decision_Making.pdf#page=246" target="_blank">Larrick, R. P., Mannes, A. E., & Soll, J. B. (2012). The Social Psychology of the Wisdom of the Crowds. In J. I. Krueger (Ed.) <i>Social Judgment and Decision Making,</i> chap. 13, (pp. 227–242). New York: Psychology Press. </a><br /><br /><a href="http://davidpfau.com/assets/generalized_bvd_proof.pdf" target="_blank">Pfau, D. (2013). A Generalized Bias-Variance Decomposition for Bregman Divergences. Unpublished manuscript.</a>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com0tag:blogger.com,1999:blog-4987609114415205593.post-42600312539498902062022-08-04T16:34:00.005+01:002022-08-04T16:36:54.236+01:00Should longtermists recommend hastening extinction rather than delaying it?<p>I'm cross-posting <a href="https://forum.effectivealtruism.org/posts/xAoZotkzcY5mvmXFY/should-longtermists-recommend-hastening-extinction-rather-1" rel="nofollow" target="_blank">this</a> from the Effective Altruism Forum. It's the latest version of my critique of longtermism that uses Lara Buchak's risk-weighted expected utility theory.</p><p>Here's the abstract: Longtermism is the view that the most urgent global priorities, and those to which we should devote the largest portion of our current resources, are those that focus on ensuring a long future for humanity, and perhaps sentient or intelligent life more generally, and improving the quality of those lives in that long future. The central argument for this conclusion is that, given a fixed amount of a resource that we are able to devote to global priorities, the longtermist’s favoured interventions have greater expected goodness than each of the other available interventions, including those that focus on the health and well-being of the current population. In this paper, I argue that, even granting the longtermist's axiology and their consequentialist ethics, we are not morally required to choose whatever option maximises expected utility, and may not be permitted to do so. Instead, if their axiology and consequentialism is correct, we should choose using a decision theory that is sensitive to risk, and allows us to give greater weight to worse-case outcomes than expected utility theory. And such decision theories do not recommend longtermist interventions. Indeed, sometimes, they recommend hastening human extinction. Many, though not all, will take this as a reductio of the longtermist's axiology or consequentialist ethics. I remain agnostic on the conclusion we should draw. </p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh8TxAywF1pMCyza11dmpxYgAWLCWbFeLMT4V4EbrVqFsbK0Y0H_IIEUsPMMP59sf6c14oFOap5eM5nk2YqDJOWSEvhOlSNYg41q1TE8VKuOhpi9RlnhaDfvedm1nGBa1TgsHNNhjnqD_ChChqWLKlD1Moaw1QxoBewpDlEFEWILvRl3kFB89Px3kjzVQ/s4032/IMG_6635.heic" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="3024" data-original-width="4032" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh8TxAywF1pMCyza11dmpxYgAWLCWbFeLMT4V4EbrVqFsbK0Y0H_IIEUsPMMP59sf6c14oFOap5eM5nk2YqDJOWSEvhOlSNYg41q1TE8VKuOhpi9RlnhaDfvedm1nGBa1TgsHNNhjnqD_ChChqWLKlD1Moaw1QxoBewpDlEFEWILvRl3kFB89Px3kjzVQ/s320/IMG_6635.heic" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">A kitten playing the long game<br /></td></tr></tbody></table>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com1tag:blogger.com,1999:blog-4987609114415205593.post-61818260260074887432022-07-27T11:03:00.001+01:002022-07-27T11:03:14.107+01:00Self-recommending decision theories for imprecise probabilities<p>A PDF version of this post is available <a href="https://drive.google.com/file/d/1hc16Xpw3KxjTaO3p6svXc80qzD54p9mp/view?usp=sharing" rel="nofollow" target="_blank">here</a>.</p><p>The question of this blogpost is this: Take the various decision theories that have been proposed for individuals with imprecise probabilities---do they recommend themselves? It is the final post in a trilogy on the topic of self-recommending decision theories (the others are <a href="https://m-phi.blogspot.com/2022/07/self-recommending-decision-theories.html" rel="nofollow" target="_blank">here</a> and <a href="https://m-phi.blogspot.com/2022/07/more-on-self-recommending-decision.html" rel="nofollow" target="_blank">here</a>). <br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhbFCh4HKyvmgpIDXYjjnMMFPATVuAUUW3tAddB3fVH68bdXnuTq-064fok9z82Yx4hX5FcSauW3E1OQwsc_35ZZEIiQegsYs_m1Vy7SIm117nU6hcCDBltSCv6tcp_ZbwT5LGI7BjVZCVE-ZyBIwnC3LLlalDzUDaJWZyEDdI_3vVnloBq2xs5Dg6CbA/s3024/IMG_5854.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="3024" data-original-width="3024" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhbFCh4HKyvmgpIDXYjjnMMFPATVuAUUW3tAddB3fVH68bdXnuTq-064fok9z82Yx4hX5FcSauW3E1OQwsc_35ZZEIiQegsYs_m1Vy7SIm117nU6hcCDBltSCv6tcp_ZbwT5LGI7BjVZCVE-ZyBIwnC3LLlalDzUDaJWZyEDdI_3vVnloBq2xs5Dg6CbA/s320/IMG_5854.jpg" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">One precise kitten and one imprecise kitten<br /></td></tr></tbody></table><span></span></p><a name='more'></a>Let's begin by unpacking the question. <br /><p></p><p>First, imprecise probabilities (sometimes known as mushy credences; for an overview, see Seamus Bradley's SEP entry <a href="https://plato.stanford.edu/entries/imprecise-probabilities/" rel="nofollow" target="_blank">here</a>). For various reasons, some formal epistemologists think we should represent an individual's beliefs not by a precise probability function, which assigns to each proposition about which they have an option a single real number between 0 and 1, but rather a set of such functions. Some think that, whatever rationality requires of them, most individuals simply don't make sufficiently strong and detailed probabilistic judgments to pick out a single probability function. Others think that, at least when the individual's evidence is very complex or very vague or very sparse, rationality in fact requires them not to make judgments that pick out just one function. Whatever the reason, many think we should represent an individual's beliefs by a set $P$ of probability functions, which we call their <i>representor</i>, following <a href="https://philpapers.org/rec/VANFIA-3" rel="nofollow" target="_blank">van Fraassen</a>. One way to understand a representor is like this: $P$ contains all the probability functions that respect the probabilistic judgments that the individual makes. For instance, if they judge that proposition $A$ is more likely than $B$, then every function in $P$ should assign higher probability to $A$ than to $B$; if they judge $A$ is more likely than not, then every function in $P$ should assign higher probability to $A$ than to its negation. If these are the only two probabilistic judgments they make, then their representor will be $P = \{p : p(A) > p(B)\ \&\ p(A) > p(\overline{A})\}$. And this clearly contains more than one probability function!<br /></p><p>Second, decision theories for an individual with imprecise probabilities. Suppose you have opinions only about two possible worlds $w_1$ and $w_2$, which are exclusive and exhaustive. And suppose your representor is$$P = \{(x, 1-x) : 0.3 \leq x \leq 0.4\}$$where $(x, 1-x)$ is the probability function that assigns probability $x$ to $w_1$ and $1-x$ to $w_2$. That is, you think $w_1$ is between 30% and 40% likely, and $w_2$ is between 60% and 70% likely, but you make no stronger judgment than that. And now suppose you face a decision problem that consists of a choice between two options, $a$ and $b$, with the following payoff table:$$\begin{array}{r|cc}& w_1 & w_2 \\ \hline a & 13 & 0 \\ b & 0 & 7 \end{array}$$Then, if you take the probability function $(0.3, 0.7)$, which takes $w_1$ to be 30% likely and $w_2$ to be 70% likely, then it expects $b$ to be better than $a$, but if you take the probability function $(0.4, 0.6)$, which takes $w_1$ to be 40% likely and $w_2$ to be 60% likely, then it expects $a$ to be better than $b$. How, then, should you choose? It turns out that there are many possibilities! I'll list what I take to be the main contenders below before asking whether any of them recommend themselves.</p><p>Third, self-recommending decision theories. Suppose you are unsure what decision problem you're about to face. Indeed you think each possible decision problem is equally likely. Then you can use any decision theory to pick between the available decision theories that you might use when faced with whichever decision problem arises. A decision theory is self-recommending if it says that it's permissible to pick itself.<br /></p><p>Let's meet the decision theories for imprecise probabilities. This list may well not be comprehensive, but I've tried to identify the main ones (thanks to Jason Konek for an impromptu tutorial on $E$-admissibility and Maximality!). I state them in terms of impermissibility, but nothing hangs on that.<br /></p><p>Suppose $P$ is a set of probability functions and $O$ is a set of options. Following <a href="http://brian.weatherson.org/vdt.pdf" rel="nofollow" target="_blank">Brian Weatherson</a>, given a particular option $o$, we let $$l_o = \min_{p \in P} \mathrm{Exp}_p(o) \hspace{10mm} \text{ and } \hspace{10mm} u_o = \max_{p \in P} \mathrm{Exp}_p(o)$$</p><p><i>Global Dominance</i> $o$ in $O$ is impermissible iff there is $o'$ in $O$ such that $u_o < l_{o'}$.</p><p><i>$\Gamma$-Maximin</i> $o$ in $O$ is impermissible iff there is $o'$ in $O$ such that $l_o < l_{o'}$.</p><p><i>$\Gamma$-Maxi</i> $o$ in $O$ is impermissible iff there is $o'$ in $O$ such that one of the following hold:</p><ul style="text-align: left;"><li>$l_o < l_{o'}$ and $u_o < u_{o'}$</li><li>$l_o < l_{o'}$ and $u_o = u_{o'}$</li><li>$l_o = l_{o'}$ and $u_o < u_{o'}$</li></ul><p><i>$\Gamma$-Hurwicz$_\lambda$</i> $o$ in $O$ is permissible iff there is $o'$ in $O$ such that $$\lambda l_o + (1-\lambda) u_o < \lambda l_{o'} + (1-\lambda) u_{o'}$$</p><p><i>$E$-Admissibility</i> $o$ in $O$ is impermissible iff for all $p$ in $P$ there is $o'$ in $O$ such that$$\mathrm{Exp}_p(o) < \mathrm{Exp}_p(o')$$</p><p><i>Maximality</i> $o$ in $O$ is impermissible iff there is $o'$ in $O$ such that, for all $p$ in $P$,$$\mathrm{Exp}_p(o) < \mathrm{Exp}_p(o')$$<br /></p><p>Notice that the difference between $E$-Admissibility and Maximality is in the order of the quantifiers.</p><p>Next, let me specify the situation in which we're testing for self-recommendation a little more precisely. We imagine that there's a maximum utility that you might receive in the decision problem, let's say $n$; and the utilities you might receive come from $\{0, 1, 2, \ldots, n\}$. And we imagine that the decision problem will consist of either three available options, $a, b, c$, each defined on the two worlds $w_1$ and $w_2$. So each decision problem has a payoff table like this where $a_1, a_2, b_1, b_2, c_1, c_2$ come from $\{0, 1, 2, \ldots, n\}$:$$\begin{array}{r|cc}& w_1 & w_2 \\ \hline a & a_1 & a_2 \\ b & b_1 & b_2 \\ c & c_1 & c_2 \end{array}$$And we assume that each such decision problem is equally probable; and we assume that which decision problem you face is independent of which world you inhabit. So the probability of being at world $w_1$ and facing the decision problem $(a_1, a_2, b_1, b_2, c_1, c_2)$ is the probability of being at world $w_1$ multiplied by the probability of facing $(a_1, a_2, b_1, b_2, c_1, c_2)$, which is $\frac{1}{n^6}$.</p><p>For all of the decision theories we've mentioned, they will sometimes permit more than one option: for instance, Global Dominance permits an option $o$ if, for any other option $o'$, $u_o > l_{o'}$. In these cases, we assume that the individual picks at random between the permissible options.<br /></p><p>Now, let $P = \{(x, 1-x) : 0.3 \leq x \leq 0.4\}$. Then all of the decision theories listed above are not self-recommending. Indeed, every one of them prefers using expected utility with probability function $m = (0.35, 0.65)$ to using themselves. That is:</p><ul style="text-align: left;"><li>the maximum expected utility of using Global Dominance with $P$ is less than the minimum expected utility of using Expected Utility with $m$;</li><li>the minimum expected utility of using $\Gamma$-Maximin with $P$ is less than the minimum expected utility of using Expected Utility with $m$;</li><li>the minimum expected utility of using $\Gamma$-Maxi is less than the minimum expected utility of using Expected Utility with $m$ and the maximum expected utility of using $\Gamma$-Maxi is less than the maximum expected utility of using Expected Utility with $m$; </li><li>the weighted average of the minimum and maximum expected utilities of using $\Gamma$-Hurwicz$_\lambda$ with $P$ is less than the weighted average of the minimum and maximum expected utilities of using Expected Utility with $m$;</li><li>for all $p$ in $P$, the expected utility of using $E$-admissibility with $P$ is less than the expected utility of using Expected Utility with $m$;</li><li>for all $p$ in $P$, the expected utility of using Maximality with $P$ is less than the expected utility of using Expected Utility with $m$.</li></ul><p></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com0tag:blogger.com,1999:blog-4987609114415205593.post-57791726400936898032022-07-19T12:41:00.002+01:002022-07-19T12:53:17.729+01:00More on self-recommending decision theories<p>A PDF of this blogpost can be found <a href="https://drive.google.com/file/d/1LHjKo_MVKjV2FJLEBWxXO-B56_Yc5miZ/view?usp=sharing" rel="nofollow" target="_blank">here</a>.</p><p>Last week, I wrote about how we might judge a decision theory by its own lights. I suggested that we might ask the decision theory whether it would choose to adopt itself as a decision procedure if it were uncertain about which decisions it would face. And I noted that many instances of Lara Buchak's <a href="https://philpapers.org/rec/BUCRAR-5" rel="nofollow" target="_blank"><i>risk-weighted expected utility theory</i> (<i>REU</i>)</a> do not recommend themselves when asked this question. In this post, I want to give a little more detail about that case, and also note a second decision theory that doesn't recommend itself, namely, <a href="https://www.cmu.edu/dietrich/philosophy/docs/seidenfeld/A%20contrast%20between%20two%20decision%20rules.pdf" rel="nofollow" target="_blank"><i>$\Gamma$-Maximin </i>(<i>MM</i>)</a>, a decision theory designed to be used when uncertainty is modeled by imprecise probabilities.</p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgIw5yY-UW6qzL6vkml-PPNNFnUzy1qHA4f2NpZm3jl9feIkW8N3K9b_6GKS6BdEyiVEdkljAQLVlClEWXzbLrPr7sMsGYlBXGqmt61yExIhUH2eL_1H17K_Ko0qZrb4Pzl1TzJ-I0j8fkQ-fL7Su6fYx6gaojygsocoo7iLNvYJy-hh9jWKy37jv9OUg/s3696/IMG_1659.jpg" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="3696" data-original-width="3696" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgIw5yY-UW6qzL6vkml-PPNNFnUzy1qHA4f2NpZm3jl9feIkW8N3K9b_6GKS6BdEyiVEdkljAQLVlClEWXzbLrPr7sMsGYlBXGqmt61yExIhUH2eL_1H17K_Ko0qZrb4Pzl1TzJ-I0j8fkQ-fL7Su6fYx6gaojygsocoo7iLNvYJy-hh9jWKy37jv9OUg/s320/IMG_1659.jpg" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">A cat judging you...harshly<br /></td></tr></tbody></table> <span></span><p></p><a name='more'></a><p></p><p>The framework is this. For the sake of the calculations I'll present here, we assume that you'll face a decision problem with the following features:</p><ul style="text-align: left;"><li>there will be two available options, $a$ and $b$;</li><li>each option is defined for two exhaustive and exclusive possibilities, which we'll call worlds, $w_1$ and $w_2$; so, each decision problem is determined by a quadruple $(a_1, a_2, b_1, b_2)$, where $a_i$ is the utility of option $a$ at $w_i$, and $b_i$ is the utility of $b$ at $w_i$; <br /></li><li>all of the utilities will be drawn from the set $\{0, 1, 2, \ldots, 20\}$; so, there are $21^4 = 194,481$ possible decision problems.</li></ul><p>This is all you know about the decision problem you'll face. You place a uniform distribution over the possible decision problems you'll face. You take each to have probability $\frac{1}{194,481}$.</p><p>You also assign probabilities to $w_1$ and $w_2$. </p><ul style="text-align: left;"><li>In the case of REU, you have a credence function $(p, 1-p)$ over $w_1$ and $w_2$, so that $p$ is your credence in $w_1$ and $1-p$ is your credence in $w_2$. </li><li>In the case of MM, you represent your uncertainty by a set of such credence functions $\{(x, 1-x) : p \leq x \leq q\}$.</li></ul>In both cases, you take the worlds to be independent from the decision problem you'll face. <br /><div><p>In the case of REU, you also have a risk function $r$. In this post, I'll only consider risk functions of the form $r(x) = r^k$. I'll write $r_k$ for that function.<br /></p><p>With all of that in place, we can ask our question about REU. Fix your credence function $(p, 1-p)$. Now, given a particular risk function $r$, does REU-with-$r$ judge itself to be the best decision theory available? Or is there an alternative decision theory---possibly REU-with-a-different-risk-function, but not necessarily---such that the risk-weighted expected utility from the point of view of $r$ of REU-with-$r$ is less that the risk-weighted expected utility from the point of view of $r$ of this alternative? And the answer is that, for many natural choices of $r$, there is.</p><p>Let's see this in action. Let $p = \frac{1}{2} = 1-p$. Let's consider the risk functions $r_k(x) = x^k$ for $0.5 \leq k \leq 2$. Then the following table gives some results. A particular entry, (row $k$, column $k'$), gives the risk-weighted expected utility from the point of view of $r_{k'}$ of using risk-weighted expected utility from the point of view of $r_k$ to make your decisions. In each column $k'$, the entry in green is what the risk function $r_{k'}$ thinks of itself; the entries in blue indicate the risk functions $r_k$ that $r_{k'}$ judges to be better than itself; and the entry in red is the risk function $r_k$ that $r_{k'}$ judges to be best.</p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh8N-B-uT0f96_fVee_g_GaanocMxZdV6Ruks1qyhrjK7zieHxQG51TUOrZxLopzrJqilRt7vPifjrDkJfjzycdMbhRv7Abp72mgKUc4K_SJRUkdXO5-P5vbV9iVb5I9JRQxAZa8fMFX0M9igBVpfhSVUU6X6J2EiGT2rcYekUs5sn-2NfqJzXsRe78iw/s1337/Screenshot%202022-07-19%20at%2010.32.14.png" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="447" data-original-width="1337" height="214" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh8N-B-uT0f96_fVee_g_GaanocMxZdV6Ruks1qyhrjK7zieHxQG51TUOrZxLopzrJqilRt7vPifjrDkJfjzycdMbhRv7Abp72mgKUc4K_SJRUkdXO5-P5vbV9iVb5I9JRQxAZa8fMFX0M9igBVpfhSVUU6X6J2EiGT2rcYekUs5sn-2NfqJzXsRe78iw/w640-h214/Screenshot%202022-07-19%20at%2010.32.14.png" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">How risk-weighted expected utility theories judge each other<br /></td></tr></tbody></table><br />There are a few trends to pick out:<p></p><ul style="text-align: left;"><li>Risk-inclined risk functions ($r_{k'}(x) = x^{k'}$ for $0.5 \leq k' < 1$) judge less risk-inclined ones to the better than themselves;</li><li>The more risk-inclined, the further away the risk functions can be and still count as better, so $r_{0.5}$ judges $r_{0.9}$ to be better than itself, but $r_{0.7}$ doesn't judge $r_1$ to be better; <br /></li><li>And similarly, mutatis mutandis, for risk-averse risk functions ($r_{k'}(x) = x^{k'}$ for $1 < k' \leq 2$). Each judges less risk-averse risk functions to be better than themselves;</li><li>And the more risk-averse, the further away a risk function can be and still be judged better.</li><li>It might look like $r_{0.9}$ and $r_{1.1}$ are self-recommending, but that's just because we haven't consider more fine-grained possibilities between them and $r_1$. When we do, we find they follow the pattern above.</li><li>The risk-neutral risk function $r_1$ is genuinely self-recommending. REU with this risk function is just expected utility theory.</li></ul><p>So much for REU. Let's turn now to MM. First, let me describe this decision rule. Suppose you face a decision problem between $a = (a_1, a_2)$ and $b = (b_1, b_2)$. Then, first, you decide how much you value each option. You take this to be the minimum expected utility it gets from credence functions in the set of credence functions that represent your uncertainty. That is, you take each $(x, 1-x)$ from your set of credence functions, so that $p \leq x \leq q$, you calculate the expected utility of $a$ relative to that credence function, and you value $a$ by the minimum expectation you come across. Then you pick whichever of the two options you value most. That is, you pick the one whose minimum expected utility is greatest.<br /></p><p>Let's turn to asking how the theory judges itself. Here, we don't have different versions of the theory specified by different risk-functions. But let me consider different sets of credences that might represent our uncertainty. I'll ask how the theory judges itself, and also how it judges the version of expected utility theory (EU) where you use the precise credence function that sits at the midpoint of the credence functions in the set that represents your uncertainty. So, for instance, if $p = 0.3$ and $q = 0.4$ and the representor is $\{(x, 1-x) : 0.3 \leq x \leq 0.4\}$, I'll be comparing how MM thinks of itself and how it thinks of the version of expected utility theory that uses the precise credence function $(0.35, 0.65)$. In the first column here, we have the values of $p$ and $q$; in the second, we have the minimum expected utility for MM relative to each of these pairs of values; in the final column, we have the minimum expected utility for EU relative to those pairs.<br /></p><p>$$\begin{array}{r|cc} & \text{MM} & \text{EU} \\ \hline (0.3, 0.4) & 12.486 & 12.490 \\ (0.3, 0.6) & 12.347 & 12.377 \\ (0.3, 0.9) & 12.690 & 12.817 \\ (0.1, 0.2) & 12.870 & 12.873\end{array}$$</p><p>Again, some notable features:</p><ul style="text-align: left;"><li>in each case, MM judges EU to be better than itself (I suspect this is connected to the fact that ther is no strictly proper scores for imprecise credences, but I'm not sure quite how yet! For treatments of that, see <a href="https://www.sciencedirect.com/science/article/pii/S0888613X12000941" rel="nofollow" target="_blank">Seidenfeld, Schervish, & Kadane</a>, <a href="https://philpapers.org/rec/SCHTAA-45" rel="nofollow" target="_blank">Schoenfield</a>, <a href="https://philarchive.org/rec/MAYSIC-2" rel="nofollow" target="_blank">Mayo-Wilson & Wheeler</a>, and <a href="https://proceedings.mlr.press/v103/konek19a.html" rel="nofollow" target="_blank">Konek</a>.)<br /></li><li>greater uncertainty (which is represented by a broader range of credence functions) leads to a bigger difference between MM and EU;</li><li>having a midpoint that lies further from the centre also seems to lead to a bigger difference.</li></ul><p>At some point, I'll try to write up some thoughts about the consequences of these facts. Could a decision theory that does not recommend itself be rationally adopted? But frankly it's far too hot to think about that today.<br /></p></div>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com0tag:blogger.com,1999:blog-4987609114415205593.post-57475628910843270522022-07-11T10:54:00.002+01:002022-07-27T19:07:08.340+01:00Self-recommending decision theories<p>A PDF of this blogpost is available <a href="https://drive.google.com/file/d/1L6zOO9MO4DAGN7XkUprY4GazIsYmy227/view?usp=sharing" rel="nofollow" target="_blank">here</a>.<br /><br />Once again, I find myself stumbling upon a philosophical thought that seems so natural that I feel reasonably confident it must have been explored before, but I can't find where. So, in this blogpost, I'll set it out in the hope that a kind reader will know where to find a proper version already written up fully.* <br /><br />I'd like to develop a type of objection that might be raised against a theory of rational decision-making. Here, I'll raise it against Lara Buchak's <a href="https://philpapers.org/rec/BUCRAR-5" rel="nofollow" target="_blank"><i>risk-weighted expected utility theory</i></a>, in particular, but there will be many other theories to which it applies. <br /><br />In brief, the objection applies to decision theories that are not self-recommending. That is, it applies to a decision theory if there is a particular instance of that theory that recommends that you use some alternative decision theory to make your decision; if you were to use this decision theory to choose which decision theory to use to make your choices, it would tell you to choose a different one, and not itself. We might naturally say that a decision theory that is not self-recommending in this sense is not a coherent means by which to make decisions, and that seems to be a strong strike against it.</p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEivrMYTrbLKUHV06sfhHcE5DDIuqQ3lBTedARfcwb4687hbiTFkaTZZNJ0--BOeApydm9AEki_tFCFpQeBLum3-c05X-_kVG5BETXJ9_6sm1DLYJMWkI2YXpSnCRWOtku82Iqhc8U2DcR4CZLjNV5MAOHSaVjvr3ixs_Mt8he89wOTOmH0NvAOsKyRNYA/s1024/IMG_5301.JPG" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="436" data-original-width="1024" height="136" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEivrMYTrbLKUHV06sfhHcE5DDIuqQ3lBTedARfcwb4687hbiTFkaTZZNJ0--BOeApydm9AEki_tFCFpQeBLum3-c05X-_kVG5BETXJ9_6sm1DLYJMWkI2YXpSnCRWOtku82Iqhc8U2DcR4CZLjNV5MAOHSaVjvr3ixs_Mt8he89wOTOmH0NvAOsKyRNYA/s320/IMG_5301.JPG" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">A self-recommending Timothy Dalton<br /></td></tr></tbody></table><p><span></span></p><a name='more'></a>On what basis might we criticize a theory of rational decision making? One popular way is to show that any individual who adopts the theory is exploitable; that is, there are decisions they might face in response to which the theory will lead them to choose certain options when there are alternative options that are guaranteed to be better; that is, in the jargon of decision theory, there are alternatives that dominate the recommendations of the decision theory in question. This is the sort of objection that <a href="https://philpapers.org/rec/GUSMA" rel="nofollow" target="_blank">money pump arguments</a> raise against their targets. For instance, suppose my decision theory permits cyclical preferences, so that I may prefer $a$ to $b$, $b$ to $c$, and $c$ to $a$. Then, if I have those preferences and choose in line with them, the money pump argument notes that I will choose $b$ over $a$, when presented with a choice between them, then pay money to swap $b$ for $c$, when presented with that option, and then pay money again to swap $c$ for $a$, when presented with that possibility. However, I might simply have chosen $a$ in the first place and then refrained from swapping thereafter, and I would have ended up better off for sure. The second sequence of choices dominates the first. So, the exploitability argument concludes, cyclical preferences are irrational and so any decision theory that permits them is flawed.<br /><br />This sort of objection is also often raised against decision theories that permit sensitivity to risk. For instance, take the most extreme risk-averse decision theory available, namely Abraham Wald's <a href="https://www.jstor.org/stable/1969022" rel="nofollow" target="_blank"><i>Maximin</i></a>. This doesn't just permit sensitivity to risk---it demands it. It says that, in any decision problem, you should choose the option whose worst-case outcome is best. So, suppose you are faced with the choice between $a$ and $b$, both of which are defined at two possible worlds, $w_1$ and $w_2$:<br />$$\begin{array}{r|cc}& w_1 & w_2 \\ \hline a & 3 & 0 \\ b & 1 & 1 \end{array}$$<br />Then Maximin says that you should choose $b$, since its worst-case outcome gives 1 utile, while the worst-case outcome of $a$ gives 0 utiles. <br /><br />Now, after facing that first decision, and choosing $b$ in line with Maximin, suppose you're now faced with the choice between $c$ and $d$:<br />$$\begin{array}{r|cc} & w_1 & w_2 \\ \hline a' & 0 & 3 \\ b' & 1 & 1 \end{array}$$ <br />You choose $b'$, just as Maximin requires. But it's easy to see that $a$ and $a'$, taken together, dominate $b$ and $b'$. $a + a'$ gives 3 utiles for sure, while $b + b'$ gives 2 utiles for sure.<br />$$\begin{array}{r|cc} & w_1 & w_2 \\ \hline a+a' & 3 & 3 \\ b+b' & 2 & 2 \end{array} $$<br />So Maximin is exploitable.<br /><br />I've argued in various places that I don't find exploitability arguments compelling.** They show only that the decision rule will lead to a bad outcome when the decision maker is faced with quite a specific series of decision problems. But that tells me little about the performance of the decision rule over the vast array of possible decisions I might face. Perhaps an exploitable rule compensates for its poor performance in those particular cases by performing extremely well in other cases. For all the exploitability objection tells me, that could well be the case.<br /><br />Recognising this problem, you might instead ask: how does this decision rule perform on average over all decision problems you might face? And indeed it's easy to show that decision theories that disagree with expected utility theory will perform worse on average than expected utility theory itself. But that's partly because we've stacked the deck in favour of expected utility theory. After all, looking at the average performance over all decision problems is just looking at the expected performance from the point of view of a credence function that assigns equal probability to all possible decision problems. And, as we'll show explicitly below, expected utility theory judges itself to be the best decision theory to use; that is, it does best in expectation; that is, it does best on average.<br /><br />But while this argument begs the question against non-expected utility theories, it does suggest a different way to test a decision theory: ask not whether it does best on average, and thus by the lights of expected utility theory; ask rather whether it does best by its own lights; ask whether it judges itself to be the best decision theory. Of course, this is a coherence test, and like all coherence tests, passing it is not sufficient for rationality. But it does seem that failing is sufficient for irrationality. It is surely irrational to use a method for selecting the best means to your ends that does not think it is the best method for selecting the best means to your ends.<br /><br />Let's begin by seeing a couple of theories that pass the test. Expected utility theory is the obvious example, and running through that will allow us to set up the formal framework. We begin with the space of possible states. There are two components to these states:<p></p><p></p><ul style="text-align: left;"><li>Let $W$ be the set of possible worlds grained finely enough to determine the utilities of all the options between which you will pick;</li><li>Let $D$ be the set of decision problems you might face.</li></ul><p>Then a state is a pair $(d, w)$ consisting of a decision problem $d$ from $D$ and a possible world $w$. And now we define your credence function $p : W \times D \rightarrow [0, 1]$. So $p(w\ \&\ d)$ is your credence that you're at world $w$ and will face decision problem $d$.<br /><br />Then expected utility theory says that, faced with a decision problem $d$ from $D$, you should pick an option $a$ from among those in $d$ that have maximal expected utility from the point of view of $p$. Given option $a$ and world $w$, we write $a(w)$ for the utility of $a$ at $w$. So the expected utility of $a$ is<br />$$<br />\sum_{w \in W} p(w | d)a(w)<br />$$<br />Now, let's ask how expected utility theory judges itself. Given a decision theory $R$ and a decision problem $d$, let $R(d)$ be the option in $d$ that $R$ requires you to take; so $R(d)(w)$ is the utility at world $w$ of the option from decision problem $d$ that decision theory $R$ requires you to take. Note that, in order for this to be well-defined for theories like $EU$ in which there might be multiple options with maximal expected utility, we must supplement those theories with a mechanism for breaking ties; but that is easily done. So $EU(d)$ is one of the acts in $d$ that maximises expected utility. Then expected utility theory assigns the following value or choiceworthiness to a decision theory $R$:<br />$$<br />EU(R) = \sum_{d \in D} \sum_{w \in W} p(d\ \&\ w)R(d)(w)<br />$$<br />Now, suppose $R$ is a decision theory and $d$ is a decision problem. Then<br />$$<br />\sum_{w \in W} p(w | d)EU(d)(w) \geq \sum_{w \in W} p(w | d)R(d)(w)<br />$$<br />with strict inequality if $R$ picks an option that doesn't maximise expected utility and $p$ assigns positive credence to a world $w$ at which this option differs from the one that does. But then<br />$$<br />\sum_{d \in D} p(d) \sum_{w \in W} p(w | d)EU(d)(w) \geq \sum_{d \in D} p(d) \sum_{w \in W} p(w | d)R(d)(w)<br />$$<br />So<br />$$<br />EU(EU) = \sum_{d \in D} \sum_{w \in W} p(d\ \&\ w)EU(d)(w) \geq \sum_{d \in D}\sum_{w \in W} p(d\ \&\ w)R(d)(w) = EU(R)<br />$$<br />with strict inequality if $p$ assigns positive credence to $d\ \&\ w$ where $R$ chooses an option that doesn't maximise expected utility and at world the utility of that option is different from the utility of the option that does. So, expected utility recommends itself.<br /><br />Maximin, which we met above, is another self-recommending decision theory. What is the value or choiceworthiness that Maximin assigns to a decision theory $R$? It is the lowest utility you can obtain from using $R$ to make a decision:<br />$$<br />M(R) = \min_{\substack{d \in D \\ w \in W}} \{R(d)(w)\}<br />$$<br />Now, suppose $M$ judges $R$ to be better than it judges itself to be. That is, $$M(R) > M(M)$$Then pick a decision problem $d^\star$ and a world $w^\star$ at which $M$ obtains its minimum. That is, $M(d^\star)(w^\star) = M(M)$. And suppose $R$ picks option $a$ from $d^\star$. Then, for each world $w$, $a(w) > M(d^\star)(w^\star)$. If this were not the case, then $R$ would achieve as low a minimum as $M$. But then $M$ would have recommended $a$ instead of $M(d^\star)$ when faced with $d^\star$, since the worst-case of $a$ is better than the worst-case of $M(d^\star)$, which occurs at $w^\star$. So, Maximin is a self-recommending decision theory.<br /><br />Now, Maximin is usually rejected as a reasonable decision rule for other reasons. For one thing, without further supplementary principles, it permits choices that are weakly dominated---that is, in some decision problems, it will declare one option permissible when there is another that is at least as good at all worlds and better at some. And since it pays no attention to the probabilities of the outcomes, it also permits choices that are stochastically dominated---that is, in some decision problems, it will declare one option permissible when there is another with the same possible outcomes, but higher probabilities for the better of those outcomes and lower probabilities for the worse. For another thing, Maximin just seems too extreme. It demands that you to take £1 for sure instead of a 1% chance of 99p and 99% chance of £10trillion.<br /><br />An alternative theory of rational decision making that attempts to accommodate less extreme attitudes to risk is Lara Buchak's risk-weighted expected utility theory. This theory encodes your attitudes to risk in a function $r : [0, 1] \rightarrow [0, 1]$ that is (i) continuous, (ii) strictly increasing, and (iii) assigns $r(0) = 0$ and $r(1) = 1$. This function is used to skew probabilities. For risk-averse agents, the probabilities are skewed in such a way that worse-case outcomes receive more weight than expected utility theory gives them, and best-case outcomes receive less weight. For risk-inclined agents, it is the other way around. For risk-neutral agents, $r(x) = x$, the probabilities aren't skewed at all, and the theory agrees exactly with expected utility theory.<br /><br />Now, suppose $W = \{w_1, \ldots, w_n\}$. Given a credence function $p$ and a risk function $r$ and an option $a$, if $a(w_1) \leq a(w_2) \leq \ldots \leq a(w_n)$, the risk-weighted expected utility of $a$ is<br />\begin{eqnarray*}<br />& & REU_{p, r}(a) \\ <br />& = & a(w_1) + \sum^{n-1}_{i=1} r(p(w_{i+1}) + \ldots + p(w_n))(a(w_{i+1}) - a(w_i)) \\<br />& = & \sum^{n-1}_{i=1} [r(p(w_i) + \ldots + p(w_n)) - r(p(w_{i+1}) + \ldots + p(w_n))]a(w_i) + r(p(w_n))a(w_n)<br />\end{eqnarray*}<br />So the risk-weighted expected utility of $a$, like the expected utility of $a$, is a weighted sum of the various utilities that $a$ can take at the different worlds. But, whereas expected utility theory weights the utility of $a$ at $w_i$ by the probability of $w_i$, risk-weighted utility theory weights it by the difference between the skewed probability that you will receive at least $a(w_i)$ from choosing $a$ and the skewed probability that you will receive more than $a(w_i)$ from choosing $a$.<br /><br />Here is an example to illustrate. The decision is between $a$ and $b$:<br />$$\begin{array}{r|cc} & w_1 & w_2 \\ \hline a & 1 & 4 \\ b & 2 & 2 \end{array}$$ <br />And suppose $p(w_1) = p(w_2) = 0.5$ and $r(x) = x^2$, for all $0 \leq x \leq 1$. Then:<br />$$<br />REU_{p, r}(a) = 1 + r(w_2)(4-1) = 1 + \frac{1}{2}^23 = \frac{7}{4}<br />$$<br />while<br />$$<br />REU_{p, r}(b) = 2 + r(w_2)(2-2) = 2 = \frac{8}{4}<br />$$<br />So, while the expected utility of $a$ (i.e. 2.5) exceeds the expected utility of $b$ (i.e. 2), and so expected utility theory demands you pick $a$ over $b$, the risk-weighted expected utility of $b$ (i.e. 2) exceeds the risk-weighted expected utility of $a$ (i.e. 1.75), and so risk-weighted expected utility demands you pick $b$ over $a$.<br /><br />Now we're ready to ask the central question of this post: does risk-weighted utility theory recommend itself? And we're ready to give our answer, which is that it doesn't.<br /><br />It's tempting to think it does, and for the same reason that expected utility theory does. After all, if you're certain that you'll face a particular decision problem, risk-weighted expected utility theory recommends using it to make the decision. How could it not? After all, it recommends picking a particular option, and therefore recommends any theory that will pick that option, since using that theory will have the same utility as picking the option at every world. So, you might expect, it will also recommend itself when you're uncertain which decision you'll face. But risk-weighted expected utility theory doesn't work like that.<br /><br />Let me begin by noting the simplest case in which it recommends something else. This is the case in which there are two decision problems, $d$ and $d'$, and you're certain that you'll face one or the other, but you're unsure which.<br />$$<br />\begin{array}{r|cc}<br />d & w_1 & w_2 \\<br />\hline<br />a & 3 & 6 \\<br />b & 2 & 8<br />\end{array}\ \ \ <br />\begin{array}{r|cc}<br />d' & w_1 & w_2 \\<br />\hline<br />a' & 4 & 19 \\<br />b' & 7 & 9<br />\end{array}<br />$$ <br />You think each is equally likely, you think each world is equally likely, and you think the worlds and decision problems are independent. So, $$p(d\ \&\ w_1) = p(d\ \&\ w_2)=p(d'\ \&\ w_1) = p(d'\ \&\ w_2) = \frac{1}{4}$$Then:<br />$$<br />REU(a) = 3 + \frac{1}{2}^2(6-3) = 3.75 > 3.5 = 2 + \frac{1}{2}^2(8-2) = REU(b)<br />$$<br />$$<br />REU(a') = 4 + \frac{1}{2}^2(19-4) = 7.75 > 7.5 = 7 + \frac{1}{2}^2(9-7) = REU(b')<br />$$<br />So REU will tell you to choose $a$ when faced with $d$ and $a'$ when faced with $d'$. Now compare that with a decision rule $R$ that tells you to pick $b$ and $b'$ respectively.<br />$$<br />REU(REU) = 3 + \frac{3}{4}^2(4-3) + \frac{1}{2}^2(6-4) + \frac{1}{4}^2(19-6) = 4.875<br />$$<br />and<br />$$<br />REU(R) = 2 + \frac{3}{4}^2(7-2) + \frac{1}{2}^2(8-7) + \frac{1}{4}^2(9-8) = 5.125<br />$$<br />So risk-weighted utility theory does not recommend itself in this situation. Yet it doesn't seem fair to criticize it on this basis. After all, perhaps it redeems itself by its performance in the face of other decision problems. In exploitability arguments, we only consider one series of decisions. Here, we only consider a pair of possible decisions. What happens when we have much much more limited information about the decision problems we'll face? <br /><br />Let's suppose that there is a finite set of utilities, $\{0, \ldots, n\}$.*** And suppose you consider every two-world, two-option decision problem with utilities from that set is possible and equally likely. That is, the following decision problems are equally likely: decision problems $d$ in which there are exactly two available options, $a$ and $b$, which are defined only on mutually exclusive and exhaustive worlds $w_1$ and $w_2$, and where the utilities $a(w_1), a(w_2), b(w_1), b(w_2)$ lie in $\{0, \ldots, n\}$. <br /><br />Here are some results: Set $n = 22$, and let $p(w_1) = p(w_2) = 0.5$. And consider risk functions of the form $r_k(x) = x^k$. For $k > 1$, $r_k$ is risk-averse; for $k = 1$, $r_k$ is risk-neutral and risk-weighted expected utility theory agrees with expected utility theory; and for $k < 1$, $r_k$ is risk-seeking. We say that $REU_{p, r_k}$ judges $REU_{p, r_{k'}}$ better than it judges itself if $$REU_{p, r_k}(REU_{p, r_k}) < REU_{p, r_k}(REU_{p, r_{k'}})$$And we write $REU_{p, r_k} \rightarrow REU_{p, r_{k'}}$. Then we have the following results:<br />$$<br />REU_{p, r_2} \rightarrow REU_{p, r_{1.5}} \rightarrow REU_{p, r_{1.4}} \rightarrow REU_{p, r_{1.3}} \rightarrow REU_{p, r_{1.2}} \rightarrow REU_{p, r_{1.1}}<br />$$<br />and<br />$$<br />REU_{p, r_{0.5}} \rightarrow REU_{p, r_{0.6}} \rightarrow REU_{p, r_{0.7}} \rightarrow REU_{p, r_{0.8}} \rightarrow REU_{p, r_{0.9}}<br />$$<br />So, for many natural risk-averse and risk-seeking risk functions, risk-weighted utility theory isn't self-recommending. And this, it seems to me, is a problem for these versions of the theory.<br /><br />Now, for all my current results say, it's possible that there is a risk function other than $r(x) = x$ for which the theory recommends itself. But my conjecture is that this doesn't happen. The present results suggest that, for each risk-averse function, there is a less risk-averse one that it judges better, and for risk-seeking ones, there is a less risk-seeking one that it judges better. But even if there were a risk function for which the theory is self-recommending, that would surely limit the versions of risk-weighted expected utility theory that are tenable. That in itself would be an interesting result.</p><p><br /></p><p>* The work of which I'm aware that comes closest to what interests me here is Catrin Campbell-Moore and Bernhard Salow's <a href="https://philpapers.org/rec/CAMAUF" rel="nofollow" target="_blank">exploration of proper scoring rules for risk-sensitive agents in Buchak's theory</a>. But it's not quite the same issue. And the idea of judging some part or whole of our decision-making apparatus by looking at its performance over all decision problems we might face I draw from <a href="https://projecteuclid.org/journals/annals-of-statistics/volume-17/issue-4/A-General-Method-for-Comparing-Probability-Assessors/10.1214/aos/1176347398.full" rel="nofollow" target="_blank">Mark Schervish's</a> and <a href="https://philpapers.org/rec/LEVAPG" rel="nofollow" target="_blank">Ben Levinstein's</a> work. But again, they are interested in using decision theories to judge credences, not using decision theories to judge themselves.<br /></p><p>** In Section 13.7 of my <a href="https://philpapers.org/rec/PETCFC-2" rel="nofollow" target="_blank"><i>Choosing for Changing Selves</i></a> and Chapter 6 of my <a href="https://philpapers.org/rec/PETDBA-2" rel="nofollow" target="_blank"><i>Dutch Book Arguments</i></a>.</p><p>*** I make this restriction because it's the one for which I have some calculations; there's no deeper motivation. From fiddling with the calculations, it looks to me as if this restriction is inessential.</p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com0tag:blogger.com,1999:blog-4987609114415205593.post-78155984560699321832022-06-23T09:04:00.004+01:002022-07-27T19:07:26.892+01:00Aggregating for accuracy: another accuracy argument for linear pooling<p> A PDF of this blogpost is available <a href="https://drive.google.com/file/d/1H9MTwjN9mzKwpZ7StE5LqL8lNFPqQSrC/view?usp=sharing" rel="nofollow" target="_blank">here</a>. <br /></p><p>I don't have an estimate for how long it will be before the Greenland ice sheet collapses, and I don't have an estimate for how long it will be before the average temperature at Earth's surface rises more than 3C above pre-industrial levels. But I know a bunch of people who do have such estimates, and I might hope that learning theirs might help me set mine. Unfortunately, each of these people has a different estimate for each of these two quantities. What should I do? Should I pick one of them at random and adopt their estimates as mine? Or should I pick some compromise between them? If the latter, which compromise?</p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3WqjvJRMLAJqxsKPOjN859PCPOQKtXcjMliYcoKDzTpb6k6_PepzhtFPY5CpJTNHRezWxV_e907DWdOFtnZJHHNF7k4MclC7__nSq33uEg5B7X6hk3g1Pwi3vWz34oKCo3H-TZPzEypVhWrOyXrzup8QoNCOjtCxFtQeLLufmjLkCdwu-hBknqE3F9A/s2886/bluebell-fall.jpg" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="2886" data-original-width="2886" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3WqjvJRMLAJqxsKPOjN859PCPOQKtXcjMliYcoKDzTpb6k6_PepzhtFPY5CpJTNHRezWxV_e907DWdOFtnZJHHNF7k4MclC7__nSq33uEg5B7X6hk3g1Pwi3vWz34oKCo3H-TZPzEypVhWrOyXrzup8QoNCOjtCxFtQeLLufmjLkCdwu-hBknqE3F9A/s320/bluebell-fall.jpg" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Cat inaccurately estimates width of step<br /></td></tr></tbody></table><span><a name='more'></a></span>The following fact gives a hint. An estimate of a quantity, such as the number of years until an ice sheet collapses or the number of years until the temperature rises by a certain amount, is better the closer it lies to the true value of the quantity and worse the further it lies from this. There are various ways to measure the distance between estimate and true value, but we'll stick with a standard one here, namely, <i>squared error</i>, which takes the distance to be the square of the difference between the two values. Then the following is simply a mathematical fact: taking the straight average of the group's estimates of each quantity as your estimate of that quantity is guaranteed to be better, in expectation, than picking a member of the group at random and simply deferring to them. This is sometimes called the Diversity Prediction Theorem, or it's a corollary of what goes by that name. <p></p><p>The result raises a natural question: Is it only by taking the straight average of the group's estimate as your own that you can be guaranteed to do better, in expectation, than by picking at random? Or is there another method for aggregating the estimates that also has this property? As I'll show, only straight averaging has this property. If you combine the group's estimates in any other way to give your own, there is a possible set of true values that will lie further from your estimate than you would lie, in expectation, were you to pick at random. The question is natural, and the answer is not difficult to prove, so I'm pretty confident this has been asked and answered before; but I haven't been able to find it, so I'd be grateful for a reference if anyone has one.</p><p>Let's make all of this precise. We have a group of $m$ individuals; each of them has an estimate for each of the quantities $Q_1, \ldots, Q_n$. We represent individual $j$ by the sequence $X_j = (x_{j1}, \ldots, x_{jn})$ of their estimates of these quantities. So $x_{ji}$ is the estimate of quantity $Q_i$ by individual $j$. Suppose $T = (t_1, \ldots, t_n)$ is the sequence of true values of these quantities. So $t_i$ is the true value of $Q_i$. Then the disvalue or badness of individual $j$'s estimates, as measured by squared error, is:$$(x_{j1} - t_1)^2 + \ldots + (x_{jn} - t_n)^2$$The disvalue or badness of individual $j$'s estimate of quantity $Q_i$ is $(x_{ji} - t_i)^2$, and the disvalue or badness of their whole set of estimates is the sum of the disvalue or badness of their individual estimates. We write $\mathrm{SE}(X, T)$ for this sum. That is,$$\mathrm{SE}(X_j, T) = \sum_i (x_{ji} - t_i)^2$$Then the Diversity Prediction Theorem says that, for any $X_1, \ldots, X_m$ and any $T$,$$\mathrm{SE}\left (\frac{1}{m}X_1 + \ldots + \frac{1}{m}X_m, T \right ) < \frac{1}{m}\mathrm{SE}(X_1, T) + \ldots + \frac{1}{m}\mathrm{SE}(X_m, T)$$And we wish to prove a sort of converse, namely, if $V \neq \frac{1}{m}X_1 + \ldots + \frac{1}{m}X_m$, then there is a possible set of true values $T = (t_1, \ldots, t_n)$ such that$$\mathrm{SE}(V, T) > \frac{1}{m}\mathrm{SE}(X_1, T) + \ldots + \frac{1}{m}\mathrm{SE}(X_m, T)$$I'll give the proof below.</p><p>Why is this interesting? One question at the core of those parts of philosophy that deal with collectives and their attitudes is this: How should you aggregate the opinions of a group of individuals to give a single set of opinions? When the opinions come in numerical form, such as when they are estimates of quantities or when they are probabilities, there are a number of proposals. Taking the straight arithmetic average as we have done here is just one. How are we to decide which to use? Standard arguments proceed by identifying a set of properties that only one aggregation method boasts, and then arguing that the properties in the set are desirable given your purpose in doing the aggregation in the first place. The result we have just noted might be used to mount just such an argument: when we aggregate estimates, we might well want a method that is guaranteed to produce aggregate estimates that are better, in expectation, than picking at random, and straight averaging is the only method that does that. </p><p>Finally, here's a slightly more general version of the result, which considers not just straight averages but also weighted averages; the proof is also given.</p><p><b>Proposition</b> Suppose $\lambda_1, \ldots, \lambda_m$ is a set of weights, so that $0 \leq \lambda_j \leq 1$ and $\sum_j \lambda_j = 1$. Then, if $V \neq \lambda_1 X_1 + \ldots + \lambda_mX_m$, then there is a possible set of true values $T = (t_1, \ldots, t_n)$ such that$$\mathrm{SE}(V, T) > \lambda_1\mathrm{SE}(X_1, T) + \ldots + \lambda_m\mathrm{SE}(X_m, T)$$</p><p><i>Proof.</i> The left-hand side of the inequality is<br />$$<br />\mathrm{SE}(V, T) = \sum_i (v_i - t_i)^2 = \sum_i v_i^2 - 2\sum_i v_it_i + \sum_i t^2_i<br />$$The right-hand side of the inequality is<br />\begin{eqnarray*}<br />\sum_j \lambda_j \mathrm{SE}(X_j, T) & = & \sum_j \lambda_j \sum_i (x_{ji} - t_i)^2 \\<br />& = & \sum_j \lambda_j \sum_i \left ( x^2_{ji} - 2x_{ji}t_i + t_i^2 \right ) \\<br />& = & \sum_{i,j} \lambda_j x^2_{ji} - 2\sum_{i,j} \lambda_j x_{ji}t_i + \sum_i t_i^2 <br />\end{eqnarray*}<br />So $\mathrm{SE}(V, T) > \sum_j \lambda_j \mathrm{SE}(X_j, T)$ iff$$<br />\sum_i v_i^2 - 2\sum_i v_it_i > \sum_{i,j} \lambda_j x^2_{ji} - 2\sum_{i,j} \lambda_j x_{ji}t_i<br />$$iff$$<br /> 2\left ( \sum_i \left ( \sum_j \lambda_j x_{ji}- v_i \right) t_i \right ) > \sum_{i,j} \lambda_j x^2_{ji} - \sum_i v_i^2 <br />$$And, if $(v_1, \ldots, v_n) \neq (\sum_j \lambda_j x_{j1}, \ldots,\sum_j \lambda_j x_{jn})$, there is $i$ such that $\sum_j \lambda_j x_{ji} - v_i \neq 0$, and so it is always possible to choose $T = (t_1, \ldots, t_n)$ so that the inequality holds, as required.<br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com0tag:blogger.com,1999:blog-4987609114415205593.post-20982599698943509492022-05-18T13:19:00.003+01:002022-07-27T19:07:44.630+01:00Should we agree? III: the rationality of groups<p>In the previous two posts in this series (<a href="https://m-phi.blogspot.com/2022/05/should-we-agree-i-arguments-for.html" rel="nofollow" target="_blank">here</a> and <a href="https://m-phi.blogspot.com/2022/05/should-we-agree-ii-new-pragmatic.html" rel="nofollow" target="_blank">here</a>), I described two arguments for the conclusion that the members of a group should agree. One was an epistemic argument and one a pragmatic argument. Suppose you have a group of individuals. Given an individual, we call the set of propositions to which they assign a credence their agenda. The group's agenda is the union of its member's agendas; that is, it includes any proposition to which some member of the group assigns a credence. The precise conclusion of the two arguments we describe is this: the group is irrational if there no single probability function defined on the group's agenda that gives the credences of each member of the group when restricted to their agenda. Following Matt Kopec, I called this norm Consensus. </p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6_QglBAUoEGpOG09V6UWe2-o0HF7V_6zq2mh54fTczHSY762ls2h8Po8ya_un5vP-1DUwRCBcm74fNUhcRnM4ZdBu1Es_7fJnZIAvqu25gP3Ax-QVgicDwwFvYp_M8ndcCDHi8pmMYLqLTOVjOWcHjiHOG5LaO0rNzpU2QuNGXXFqK3bB-5QmH0Vxqg/s910/IMG_1633.jpg" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="910" data-original-width="910" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6_QglBAUoEGpOG09V6UWe2-o0HF7V_6zq2mh54fTczHSY762ls2h8Po8ya_un5vP-1DUwRCBcm74fNUhcRnM4ZdBu1Es_7fJnZIAvqu25gP3Ax-QVgicDwwFvYp_M8ndcCDHi8pmMYLqLTOVjOWcHjiHOG5LaO0rNzpU2QuNGXXFqK3bB-5QmH0Vxqg/s320/IMG_1633.jpg" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Cats showing a frankly concerning degree of consensus<br /></td></tr></tbody></table><span><a name='more'></a></span><p></p><p>Both arguments use the same piece of mathematics, but they interpret it differently. Both appeal to mathematical functions that measure how well our credences achieve the goals that we have when we set them. There are (at least) two such goals: we aim to have credences that will guide our actions well and we aim to have credences that will represent the world accurately. In the pragmatic argument, the mathematical function measures how well our credences achieve the first goal. In particular, they measure the utility we can expect to gain by having the credences we have and choosing in line with them when faced with whatever decision problems life throws at us. In the epistemic argument, the mathematical function measures how well our credences achieve the second goal. In particular, they measure the accuracy of our credences. As we noted in the second post on this, work by <a href="https://projecteuclid.org/journals/annals-of-statistics/volume-17/issue-4/A-General-Method-for-Comparing-Probability-Assessors/10.1214/aos/1176347398.full" rel="nofollow" target="_blank">Mark Schervish</a> and <a href="https://philpapers.org/rec/LEVAPG" rel="nofollow" target="_blank">Ben Levinstein</a> shows that the functions that measure these goals have the same properties: they are both strictly proper scoring rules. The arguments then appeal to the following fact: given a strictly proper scoring rule, if the members of a group do not agree on the credences they assign in the way required by Consensus, then there are some alternative credences they might assign instead that are guaranteed to be better according to that scoring rule.<br /></p><p>I'd like to turn now to assessing these arguments. My first question is this: In the norm of Probabilism, rationality requires something of an individual, but in the norm of Consensus, rationality requires something of a group of individuals. We understand what it means to say that an individual is irrational, but what could it mean to say that a group is irrational?</p><p>Here, I follow <a href="https://philpapers.org/rec/EASTCO-3" rel="nofollow" target="_blank">Kenny Easwaran's suggestion</a> that collective entities---in his case, cities; in my case, groups---can be said quite literally to be rational or irrational. For Easwaran, a city is rational "to the extent that the collective practices of its people enable diverse inhabitants to simultaneously live the kinds of lives they are each trying to live." As I interpret him, the idea is this: a city, no less than its individual inhabitants, has an end or goal or telos. For Easwaran, for instance, the end of a city is enabling its inhabitants to live as they wish to. And a city is irrational if it does not provide---in its physical and technological infrastructure, its byelaws and governing institutions---the best means to that end among those that are available. Now, we might disagree with Easwaran's account of a city's ends. But the template he provides by which we might understand group rationality is nonetheless helpful. Following his lead, we might say that a group, no less than its individual members, has an end. For instance, its end might be maximising the total utility of its members, or it might be maximizing the total epistemic value of their credences. And it is then irrational if it does not provide the best means to that end among those available. So, for instance, as long as agreement between members is available, our pragmatic and epistemic arguments for Consensus seem to show that a group whose ends are as I just described does not provide the best means to its ends if it does not deliver such agreement.</p><p>Understanding group rationality as Easwaran does helps considerably. As well as making sense of the claim that the group itself can be assessed for rationality, it also helps us circumscribe the scope of the two arguments we've been exploring, and so the scope of the version of Consensus that they justify. After all, it's clear on this conception that these arguments will only justify Consensus for a group if</p><ol style="text-align: left;"><li>that group has the end of maximising total expected pragmatic utility or total epistemic utility, i.e., maximising the quantities measured by the mathematical functions described above;</li><li>there are means available to it to achieve Consensus.</li></ol><p>So, for instance, a group of sworn enemies hellbent of thwarting each other's plans is unlikely to have as its end maximising total utility, while a group composed of randomly selected individuals from across the globe is unlikely to have as its end maximising total epistemic utility, and indeed a group so disparate might lack any ends at all.</p><p>And we can easily imagine situations in which there are no available means by which the group could achieve Consensus, perhaps because it would be impossible to set up reliable lines of communication.</p><p>This allows us to make sense of two of the conditions that <a href="https://philpapers.org/rec/GILIPA" rel="nofollow" target="_blank">Donald Gillies</a> places on the groups to which he takes his sure loss argument to apply (this is the first version of the pragmatic argument for Consensus; the one I presented in the first post and then abandoned in favour of the second version in the second post). He says (i) the members of the group must have a shared purpose, and (ii) there must be good lines of communication between them. Let me take these in turn to understand their status more precisely.</p><p>It's natural to think that, if a group has a shared purpose, it will have as its end maximising the total utility of the members of the group. And indeed in some cases this is almost certainly true. Suppose, for instance, that every member of a group cares only about the amount of biodiversity in a particular ecosystem that is close to their hearts. Then they will have the same utility function, and it is natural to say that maximising that shared utility is the group's end. But of course maximising that shared utility is equivalent to maximising the group's total utility, since the total utility is simply the shared utility scaled up by the number of members of the group.</p><p>However, it is also possible for a group to have a shared purpose without its end being to maximise total utility. After all, a group can have a shared purpose without each member taking that purpose to be the one and only valuable end. Imagine a different group: each member cares primarily about the level of biodiversity in their preferred area, but each also cares deeply about the welfare of their family. In this case, you might take the group's end to be maximising biodiversity in the area in question, particularly if it was this shared interest that brought them together as a group in the first place, but maximising this good might require the group not to maximise total utility, perhaps because some members of the group have family who are farmers and who will be adversely affected by whatever is the best means to the end of greater biodiversity.</p><p>What's more, it's possible for a group to have as its end maximising total utility without having any shared purpose at all. For instance, a certain sort of utilitarian might say that the group of all sentient beings has as its end the maximisation of the total utility of its members. But that group does not have any shared purpose.</p><p>So I think we can use the pragmatic and epistemic arguments to determine the groups to which the norm of Consensus applies, or at least the groups for which our pragmatic and epistemic arguments can justify its application. It is those groups that have as their end either maximising the total pragmatic utility of the group, or maximising their total epistemic utility, or maximising some weighted average of the two---after all, the weighted average of two strictly proper scoring rules, one measuring epistemic utility and one measuring pragmatic utility, is itself a strictly proper scoring rule. Of course, this requires an account of when a group has a particular end. This, like all questions about when collectives have certain attitudes, is delicate. I won't say anything more about it here.</p><p>Let's turn next to Gillies' claim that Consensus applies only to groups between whose members there are reliable lines of communication. In fact, I think our versions of the arguments show that this condition lives a strange double life. On the one hand, if such lines of communcation are necessary to achieve agreement across the group, then the norm of Consensus simply does not apply to a group when these lines of communication are impossible, perhaps because of geographical, social, or technological barriers. A group cannot be judged irrational for failing to achieve something it could not possibly achieve, however much closer it would get to its goal if it could achieve that. </p><p>On the other hand, if such lines of communication are available, and if they increase the chance of agreement among members of the group, then our two arguments for Consensus are equally arguments for establishing such lines of communication, providing that the cost of doing so is outweighed by the gain in pragmatic or epistemic utility that comes from achieving agreement.</p><p>But these arguments do something else as well. They lend nuance to Consensus. In some cases in which some lines of communication are available but others aren't, or are too costly, our arguments still provide norms. Take, for instance, a case in which some central planner is able to communicate a single set of prior credences that each member of the group should have, but after the members start receiving evidence, this central planner can no longer coordinate their credences. And suppose we know that the members will receive different evidence: they'll be situated in different places, and so they'll see different things, have access to different information sources, and so on. So we know that, if they update on the evidence they receive in the standard way, they'll end up having different credences from one another and therefore violating Consensus. You might think, from looking at Consensus, that the group would do better, both pragmatically and epistemically, if each of its members were to ignore whatever evidence were to come in and to stick with their prior regardless in order to be sure that they remain in agreement and satisfy Consensus both in their priors and their posteriors.</p><p>In fact, however, this isn't the case. Let's take an extremely simple example. The group has just two members, Ada and Baz. Each has opinions only about the outcomes of two independent tosses of a fair coin. So the possible worlds are HH, HT, TH, TT. Ada will learn the outcome of the first, and Baz will learn the outcome of the second. A central planner can communicate to them a prior they should adopt, but that central planner can't receive information from them, and so can't receive their evidence and pool it and communicate a shared posterior to them. How should Ada and Baz proceed? How should they pick their priors, and what strategies should each adopt for updating when the evidence comes in? The entity we're assessing for rationality is the quadruple that contains Ada's prior together with her plan for updating, and Baz's prior together with his plan for updating. Which of these are available? Well, nothing constrains Ada's priors and nothing constrain's Baz's. But there are constraints on their updating rules. Ada's updating rule must give the same recommendation at any two worlds at which her evidence is the same---so, for instance, it must give the same recommendation at HH as at HT, since all she learns at both is that the first coin landed heads. And Baz's updating rule must give the same recommendation at any two worlds at which his evidence is the same---so, for instance, it must give the same recommendation at HH as at TH. Then consider the following norm:</p><p><b>Prior Consensus</b> Ada and Baz should have the same prior and both should plan to update on their private evidence by conditioning on it.</p><p>And the argument for this is that, if they don't, there's a quadruple of their priors and plans that (i) satisfy the constraint outlined above and (ii) together have greater total epistemic utility at each possible world; and there's a quadruple of their priors and plans that (i) satisfy the constraint outlined above and (ii) together have greater total expected pragmatic utility at each possible world. This is a corollary of an argument that <a href="https://philpapers.org/rec/BRIAAA-11" rel="nofollow" target="_blank">Ray Briggs and I</a> gave, and that <a href="https://philpapers.org/rec/NIEAAC" rel="nofollow" target="_blank">Michael Nielsen</a> corrected and improved on. So, if Ada and Baz are in agreement on their prior, and plan to stick with it rather than update on their evidence because that way they'll retain agreement, then they're be accuracy dominated and pragmatically dominated.</p><p>You might wonder how this is possible. After all, whatever evidence Ada and Baz each receive, Prior Consensus requires them to update on it in a way that leads them to disagree, and we know that they are then accuracy and pragmatically dominated. This is true, and it would tell against the priors + updating plans recommended by Prior Consensus if there were some way for Ada and Baz to communicate after their evidence came in. It's true that, for each possible world, there is some credence function such that if, at each world, Ada and Baz were to have that credence function rather than the ones they obtain by updating their shared prior on their private evidence, then they'd end up with greater total accuracy and pragmatic utility. But, without the lines of communication, they can't have that.</p><p>So, by looking in some detail at the arguments for Consensus, we come to understand better the groups to which it applies and the norms that apply to those groups to which it doesn't apply in its full force. <br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com1tag:blogger.com,1999:blog-4987609114415205593.post-43468244280791864632022-05-06T10:08:00.002+01:002022-07-27T19:08:02.431+01:00Should we agree? II: a new pragmatic argument for consensus<p>There is a PDF version of this blogpost available <a href="https://drive.google.com/file/d/1X0a65SJpW4J-oiw4L_byBMoN3uEf7wZf/view?usp=sharing" target="_blank">here</a>.</p><p>In <a href="https://m-phi.blogspot.com/2022/05/should-we-agree-i-arguments-for.html" target="_blank">the previous post</a>, I introduced the norm of Consensus. This is a claim about the rationality of groups. Suppose you've got a group of individuals. For each individual, call the set of propositions to which they assign a credence their agenda. They might all have quite different agendas, some of them might overlap, others might not. We might say that the credal states of these individual members cohere with one another if there is a some probability function that is defined for any proposition that appears in any member's agenda, and the credences each member assigns to the propositions in their agenda match those assigned by this probability function to those propositions. Then Consensus says that a group is irrational if it does not cohere.</p><p></p><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto;"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOVZ9sytsZsNMQfil3puZgn03ZtNd75uiOLsLcKmmPH5HWYuFcKKBzBUqsVgBFoKsd_jLkRKAxJvjLwefMq6IYbxqm5S5poKdKuNG2_zzto3PjfmPSyCnROGne364F0prJkyRPEST55cRCcqsMGp4SiW_LVtTU6YanOkV8XfGiK2ZGdMhGqLXf_LV7jw/s3151/IMG_2162.jpg" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="2671" data-original-width="3151" height="271" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOVZ9sytsZsNMQfil3puZgn03ZtNd75uiOLsLcKmmPH5HWYuFcKKBzBUqsVgBFoKsd_jLkRKAxJvjLwefMq6IYbxqm5S5poKdKuNG2_zzto3PjfmPSyCnROGne364F0prJkyRPEST55cRCcqsMGp4SiW_LVtTU6YanOkV8XfGiK2ZGdMhGqLXf_LV7jw/s320/IMG_2162.jpg" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><i>A group coming to consensus</i><br /></td></tr></tbody></table><span><a name='more'></a></span>In that post, I noted that there are two sorts of argument for this norm: a pragmatic argument and an epistemic argument. The pragmatic argument is a sure loss argument. It is based on the fact that, if the individuals in the group don't agree, there is a series of bets that their credences require them to accept that will, when taken together, lose the group money for sure. In this post, I want to argue that there is a problem with the sure loss argument for Consensus. It isn't peculiar to this argument, and indeed applies equally to any argument that tries to establish a rational requirement by showing that someone who violates it is exploitable. Indeed, I've raised it elsewhere against the sure loss argument for Probabilism <a href="https://www.cambridge.org/core/books/dutch-book-arguments/3A54DE6AB1006E159A85FC17922D5779" target="_blank">(Section 6.2, Pettigrew 2020)</a> and the money pump argument against non-exponential discounting and changing preferences in general <a href="https://global.oup.com/academic/product/choosing-for-changing-selves-9780198814962" target="_blank">(Section 13.7.4, Pettigrew 2019)</a>. I'll describe the argument here, and then offer a solution based on work by <a href="http://deanfoster.net/teaching/data_mining/schervish.pdf" target="_blank">Mark Schervish (1989)</a> and <a href="https://philpapers.org/rec/LEVAPG" target="_blank">Ben Levinstein (2017)</a>. I've described this sort of solution before <a href="https://www.cambridge.org/core/books/dutch-book-arguments/3A54DE6AB1006E159A85FC17922D5779" target="_blank">(Section 6.3, Pettigrew 2020)</a>, and <a href="https://philpapers.org/rec/KONDOI" target="_blank">Jason Konek (ta)</a> has recently put it to interesting work addressing an issue with <a href="https://global.oup.com/academic/product/unsettled-thoughts-9780198833710" target="_blank">Julia Staffel's (2020)</a> account of degrees of incoherence.<br /><br />Sure loss and money pump arguments judge the rationality of attitudes, whether credences or preferences, by looking at the quality of the choices they require us to make. As Bishop Butler said, probability is the very guide of life. These arguments evaluate credences by exactly how well they provide that guide. So they are teleological arguments: they attempt to derive facts about the epistemic right---namely, what is rationally permissible---from facts about the epistemic good---namely, leading to pragmatically good choices.<br /><br />Say that one sequence of choices <i>dominates</i> another if, taken together, the first leads to better outcomes for sure. Say that a collection of attitudes is <i>exploitable</i> if there is a sequence of decision problems you might face such that, if faced with them, these attitudes will require you to make a dominated sequence of choices.<br /><br />For instance, take the sure loss argument for Probabilism: if you violate Probabilism because you believe $A\ \&\ B$ more strongly than you believe $A$, your credence in the former will require you to pay some amount of money for a bet that pays out a pound if $A\ \&\ B$ true and nothing if it's false, and your credence in the latter will require you to sell for less money a bet that pays out a pound if $A$ is true and nothing if it's false; yet you'd be better off for sure rejecting both bets. So rejecting both bets dominates accepting both; your credences require you to accept both; so your credences are exploitable. Or take the money pump argument against cyclical preferences: if you prefer $A$ to $B$ and $B$ to $C$ and $C$ to $A$, then you'll choose $B$ when offered a choice between $B$ and $C$, you'll then pay some amount to swap to $A$, and you'll then pay some further amount to swap to $C$; yet you'd be better off for sure simply choosing $C$ in the first place and not swapping either time that possibility was offered. So choosing $C$ and sticking with it dominates the sequence of choices your preferences require; so your preferences are exploitable.<br /><br />But, I contend, the existence of a sequence of decision problems in response to which your attitudes require you to make a dominated series of choices does not on its own render those attitudes irrational. After all, it is just one possible sequence of decision problems you might face. And there are many other sequences you might face instead. The argument does not consider how your attitudes will require you to choose when faced with those alternative sequences, and yet surely that is relevant to assessing those attitudes, for it might be that however bad is the dominated sequences of choices the attitudes require you to make when faced with the sequence of decision problems described in the argument for exploitability, there is another sequence of decision problems where those same attitudes require you to make a series of choices that are very good; indeed, they might be so good that they somehow outweigh the badness of the dominated sequence. So, instead of judging your attitudes by looking only at the outcome of choosing in line with them when faced with a single sequence of decision problems, we should rather judge them by looking at the outcome of choosing in line with them when faced with any decision problem that might come your way, weighting each by how likely you are to face it, to give a balanced view of the pragmatic benefits of having those credences. That's the approach I'll present now, and I'll show that it leads to a new and better pragmatic argument for Probabilism and Consensus.<br /><br />As I presented them, the sure loss arguments for Probabilism and Consensus both begin with a principle that I called Ramsey's Thesis. This is a claim about the prices that an individual's credence in a proposition requires her to pay for a bet on that proposition. It says that, if $p$ is your credence in $A$ and $x < pS$, then you are required to pay $£x$ for a bet that pays out $£S$ if $A$ is true and $£0$ if $A$ is false. Now in fact this is a particular consequence of a more general norm about how our credences require us to choose. Let's call the more general norm Extended Ramsey's Thesis. It says how our credence in a proposition requires us to choose when faced with a series of options, all of whose payoffs depend only on the truth or falsity of that proposition. Given a proposition $A$, let's say that an option is an $A$-option if its payoffs at any two worlds at which $A$ is true are the same, and its payoffs at any two worlds at which $A$ is false are the same. Then, given a credence $p$ in $A$ and an $A$-option $a$, we say that the expected payoff of $a$ by the lights of $p$ is<br />$$<br />p \times \text{payoff of $a$ when $A$ is true} + (1-p) \times \text{payoff of $a$ when $A$ is false}<br />$$Now suppose you face a decision problem in which all of the available options are $A$-options. Then Extended Ramsey's Thesis says that you are required to pick an option whose expected payoff by the lights of your credence in $A$ is maximal.*<br /><br />Next, we make a move that is reminiscent of the central move in I. J. Good's argument for Carnap's Principle of Total Evidence <a href="https://philpapers.org/rec/GOOOTP" target="_blank">(Good 1967)</a>. We say what we take the payoff to be of having a particular credence in a particular proposition given a particular way the world is and when faced with a particular decision problem. Specifically, we define the payoff of having credence $p$ in the proposition $A$ when that proposition is true, and when you're faced with a decision problem $D$ in which all of the options are $A$-options, to be the payoff when $A$ is true of whichever $A$-option available in $D$ maximises expected payoff by the lights of $p$. And we define the payoff of having credence $p$ in the proposition $A$ when that proposition is false, and when you're faced with a decision problem $D$ in which all of the options are $A$-options, to be the payoff when $A$ is false of whichever $A$-option available in $D$ maximises expected payoff by the lights of $p$. So the payoff of having a credence is the payoff of the option you're required to pick using that credence.<br /><br />Finally, we make the move that is central to Schervish's and Levinstein's work. We now know the payoff of having a particular credence in propositiojn $A$ when you face a decision problem in which all options are $A$-options. But of course we don't know which such decision problems we'll face. So, when we evaluate the payoff of having a credence in $A$ when $A$ is true, for instance, we look at all the decision problems populated by $A$-options we might face and weight them by how likely we are to face them and then take the payoff of having that credence when $A$ is true to be the expected payoff of the $A$-options it would leave us to choose faced with the decision problems we'll face. And then we note, as Schervish and Levinstein themselves note: if we make certain natural assumptions about how likely we are to face different decisions, then this resulting measure of the pragmatic payoff of having credence $p$ in proposition $A$ is a continuous and strictly proper scoring rule. That is, mathematically, the functions we use to measure the pragmatic value of a credence function are identical to the functions we use to evaluate the epistemic value of a credence that we use in the epistemic utility argument for Probabilism and Consensus.**<br /><br />With this construction in place, we can piggyback on the theorems stated in the previous post to give new pragmatic arguments for Probabilism and Consensus. First: Suppose your credences do not obey Probabilism. Then there are alternative ones you might have instead that do obey that norm and, at any world, if we look at each decision problem you might face and ask what payoff you'd receive at that world were you to choose from the options in that decision problem as the two different sets of credences require, and then weight those payoffs by how likely they are to face that decision to give their expected payoff, then the alternatives will always have the greater expected payoff. This gives strong reason to obey Probabilism.<br /><br />Second: Take a group of individuals. Now suppose the group's credences do not obey Consensus. Then there are alternative credences each member might have instead such that, if they were to have them, the group would obey Consensus and, at any world, if we look at each decision problem each member might face and ask what payoff that individual would receive at that world were they to choose from the options in that decision problem as the two different sets of credences require, and then weight those payoffs by how likely they are to face that decision to give their expected payoff, then the alternatives will always have the greater expected total payoff when this is summed across the whole group.<br /><br />So that is our new and better pragmatic argument for Consensus. The sure loss argument points out a single downside to a group that violates the norm. Such a group is vulnerable to exploitation. But it remains silent on whether there are upsides that might balance out that downside. The present argument addresses that problem. It finds that, if a group violates the norm, there are alternative credences they might have that are guaranteed to serve them better in expectation as a basis for decision making.<br /><br />* Notice that, if $x < pS$, then the expected payoff of a bet that pays $S$ if $A$ is true and $0$ if $A$ is false is<br />$$<br />p(-x + S) + (1-p)(-x) = pS- x<br />$$<br />which is positive. So, if the two options are accept or reject the bet, accepting maximises expected payoff by the lights of $p$, and so it is required, as Ramsey's Thesis says.<p></p><p>** <a href="https://philpapers.org/archive/KONDOI.pdf" target="_blank">Konek (ta)</a> gives a clear formal treatment of this solution. For those who want the technical details, I'd recommend the Appendix of that paper. I think he presents it better than I did in <a href="https://www.cambridge.org/core/books/dutch-book-arguments/3A54DE6AB1006E159A85FC17922D5779" target="_blank">(Pettigrew 2020)</a>.</p><p></p><h2 style="text-align: left;">References</h2><p style="text-align: left;"> </p>Good, I. J. (1967). On the Principle of Total Evidence. <i>The British Journal for the Philosophy of Science</i>, 17, 319–322. <br /><br />Konek, J. (ta). Degrees of incoherence, dutch bookability & guidance value. <i>Philosophical Studies</i>. <br /><br />Levinstein, B. A. (2017). A Pragmatist’s Guide to Epistemic Utility. <i>Philosophy of Science</i>, 84(4), 613–638. <br /><br />Pettigrew, R. (2019). <i>Choosing for Changing Selves</i>. Oxford, UK: Oxford University Press. <br /><br />Pettigrew, R. (2020). <i>Dutch Book Arguments</i>. Elements in Decision Theory and Philosophy. Cambridge, UK: Cambridge University Press. <br /><br />Schervish, M. J. (1989). A general method for comparing probability assessors. <i>The Annals of Statistics</i>, 17, 1856–1879. <br /><br />Staffel, J. (2020). <i>Unsettled Thoughts</i>. Oxford University Press. <p><br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com0tag:blogger.com,1999:blog-4987609114415205593.post-8456054485020270232022-05-05T07:50:00.001+01:002022-07-27T19:08:16.636+01:00Should we agree? I: the arguments for consensus<p>You can find a PDF of this blogpost <a href="https://drive.google.com/file/d/1Z-1IYfOivGOdYCoLo7hYxL28ZcMO2KZl/view?usp=sharing" target="_blank">here</a>.<br /><br />Should everyone agree with everyone else? Whenever two members of a group have an opinion about the same claim, should they both be equally confident in it? If this is sometimes required of groups, of which ones is it required and when? Whole societies at any time in their existence? Smaller collectives when they're engaged in some joint project?<br /><br />Of course, you might think these are purely academic questions, since there's no way we could achieve such consensus even if we were to conclude that it is desirable, but that seems too strong. Education systems and the media can be deployed to push a population towards consensus, and indeed this is exactly how authoritarian states often proceed. Similarly, social sanctions can create incentives for conformity. So it seems that a reasonable degree of consensus might be possible.<br /><br />But is it desirable? In this series of blogposts, I want to explore two formal arguments. They purport to establish that groups should be in perfect agreement; and they explain why getting closer to consensus is better, even if perfect agreement isn't achieved---in this case, a miss is not as good as a mile. It's still a long way from their conclusions to practical conclusions about how to structure a society, but they point sufficiently strongly in a surprising direction that it is worth exploring them. In this first post, I set out the arguments as they have been given in the literature and polish them up a bit so that they are as strong as possible.<br /><br />Since they're formal arguments, they require a bit of mathematics, both in their official statement and in the results on which they rely. But I want to make the discussion as accessible as possible, so, in the main body of the blogpost, I state the arguments almost entirely without formalism. Then, in the technical appendix, I sketch some of the formal detail for those who are interested.<br /></p><h2 style="text-align: left;"><span><a name='more'></a></span>Two sorts of argument for credal norms</h2><p>There are two sorts of argument we most often use to justify the norms we take to govern our credences: there are pragmatic arguments, of which the betting arguments are the most famous; and there are epistemic arguments, of which the epistemic utility arguments are the most well known. <br /><br />Take the norm of Probabilism, for instance, which says that your credences should obey the axioms of the probability calculus. The betting argument for Probabilism is sometimes known as the <i>Dutch Book</i> or <i>sure loss argument</i>.* It begins by claiming that the maximum amount you are willing to pay for a bet on a proposition that pays out a certain amount if the proposition is true and nothing if it is false is proportional to your credence in that proposition. Then it shows that, if your credences do not obey the probability axioms, there is a set of bets each of which they require you to accept, but which when taken together lose you money for sure; and if your credences do obey those axioms, there is no such set of bets.<br /><br />The epistemic utility argument for Probabilism, on the other hand, begins by claiming that any measure of the epistemic value of credences must have certain properties.** It then shows that, by the lights of any epistemic utility function that does have those properties, if your credences do not obey the probability axioms, then there are alternatives that are guaranteed to be have greater epistemic utility than yours; and if they do obey those axioms, there are no such alternatives.<br /><br />Bearing all of this in mind, consider the following two facts. <br /></p><p>(I) Suppose we make the same assumptions about which bets an individual's credences require them to accept that we make in the betting argument for Probabilism. Then, if two members of a group assign different credences to the same proposition, there is a bet the first should accept and a bet the second should accept that, taken together, leave the group poorer for sure (<a href="https://philpapers.org/rec/RYDCOA" target="_blank">Ryder 1981</a>, <a href="https://philpapers.org/rec/GILIPA" target="_blank">Gillies 1991</a>). </p><p>(II) Suppose we measure the epistemic value of credences using an epistemic utility function that boasts the properties required of it by the epistemic utility argument for Probabilism. Then, if two members of a group assign different credences to the same proposition, there is a single credence such that the group is guaranteed to have greater total epistemic utility if every member adopts that single credence in that proposition (<a href="https://philpapers.org/rec/KOPWOT" target="_blank">Kopec 2012</a>).<br /><br />Given the epistemic utility and betting arguments for Probabilism, neither (I) nor (II) is very surprising. After all, one consequence of Probabilism is that an individual must assign the same credence to two propositions that have the same truth value as a matter of logic. But from the point of view of the betting argument or the epistemic utility argument, this is structurally identical to the requirement that two different people assign the same credence to the same proposition, since obviously a single proposition necessarily has the same truth value as itself! However we construct the sure loss bets against the individual who violates the consequence of Probabilism, we can use an analogous strategy to construct the sure loss bets against the pair who disagree in the credences they assign. And however we construct the alternative credences that are guaranteed to be more accurate than the ones that violate the consequence of Probabilism, we can use an analogous strategy to construct the alternative credence that, if adopted by all members of the group that contains two individuals who currently disagree, would increase their total epistemic utility for sure.<br /><br />Just as a betting argument and an epistemic utility argument aim to establish the individual norm of Probabilism, we might ask whether there is a group norm for which we can give a betting argument and an epistemic utility argument by appealing to (I) and (II)? That is the question I'd like to explore in these posts. In the remainder of this post, I'll spell out the details of the epistemic utility argument and the betting argument for Probabilism, and then adapt those to give analogous arguments for Consensus.</p><h2 style="text-align: left;">The Epistemic Utility Argument for Probabilism</h2><p>Two small bits of terminology first:</p><ul style="text-align: left;"><li>Your <i>agenda</i> is the set of propositions about which you have an opinion. We'll assume throughout that all individuals have finite agendas.</li><li>Your <i>credence function</i> takes each proposition in your agenda and returns your credence in that proposition.</li></ul><p>With those in hand, we can state Probabilism</p><p><b>Probabilism</b> Rationality requires of an individual that their credence function is a probability function. </p><p>What does it mean to say that a credence function is a probability function? There are two cases to consider.<br /><br />First, suppose that, whenever a proposition is in your agenda, its negation is as well; and whenever two propositions are in your agenda, their conjunction and their disjunction are as well. When this holds, we say that your agenda is a <i>Boolean algebra</i>. And in that case your credence function is a probability function if two conditions hold: first, you assign the minimum possible credence, namely 0, to any contradiction and the maximum possible credence, namely 1, to any tautology; second, your credence in a disjunction is the sum of your credences in the disjuncts less your credence in their conjunction (just like the number of people in two groups is the number in the first plus the number in the second less the number in both).<br /><br />Second, suppose that your agenda is not a Boolean algebra. In that case, your credence function is a probability function if it is possible to extend it to a probability function on the smallest Boolean algebra that contains your agenda. That is, it's possible to fill out your agenda so that it's closed under negation, conjunction, and disjunction, and then extend your credence function so that it assign credences to those new propositions in such a way that the result is a probability function on the expanded agenda. Defining probability functions on agendas that are not Boolean algebras allows us to say, for instance, that, if your agenda is just <i>It will be windy tomorrow</i> and <i>It will be windy and rainy tomorrow</i>, and you assign credence 0.6 to <i>It will be windy</i> and 0.8 to <i>It will be windy and rainy</i>, then you violate Probabilism because there's no way to assign credences to <i>It won't be windy</i>, <i>It will be windy or rainy</i>, <i>It won't be rainy</i>, etc in such a way that the result is a probability function.<br /><br />The Epistemic Utility Argument for Probabilism begins with three claims about how to measure the epistemic value of a whole credence function. The first is Individual Additivity, which says that the epistemic utility of a whole credence function is simply the sum of the epistemic utilities of the individual credences it assigns. The second is Continuity, which says that, for any proposition, the epistemic utility of a credence in that proposition is a continuous function of that credence. And the third is Strict Propriety, which says that, for any proposition, each credence in that proposition should expect itself to be have greater epistemic utility than it expects any alternative credence in that proposition to have. With this account in hand, the argument then appeals to a mathematical theorem, which tells us two consequences of measuring epistemic value using an epistemic utility function that has the three properties just described, namely, Individual Additivity, Continuity, and Strict Propriety.</p><p>(i) For any credence function that violates Probabilism, there is a credence function defined on the same agenda that satisfies it and that has greater epistemic utility regardless of how the world turns out. In this case, we say that the alternative credence function <i>dominates</i> the original one. </p><p>(ii) For any credence function that is a probability function, there is no credence function that dominates it. Indeed, there is no alternative credence function that is even as good as it at every world. For any alternative, there will be some world where that alternative is strictly worse.</p><p>The argument concludes by claiming that an option is irrational if there is some alternative that is guaranteed to be better and no option that is guaranteed to be better than that alternative.</p><h2 style="text-align: left;">The Epistemic Utility Argument for Consensus</h2><p>As I stated it above, and as it is usually stated in the literature, Consensus says that, whenever two members of a group assign credences to the same proposition, they should assign the same credence. But in fact the epistemic argument in its favour establishes something stronger. Here it is:<b> </b></p><p><b>Consensus</b> Rationality requires of a group that there is a single probability function defined on the union of the agendas of all of the members of the group such that the credence function of each member assigns the same credence to any proposition in their agenda as this probability function does.<br /><br />This goes further than simply requiring that all agents agree on the credence they assign to any proposition to which they all assign credences. Indeed, it would place constraints even on a group whose members' agendas do not overlap at all. For instance, if you have credence 0.6 that it will be rainy tomorrow, while I have credence 0.8 that it will be rainy and windy, the pair of us will jointly violate Consensus, even though we don't assign credences to any of the same propositions, since no probability function assigns 0.6 to one proposition and 0.8 to the conjunction of that proposition with another one. In these cases, we say that the group's credences don't cohere.<br /><br />One notable feature of Consensus is that it purports to govern groups, not individuals, and we might wonder what it could mean to say that a group is irrational. I'll return to that in a later post. It will be useful to have the epistemic utility and betting arguments for Consensus to hand first. <br /><br />The Epistemic Utility Argument for Consensus begins, as the epistemic argument for Probabilism does, with Individual Additivity, Continuity, and Strictly Propriety. And it adds to those Group Additivity, which says that group's epistemic utility is the sum of the epistemic utilities of the credence functions of its members. With this account of group epistemic value in hand, the argument then appeals again to a mathematical theorem, but a different one, which tells us two consequences of Group and Individual Additivity, Continuity, and Strict Propriety:***</p><p>(i) For any group that violates Consensus, there is, for each individual, an alternative credence function defined on their agenda that they might adopt such that, if all were to adopt these, the group would satisfy Consensus and it would be more accurate regardless of how the world turns out. In this case, we say that the alternative credence functions <i>collectively dominate</i> the original ones.</p><p>(ii) For any group that satisfies Consensus, there are no credence functions the group might adopt that collectively dominate it.</p><p>The argument concludes by assuming again the norm that an option is irrational if there is some alternative that is guaranteed to be better.</p><h2 style="text-align: left;">The Sure Loss Argument for Probabilism</h2><p>The Sure Loss Argument for Probabilism begins with a claim that I call <i>Ramsey's Thesis</i>. It tells you the prices at which your credences require you to buy and sell bets. It says that, if your credence in $A$ is $p$, and $£x < £pS$, then you should be prepared to pay $£x$ for a bet that pays out $£S$ if $A$ is true and $£0$ if $A$ is false. And this is true for any stakes $S$, whether positive, negative, or zero. Then it appeals to a mathematical theorem, which tells us two consequences of Ramsey's Thesis.</p><p>(i) For any credence function that violates Probabilism, there is a series of bets, each of which your credences require you to accept, that, taken together, lose you money for sure.</p><p>(ii) For any credence function satisfies Probabilism, there is no such series of bets.</p><p>The argument concludes by assuming a norm that says that it is irrational to have credences that require you to make a series of choices when there is an alternative series of choices you might have made that would be better regardless of how the world turns out.</p><h2 style="text-align: left;">The Sure Loss Argument for Consensus</h2><p>The Sure Loss Argument for Consensus also begins with Ramsey's Thesis. It appeals to a mathematical theorem that tells us two consequences of Ramsey's Thesis.</p><p>(i) For any group that violates Consensus, there is a series of bets, each offered to a member of the group whose credences require that they accept it, that, taken together, lose the group money for sure.</p><p>(ii) For any group that satisfies Consensus, there is no such series of bets.</p><p>And it concludes by assuming that it is irrational for the members of a group to have credences that require them to make a series of choices when there is an alternative series of choices they might have made that would be better for the group regardless of how the world turns out.<br /><br />So now we have the Epistemic Utility and Sure Loss Arguments for Consensus. In fact, I think the Sure Loss Argument doesn't work. So in the next post I'll say why and provide a better alternative based on work by <a href="https://projecteuclid.org/journals/annals-of-statistics/volume-17/issue-4/A-General-Method-for-Comparing-Probability-Assessors/10.1214/aos/1176347398.full">Mark Schervish</a> and <a href="https://philpapers.org/rec/LEVAPG" target="_blank">Ben Levinstein</a>. But in the meantime, here's the technical appendix.</p><h2 style="text-align: left;">Technical appendix</h2><p>First, note that Probabilism is the special case of Consensus when the group has only one member. So we focus on establishing Consensus.<br /><br />Some definitions to begin:</p><ul style="text-align: left;"><li>If $c$ is a credence function defined on the agenda $\mathcal{F}_i = \{A^i_1, \ldots, A^i_{k_i}\}$, represent it as a vector as follows:$$c = \langle c(A^i_1), \ldots, c(A^i_{k_i})\rangle$$</li><li>Let $\mathcal{C}_i$ be the set of credence functions defined on $\mathcal{F}_i$, represented as vectors in this way.</li><li>If $c_1, \ldots, c_n$ are credence functions defined on $\mathcal{F}_1, \ldots, \mathcal{F}_n$ respectively, represent them collectively as a vector as follows:<br />$$<br />c_1 \frown \ldots \frown c_n = \langle c_1(A^1_1), \ldots, c_1(A^1_{k_1}), \ldots, c_n(A^n_1), \ldots, c_n(A^n_{k_n}) \rangle<br />$$</li><li>Let $\mathcal{C}$ be the set of sequences of credence functions defined on $\mathcal{F}_1, \ldots, \mathcal{F}_n$ respectively, represented as vectors in this way. </li><li>If $w$ is a classically consistent assignment of truth values to the propositions in $\mathcal{F}_i$, represent it as a vector $$w = \langle w(A^i_1), \ldots, w(A^i_{k_i})\rangle$$ where $w(A) = 1$ if $A$ is true according to $w$, and $w(A) = 0$ if $A$ is false according to $w$.</li><li>Let $\mathcal{W}_i$ be the set of classically consistent assignments of truth values to the propositions in $\mathcal{F}_i$, represented as vectors in this way.</li><li>If $w$ is a classically consistent assignment of truth values to the propositions in $\mathcal{F} = \bigcup^n_{i=1} \mathcal{F}_i$, represent the restriction of $w$ to $\mathcal{F}_i$ by the vector $$w_i = \langle w(A^i_1), \ldots, w(A^i_{k_i})\rangle$$So $w_i$ is in $\mathcal{W}_i$. And represent $w$ as a vector as follows:<br />$$<br />w = w_1 \frown \ldots \frown w_n = \langle w(A^1_1), \ldots, w(A^1_{k_1}), \ldots, w(A^n_1), \ldots, w(A^n_{k_n})\rangle<br />$$</li><li>Let $\mathcal{W}$ be the set of classical consistent assignments of truth values to the propositions in $\mathcal{F}$, represented as vectors in this way.</li></ul><p>Then we have the following result, which generalizes a result due to <a href="https://philpapers.org/rec/DEFTOP" target="_blank">de Finetti (1974)</a>:<br /><br /><b>Proposition 1</b> A group of individuals with credence functions $c_1, \ldots, c_n$ satisfy Consensus iff $c_1 \frown \ldots \frown c_n$ is in the closed convex hull of $\mathcal{W}$.<br /><br />We then appeal to two sets of results. First, concerning epistemic utility measures, which generalizes a result to <a href="https://philpapers.org/rec/PREPCA">Predd, et al. (2009)</a>:<br /></p><p><b>Theorem 1<br /></b></p><p>(i) Suppose $\mathfrak{A}_i : \mathcal{C}_i \times \mathcal{W}_i \rightarrow [0, 1]$ is a measure of epistemic utility that satisfies Individual Additivity, Continuity, and Strict Propriety. Then there is a Bregman divergence $\mathfrak{D}_i : \mathcal{C}_i \times \mathcal{C}_i \rightarrow [0, 1]$ such that $\mathfrak{A}_i(c, w) = -\mathfrak{D}_i(w, c)$.</p><p>(ii) Suppose $\mathfrak{D}_1, \ldots, \mathfrak{D}_n$ are Bregman divergences defined on $\mathcal{C}_1, \ldots, \mathcal{C}_n$, respectively. And suppose $\mathcal{X}$ is a closed convex subset of $\mathcal{C}$. And suppose $c_1 \frown \ldots \frown c_n$ is not in $\mathcal{X}$. Then there is $c^\star_1 \frown \ldots \frown c^\star_n$ in $\mathcal{Z}$ such that, for all $z_1 \frown \ldots \frown z_n$ in $\mathcal{Z}$,<br />$$<br />\sum^n_{i=1} \mathfrak{D}_i(z_i, c^\star_i) < \sum^n_{i=1} \mathfrak{D}_i(z_i, c_i)<br />$$<br /><br />So, by Proposition 1, if a group $c_1, \ldots, c_n$ does not satisfy Consensus, then $c_1 \frown \ldots \frown c_n$ is not in the closed convex hull of $\mathcal{W}$, and so by Theorem 1 there is $c^\star_1 \frown \ldots \frown c^\star_n$ in the closed convex hull of $\mathcal{W}$ such that, for all $w$ in $\mathcal{W}$, $$\mathfrak{A}_i(c, w) < \mathfrak{A}(c^\star, w)$$ as required.<br /><br />Second, concerning bets, which is a consequence of the Separating Hyperplane Theorem:</p><p><b>Theorem 2</b><br />Suppose $\mathcal{Z}$ is a closed convex subset of $\mathcal{C}$. And suppose $c_1 \frown \ldots \frown c_n$ is not in $\mathcal{Z}$. Then there are vectors<br />$$<br />x = \langle x^1_1, \ldots, x^1_{k_1}, \ldots, x^n_1, \ldots, x^n_{k_n}\rangle<br />$$<br />and<br />$$<br />S = \langle S^1_1, \ldots, S^1_{k_1}, \ldots, S^n_1, \ldots, S^n_{k_n}\rangle<br />$$<br />such that, for all $x^i_j$ and $S^i_j$,<br />$$<br />x^i_j < c_i(A^i_j)S^i_j<br />$$<br />and, for all $z$ in $\mathcal{Z}$,<br />$$<br />\sum^n_{i=1} \sum^{k_i}_{j = 1} x^i_j > \sum^n_{i=1} \sum^{k_i}_{j=1} z^i_jS^i_j<br />$$<br /><br />So, by Proposition 1, if a group $c_1, \ldots, c_n$ does not satisfy Consensus, then $c_1 \frown \ldots \frown c_n$ is not in the closed convex hull of $\mathcal{W}$, and so, by Theorem 2, there is $x = \langle x^1_1, \ldots, x^1_{k_1}, \ldots, x^n_1, \ldots, x^n_{k_n}\rangle$ and $S = \langle S^1_1, \ldots, S^1_{k_1}, \ldots, S^n_1, \ldots, S^n_{k_n}\rangle$ such that (i) $x^i_j < c_i(A^i_j)S^i_j$ and (ii) for all $w$ in $\mathcal{W}$, <br />$$\sum^n_{i=1} \sum^{k_i}_{j = 1} x^i_j > \sum^n_{i=1} \sum^{k_i}_{j=1} w(A^i_j)S^i_j$$<br />But then (i) says that the credences of individual $i$ require them to pay $£x^i_j$ for a bet on $A^i_j$ that pays out $£S^i_j$ if $A^i_j$ is true and $£0$ if it is false. And (ii) says that the total price of these bets across all members of the group---namely, $£\sum^n_{i=1} \sum^{k_i}_{j = 1} x^i_j$---is greater than the amount the bets will payout at any world---namely, $£\sum^n_{i=1} \sum^{k_i}_{j=1} w(A^i_j)S^i_j$.</p><p></p><p>* This was introduced independently by <a href="https://philpapers.org/rec/RAMTAP" target="_blank">Frank P. Ramsey (1931)</a> and <a href="https://philpapers.org/rec/DEFLPS">Bruno de Finetti (1937)</a>. For overviews, see (<a href="https://philpapers.org/rec/GENDBA" target="_blank">Hajek 2008</a>, <a href="https://philpapers.org/rec/VINDBA" target="_blank">Vineberg 2016</a>, <a href="https://philpapers.org/rec/PETTDB">Pettigrew 2020</a>).</p><p>**Much of the discussion of these arguments in the literature focusses on versions on which the epistemic value of a credence is taken to be its accuracy. This literature begins with <a href="https://philpapers.org/rec/ROSFAA-3" target="_blank">Rosenkrantz (1981)</a> and <a href="https://philpapers.org/rec/JOYANV" target="_blank">Joyce (1998)</a>. But, following <a href="https://philpapers.org/rec/JOYAAC" target="_blank">Joyce (2009)</a> and <a href="https://philpapers.org/rec/PREPCA" target="_blank">Predd (2009)</a>, it has been appreciated that we need not necessarily assume that accuracy is the only source of epistemic value in order to get the argument going. <br /></p><p>*** <a href="https://philpapers.org/rec/KOPWOT" target="_blank">Matthew Kopec (2012)</a> offers a proof of a slightly weaker result. It doesn't quite work because it assumes that all strictly proper measures of epistemic value are convex, when they are not---the spherical scoring rule is not. I offer an alternative proof of this stronger result in the technical appendix below.</p><p></p><h1 style="text-align: left;">References</h1><p style="text-align: left;"> </p><div class="page" title="Page 9"> <div class="layoutArea"><div class="column">de Finetti, B. (1937 [1980]). Foresight: Its Logical Laws, Its Subjective Sources. In H. E. Kyburg, & H. E. K. Smokler (Eds.) <i>Studies in Subjective Probability</i>. Huntingdon, N. Y.: Robert E. Kreiger Publishing Co. <br /><br />de Finetti, B. (1974). <i>Theory of Probability</i>, vol. I. New York: John Wiley & Sons.<br /><br />Gillies, D. (1991). Intersubjective probability and confirmation theory. <i>The British Journal for the Philosophy of Science</i>, 42(4), 513–533. <br /><br />Hájek, A. (2008). Dutch Book Arguments. In P. Anand, P. Pattanaik, & C. Puppe (Eds.) <i>The Oxford Handbook of Rational and Social Choice</i>, (pp. 173–195). Oxford: Oxford University Press. <br /><br />Joyce, J. M. (1998). A Nonpragmatic Vindication of Probabilism. <i>Philosophy of Science</i>, 65(4), 575–603.<br /><br />Joyce, J. M. (2009). Accuracy and Coherence: Prospects for an Alethic Epistemology of Partial Belief. In F. Huber, & C. Schmidt-Petri (Eds.) <i>Degrees of Belief</i>. Dordrecht and Heidelberg: Springer. <br /><br />Kopec, M. (2012). We ought to agree: A consequence of repairing Goldman’s group scoring rule. <i>Episteme</i>, 9(2), 101–114.<br /><br />Pettigrew, R. (2020). <i>Dutch Book Arguments</i>. Cambridge University Press. <br /><br />Predd, J., Seiringer, R., Lieb, E. H., Osherson, D., Poor, V., & Kulkarni, S. (2009). Probabilistic Coherence and Proper Scoring Rules. <i>IEEE Transactions of Information Theory</i>, 55(10), 4786–4792.<br /><br />Ramsey, F. P. (1926 [1931]). Truth and Probability. In R. B. Braithwaite (Ed.) <i>The Foundations of Mathematics and Other Logical Essays</i>, chap. VII, (pp. 156–198). London: Kegan, Paul, Trench, Trubner & Co. <br /><br />Rosenkrantz, R. D. (1981). <i>Foundations and Applications of Inductive Probability</i>. Atascadero, CA: Ridgeview Press.<br /><br />Ryder, J. (1981). Consequences of a simple extension of the Dutch Book argument. <i>The British Journal for the Philosophy of Science</i>, 32(2), 164–167. <br /><br />Vineberg, S. (2016). Dutch Book Arguments. In E. N. Zalta (Ed.) <i>Stanford Encyclopedia of Philosophy</i>. Metaphysics Research Lab, Stanford University. </div></div></div><div class="page" title="Page 10"><div class="layoutArea"> </div> </div> <p><br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com0tag:blogger.com,1999:blog-4987609114415205593.post-42830170852517456922021-04-13T11:14:00.003+01:002022-07-27T19:09:22.014+01:00What we together risk: three vignettes in search of a theory<p>For a PDF version of this post, see <a href="https://drive.google.com/file/d/10fQudSUQKI2Ut59tEJFjg03aLpusi5MY/view?usp=sharing" target="_blank">here</a>.<br /><br />Many years ago, I was climbing Sgùrr na Banachdich with my friend Alex. It's a mountain in the Black Cuillin, a horseshoe of summits that surround Loch Coruisk at the southern end of the Isle of Skye. It's a Munro---that is, it stands over 3,000 feet above sea level---but only just---it measures 3,166 feet. About halfway through our ascent, the mist rolled in and the rain came down heavily, as it often does near these mountains, which attract their own weather system. At that point, my friend and I faced a choice: to continue our attempt on the summit or begin our descent. Should we continue, there were a number of possible outcomes: we might reach the summit wet and cold but not injured, with the mist and rain gone and in their place sun and views across to Bruach na Frìthe and the distinctive teeth-shaped peaks of Sgùrr nan Gillean; or we might reach the summit without injury, but the mist might remain, obscuring any view at all; or we might get injured on the way and either have to descend early under our own steam or call for help getting off the mountain. On the other hand, should we start our descent now, we would of course have no chance of the summit, but we were sure to make it back unharmed, for the path back is good and less affected by rain.</p><p><span></span></p><a name='more'></a>Alex and I had climbed together a great deal that summer and the summer before. We had talked at length about what we enjoyed in climbing and what we feared. To the extent that such comparisons make sense and can be known, we both knew that we both gained exactly the same pleasure from reaching a summit, the same additional pleasure if the view was clear; we gained the same displeasure from injury, the same horror at the thought of having to call for assistance getting off a mountain. What's more, we both agreed exactly on how likely each possible outcome was: how likely we were to sustain an injury should we persevere; how likely that the mist would clear in the coming few hours; and so on. Nonetheless, I wished to turn back, while Alex wanted to continue. <br /><br />How could that be? We both agreed how good or bad each of the options was, and both agreed how likely each would be were we to take either of the courses of action available to us. Surely we should therefore have agreed on which course of action would maximise our expected utility, and therefore agreed which would be best to undertake. Yes, we did agree on which course of action would maximise our expected utility. However, no, we did not therefore agree on which was best, for there are theories of rational decision-making that do not demand that you must rank options by their expected utility. These are the risk-sensitive decision theories, and they include John Quiggin's rank-dependent decision theory and Lara Buchak's risk-weighted expected utility theory. According to Quiggin's and Buchak's theories, what you consider best is not determined only by your utilities and your probabilities, but also by your attitudes to risk. The more risk-averse will give greater weight to the worst-case scenarios and less to the best-case ones than expected utility demands; the more risk-inclined will give greater weight to the best outcomes and less to the worst than expected utility does; and the risk-neutral person will give exactly the weights prescribed by expected utility theory. So, perhaps I preferred to begin our descent from Sgùrr na Banachdich while Alex preferred to continue upwards because I was risk-averse and he was risk-neutral or risk-seeking, or I was risk-neutral and he was risk-seeking. In any case, he must have been less risk-averse than I was. <br /><br />Of course, as it turned out, we sat on a mossy rock in the rain and discussed what to do. We decided to turn back. Luckily, as it happened, for a thunderstorm hit the mountains an hour later at just the time we'd have been returning from the summit. But suppose we weren't able to discuss the decision. Suppose we'd roped ourselves together to avoid getting separated in the mist, and he'd taken the lead, forcing him to make the choice on behalf of both of us. In that case, what should he have done?<br /><br />As I will do throughout these reflections, let me simply report by own reaction to the case. I think, in that case, Alex should have chosen to descend (and not only because that was my preference---I'd have thought the same had it been he who wished to descend and me who wanted to continue!). Had he chosen to continue---even if all had turned out well and we'd reached the summit unharmed and looked over the Cuillin ridge in the sun---I would still say that he chose wrongly on our behalf. This suggests the following principle (in joint work, Ittay Nissan Rozen and Jonathan Fiat argue for a version of this principle that applies in situations in which the individuals do not assign the same utilities to the outcomes):<p></p><span><!--more--></span><span><!--more--></span><p><b>Principle 1</b> Suppose two people assign the same utilities to the possible outcomes, and assign the same probabilities to the outcomes conditional on choosing a particular course of action. And suppose that you are required to choose between those courses of action on their behalf. Then you must choose whatever the more risk-averse of the two would choose.<br /><br />However, I think the principle is mistaken. A few years after our unsuccessful attempt on Sgùrr na Banachdich, I was living in Bristol and trying to decide whether to take up a postdoctoral fellowship there or a different one based in Paris (a situation that seems an unimaginable luxury and privilege when I look at today's academic job market). Staying in Bristol was the safe bet; moving to Paris was a gamble. I already knew what it would be like to live in Bristol and what the department was like. I knew I'd enjoy it a great deal. I'd visited Paris, but I didn't know what it would be like to live there, and I knew the philosophical scene even less. I knew I'd enjoy living there, but I didn't know how much. I figured I might enjoy it a great deal more than Bristol, but also I might enjoy it somewhat less. The choice was complicated because my partner at the time would move too, if that's what we decided to do. Fortunately, just as Alex and I agreed on how much we valued the different outcomes that faced us on the mountain, so my partner and I agreed on how much we'd value staying in Bristol, how much we'd value living in Paris under the first, optimistic scenario, and how much we'd value living there under the second, more pessimistic scenario. We also agreed how likely the two Parisian scenarios were---we'd heard the same friends describing their experiences of living there, and we'd drawn the same conclusions about how likely we were to value the experience ourselves to different extents. Nonetheless, just as Alex and I had disagreed on whether or not to start our descent despite our shared utilities and probabilities, so my partner and I disagreed on whether or not to move to Paris. Again the more risk-averse of the two, I wanted to stay in Bristol, while he wanted to move to Paris. Again, of course, we sat down to discuss this. But suppose that hadn't been possible. Perhaps my partner had to make the decision for both of us at short notice and I was not available to consult. How should he have chosen?<br /><br />In this case, I think either choice would have been permissible. My partner might have chosen Paris or he might have chosen Bristol and either of these would have been allowed. But of course this runs contrary to Principle 1. <br /><br />So what is the crucial difference between the decision on Sgùrr na Banachdich and the decision whether to move cities? In each case, there is an option---beginning our descent or staying in Bristol---that is certain to have a particular level of value; and there is an alternative option---continuing to climb or moving to Paris---that might give less value than the sure thing, but might give more. And, in each case, the more risk-averse person prefers the sure thing to the gamble, while the more risk-inclined prefers the gamble. So why must someone choosing for me and Alex in the first case choose to descend, while someone choosing for me and my partner in the second case choose either Bristol or Paris? <br /><br />Here's my attempt at a diagnosis: in the choice of cities, there is no risk of harm, while in the decision on the mountain, there is. In the first case, the gamble opens up a possible outcome in which we're harmed---we are injured, perhaps quite badly. In the second case, the gamble doesn't do that---we countenance the possibiilty that moving to Paris might not be as enjoyable as remaining in Bristol, but we are certain it won't harm us! This suggests the following principle:</p><p><b>Principle 2</b> Suppose two people assign the same utilities to the possible outcomes, and assign the same probabilities to the outcomes conditional on choosing a particular course of action. And suppose that you are required to choose between those courses of action on their behalf. Then there are two cases: if one of the available options opens the possibility of a harm, then you must choose whatever the more risk-averse of the two would choose; if neither of the available options opens the possibility of a harm, then you may choose an option if at least one of the two would choose it. </p><p>So risk-averse preferences do not always take precedence, but they do when harms are involved. Why might that be? <br /><br />A natural answer: to expose someone to the risk of a harm requires their consent. That is, when there is an alternative option that opens no possibility of harm, you are only allowed to choose an option that opens up the possibility of a harm if everyone affected would consent to being subject to that risk. So Alex should only choose to continue our ascent and expose us to the risk of injury if I would consent to that, and of course I wouldn't, since I'd prefer to descend. But my partner is free to choose the move to Paris even though I wouldn't choose that, because it exposes us to no risk of harm. <br /><br />A couple of things to note: First, in our explanation, reference to risk-aversion, risk-neutrality, and risk-inclination have dropped out. What is important is not who is more averse to risk, but who consents to what. Second, our account will only work if we employ an absolute notion of harm. That is, I must say that there is some threshold and an option harms you if it causes your utility to fall below that threshold. We cannot use a relative notion of harm on which an option harms you if it merely causes your utility to fall. After all, using a relative notion of harm, the move to Paris will harm you should it turn out to be worse than staying in Bristol. <br /><br />The problem with Principle 2 and the explanation we have just given is that it does not generalise to cases in which more than two people are involved. That is, the following principle seems false:</p><p><b>Principle 3</b> Suppose each member of a group of people assign the same utilities to the possible outcomes, and assign the same probabilities to the outcomes conditional on choosing a particular course of action. And suppose that you are required to choose between those courses of action on their behalf. Then there are two cases: if one of the available options opens the possibility of a harm, then you must choose whatever the most risk-averse of them would choose; if neither of the available options opens the possibility of a harm, then you may choose an option if at least one member of the group would choose it.</p><p>A third vignette might help to illustrate this.<br /><br />I grew up between two power stations. My high school stood in the shadow of the coal-fired plant at Cockenzie, while the school where my mother taught stood in the lee of the nuclear plant at Torness Point. And I was born two years after the Three Mile Island accident and the Chernobyl tragedy happened as I started school. So the risks of nuclear power were somewhat prominent growing up. Now, let's imagine a community of five million people who currently generate their energy from coal-fired plants---a community like Scotland in 1964, just before its first nuclear plant was constructed. This community is deciding whether to build nuclear plants to replace its coal-fired ones. All agree that having a nuclear plant that suffered no accidents would be vastly preferable to having coal plants, and all agree that a nuclear plant that suffered an accident would be vastly worse than the coal plants. And we might imagine that they also all assign the same probability to the prospective nuclear plants suffering an accident---perhaps they all defer to a recent report from the country's atomic energy authority. But, while they agree on the utilities and the probabilities, they have don't all have the same attitudes to risk. In the end, 4.5million people prefer to build the nuclear facilities, while half a million, who are more risk-averse, prefer to retain the coal-fired alternatives. Principle 3 says that, for someone choosing on behalf of this population, the only option they can choose is to retain the coal-fired plants. After all, a nuclear accident is clearly a harm, and there are individuals who would suffer that harm who would not consent to being exposed to the risk. But surely that's wrong. Surely, despite such opposition, it would be acceptable to build the nuclear plant.<br /><br />So, while Principle 2 might yet be true, Principle 3 is wrong. And I think my attempt to explain the basis of Principle 2 must be wrong as well, for if it were right, it would also support Principle 3. After all, in no other case I can think of in which a lack of consent is sufficient to block an action does that block disappear if there are sufficiently many people in favour of the action. <br /><br />So what general principles underpin our reactions to these three vignettes? Why do the preferences of the more risk-averse individuals carry more weight when one of the outcomes involves a harm than when they don't, but not enough weight to overrule a significantly greater number of more risk-inclined individuals? That's the theory I'm in search of here.<br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com60tag:blogger.com,1999:blog-4987609114415205593.post-37717036127288502112021-04-06T08:04:00.003+01:002022-07-27T19:09:46.461+01:00Believing is said of groups in many ways<p style="text-align: left;">For a PDF version of this post, see <a href="https://drive.google.com/file/d/1x1LgnTfosyWO6BUpPHU1W9xCJ5Hl0Lqe/view?usp=sharing" target="_blank">here</a>. <br /></p><h2 style="text-align: left;">In defence of pluralism <br /></h2><p>Recently, after a couple of hours discussing a problem in the philosophy of mathematics, a colleague mentioned that he wanted to propose a sort of pluralism as a solution. We were debating the foundations of mathematics, and he wanted to consider the claim that there might be no single unique foundation, but rather many different foundations, no one of them better than the others. Before he did so, though, he wanted to preface his suggestion with an apology. Pluralism, he admitted, is unpopular wherever it is proposed as a solution to a longstanding philosophical problem. </p><p><span></span></p><a name='more'></a>I agree with his sociological observation. Philosophers tend to react badly to pluralist solutions. But why? And is the reaction reasonable? This is pure speculative generalisation based on my limited experience, but I've found that the most common source of resistance is a conviction that there is a particular special role that the concept in question must play; and moreover, in that role, whether or not something falls under the concept determines some important issue concerning it. So, in the philosophy of mathematics, you might think that a proof of a mathematical proposition is legitimate just in case it can be carried out in the system that provides the foundation for mathematics. And, if you allow a plurality of foundations of differing logical strength, the legitimacy of certain proof becomes indeterminate---relative to some foundations, they're legit; relative to others, they aren't. Similarly, you might think that a person who accidentally poisons another person is innocent of murder if, and only if, they were justified in their belief that the liquid they administered was not poisonous. And, if you allow a plurality of concepts of justification, then whether or not the person is innocent might become indeterminate. <p></p><p>I tend to respond to such concerns in two ways. First, I note that, while the special role that my interlocutor picks out for the concept we're discussing is certainly among the roles that this concept needs to play, it isn't the only one; and it is usually not clear why we should take it to be the most important one. One role for a foundation of mathematics is to test the legitimacy of proofs; but another is to provide a universal language that mathematicians might use, and that might help them discover new mathematical truths (see <a href="https://philpapers.org/rec/MARCTA" target="_blank">this paper</a> by Jean-Pierre Marquis for a pluralist approach that takes both of these roles seriously).</p><p>Second, I note that we usually determine the important issues in question independently of the concept and then use our determinations to test an account of the concept, not the other way around. So, for instance, we usually begin by determining whether we think a particular proof is legitimate---perhaps by asking what it assumes and whether we have good reason for believing that those assumptions are true---and then see whether a particular foundation measures up by asking whether the proof can be carried out within it. We don't proceed the other way around. And we usually determine whether or not a person is innocent independently of our concept of justification---perhaps just by looking at the evidence they had and their account of the reasoning they undertook---and then see whether a particular account of justification measures up by asking whether the person is innocent according to it. Again, we don't proceed the other way around.</p><p>For these two reasons, I tend not to be very moved by arguments against pluralism. Moreover, while it's true that pluralism is often greeted with a roll of the eyes, there are a number of cases in which it has gained wide acceptance. We no longer talk of the <i>probability</i> of an event but distinguish between its <i>chance</i> of occurring, a particular individual's <i>credence</i> in it occurring, and perhaps even it's <i>evidential probability</i> relative to a body of evidence. That is, we are pluralists about probability. Similarly, we no longer talk of a particular belief being <i>justified simpliciter</i>, but distinguish between <i>propositional</i>, <i>doxastic</i>, and <i>personal justification</i>. We are, along some dimensions at least, pluralists about justification. We no longer talk of a person having a <i>reason</i> to choose one thing rather than another, but distinguish between their <i>internal</i> and <i>external reasons</i>. </p><p>I want to argue that we should extend pluralism to so-called <i>group beliefs</i> or<i> collective beliefs</i>. Britain believes lockdowns are necessary to slow the virus. Scotland believes it would fare well economically as an independent country. The University believes the pension fund has been undervalued and requires no further increase in contributions in the near future to meet its obligations in the further future. In 1916, Russia believed Rasputin was dishonest. In each of these sentences, we seem to ascribe a belief to a group or collective entity. When is it correct to do this? I want to argue that there is no single answer. Rather, as Aristotle said of being, believing is said of groups in many ways---that is, a pluralist account is appropriate.</p><p>I've been thinking about this recently because I've been reading Jennifer Lackey's fascinating new book, <i><a href="https://global.oup.com/academic/product/the-epistemology-of-groups-9780199656608" target="_blank">The Epistemology of Groups</a></i> (all page numbers in what follows refer to that). In it, Lackey offers an account of group belief, justified group belief, group knowledge, and group assertion. I'll focus here only on the first.</p><h2 style="text-align: left;">Lackey's treatment of group belief</h2><h3 style="text-align: left;">Three accounts of group belief <br /></h3><p style="text-align: left;">Lackey considers two existing accounts of group belief as well as her own proposal. </p><p style="text-align: left;">The first, due to <a href="https://philpapers.org/rec/GILOSF-2" target="_blank">Margaret Gilbert</a> and with amendments by <a href="https://philpapers.org/rec/TUOGB" target="_blank">Raimo Tuomela</a>, is a non-summative account that treats groups as having 'a mind of their own'. Lackey calls it the <i>Joint Acceptance Account</i> (<i>JAA</i>). I'll stick with the simpler Gilbert version, since the points I'll make don't rely on Tuomela's more involved amendment (24):</p><p style="text-align: left;"><b>JAA</b> A group $G$ believes that $p$ iff it is common knowledge in $G$ that the members of $G$ individually have intentionally and openly expressed their willingness jointly to accept that $p$ with the other members of $G$.</p><p style="text-align: left;">The second, due to Philip Pettit, is a summative account that treats group belief as strongly linked to individual belief. Lackey calls it the <i>Premise-Based Aggregation Account</i> (PBAA) (29). Here's a rough paraphrase:</p><p style="text-align: left;"><b>PBAA</b> A group $G$ believes that $p$ iff there is some collection of propositions $q_1, \ldots, q_n$ such that (i) it is common knowledge among the operative members of $G$ that $p$ is true iff each $q_i$ is true, (ii) for each operative member of $G$, they believe $p$ iff they believe each $q_i$, and (iii) for each $q_i$, the majority of operative members of $G$ believe $q_i$.</p><p style="text-align: left;">Lackey's own proposal is the <i>Group Agent Account</i> (GAA) (48-9):</p><p style="text-align: left;"><b>GAA</b> A group $G$ believes that $p$ iff (i) there is a significant percentage of $G$'s operative members who believe that $p$, and (ii) are such that adding together the bases of their beliefs that $p$ yields a belief set that is not substantively incoherent.</p><h3 style="text-align: left;">Group lies (and bullshit) and judgment fragility: two desiderata for accounts of group belief<br /></h3><p style="text-align: left;">To distinguish between these three accounts, Lackey enumerates four desiderata for accounts of group belief that she takes to tell against JAA and PBAA and in favour of GAA. The first three are related to an objection to Gilbert's account of group belief that was developed by <a href="https://philpapers.org/rec/WRACBA-2" target="_blank">K. Brad Wray</a>, <a href="https://philpapers.org/rec/MEICAA-2" target="_blank">A. W. M. Meijers</a>, and <a href="https://philpapers.org/rec/HAKOTP" target="_blank">Raul Hakli</a> in the 2000s. According to this, JAA makes it too easy for groups to actively, consciously, and intentionally choose what they believe: all they need to do is intentionally and openly express their willingness jointly to accept the proposition in question. Lackey notes two consequences of this: (a) on such an account, it is difficult to give a satisfactory account of group lies (or group bullshit, though I'll focus on group lies); (b) on such an account, whether or not a group believes something at a particular time is sensitive to the group's situation at that time in a way that beliefs should not be sensitive.</p><p style="text-align: left;">So Lackey's first desideratum for an account of group belief is that it must be able to accommodate a plausible account of group lies (and the second that it accommodate group bullshit, but as I said I'll leave that for now). Suppose each member of a group strongly believes $p$ on the basis of excellent evidence that they all share, but they also know that the institution will be culpable of a serious crime if it is taken to believe $p$. Then they might jointly agree to accept $\neg p$. And, if they do, Gilbert must say that they do believe $\neg p$. But were they to assert $\neg p$, we would take the group to have lied, which would require that it believes $p$. The point is that, if a group's belief is so thorougly within its voluntary control, it can manipulate it whenever it likes in order to avoid ever lying in situations in which dishonesty would be subject to censure. </p><p style="text-align: left;">Lackey's third desideratum for an account of group belief is that such belief should not be rendered sensitive in certain ways to the situation in which the group formed it. Suppose that, on the basis of the same shared evidence, a substantial majority of members of a group judge the horse Cisco most likely to win the race, the horse Jasper next most likely, and the horse Whiskey very unlikely to win. But, again on the basis of this same shared body of evidence, the remaining minority of members judge Whiskey most likely to win, Jasper next most likely, and Cisco very unlikely to win. The group would like a consensus before it reports its opinion, but time is short---the race is about to begin, say, and the group has been asked for its opinion before the starting gates open. So, in order to achieve something close to a consensus, it unanimously agrees to accept that Jasper will win, even though he is everyone's second favourite. Yet we might also assume that, had time not been short, the majority would have been able to persuade the minority of Cisco's virtues; and, in that case, they'd unanimously agree to accept that Cisco will win. So, according to Gilbert's account, under time pressure, the group believes Jasper will win, while with world enough and time, they would have believed that Cisco will win. Lackey holds that no account of group belief should make it sensitive to the situation in which it is formed in this way, and thus rejects JAA.</p><p style="text-align: left;">Lackey argues that any account of group belief must satisfy the two desiderata we've just considered. I agree that we need at least one account of group belief that satisfies the first desideratum, but I'm not convinced that all need do this---but I'll leave that for later, when I try to motivate pluralism. For now, I'd like to explain why I'm not convinced that any account needs to satisfy the second desideratum. After all, we know from various empirical studies in social psychology, as well as our experience as thinkers and reasoners and believers, that our ordinary beliefs as individuals are sensitive to the situation in which they're formed in just the sort of way that Lackey wishes to rule out for the beliefs of groups. One of the central theses of Amos Tversky and Daniel Kahneman's work is that we use a different reasoning system when we are forced to make a judgment under time pressure from the one we use when more time is available. So, when my implicit biases are mobilised under time pressure, I might come to believe that a particular job candidate is incompetent, while I might judge them to be competent were I to have more time to assess their track record and override my irrational hasty judgment. And, whenever we are faced with a complex body of evidence that, on the face of it, seems to point in one direction, but which, under closer scrutiny, points in the opposite direction, we will form a different belief if we must do so under time pressure than if we have greater leisure to unpick and balance the different components of the evidence. If individual beliefs can be sensitive to the situation in which they're formed in this way, I see no reason why group beliefs might not also be sensitive in this way.</p><p style="text-align: left;">Before moving on, I'd like to consider whether the PBAA---Pettit's premise-based aggregation account---satisfies Lackey's first desideratum. If it doesn't, it can't be for the same reason that Gilbert's JAA doesn't. After all, according to the PBAA, the group's belief is no more under its voluntary control than the beliefs of its individual members. If, for each $q_i$, a majority believes $q_i$, then the group believes $p$. The only way a group could manipulate its belief is by manipulating the beliefs of its members. But if that sort of manipulation rules out a group belief, Lackey's account is just as vulnerable.</p><p style="text-align: left;">So why does Lackey think that PBAA cannot adequately account for group lies. She considers a case in which the three board members of a tobacco company know that smoking is safe to health iff it doesn't cause lung cancer and it doesn't cause emphysema and it doesn't cause heart disease. The first member believes it doesn't cause lung cancer or heart disease, but believes it does cause emphysema, and so believes it is not safe to health; the second believes it doesn't cause emphysema or heart disease, but it does cause lung cancer, and so believes it is not safe to health; and the third believes it doesn't cause lung cancer or emphysema, but it does cause heart disease, and so believes it is not safe to health. The case is illustrated in Table 1. </p><p style="text-align: left;"></p><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-YHIjnYssXPM/YGs0NjHQ1NI/AAAAAAAAFIQ/KuXsSeEASn4o8QovUJIdQqZMsbfAfNr1QCLcBGAsYHQ/s2048/IMG_1244.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1389" data-original-width="2048" height="271" src="https://1.bp.blogspot.com/-YHIjnYssXPM/YGs0NjHQ1NI/AAAAAAAAFIQ/KuXsSeEASn4o8QovUJIdQqZMsbfAfNr1QCLcBGAsYHQ/w400-h271/IMG_1244.jpg" width="400" /></a></div><br />Then each board member believes it is not safe to health, but PBAA says that it is, because a majority (first and third) believe it doesn't cause lung cancer, a majority (second and third) believe it doesn't cause emphysema, and a majority (first and second) believe it doesn't cause heart disease. If the company then asserts that it is safe to health, then Lackey claims that it lies, while PBAA says that it believes the proposition it asserts and so does not lie.<p></p><p>I think this case is a bit tricky. I suspect our reaction to it is influenced by our knowledge of how the real-world version played out and the devastating effect it has had. So let us imagine that this group of three is not the board of a tobacco company, but the scientific committee of a public health organisation. The structure of the case will be exactly the same, and the nature of the organisation should not affect whether or not belief is present. Now suppose that, since the stakes are so high, each member would only come to believe of a specific putative risk that it is not present if their credence that it is not present is above 95%. That is, there is some pragmatic encroachment here to the extent that the threshold for belief is determined in part by the stakes involved. And suppose further that the first member of the scientific committee has credence 99% that smoking doesn't cause lung cancer, 99% that it doesn't cause heart disease, and 93% that it doesn't cause emphysema. And let's suppose that, by a tragic bout of bad luck that has bestowed on them very misleading evidence, the evidence available to them supports these credences. Then their credence that smoking is safe to health must be at most 93%---since the probability of a conjunction must be at most the probability of any of the conjuncts---and thus below 95%. So the first member doesn't believe it is safe to health. And suppose the same for the other two members of the committee, but for the other combinations of risks. So the second is 99% sure it doesn't cause emphysema and 99% sure it doesn't cause heart disease, but only 93% sure it doesn't cause lung cancer. And the third is 99% sure it doesn't cause lung cancer and 99% sure it doesn't cause emphysema, but only 93% sure it doesn't cause heart disease. So none of the three believe that smoking is safe to health. The case is illustrated in Table 2. </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-FZhr93B6w1s/YGs0jcs0hII/AAAAAAAAFIY/FRWSozNrqfUiYhEQBqPOuTuBIVBON82rgCLcBGAsYHQ/s2048/IMG_1243.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1384" data-original-width="2048" height="270" src="https://1.bp.blogspot.com/-FZhr93B6w1s/YGs0jcs0hII/AAAAAAAAFIY/FRWSozNrqfUiYhEQBqPOuTuBIVBON82rgCLcBGAsYHQ/w400-h270/IMG_1243.jpg" width="400" /></a></div><br /> However, just averaging the group's credences in each of the three specific risks, we might say that it is 97% sure that smoking doesn't cause lung cancer, 97% sure it doesn't cause emphysema, and 97% sure it doesn't cause heart disease ($\frac{0.99 + 0.99 + 0.93}{3} = 0.97$). And it is then possible that the group assigns a higher than 95% credence to the conjunction of these three. And, if it does, it seems to me, the PBAA may well get things right, and the group does not lie if it says that smoking carries no health risks. <p></p><p style="text-align: left;">Nonetheless, I think the PBAA cannot be right. In the example I just described, I noted that, just taking a straight average gives, for each specific risk, a credence of 97% that it doesn't exist. And I noted that it's then possible that the group credence that smoking is safe to health is above 95%. But of course, it's also possible that it's below 95%. This would happen, for instance, if the group were to take the three risks to be independent. Then the group credence that smoking is safe to health would be a little over 91%---too low for the group to believe it given the stakes. But PBAA would still say that the group believes that smoking is safe to health. The point is that PBAA is not sufficiently sensitive to the more fine-grained attitudes to the propositions that lie behind the beliefs in those propositions. Simply knowing what each member believes about the three putative risks is not sufficient to determine what the group thinks about them. You also need to look to their credences.<br /></p><p style="text-align: left;">Of course, there are lots of reasons to dislike straight averaging as a means for pooling credences---it can't preserve judgments of independence, for instance---and lots of reasons to dislike the naive application of a threshold or Lockean view of belief that is in the background here---it gives rise to the lottery paradox. But it seems that, for any reasonable method of probablistic aggregation and any reasonable account of the relationship between belief and credence, there will be cases like this in which the PBAA says the group believes a proposition when it shouldn't. So I agree with Lackey that the PBAA sometimes gets things wrong, but I disagree about exactly when.<br /></p><h3 style="text-align: left;">Base fragility: a further desideratum</h3><p style="text-align: left;">Consider an area of science in which two theories vie for precedence, $T_1$ and $T_2$. Half of the scientists working in this area believe the following:</p><ul style="text-align: left;"><li>($A_1$) $T_1$ is simpler than $T_2$,</li><li>($B_1$) $T_2$ is more explanatory than $T_1$,</li><li>($C_1$) simplicity always trumps explanatory power in theory choice.</li></ul><p style="text-align: left;">These scientists consequently believe $T_1$. The other half of the scientists believe the following: </p><ul style="text-align: left;"><li>($A_2$) $T_2$ is simpler than $T_1$,</li><li>($B_2$) $T_1$ is more explanatory than $T_2$,</li><li>($C_2$) explanatory power always trumps simplicity in theory choice.</li></ul><p style="text-align: left;">These scientists consequently believe $T_1$. So all scientists believe $T_1$. But they do so for diametrically opposed reasons. Indeed, all of their beliefs about the comparisons between $T_1$ and $T_2$ are in conflict, but because their views about theory choice are also in conflict, they end up believing the same theory. Does the scientific community believe $T_1$? Lackey says no. In order for a group to believe a proposition, the bases of the members' beliefs must not be substantively incoherent. In our example, for half of the members, the basis of their belief in $T_1$ is $A_1\ \&\ B_1\ \&\ C_1$, while for the other half, it's $A_2\ \&\ B_2\ \&\ C_2$. And $A_1$ contradicts $A_2$, $B_1$ contradicts $B_2$, and $C_1$ contradicts $C_2$. The bases are about as incoherent as can be. </p><p style="text-align: left;">Is Lackey correct to say that the scientific community does not believe in this case? I'm not so sure. For one thing, attributing belief in $T_1$ would help to explain a lot of the group's behaviour. Why does the scientific community fund and pursue research projects that are of interest only if $T_1$ is true? Why does the scientific community endorse and teach from textbooks that give much greater space to expounding and explaining $T_1$? Why do departments in this area hire those with the mathematical expertise required to understand $T_1$ when that expertise is useless for understanding $T_2$? In each case, we might say: because the community believes $T_1$.</p><p style="text-align: left;">Lackey raises two worries about group beliefs based in incoherent bases: (i) they cannot be subject to rational evaluation; (ii) they cannot coherently figure in accounts of collective deliberation. On (ii), it seems to me that the group belief could figure in deliberation. Suppose the community is deliberating about whether to invite a $T_1$-theorist or a $T_2$-theorist to give the keynote address at the major conference in the area. It seems that the group's belief in the superiority of $T_1$ could play a role in the discussions: 'Yes, we want the speaker who will pose the greatest challenge intellectually, but we don't want to hear a string of falsehoods, so let's go with the $T_1$-theorist,' they might reason.</p><p style="text-align: left;">On (i): Lackey asks what we would say if the group were to receive new evidence that $T_1$ has greater simplicity and less explanatory power than we initially thought. For the first half of the group, this would make their belief in $T_1$ more justified; for the second half, it would make their belief less justified. What would it do to the group's belief? Without an account of justification for group belief, it's hard to say. But I don't think the incoherent bases rule out an answer. For instance, we might be reliabilists about group justification. And if we are, then we look at all the times that the members of the group have made judgments about simplicity and explanatory power that have the same pattern as they have time---that is, half one way, half the other---and we look at the proportion of those times that the group belief---formed by whatever aggregation method we favour---has been true. If it's high, then the belief is justified; if it's not, it's not. And we can do that for the group before and after this new evidence comes in. And by doing that, we can compare the level of justification for the group belief.</p><p style="text-align: left;">Of course, this is not to say that reliabilism is the correct account of justification for group beliefs. But it does suggest that incoherent bases don't create a barrier to such accounts.<br /></p><h2 style="text-align: left;">Varieties of group belief</h2><p style="text-align: left;">One thing that is striking when we consider different proposed accounts of group belief is how large the supervenience base might be; that is, how many different features of a group $G$ might partially determine whether or not it believes a proposition $p$. Here's a list, though I don't pretend that it's exhaustive:</p><p style="text-align: left;">(1) <i>The beliefs of individual members of the group</i></p><p style="text-align: left;">(1a) Some accounts are concerned only with individual members' beliefs in $p$; others are interested in members' beliefs beyond that. For instance, a simple majoritarian account is interested only in members' beliefs in $p$. But Pettit's PBAA is interested instead in members' beliefs in each proposition from a set $q_1, \ldots, q_n$ whose conjunction is equivalent to $p$. And Lackey's GAA is interested in the members' beliefs in $p$ as well as the members' beliefs that form the bases for their belief in $p$ when they do believe $p$.</p><p style="text-align: left;">(1b) Some accounts are concerned with the individual beliefs of <i>all</i> members of the group, some only with so-called <i>operative members</i>. For instance, some will say that what determines whether a company believes $p$ is only whether or not members of their board believe $p$, while others will say that all employees of the company count.</p><p style="text-align: left;">(2) <i>The credences of individual members of the group</i></p><p style="text-align: left;">There are distinctions corresponding to (1a) and (1b) here as well.<br /></p><p style="text-align: left;">(3) <i>The outcomes of discussions between the members of the group</i></p><p style="text-align: left;">(3a) Some will say that only discussions that actually take place make a difference---you might say that, before a discussion takes place, the members of the group each believe $p$, but after they discuss it and retain those beliefs, you can say that the group believes $p$; others will say that hypothetical discussions can also make a difference---if individual members would dramatically change their beliefs were they to discuss the matter, that might mean the group does not believe, even if all members do.</p><p style="text-align: left;">(3b) Some will say that it is not the individual members' beliefs after discussion that is important, but their joint decision to accept $p$ as the group's belief. (Margaret Gilbert's JAA is such an account.)<br /></p><p style="text-align: left;">(4) <i>Belief-forming structures within the group</i></p><p style="text-align: left;">(4a) Some groups are extremely highly structured, and some of these structures relate to group belief formation. Some accounts of group belief acknowledge this by talking of 'operative members' of groups, and taking their attitudes to have greater weight in determining the group's attitude. For instance, it is common to say that the operative members of a company are its board members; the operative members of a British university might be its senior management team; the operative members of a trade union might be its executive committee. But of course many groups have much more complex structures than these. For instance, many large organisations are concerned with complex problems that break down into smaller problems, each of which requires a different sort of expertise to understand. The World Health Organization (WHO) might be such an example, or the Intergovernmental Panel on Climate Change (IPCC), or Médecins san Frontières (MSF). In each case, there might be a rigid reporting structure whereby subcommittees report their findings to the main committee, but each subcommittee might form its own subcommittees that report to them; and there might be strict rules about how the findings of a subcommittee must be taken into account by the committee to which it reports before that committee itself reports upwards. In such a structure, the notion of operative members and their beliefs is too crude to capture what's necessary. <br /></p><p style="text-align: left;">(5) <i>The actions of the group </i></p><p style="text-align: left;">(5a) Some might say that a group has a belief just in case it acts in a way that is best explained by positing a group belief. Why does the scientific community persist in appointing only $T_1$-theorists and no $T_2$-theorists? Answer: It believes $T_1$. (I think Kenny Easwaran and Reuben Stern take this view in their recent joint work.)</p><p style="text-align: left;">So, in the case of group beliefs, the disagreement between different accounts does not concern only the conditions on an agree supervenience base; it also concerns the extent of the supervenience base itself. Now, this might soften us up for pluralism, but it is hardly an argument. To give an argument, I'd like to consider a range of possible accounts and, for each, describe a role that group beliefs are typically taken to play and for which this account is best suited.</p><h3 style="text-align: left;">Group beliefs as summaries</h3><p style="text-align: left;">One thing we do when we ascribe beliefs to groups is simply to summarise the views of the group. If I say that, in 1916, Russia believed that Rasputin was dishonest, I simply give a summary of the views of people who belong to the group to which 'Russia' refers in this sentence, namely, Russians alive in 1916. And I say roughly that a substantial majority believed that he was dishonest. </p><p style="text-align: left;">For this role, a simple majoritarian account (SMA) seems best:</p><p style="text-align: left;"><b>SMA</b> A group $G$ believes $p$ iff a substantial majority of members of $G$ believes $p$.</p><p style="text-align: left;">There is an interesting semantic point in the background here. Consider the sentence: 'At the beginning of negotiations at Brest-Litovsk in 1917-8, Russia believed Germany's demands would be less harsh than they turned out to be.' We might suppose that, in fact, this belief was not widespread in Russia, but it was almost universal among the Bolshevik government. Then we might nonetheless say that the sentence is true. At first sight, it doesn't seem that SMA can account for this. But it might do if 'Russia' refers to different groups in the two different sentences: to the whole population in 1916 in the first sentence; to the members of the Bolshevik government in the second. </p><p style="text-align: left;">I'm tempted to think that this happens a lot when we discuss group beliefs. Groups are complex entities, and the name of a group might be used in one sentence to pick out some subset of its structure---just its members, for instance---and in another sentence some other subset of its structure---its members as well as its operative group, for instance---and in another sentence yet some further subset of its structure---its members, its operative group, and the rules by which the operative group abide when they are debating an issue.</p><p style="text-align: left;">Of course, this might look like straightforward synecdoche, but I'm inclined to think it's not, because it isn't clear that there is one default referent of the term 'Russia' such that all other terms are parasitic on that. Rather, there are just many many different group structures that might be picked out by the term, and we have to hope that context determines this with sufficient precision to evaluate the sentence.<br /></p><h3 style="text-align: left;">Group beliefs as attitudes that play a functional role</h3><p style="text-align: left;">An important recent development in our understanding of injustice and oppression has been the recognition of structural forms of racism, sexism, ableism, homophobia, transphobia, and so on. The notion is contested and there are many competing definitions, but to illustrate the point, let me quote from <a href="https://www.nejm.org/doi/full/10.1056/NEJMms2025396" target="_blank">a recent article</a> in the New England Journal of Medicine that considers structural racism in the US healthcare system:</p><blockquote><p style="text-align: left;">All definitions [of structural racism] make clear that racism is not simply the result of private prejudices held by individuals, but is also produced and reproduced by laws, rules, and practices, sanctioned and even implemented by various levels of government, and embedded in the economic system as well as in cultural and societal norms (Bailey, et al. 2021).</p></blockquote><p>The point is that a group---a university, perhaps, or an entire healthcare system, or a corporation---might act as if it holds racist or sexist beliefs, even though no majority of its members holds those beliefs. A university might pay academics who are women less, promote them less frequently, and so on, even while few individuals within the organisation, and certainly not a majority, believe that women's labour is worth less, and that women are less worthy of promotion. In such a case, we might wish to ascribe those beliefs to the institution as a whole. After all, on certain functionalist accounts of belief, to have a belief simply is to be in a state that has certain casual relationships with other states, including actions. And the state of a group is determined not only by the state of the individuals within it but also by the other structural features of the group, such as its laws, rules and practices. And if the states of the individuals within the group, combined with these laws, rules and practices give rise to the sort of behaviour that we would explain in a individual by positing a belief, it seems reasonable to do so in the group case as well. What's more, doing so helps to explain group behaviour in just the same way that ascribing beliefs to individuals helps to explain their behaviour. (As mentioned above, I take it that Kenny Easwaran and Reuben Stern take something like this view of group belief.)<br /></p><h3 style="text-align: left;">Group beliefs as ascriptions that have legal standing</h3><p style="text-align: left;">In her book, Lackey pays particular attention to cases of group belief that are relevant to corporate culpability and liability. In the 1970s, did the tobacco company Philip Morris believe that their product is hazardous to health, even while they repeatedly denied it? Between 1998 and 2014, did Volkswagen believe that their diesel emissions reports were accurate? In 2003, did the British government believe that Iraq could deploy biological weapons within forty-five minutes of an order to do so? Playing this role well is an important job for an account of group belief. It can have very significant real world consequences: Do those who trusted the assertions of tobacco companies and became ill as a result receive compensation? Do governments have a case against car manufacturers? Should a government stand down?</p><p style="text-align: left;">In fact, I think the consequences are often so large and, perhaps more importantly, so varied that the decision whether or not to put them in train should not depend on the applicability of a single concept with a single precise definition. Consider cases of corporate culpability. There are many ways in which this might be punished. We might fine the company. We might demand that it change certain internal policies or rules. We might demand that it change its corporate structure. We might do many things. Some will be appropriate and effective if the company believes a crucial proposition in one sense; some appropriate if it believes that proposition in some other sense. For instance, a fine does many things, but among them is this: it affects the wealth of the company's shareholders, who will react by putting pressure on the company's board. Thus, it might be appropriate to impose a fine if we think that the company believed the proposition that it denied in its public assertions in the sense that a substantial majority of its board believed it. On the other hand, demanding that the company change certain internal policies or rules would be appropriate if the company believes the proposition that it publicly denied in the sense that it is the outcome of applying its belief-forming rules and policies (such as, for instance, the nested set of subcommittees that I imagined for the WHO or the IPPC or MSF above).</p><p style="text-align: left;">The point is that our purpose in ascribing culpability and liability to a group is essentially pragmatic. We do it in order to determine what sort of punishment we might mete out. This is perhaps in contrast to cases of individual culpability and liability, where we are interested also in the moral status of the individual's action independent of how we respond to it. But, in many cases, such as when a corporation has lied, which punishment is appropriate depends on which of the many ways in which a group can believe the company believed the negation of the proposition it asserted in its lie.</p><p style="text-align: left;">So it seems to me that, even if this role were the only role that our concept of group belief had to play, pluralism would be appropriate. Groups are complex entities and there are consequently many ways in which we can seek to change them in order to avoid the sorts of harms that arise when they behave badly. We need different concepts of group belief in order to identify which is appropriate in a given case.</p><p style="text-align: left;">It's perhaps worth noting that, while Lackey's opens her book with cases of corporate culpability, and this is a central motivation for her emphasis on group lying, it isn't clear to me that her group agent account (GAA) can accommodate all cases of corporate lies. Consider the following situation. The board of a tobacco company is composed of eleven people. Each of them believes that tobacco is hazardous to health. However, some believe it for very different reasons from the others. They have all read the same scientific literature on the topic, but six of them remember it correctly and the other five remember it incorrectly. The six who remember it correctly remember that tobacco contains chemical A and remember that when chemical A comes into contact with tissue X in the human body, it causes cancer in that tissue; and they also remember that tobacco does not contain chemical B and they remember that, when chemical B comes into contact with tissue Y in the human body, it does not cause cancer in that tissue. The five who remember the scientific literature incorrectly believe that tobacco contains chemical B and believe that when chemical B comes into contact with tissue Y in the human body, it causes cancer in that tissue; and they also believe that tobacco does not contain chemical A and they believe that, when chemical A comes into contact with tissue X in the human body, it does not cause cancer in that tissue. So, all board members believe that smoking causes cancer. However, the bases of their beliefs forms an incoherent set. The two propositions on which the six base their belief directly contradict the two propositions on which the five base theirs. The board then issues a statement saying that tobacco does not cause cancer. The board is surely lying, but according to GAA, they are not because the bases of their beliefs conflict and so they do not believe that tobacco does cause cancer.<br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com28tag:blogger.com,1999:blog-4987609114415205593.post-67712133467929694302021-03-14T12:31:00.002+00:002022-07-27T19:24:44.051+01:00Permissivism and social choice: a response to Blessenohl<p>In <a href="https://www.journals.uchicago.edu/doi/abs/10.1086/708011" target="_blank">a recent paper</a> discussing Lara Buchak's risk-weighted expected utility theory, Simon Blessenohl notes that the objection he raises there to Buchak's theory might also tell against permissivism about rational credence. I offer a response to the objection here.</p><p><span></span></p><a name='more'></a>In his objection, Blessenohl suggests that credal permissivism gives rise to an unacceptable tension between the individual preferences of agents and the collective preferences of the groups to which those agents belong. He argues that, whatever brand of permissivism about credences you tolerate, there will be a pair of agents and a pair of options between which they must choose such that both agents will prefer the first to the second, but collectively they will prefer the second to the first. He argues that this consequence tells against permissivism. I respond that this objection relies on an equivocation between two different understandings of collective preferences: on the first, they are an attempt to summarise the collective view of the group; on the second, they are the preferences of a third-party social chooser tasked with making decisions on behalf of the group. I claim that, on the first understanding, Blessenohl's conclusion does not follow; and, on the second, it follows but is not problematic.<br /><p></p><p>It is well known that, if two people have difference credences in a given proposition, there is a sense in which the pair of them, taken together, is vulnerable to a sure loss set of bets.* That is, there is a bet that the first will accept and a bet that the second will accept such that, however the world turns out, they'll end up collectively losing money. Suppose, for instance, that Harb is 90% confident that Ladybug will win the horse race that is about to begin, while Jay is only 60% confident. Then Harb's credences should lead him to buy a bet for £80 that will pay out £100 if Ladybug wins and nothing if she loses, while Jay's credences should lead him to sell that same bet for £70 (assuming, as we will throughout, that the utility of £$n$ is $n$). If Ladybug wins, Harb ends up £20 up and Jay ends up £30 down, so they end up £10 down collectively. And if Ladybug loses, Harb ends up £80 down while Jay ends up £70 up, so they end up £10 down as a pair.<br /><br />So, for individuals with different credences in a proposition, there seems to be a tension between how they would choose as individuals and how they would choose as a group. Suppose they are presented with a choice between two options: on the first, $A$, both of them enter into the bets just described; on the second, $B$, neither of them do. We might represent these two options as follows, where we assume that Harb's utility for receiving £$n$ is $n$, and the same for Jay:$$A = \begin{pmatrix}<br />20 & -80 \\<br />-30 & 70<br />\end{pmatrix}\ \ \ <br />B = \begin{pmatrix}<br />0 & 0 \\<br />0 & 0<br />\end{pmatrix}$$The top left entry is Harb's winnings if Ladybug wins, the top right is Harb's winnings if she loses; the bottom left is Jay's winnings if she wins, and the bottom left is Jay's winnings if she loses. So, given a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, each row represents a <i>gamble</i>---that is, an assignment of utilities to each state of the world---and each column represents a <i>utility distribution</i>---that is, an assignment of utilities to each individual. So $\begin{pmatrix} a & b \end{pmatrix}$ represents the gamble that the option bequeaths to Harb---$a$ if Ladybug wins, $b$ if she loses---while $\begin{pmatrix} c & d \end{pmatrix}$ represents the gamble bequeathed to Jay---$c$ if she wins, $d$ if she loses. And $\begin{pmatrix} a \\ c \end{pmatrix}$ represents the utility distribution if Ladybug wins---$a$ to Harb, $c$ to Jay---while $\begin{pmatrix} b \\ d \end{pmatrix}$ represents the utility distribution if she loses---$b$ to Harb, $d$ to Jay. Summing the entries in the first column gives the group's collective utility if Ladybug wins, and summing the entries in the second column gives their collective utility if she loses.<br /><br />Now, suppose that Harb cares only for the utility that he will gain, and Jay cares only his own utility; neither cares at all about the other's welfare. Then each prefers $A$ to $B$. Yet, considered collectively, $B$ results in greater total utility for sure: for each column, the sum of the entries in that column in $B$ (that is, $0$) exceeds the sum in that column in $A$ (that is, $-10$). So there is a tension between what the members of the group unanimously prefer and what the group prefers.<br /><br />Now, to create this tension, I assumed that the group prefers one option to another if the total utility of the first is sure to exceed the total utility of the second. But this is quite a strong claim. And, as Blessenohl notes, we can create a similar tension by assuming something much weaker.<br /><br />Suppose again that Harb is 90% confident that Ladybug will win while Jay is only 60% confident that she will. Now consider the following two options:$$A' = \begin{pmatrix}<br />20 & -80 \\<br />0 & 0<br />\end{pmatrix}\ \ \ <br />B' = \begin{pmatrix}<br />5 & 5 \\<br />25 & -75<br />\end{pmatrix}$$In $A'$, Harb pays £$80$ for a £$100$ bet on Ladybug, while in $B'$ he receives £$5$ for sure. Given his credences, he should prefer $A'$ to $B'$, since the expected utility of $A'$ is $10$, while for $B'$ it is $5$. And in $A'$, Jay receives £0 for sure, while in $B'$ he pays £$75$ for a £$100$ bet on Ladybug. Given his credences, he should prefer $A'$ to $B'$, since the expected utility of $A'$ is $0$, while for $B'$ it is $-15$. But again we see that $B'$ will nonetheless end up producing greater total utility for the pair---$30$ vs $20$ if Ladybug wins, and $-70$ vs $-80$ if Ladybug loses. But we can argue in a different way that the group should prefer $B'$ to $A'$. This different way of arguing for this conclusion is the heart of Blessenohl's result.<br /><br />In what follows, we write $\preceq_H$ for Harb's preference ordering, $\preceq_J$ for Jay's, and $\preceq$ for the group's. First, we assume that, when one option gives a particular utility $a$ to Harb for sure and a particular utility $c$ to Jay for sure, then the group should be indifferent between that and the option that gives $c$ to Harb for sure and $a$ to Jay for sure. That is, the group should be indifferent between an option that gives the utility distribution $\begin{pmatrix} a \\ c\end{pmatrix}$ for sure and an option that gives $\begin{pmatrix} c \\ a\end{pmatrix}$ for sure. Blessenohl calls this <i>Constant Anonymity</i>:</p><p><b>Constant Anonymity</b> For any $a, c$,$$\begin{pmatrix}<br />a & a \\<br />c & c<br />\end{pmatrix} \sim<br />\begin{pmatrix}<br />c & c \\<br />a & a <br />\end{pmatrix}$$This allows us to derive the following:$$\begin{pmatrix}<br />20 & 20 \\<br />0 & 0<br />\end{pmatrix} \sim<br />\begin{pmatrix}<br />0 & 0 \\<br />20 & 20 <br />\end{pmatrix}\ \ \ \text{and}\ \ \ <br />\begin{pmatrix}<br />-80 & -80 \\<br />0 & 0<br />\end{pmatrix} \sim<br />\begin{pmatrix}<br />0 & 0 \\<br />-80 & -80 <br />\end{pmatrix}$$And now we can introduce our second principle:<br /><br /><b>Preference Dominance</b> For any $a, b, c, d, a', b', c', d'$, if$$\begin{pmatrix}<br />a & a \\<br />c & c<br />\end{pmatrix} \preceq<br />\begin{pmatrix}<br />a' & a' \\<br />c' & c'<br />\end{pmatrix}\ \ \ \text{and}\ \ \ <br />\begin{pmatrix}<br />b & b \\<br />d & d<br />\end{pmatrix} \preceq<br />\begin{pmatrix}<br />b' & b' \\<br />d' & d'<br />\end{pmatrix}$$then$$\begin{pmatrix}<br />a & b \\<br />c & d<br />\end{pmatrix} \preceq<br />\begin{pmatrix}<br />a' & b' \\<br />c' & d' <br />\end{pmatrix}$$Preference Dominance says that, if the group prefers obtaining the utility distribution $\begin{pmatrix} a \\ c\end{pmatrix}$ for sure to obtaining the utility distribution $\begin{pmatrix} a' \\ c'\end{pmatrix}$ for sure, and prefers obtaining the utility distribution $\begin{pmatrix} b \\ d\end{pmatrix}$ for sure to obtaining the utility distribution $\begin{pmatrix} b' \\ d'\end{pmatrix}$ for sure, then they prefer obtaining $\begin{pmatrix} a \\ c\end{pmatrix}$ if Ladybug wins and $\begin{pmatrix} b \\ d\end{pmatrix}$ if she loses to obtaining $\begin{pmatrix} a' \\ c'\end{pmatrix}$ if Ladybug wins and $\begin{pmatrix} b' \\ d'\end{pmatrix}$ if she loses.<br /><br />Preference Dominance, combined with the indifferences that we derived from Constant Anonymity, gives$$\begin{pmatrix}<br />20 & -80 \\<br />0 & 0<br />\end{pmatrix} \sim<br />\begin{pmatrix}<br />0 & 0 \\<br />20 & -80 <br />\end{pmatrix}$$And then finally we introduce a closely related principle: </p><p><b>Utility Dominance</b> For any $a, b, c, d, a', b', c', d'$, if $a < a'$, $b < b'$, $c < c'$, and $d < d'$, then$$\begin{pmatrix}<br />a & b \\<br />c & d<br />\end{pmatrix} \prec<br />\begin{pmatrix}<br />a' & b' \\<br />c' & d' <br />\end{pmatrix}$$</p><p>This simply says that if one option gives more utility than another to each individual at each world, then the group should prefer the first to the second. So$$\begin{pmatrix}<br />0 & 0 \\<br />20 & -80 <br />\end{pmatrix} \prec<br />\begin{pmatrix}<br />5 & 5 \\<br />25 & -75<br />\end{pmatrix}$$Stringing these together, we have$$A' = \begin{pmatrix}<br />20 & -80 \\<br />0 & 0<br />\end{pmatrix} \sim<br />\begin{pmatrix}<br />0 & 0 \\<br />20 & -80 <br />\end{pmatrix} \prec<br />\begin{pmatrix}<br />5 & 5 \\<br />25 & -75<br />\end{pmatrix} = B'$$And thus, assuming that $\preceq$ is transitive, while Harb and Jay both prefer $A'$ to $B'$, the group prefers $B'$ to $A'$.<br /><br />More generally, Blessenohl proves an impossibility result. Add to the principles we have already stated the following:</p><p><b>Ex Ante Pareto</b> If $A \preceq_H B$ and $A \preceq_J B$, then $A \preceq B$.</p><p>And also:</p><p><b>Egoism</b> For any $a, b, c, d, a', b', c', d'$,$$\begin{pmatrix}<br />a & b \\<br />c & d<br />\end{pmatrix} \sim_H \begin{pmatrix}<br />a & b \\<br />c' & d'<br />\end{pmatrix}\ \ \ \text{and}\ \ \ <br />\begin{pmatrix}<br />a & b \\<br />c & d<br />\end{pmatrix} \sim_J \begin{pmatrix}<br />a' & b' \\<br />c & d<br />\end{pmatrix}$$That is, Harb cares only about the utilities he obtains from an option, and Jay cares only about the utilities that he obtains. And finally:</p><p><b>Individual Preference Divergence</b> There are $a, b, c, d$ such that$$\begin{pmatrix}<br />a & b \\<br />a & b<br />\end{pmatrix} \prec_H \begin{pmatrix}<br />c & d \\<br />c & d<br />\end{pmatrix}\ \ \ \text{and}\ \ \ <br />\begin{pmatrix}<br />a & b \\<br />a & b<br />\end{pmatrix} \succ_J \begin{pmatrix}<br />c & d \\<br />c & d<br />\end{pmatrix}$$Then Blessenohl shows that there are no preferences $\preceq_H$, $\preceq_J$, and $\preceq$ that satisfy Individual Preference Divergence, Egoism, Ex Ante Pareto, Constant Anonymity, Preference Dominance, and Utility Dominance.** And yet, he claims, each of these is plausible. He suggests that we should give up Individual Preference Divergence, and with it permissivism and risk-weighted expected utility theory.<br /><br />Now, the problem that Blessenohl identifies arises because Harb and Jay have different credences in the same proposition. But of course impermissivists agree that two rational individuals can have different credences in the same proposition. So why is this a problem specifically for permissivism? The reason is that, for the impermissivist, if two rational individuals have different credences in the same proposition, they must have different evidence. And for individuals with different evidence, we wouldn't necessarily want the group preference to preserve unanimous agreement between the individuals. Instead, we'd want the group to choose using whichever credences are rational in the light of the joint evidence obtained by pooling the evidence held by each individual in the group. And those might render one option preferable to the other even though each of the individuals, with their less well informed credences, prefer the second option to the first. So Ex Ante Pareto is not plausible when the individuals have different evidence, so impermissivism is safe.<br /><br />To see this, consider the following example: There are two medical conditions, $X$ and $Y$, that affect racehorses. If they have $X$, they're 90% likely to win the race; if they have $Y$, they're 60% likely; if they have both, they're 10% likely to win. Suppose Harb knows that Ladybug has $X$, but has no information about whether she has $Y$; and suppose Jay knows Ladybug has $Y$ and no information about $X$. Then both are rational. And both prefer $A$ to $B$ from above. But we wouldn't expect the group to prefer $A$ to $B$, since the group should choose using the credence it's rational to have if you know both that Ladybug has $X$ and that she has $Y$; that is, the group should choose by pooling the individual's evidence to give the group evidence, and then choose using the probabilities relative to that. And, relative to that evidence, $B$ is preferable to $A$.<br /><br />The permissivist, in contrast, cannot make this move. After all, for them it is possible for two rational individuals to disagree even though they have exactly the same evidence, and therefore the same pooled evidence. Blessenohl considers various ways the permissivist or the risk-weighted expected utility theorist might answer his objection, either by denying Ex Ante Pareto or Preference or Utility Dominance. He considers each response unsuccessful, and I tend to agree with his assessments. However, oddly, he explicitly chooses not to consider the suggestion that we might drop Constant Anonymity. I'd like to suggest that we should consider doing exactly that.<br /><br />I think Blessenohl's objection relies on an ambiguity in what the group preference ordering $\preceq$ represents. On one understanding, it is no more than an attempt to summarise the collective view of the group; on another, it represents the preferences of a third party brought in to make decisions on behalf of the group---the social chooser, if you will. I will argue that Ex Ante Pareto is plausible on the first understanding, but Constant Anonymity isn't; and Constant Anonymity is plausible on the second understanding, but Ex Ante Pareto isn't. <br /><br />Let's treat the first understanding of $\preceq$. On this, $\preceq$ represents the group's collective opinions about the options on offer. So just as we might try to summarise the scientific community's view on the future trajectory of Earth's average surface temperate or the mechanisms of transmission for SARS-CoV-2 by looking at the views of individual scientists, so might we try to summarise Harb and Jay's collective view of various options by looking at their individual views. Understood in this way, Constant Anonymity does not look plausible. Its motivation is, of course, straightforward. If $a < b$ and$$\begin{pmatrix}<br />a & a \\<br />b & b<br />\end{pmatrix} \prec <br />\begin{pmatrix}<br />b & b \\<br />a & a <br />\end{pmatrix}$$then the group's collective view unfairly and without justification favours Harb over Jay. And if$$\begin{pmatrix}<br />a & a \\<br />b & b<br />\end{pmatrix} \succ <br />\begin{pmatrix}<br />b & b \\<br />a & a <br />\end{pmatrix}$$then it unfairly and without justification favours Jay over Harb. So we should rule out both of these. But this doesn't entail that the group preference should be indifferent between these two options. That is, it doesn't entail that we should have$$\begin{pmatrix}<br />a & a \\<br />b & b<br />\end{pmatrix} \sim <br />\begin{pmatrix}<br />b & b \\<br />a & a <br />\end{pmatrix}$$After all, when you compare two options $A$ and $B$, there are four possibilities:</p><ol style="text-align: left;"><li>$A \preceq B$ and $B \preceq A$---that is, $A \sim B$;</li><li>$A \preceq B$ and $B \not \preceq A$---that is, $A \prec B$;</li><li>$A \not \preceq B$ and $B \preceq A$---that is, $A \succ B$;</li><li>$A \not \preceq B$ and $B \not \preceq A$---that is, $A$ and $B$ and not compatible.</li></ol><p>The argument for Constant Anonymity rules out (2) and (3), but it does not rule out (4). What's more, it's easy to see that, if we weaken Constant Anonymity so that it requires (1) or (4) rather than requiring (1), then we see that all of the principles are consistent with it. So introduce <i>Weak Constant Anonymity</i>:</p><p><b>Weak Constant Anonymity</b> For any $a, c$, then either$$\begin{pmatrix}<br />a & a \\<br />c & c<br />\end{pmatrix} \sim<br />\begin{pmatrix}<br />c & c \\<br />a & a <br />\end{pmatrix}$$or$$\begin{pmatrix}<br />a & a \\<br />c & c<br />\end{pmatrix}\ \ \text{and}\ \ <br />\begin{pmatrix}<br />c & c \\<br />a & a <br />\end{pmatrix}\ \ \text{are incomparable}$$<br /><br />Then define the preference ordering $\preceq^*$ as follows:$$A \preceq^* B \Leftrightarrow \left ( A \preceq_H B\ \&\ A \preceq_J B \right )$$Then $\preceq^*$ satisfies Ex Ante Pareto, Weak Constant Anonymity, Preference Dominance, and Utility Dominance. And indeed $\preceq^*$ seems a very plausible candidate for the group preference ordering understood in this first way: where Harb and Jay disagree, it simply has no opinion on the matter; it has opinions only where Harb and Jay agree, and then it shares their shared opinion. <br /><br />On the understanding of $\preceq$ as summarising the group's collective view, if $\begin{pmatrix}<br />a & a \\<br />c & c<br />\end{pmatrix} \sim<br />\begin{pmatrix}<br />c & c \\<br />a & a <br />\end{pmatrix}$ then the group collectively thinks that this option $\begin{pmatrix}<br />a & a \\<br />c & c<br />\end{pmatrix}$ is exactly as good as this option $\begin{pmatrix}<br />c & c \\<br />a & a <br />\end{pmatrix}$. But the group absolutely does not think that. Indeed, Harb and Jay both explicitly deny it, though for opposing reasons. So Constant Anonymity is false.<br /><br />Let's turn next to the second understanding. On this, $\preceq$ is the preference ordering of the social chooser. Here, the original, stronger version of Constant Anonymity seems more plausible. After all, unlike the group itself, the social chooser should have the sort of positive commitment to equality and fairness that the group definitively does not have. As we noted above, Harb and Jay unanimously reject the egalitarian assessment represented by $\begin{pmatrix}<br />a & a \\<br />c & c<br />\end{pmatrix} \sim<br />\begin{pmatrix}<br />c & c \\<br />a & a <br />\end{pmatrix}$. They explicitly both think that these two options are not equally good---if $a < c$, then Harb thinks the second is strictly better, while Jay thinks the first is strictly better. So, as we argued above, we take the group view to be that they are incomparable. But the social chooser should not remain so agnostic. She should overrule the unanimous rejection of the indifference relation between them and accept it. But, having thus overruled one unanimous view and taken a different one, it is little surprise that she will reject other unanimous views, such as Harb and Jay's unanimous view that $A'$ is better than $B'$ above. That is, it is little surprise that she should violate Ex Ante Pareto. After all, her preferences are not only informed by a value that Harb and Jay do not endorse; they are informed by a value that Harb and Jay explicitly reject, given our assumption of Egoism. This is the value of fairness, which is embodied in the social chooser's preferences in Constant Anonymity and rejected in Harb's and Jay's preferences by Egoism. If we require of our social chooser that they adhere to this value, we should not expect Ex Ante Pareto to hold.</p><p>* See Philippe Mongin's 1995 paper <a href="https://www.sciencedirect.com/science/article/abs/pii/S0022053185710447" target="_blank">'Consistent Bayesian Aggregation'</a> for wide-ranging results in this area.</p><p>** Here's the trick: if$$\begin{pmatrix}<br />a & b \\<br />a & b<br />\end{pmatrix} \prec_H \begin{pmatrix}<br />c & d \\<br />c & d<br />\end{pmatrix}\ \ \ \text{and}\ \ \ <br />\begin{pmatrix}<br />a & b \\<br />a & b<br />\end{pmatrix} \succ_J \begin{pmatrix}<br />c & d \\<br />c & d<br />\end{pmatrix}$$<br />Then let$$A' = \begin{pmatrix}<br />c & d \\<br />a & b<br />\end{pmatrix}\ \ \ \text{and}\ \ \ <br />B' = \begin{pmatrix}<br />a & b \\<br />c & d<br />\end{pmatrix}$$Then $A' \succ_H B'$ and $A' \succ_J B'$, but $A' \sim B'$. </p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com32tag:blogger.com,1999:blog-4987609114415205593.post-77478817057816758322021-01-06T14:24:00.002+00:002022-07-27T19:10:50.375+01:00Life on the edge: a response to Schultheis' challenge to epistemic permissivism about credences<p>In their 2018 paper, <a href="https://philpapers.org/rec/SCHLOT-25" target="_blank">'Living on the Edge'</a>, Ginger Schultheis issues a powerful challenge to epistemic permissivism about credences, the view that there are bodies of evidence in response to which there are a number of different credence functions it would be rational to adopt. The heart of the argument is the claim that a certain sort of situation is impossible. Schultheis thinks that all motivations for permissivism must render situations of this sort possible. Therefore, permissivism must be false, or at least these motivations for it must be wrong.</p><p><span></span></p><a name='more'></a>Here's the situation, where we write $R_E$ for the set of credence functions that it is rational to have when your total evidence is $E$. <p></p><ul style="text-align: left;"><li>Our agent's total evidence is $E$.</li><li>There is $c$ in $R_E$ that our agent knows is a rational response to $E$.</li><li>There is $c'$ in $R_E$ that our agent does not know is a rational response to $E$.</li></ul><p>Schultheis claims that the permissivist must take this to be possible, whereas in fact it is impossible. Here are a couple of specific examples that the permissivist will typically take to be possible.</p><p>Example 1: we might have a situation in which the credences it is rational to assign to a proposition $X$ in response to evidence $E$ form the interval $[0.4, 0.7]$. But we might not be sure of quite the extent of the interval. For all we know, it might be $[0.41, 0.7]$ or $[0.39, 0.71]$. Or it might be $[0.4, 0.7]$. So we are sure that $0.5$ is a rational credence in $X$, but we're not sure whether $0.4$ is a rational credence in $X$. In this case, $c(X) = 0.5$ and $c'(X) = 0.4$.</p><p>Example 2: you know that Probablism is a rational requirement on credence functions, and you know that satisfying the Principle of Indifference is rationally permitted, but you don't know whether or not it is also rationally required. In this case, $c$ is the uniform distribution required by the Principle of Indifference, but $c'$ is any other probability function.<br /></p><p>Schultheis then appeals to a principle called <i>Weak Rationality Dominance</i>. We say that one credence function $c$ <i>rationally dominates</i> another $c'$ if $c$ is rational in all worlds in which $c'$ is rational, and also rational in some worlds in which $c'$ is not rational. Weak Rationality Dominance says that it is irrational to adopt a rationally dominated credence function. The important consequence of this for Schultheis' argument is that, if you know that $c$ is rational, but you don't know whether $c'$ is, then $c'$ is irrational. As a result, in our example above, $c'$ is not rational, contrary to what the permissivist claims, because it is rationally dominated by $c$. So permissivism must be false.<br /></p><p>If Weak Rationality Dominance is correct, then, it follows that the permissivist must say that, for any body of evidence $E$ and set $R_E$ of rational responses, the agent with evidence $E$ either <i>must know of each</i> credence function in $R_E$ that it is in $R_E$, or they <i>must not know of any</i> credence function in $R_E$ that it is in $R_E$. If they <i>know of some</i> credence functions in $R_E$ that they are in $R_E$ and <i>not know of others</i> in $R_E$ that they are in $R_E$, then they clash with Weak Rationality Dominance. But, whatever your reason for being a permissivist, it seems very likely that it will entail situations in which there are some credence functions that are rational responses to your evidence and that you know are such responses, while you are unsure about other credence functions that are, in fact, rational responses whether or not they are, in fact, rational responses. This is Schultheis' challenge.</p><p>I'd like to explore a response to Schultheis' argument that takes issue with Weak Rationality Dominance (WRD). I'll spell out the objection in general to begin with, and then see how it plays out for a specific motivation for permissivism, namely, the Jamesian motivation I sketched in <a href="https://philpapers.org/rec/SCHLOT-25" target="_blank">this previous blogpost</a>. </p><p>One worry about WRD is that it seems to entail a deference principle of exactly the sort that I objected to in <a href="https://m-phi.blogspot.com/2020/12/deferring-to-rationality-does-it.html" target="_blank">this blogpost</a>. According to such deference principles, for certain agents in certain situations, if they learn of a credence function that it is rational, they should adopt it. For instance, Ben Levinstein claims that, if you are certain that you are irrational, and you learn that $c$ is rational, then you should adopt $c$ -- or at least you should have the conditional credences that would lead you to do this if you were to apply conditionalization. We might slightly strengthen Levinstein's version of the deference principle as follows: if you are unsure whether you are rational or not, and you learn that $c$ is rational, then you should adopt $c$. WRD entails this deference principle. After all, suppose you have credence function $c'$, and you are unsure whether or not it is rational. And suppose you learn that $c$ is rational (and don't thereby learn that $c'$ is as well). Then, according to Schultheis' principle, you are irrational if you stick with $c'$.</p><p>In the previous blogpost, I objected to Levinstein's deference principle, and others like it, because it relies on the assumption that all rational credence functions are better than all irrational credence functions. I think that's false. I think there are certain sorts of flaw that render you irrational, and lacking those flaws renders you rational. But lacking those flaws doesn't ensure that you're going to be better than someone who has those flaws. Consider, for instance, the extreme subjective Bayesian who justifies their position using an accuracy dominance argument of the sort pioneered by Jim Joyce. That is, they say that accuracy is the sole epistemic good for credence functions. And they say that non-probabilistic credence functions are irrational because, for any such credence function, there are probabilistic ones that accuracy dominate them; and all probabilistic credence functions are rational because, for any such credence function, there is no probabilistic one that accuracy dominates it. Now, suppose I have credence $0.91$ in $X$ and $0.1$ in $\overline{X}$. And suppose I am either sure that this is irrational, or I'm uncertain it is. I then learn that assigning credence $0.1$ to $X$ and $0.9$ to $\overline{X}$ is rational. What should I do? It isn't at all obvious to me that I should move from my credence function to the one I've learned is rational. After all, even from my slightly incoherent standpoint, it's possible to see that the rational one is going to be a lot less accurate than mine if $X$ is true, and I'm very confident that it is. </p><p>So I think that the rational deference principle is wrong, and therefore any version of WRD that entails it is also wrong. But perhaps there is a more restricted version of WRD that is right. And one that is nonetheless capable of sinking permissivism. Consider, for instance, a restricted version of WRD that applies only to agents who have no credence function --- that is, it applies to your initial choice of a credence function; it does not apply when you have a credence function and you are deciding whether to adopt a new one. This makes a difference. The problem with a version that applies when you already have a credence function $c'$ is that, even if it is irrational, it might nonetheless be better than the rational credence function $c$ in some situation, and it might be that $c'$ assigns a lot of credence to that situation. So it's hard to see how to motivate the move from $c'$ to $c$. However, in a situation in which you have no credence function, and you are unsure whether $c'$ is rational (even though it is) and you're certain that $c$ is rational (and indeed it is), WRD's demand that you should not pick $c'$ seems more reasonable. You occupy no point of view such that $c'$ is less of a depature from that point of view than $c$ is. You know only that $c$ lacks the flaws for sure, whereas $c'$ might have them. Better, then, to go for $c$, is it not? And if it is, this is enough to defeat permissivism.</p><p>I think it's not quite that simple. I noted above that Levinstein's deference principle relies on the assumption that all rational credence functions are better than all irrational credence functions. Schultheis' WRD seems to rely on something even stronger, namely, the assumption that all rational credence functions are equally good in all situations. For suppose they are not. You might then be unsure whether $c'$ is rational (though it is) and sure that $c$ is rational (and it is), but nonetheless rationally opt for $c'$ because you know that $c'$ has some good feature that you know $c$ lacks and you're willing to take the risk of having an irrational credence function in order to open the possibility of having that good feature.</p><p>Here's an example. You are unsure whether it is rational to assign $0.7$ to $X$ and $0.3$ to $\overline{X}$. It turns out that it is, but you don't know that. On the other hand, you do know that it is rational to assign 0.5 to each proposition. But the first assignment and the second are not equally good in all situations. The second has the same accuracy whether $X$ is true or false; the first, in constrast, is better than the first if $X$ is true and worse than the first if $X$ is false. The second does not open up the possibility of high accuracy that the first does; though, to compensate, it also precludes the possibility of low accuracy, which the first doesn't. Surveying the situation, you think that you will take the risk. You'll adopt the first, even though you aren't sure whether or not it is rational. And you'll do this because you want the possibility of being rational and having that higher accuracy. This seems a rational thing to do. So, it seems to me, WRD is false.<br /></p><p>Although I think this objection to WRD works, I think it's helpful to see how it might play out for a particular motivation for permissivism. Here's the motivation: Some credence functions offer the promise of great accuracy -- for instance, assigning 0.9 to $X$ and 0.1 to $\overline{X}$ will be very accurate if $X$ is true. However, those that do so also open the possibility of great inaccuracy -- if $X$ is false, the credence function just considered is very inaccurate. Other credence functions neither offer great accuracy nor risk great inaccuracy. For instance, assigning 0.5 to both $X$ and $\overline{X}$ guarantees the same inaccuracy whether or not $X$ is true. You might say that you are more risk-averse the lower is the maximum possible inaccuracy you are willing to risk. Thus, the options that are rational for you are those undominated options with maximum inaccuracy at most whatever the threshold is that you set. Now, suppose you use the Brier score to measure your inaccuracy -- so that the inaccuracy of the credence function $c(X) = p$ and $c(\overline{X}) = 1-p$ is $2(1-p)^2$ if $X$ is true and $2p^2$ if $X$ is false. And suppose you are willing to tolerate a maximum possible inaccuracy of $0.5$, which also gives you a mininum inaccuracy of $0.5$. In that case, only $c(X) = 0.5 = c(\overline{X})$ will be rational from the point of view of your risk attitudes --- since $2(1-0.5)^2 = 0.5 = 2(0.5^2)$. On the other hand, suppose you are willing to tolerate a maximum inaccuracy of $0.98$, which also gives you a minimum inaccuracy of $0.18$. In that case, any credence function $c$ with $0.3 \leq c(X) \leq 0.7$ and $c(\overline{X}) = 1-c(X)$ is rational from the point of view of your risk attitudes.</p><p>Now, suppose that you are in the sort of situation that Schultheis imagines. You are uncertain of the extent of the set $R_E$ of rational responses to your evidence $E$. On the account we're considering, this must be because you are uncertain of your own attitudes to epistemic risk. Let's say that the threshold of maximum inaccuracy that you're willing to tolerate is $0.98$, but you aren't certain of that --- you think it might be anything between $0.72$ and $1.28$. So you're sure that it's rational to assign anything between 0.4 and 0.6 to $X$, but unsure whether it's rational to assign $0.7$ to $X$ --- if your threshold turns out to be less than 0.98, then assigning $0.7$ to $X$ would be irrational, because it risks inaccuracy of $0.98$. In this situation, is it rational to assign $0.7$ to $X$? I think it is. Among the credence functions that you know for sure are rational, the ones that give you the lowest possible inaccuracy are the one that assigns 0.4 to $X$ and the one that assigns 0.6 to $X$. They have maximum inaccuracy of 0.72, and they open up the possibility of an inaccuracy of 0.32, which is lower than the lowest possible inaccuracy opened up by any others that you know to be rational. On the other hand, assigning 0.7 to $X$ opens up the possibility of an inaccuracy of 0.18, which is considerably lower. As a result, it doesn't seem irrational to assign 0.7 to $X$, even though you don't know whether it is rational from the point of view of your attitudes to risk, and you do know that assigning 0.6 is rational. </p><p>There is another possible response to Schultheis' challenge for those who like this sort of motivation for permissivism. You might simply say that, if your attitudes to risk are such that you will tolerate a maximum inaccuracy of at most $t$, then regardlesss of whether you know this fact, indeed regardless of your level of uncertainty about it, the rational credence functions are precisely those that have maximum inaccuracy of at most $t$. This sort of approach is familiar from expected utility theory. Suppose I have credences in $X$ and in $\overline{X}$. And suppose I face two options whose utility is determined by whether or not $X$ is true or false. Then, regardless of what I believe about my credences in $X$ and $\overline{X}$, I should choose whichever option maximises expected utility from the point of view of my actual credences. The point is this: if what it is rational for you to believe or to do is determined by some feature of you, whether it's your credences or your attitudes to risk, being uncertain about those features doesn't change what it is rational for you to do. This introduces a certain sort of externalism to our notion of rationality. There are features of ourselves -- our credences or our attitudes to risk -- that determine what it is rational for us to believe or do, which are nonetheless not luminous to us. But I think this is inevitable. Of course, we might might move up a level and create a version of expected utility theory that appeals not to our first-order credences but to our credences concerning those first-order credences -- perhaps you use the higher-order credences to define a higher-order expected value for the first-order expected utilities, and you maximize that. But it simply pushes the problem back a step. For your higher-order credences are no more luminous than your first-order ones. And to stop the regress, you must fix some level at which the credences at that level simply determine the expectation that rationality requires you to maximize, and any uncertainty concerning those does not affect rationality. And the same goes in this case. So, given this particular motivation for permissivism, which appeals to your attitudes to epistemic risk, it seems that there is another reason why WRD is false. If $c$ is in $R_E$, then it is rational for you, regardless of your epistemic attitude to its rationality.<br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com5tag:blogger.com,1999:blog-4987609114415205593.post-85533560114109447092021-01-04T11:53:00.001+00:002022-07-27T19:11:30.478+01:00Using a generalized Hurwicz criterion to pick your priors<p>Over the summer, I got interested in the problem of the priors again. Which credence functions is it rational to adopt at the beginning of your epistemic life? Which credence functions is it rational to have before you gather any evidence? Which credence functions provide rationally permissible responses to the empty body of evidence? As is my wont, I sought to answer this in the framework of epistemic utility theory. That is, I took the rational credence functions to be those declared rational when the appropriate norm of decision theory is applied to the decision problem in which the available acts are all the possible credence functions, and where the epistemic utility of a credence function is measured by a strictly proper measure. I considered a number of possible decision rules that might govern us in this evidence-free situation: <a href="https://m-phi.blogspot.com/2020/07/hurwiczs-criterion-of-realism-and.html" target="_blank">Maximin, the Principle of Indifference, and the Hurwicz criterion</a>. And I concluded in favour of <a href="https://m-phi.blogspot.com/2020/07/a-generalised-hurwicz-criterion.html" target="_blank">a generalized version of the Hurwicz criterion</a>, which I axiomatised. I also <a href="https://m-phi.blogspot.com/2020/07/taking-risks-and-picking-priors.html" target="_blank">described</a> which credence functions that decision rule would render rational in the case in which there are just three possible worlds between which we divide our credences. In this post, I'd like to generalize the results from that treatment to the case in which there any finite number of possible worlds.</p><p><span></span></p><a name='more'></a>Here's the decision rule (where $a(w_i)$ is the utility of $a$ at world $w_i$).<br /><p></p><p><b>Generalized Hurwicz Criterion </b>Given an option $a$ and a sequence of weights $0 \leq \lambda_1, \ldots, \lambda_n \leq 1$ with $\sum^n_{i=1} \lambda_i = 1$, which we denote $\Lambda$, define the generalized Hurwicz score of $a$ relative to $\Lambda$ as follows: if $$a(w_{i_1}) \geq a(w_{i_2}) \geq \ldots \geq a(w_{i_n})$$ then $$H^\Lambda(a) := \lambda_1a(w_{i_1}) + \ldots + \lambda_na(w_{i_n})$$That is, $H^\Lambda(a)$ is the weighted average of all the possible utilities that $a$ receives, where $\lambda_1$ weights the highest utility, $\lambda_2$ weights the second highest, and so on.</p><p>The Generalized Hurwicz Criterion says that you should order options by their generalized Hurwicz score relative to a sequence $\Lambda$ of weightings of your choice. Thus, given $\Lambda$,$$a \preceq^\Lambda_{ghc} a' \Leftrightarrow H^\Lambda(a) \leq H^\Lambda(a')$$And the corresponding decision rule says that you should pick your Hurwicz weights $\Lambda$ and then, having done that, it is irrational to choose $a$ if there is $a'$ such that $a \prec^\Lambda_{ghc} a'$.</p><p>Now, let $\mathfrak{U}$ be an additive strictly proper epistemic utility measure. That is, it is generated by a strictly proper scoring rule. A <i>strictly proper scoring rule</i> is a function $\mathfrak{s} : \{0, 1\} \times [0, 1] \rightarrow [-\infty, 0]$ such that, for any $0 \leq p \leq 1$, $p\mathfrak{s}(1, x) + (1-p)\mathfrak{s}(0, x)$ is maximized, as a function of $x$, uniquely at $x = p$. And an epistemic utility measure is generated by $\mathfrak{s}$ if, for any credence function $C$ and world $w_i$,$$\mathfrak{U}(C, w_i) = \sum^n_{j=1} \mathfrak{s}(w^j_i, c_j)$$where</p><ul style="text-align: left;"><li>$c_j = C(w_j)$, and <br /></li><li>$w^j_i = 1$ if $j=i$ and $w^j_i = 0$ if $j \neq i$ </li></ul><p>In what follows, we write the sequence $(c_1, \ldots, c_n)$ to represent the credence function $C$.</p><p>Also, given a sequence $(\alpha_1, \ldots, \alpha_k)$ of numbers, let$$\mathrm{Av}((\alpha_1, \ldots, \alpha_k)) := \frac{\alpha_1 + \ldots + \alpha_k}{k}$$That is, $\mathrm{av}(A)$ is the average of the numbers in $A$. And given $1 \leq k \leq n$, let $A|_k = (a_1, \ldots, a_k)$. That is, $A|_k$ is the truncation of the sequence $A$ that omits all terms after $a_k$. Then we say that $A$ does not exceed its average if, for each $1 \leq k \leq n$,$$\mathrm{av}(A) \geq \mathrm{av}(A|_k)$$That is, at no point in the sequence does the average of the numbers up to that point exceed the average of all the numbers in the sequence. <br /></p><p><b>Theorem 1</b> Suppose $\Lambda = (\lambda_1, \ldots, \lambda_n)$ is a sequence of generalized Hurwicz weights. Then there is a sequence of subsequences $\Lambda_1, \ldots, \Lambda_m$ of $\Lambda$ such that</p><ol style="text-align: left;"><li>$\Lambda = \Lambda_1 \frown \ldots \frown \Lambda_m$</li><li>$\mathrm{av}(\Lambda_1) \geq \ldots \geq \mathrm{av} (\Lambda_m)$</li><li>each $\Lambda_i$ does not exceed its average</li></ol><p>Then, the credence function$$(\underbrace{\mathrm{av}(\Lambda_1), \ldots, \mathrm{av}(\Lambda_1)}_{\text{length of $\Lambda_1$}}, \underbrace{\mathrm{av}(\Lambda_2), \ldots, \mathrm{av}(\Lambda_2)}_{\text{length of $\Lambda_2$}}, \ldots, \underbrace{\mathrm{av}(\Lambda_m), \ldots, \mathrm{av}(\Lambda_m)}_{\text{length of $\Lambda_m$}})$$maximizes $H^\Lambda(\mathfrak{U}(-))$ among credence functions $C = (c_1, \ldots, c_n)$ for which $c_1 \geq \ldots \geq c_n$.</p><p>This is enough to give us all of the credence functions that maximise $H^\Lambda(\mathfrak{U}(-))$: they are the credence function mentioned together with any permutation of it --- that is, any credence function obtained from that one by switching around the credences assigned to the worlds.<br /></p><p><i>Proof of Theorem 1.</i> Suppose $\mathfrak{U}$ is a measure of epistemic value that is generated by the strictly proper scoring rule $\mathfrak{s}$. And suppose that $\Lambda$ is the following sequence of generalized Hurwicz weights $0 \leq \lambda_1, \ldots, \lambda_n \leq 1$ with $\sum^n_{i=1} \lambda_i = 1$. <br /><br />First, due to a theorem that originates in Savage and is stated and proved fully by Predd, et al., if $C$ is not a probability function---that is, if $c_1 + \ldots + c_n \neq 1$---then there is a probability function $P$ such that $\mathfrak{U}(P, w_i) > \mathfrak{U}(C, w_i)$ for all worlds $w_i$. Thus, since GHC satisfies Strong Dominance, whatever maximizes $H^\Lambda(\mathfrak{U}(-))$ will be a probability function. <br /><br />Now, since $\mathfrak{U}$ is generated by a strictly proper scoring rule, it is also truth-directed. That is, if $c_i > c_j$, then $\mathfrak{U}(C, w_i) > \mathfrak{U}(C, w_j)$. Thus, if $c_1 \geq c_2 \geq \ldots \geq c_n$, then$$H^\Lambda(\mathfrak{U}(C)) = \lambda_1\mathfrak{U}(C, w_1) + \ldots + \lambda_n\mathfrak{U}(C, w_n)$$This is what we seek to maximize. But notice that this is just the expectation of $\mathfrak{U}(C)$ from the point of view of the probability distribution $\Lambda = (\lambda_1, \ldots, \lambda_n)$.<br /><br />Now, Savage also showed that, if $\mathfrak{s}$ is strictly proper and continuous, then there is a differentiable and strictly convex function $\varphi$ such that, if $P, Q$ are probabilistic credence functions, then<br />\begin{eqnarray*}<br />\mathfrak{D}_\mathfrak{s}(P, Q) & = & \sum^n_{i=1} \varphi(p_i) - \sum^n_{i=1} \varphi(q_i) - \sum^n_{i=1} \varphi'(q_i)(p_i - q_i) \\<br />& = & \sum^n_{i=1} p_i\mathfrak{U}(P, w_i) - \sum^n_{i=1} p_i\mathfrak{U}(Q, w_i)<br />\end{eqnarray*}<br />So $C$ maximizes $H^\Lambda(\mathfrak{U}(-))$ among credence functions $C$ with $c_1 \geq \ldots \geq c_n$ iff $C$ minimizes $\mathfrak{D}_\mathfrak{s}(\Lambda, -)$ among credence functions $C$ with $c_1 \geq \ldots \geq c_n$. We now use the KKT conditions to calculate which credence functions minimize $\mathfrak{D}_\mathfrak{s}(\Lambda, -)$ among credence functions $C$ with $c_1 \geq \ldots \geq c_n$. <br /><br />Thus, if we write $x_n$ for $1 - x_1 - \ldots - x_{n-1}$, then <br />\begin{multline*}<br />f(x_1, \ldots, x_{n-1}) = \mathfrak{D}((\lambda_1, \ldots, \lambda_n), (x_1, \ldots, x_n)) = \\<br />\sum^n_{i=1} \varphi(\lambda_i) - \sum^n_{i=1} \varphi(x_i) - \sum^n_{i=1} \varphi'(x_i)(\lambda_i - x_i) <br />\end{multline*}<br />So<br />\begin{multline*}<br />\nabla f = \langle \varphi''(x_1) (x_1 - \lambda_1) - \varphi''(x_n)(x_n - \lambda_n), \\<br />\varphi''(x_2) (x_2 - \lambda_2) - \varphi''(x_n)(x_n - \lambda_n), \ldots \\<br />\varphi''(x_{n-1}) (x_{n-1} - \lambda_{n-1}) - \varphi''(x_n)(x_n - \lambda_n) )\rangle<br />\end{multline*}<br /><br />Let $$\begin{array}{rcccl}<br />g_1(x_1, \ldots, x_{n-1}) & = & x_2 - x_1& \leq & 0\\<br />g_2(x_1, \ldots, x_{n-1}) & = & x_3 - x_2& \leq & 0\\<br />\vdots & \vdots & \vdots & \vdots & \vdots \\<br />g_{n-2}(x_1, \ldots, x_{n-1}) & = & x_{n-1} - x_{n-2}& \leq & 0 \\<br />g_{n-1}(x_1, \ldots, x_{n-1}) & = & 1 - x_1 - \ldots - x_{n-2} - 2x_{n-1} & \leq & 0<br />\end{array}$$So,<br />\begin{eqnarray*}<br />\nabla g_1 & = & \langle -1, 1, 0, \ldots, 0 \rangle \\<br />\nabla g_2 & = & \langle 0, -1, 1, 0, \ldots, 0 \rangle \\<br />\vdots & \vdots & \vdots \\<br />\nabla g_{n-2} & = & \langle 0, \ldots, 0, -1, 1 \rangle \\<br />\nabla g_{n-1} & = & \langle -1, -1, -1, \ldots, -1, -2 \rangle \\ <br />\end{eqnarray*}<br />So the KKT theorem says that $x_1, \ldots, x_n$ is a minimizer iff there are $0 \leq \mu_1, \ldots, \mu_{n-1}$ such that$$\nabla f(x_1, \ldots, x_{n-1}) + \sum^{n-1}_{i=1} \mu_i \nabla g_i(x_1, \ldots, x_{n-1}) = 0$$That is, iff there are $0 \leq \mu_1, \ldots, \mu_{n-1}$ such that<br />\begin{eqnarray*}<br />\varphi''(x_1) (x_1 - \lambda_1) - \varphi''(x_n)(x_n - \lambda_n) - \mu_1 - \mu_{n-1} & = & 0 \\<br />\varphi''(x_2) (x_2 - \lambda_2) - \varphi''(x_n)(x_n - \lambda_n) + \mu_1 - \mu_2 - \mu_{n-1} & = & 0 \\<br />\vdots & \vdots & \vdots \\<br />\varphi''(x_{n-2}) (x_{n-2} - \lambda_{n-2}) - \varphi''(x_n)(x_n - \lambda_n) + \mu_{n-3} - \mu_{n-2} - \mu_{n-1}& = & 0 \\<br />\varphi''(x_{n-1}) (x_{n-1} - \lambda_{n-1}) - \varphi''(x_n)(x_n - \lambda_n)+\mu_{n-2} - 2\mu_{n-1} & = & 0<br />\end{eqnarray*}<br />By summing these identities, we get:<br />\begin{eqnarray*}<br />\mu_{n-1} & = & \frac{1}{n} \sum^{n-1}_{i=1} \varphi''(x_i)(x_i - \lambda_i) - \frac{n-1}{n} \varphi''(x_n)(x_n - \lambda_n) \\<br />&= & \frac{1}{n} \sum^n_{i=1} \varphi''(x_i)(x_i - \lambda_i) - \varphi''(x_n)(x_n - \lambda_n) \\<br />& = & \sum^{n-1}_{i=1} \varphi''(x_i)(x_i - \lambda_i) - \frac{n-1}{n}\sum^n_{i=1} \varphi''(x_i)(x_i - \lambda_i)<br />\end{eqnarray*}<br />So, for $1 \leq k \leq n-2$,<br />\begin{eqnarray*}<br />\mu_k & = & \sum^k_{i=1} \varphi''(x_i)(x_i - \lambda_i) - k\varphi''(x_n)(x_n - \lambda_n) - \\<br />&& \hspace{20mm} \frac{k}{n}\sum^{n-1}_{i=1} \varphi''(x_i)(x_i - \lambda_i) + k\frac{n-1}{n} \varphi''(x_n)(x_n - \lambda_n) \\<br />& = & \sum^k_{i=1} \varphi''(x_i)(x_i - \lambda_i) - \frac{k}{n}\sum^{n-1}_{i=1} \varphi''(x_i)(x_i - \lambda_i) -\frac{k}{n} \varphi''(x_n)(x_n - \lambda_n) \\<br />&= & \sum^k_{i=1} \varphi''(x_i)(x_i - \lambda_i) - \frac{k}{n}\sum^n_{i=1} \varphi''(x_i)(x_i - \lambda_i)<br />\end{eqnarray*}<br />So, for $1 \leq k \leq n-1$,<br />$$\mu_k = \sum^k_{i=1} \varphi''(x_i)(x_i - \lambda_i) - \frac{k}{n}\sum^n_{i=1} \varphi''(x_i)(x_i - \lambda_i)$$<br />Now, suppose that there is a sequence of subsequences $\Lambda_1, \ldots, \Lambda_m$ of $\Lambda$ such that<br /></p><ol style="text-align: left;"><li>$\Lambda = \Lambda_1 \frown \ldots \frown \Lambda_m$</li><li>$\mathrm{av}(\Lambda_1) \geq \ldots \geq \mathrm{av}(\Lambda_m)$</li><li>each $\Lambda_i$ does not exceed its average.<br /></li></ol><p>And let $$P = (\underbrace{\mathrm{av}(\Lambda_1), \ldots, \mathrm{av}(\Lambda_1)}_{\text{length of $\Lambda_1$}}, \underbrace{\mathrm{av}(\Lambda_2), \ldots, \mathrm{av}(\Lambda_2)}_{\text{length of $\Lambda_2$}}, \ldots, \underbrace{\mathrm{av}(\Lambda_m), \ldots, \mathrm{av}(\Lambda_m)}_{\text{length of $\Lambda_m$}})$$Then we write $i \in \Lambda_j$ if $\lambda_i$ is in the subsequence $\Lambda_j$. So, for $i \in \Lambda_j$, $p_i = \mathrm{av}(\Lambda_j)$. Then$$\frac{k}{n}\sum^n_{i=1} \varphi''(p_i)(p_i - \lambda_i) = \frac{k}{n} \sum^m_{j = 1} \sum_{i \in \Lambda_j} \varphi''(\mathrm{av}(\Lambda_j))(\mathrm{av}(\Lambda_j) - \lambda_i) = 0 $$<br />Now, suppose $k$ is in $\Lambda_j$. Then<br />\begin{multline*}<br />\mu_k = \sum^k_{i=1} \varphi''(p_i)(p_i - \lambda_i) = \\<br />\sum_{i \in \Lambda_1} \varphi''(p_i)(p_i - \lambda_i) + \sum_{i \in \Lambda_2} \varphi''(p_i)(p_i - \lambda_i) + \ldots + \\<br />\sum_{i \in \Lambda_{j-1}} \varphi''(p_i)(p_i - \lambda_i) + \sum_{i \in \Lambda_j|_k} \varphi''(p_i)(p_i - \lambda_i) = \\<br />\sum_{i \in \Lambda_j|_k} \varphi''(p_i)(p_i - \lambda_i) = \sum_{i \in \Lambda_j|_k} \varphi''(\mathrm{av}(\Lambda_j)(\mathrm{av}(\Lambda_j) - \lambda_i) <br />\end{multline*}<br />So, if $|\Lambda|$ is the length of the sequence $\Lambda$,$$\mu_k \geq 0 \Leftrightarrow |\Lambda_j|_k|\mathrm{av}(\Lambda_j) - \sum_{i \in \Lambda_j|_k} \lambda_i \geq 0 \Leftrightarrow \mathrm{av}(\Lambda_j) \geq \mathrm{av}(\Lambda_j|_k)$$But, by assumption, this is true for all $1 \leq k \leq n-1$. So $P$ minimizes $H^\Lambda(\mathfrak{U}(-))$, as required.<br /><br />We now show that there is always a series of subsequences that satisfy (1), (2), (3) from above. We proceed by induction. </p><p><i>Base Case </i> $n = 1$. Then it is clearly true with the subsequence $\Lambda_1 = \Lambda$.</p><p><i>Inductive Step</i> Suppose it is true for all sequences $\Lambda = (\lambda_1, \ldots, \lambda_n)$ of length $n$. Now consider a sequence $(\lambda_1, \ldots, \lambda_n, \lambda_{n+1})$. Then, by the inductive hypothesis, there is a sequence of sequences $\Lambda_1, \ldots, \Lambda_m$ such that <br /></p><ol style="text-align: left;"><li>$\Lambda \frown (\lambda_{n+1}) = \Lambda_1 \frown \ldots \frown \Lambda_m \frown (\lambda_{n+1})$</li><li>$\mathrm{av}(\Lambda_1) \geq \ldots \geq \mathrm{av} (\Lambda_m)$</li><li>each $\Lambda_i$ does not exceed its average.<br /></li></ol><p>Now, first, suppose $\mathrm{av}(\Lambda_m) \geq \lambda_{n+1}$. Then let $\Lambda_{m+1} = (\lambda_{n+1})$ and we're done.<br /><br />So, second, suppose $\mathrm{av}(\Lambda_m) < \lambda_{n+1}$. Then we find the greatest $k$ such that$$\mathrm{av}(\Lambda_k) \geq \mathrm{av}(\Lambda_{k+1}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1}))$$Then we let $\Lambda^*_{k+1} = \Lambda_{k+1}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1})$. Then we can show that</p><ol style="text-align: left;"><li>$(\lambda_1, \ldots, \lambda_n, \lambda_{n+1}) = \Lambda_1 \frown \Lambda_2 \frown \ldots \frown \Lambda_k \frown \Lambda^*_{k+1}$.</li><li>Each $\Lambda_1, \ldots, \Lambda_k, \Lambda^*_{k+1}$ does not exceed average.</li><li>$\mathrm{av}(\Lambda_1) \geq \mathrm{av}(\Lambda_2) \geq \ldots \geq \mathrm{av}(\Lambda_k) \geq \mathrm{av}(\Lambda^*_{k+1})$.<br /></li></ol><p>(1) and (3) are obvious. So we prove (2). In particular, we show that $\Lambda^*_{k+1}$ does not exceed average. We assume that each subsequence $\Lambda_j$ starts with $\Lambda_{i_j+1}$</p><ul style="text-align: left;"><li>Suppose $i \in \Lambda_{k+1}$. Then, since $\Lambda_{k+1}$ does not exceed average, $$\mathrm{av}(\Lambda_{k+1}) \geq \mathrm{av}(\Lambda_{k+1}|_i)$$But, since $k$ is the greatest number such that$$\mathrm{av}(\Lambda_k) \geq \mathrm{av}(\Lambda_{k+1}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1}))$$We know that$$\mathrm{av}(\Lambda_{k+2}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1})) > \mathrm{av}(\Lambda_{k+1})$$So$$\mathrm{av}(\Lambda_{k+1}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1})) > \mathrm{av}(\Lambda_{k+1})$$So$$\mathrm{av}(\Lambda_{k+1}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1})) > \mathrm{av}(\Lambda_{k+1}|_i)$$</li><li>Suppose $i \in \Lambda_{k+2}$. Then, since $\Lambda_{k+2}$ does not exceed average, $$\mathrm{av}(\Lambda_{k+2}) \geq \mathrm{av}(\Lambda_{k+2}|_i)$$But, since $k$ is the greatest number such that$$\mathrm{av}(\Lambda_k) \geq \mathrm{av}(\Lambda_{k+1}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1}))$$We know that$$\mathrm{av}(\Lambda_{k+3}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1})) > \mathrm{av}(\Lambda_{k+2})$$So$$\mathrm{av}(\Lambda_{k+1}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1})) > \mathrm{av}(\Lambda_{k+2}|_i)$$But also, from above,$$ \mathrm{av}(\Lambda_{k+1}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1})) > \mathrm{av}(\Lambda_{k+1})$$So$$\mathrm{av}(\Lambda_{k+1}\frown \ldots \frown \Lambda_m \frown (\lambda_{n+1})) > \mathrm{av}(\Lambda_{k+1} \frown \Lambda_{k+2}|_i)$$</li><li>And so on.</li></ul><p>This completes the proof. $\Box$ <br /></p><p><br /></p><p><br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com8tag:blogger.com,1999:blog-4987609114415205593.post-530923806820880722021-01-01T16:12:00.003+00:002022-07-27T19:11:42.555+01:00How permissive is rationality? Horowitz's value question for moderate permissivism<p>Rationality is good; irrationality is bad. Most epistemologists would agree with this rather unnuanced take, regardless of their view of what exactly constitutes rationality and its complement. Granted this, a good test of a thesis in epistemology is whether it can explain why these two claims are true. Can it answer <i>the value question</i>: Why is rationality valuable and irrationality not? And indeed <a href="https://philpapers.org/rec/HORIR" target="_blank">Sophie Horowitz gives an extremely illuminating appraisal</a> of different degrees of epistemic permissivism and impermissivism by asking of each what answer it might give. Her conclusion is that the extreme permissivist -- played in her paper by the extreme subjective Bayesian, who thinks that satisfying Probabilism and being certain of your evidence is necessary and sufficient for rationality -- can give a satisfying answer to this question, or, at least, an answer that is satisfying from their own point of view. And the extreme impermissivist -- played here by the objective Bayesian, who thinks that rationality requires something like the maximum entropy distribution relative to your evidence -- can do so too. But, Horowitz argues, the moderate permissivist -- played by the moderate Bayesian, who thinks rationality imposes requirements more stringent than merely Probabilism, but who does not think they're stringent enough to pick out a unique credence function -- cannot. In this post, I'd like to raise some problems for Horowitz's assessment, and try to offer my own answer to the value question on behalf of the moderate Bayesian. (Full disclosure: If I'm honest, I think I lean towards extreme permissivism, but I'd like to show that moderate permissivism can defend itself against Horowitz's objection.)</p><p><span></span></p><a name='more'></a>Let's begin with the accounts that Horowitz gives on behalf of the extreme permissivist and the impermissivist.<p></p><p>The extreme permissivist -- the extreme subjective Bayesian, recall -- can say that only by being rational can you have a credence function that is <i>immodest</i> -- where a credence function is immodest if it uniquely maximizes expected epistemic utility from its own point of view. This is because Horowitz, like others in the epistemic utility theory literature, assume that epistemic utility is measured by <i>strictly proper measures</i>, so that, every probabilistic credence function expects itself to be better than any alternative credence function. From this, we can conclude that, on the extreme permissivist view, rationality is sufficient for immodesty. It's trickier to show that it is also necessary, since it isn't clear what we mean by the expected epistemic utility of a credence function from the point of view of a non-probabilistic credence function -- the usual definitions of expectation make sense only for probabilistic credence functions. Fortunately, however, we don't have to clarify this much. We need only say that, at the very least, if one credence function is epistemically better than another at all possible worlds -- that is, in decision theory parlance, the first dominates the second -- then any credence function, probabilistic or not, will expect the first to be better than the second. We then combine this with the result that, if epistemic utility is measured by a stricty proper measure, then, for each non-probabilistic credence function, there is a probabilistic credence function that dominates it, while for each probabilistic credence function, there is no such dominator (this result traces back to <a href="https://philpapers.org/rec/SAVEOP" target="_blank">Savage's 1971 paper</a>; <a href="https://philpapers.org/rec/PREPCA" target="_blank">Predd, et al.</a> give the proof in detail when the measure is additive; <a href="https://philpapers.org/rec/PETAEW" target="_blank">I then generalised it</a> to remove the additivity assumption). This then shows that being rational is necessary for being immodest. So, according to Horowitz's answer on behalf of the extreme permissivist, being rational is good and being irrational is bad because being rational is necessary and sufficient for being immodest; and it's good to be immodest and bad to be modest. </p><p>On the other hand, the impermissivist can say that, by being rational, you are maximizing expected accuracy from the point of view of the one true rational credence function. That's their answer to the value question, according to Horowitz.<br /></p><p>We'll return to the question of whether these answers are satisfying below. But first I want to turn to Horowitz's claim that the moderate Bayesian cannot give a satisfactory answer. I'll argue that, <i>if</i> the two answers just given on behalf of the extreme permissivist and extreme impermissivist are satisfactory, <i>then</i> there is a satisfactory answer that the moderate permissivist can give. Then I'll argue that, in fact, these answers aren't very satisfying. And I'll finish by sketching my preferred answer on behalf of the moderate permissivist. This is inspired by William James' account of epistemic risks in <i>The Will to Believe</i>, which leads me to discuss <a href="https://philpapers.org/rec/HOREVA" target="_blank">another Horowitz paper</a>. <br /></p><p>Horowitz's strategy is to show that the moderate permissivist cannot find a good epistemic feature of credence functions that belongs to all that they count as rational, but does not belong to any they count as irrational. The extreme permissivist can point to immodesty; the extreme impermissivist can point to maximising expected epistemic utility from the point of view of the sole rational credence function. But, for the moderate, there's nothing. Or so Horowitz argues.<br /></p><p>For instance, Horowitz initially considers the suggestion that rational credence functions guarantee you a minimum amount of epistemic utility. As she notes, the problem with this is that either it leads to impermissivism, or it fails to include all and only the credence functions the moderate considers rational. Let's focus on the case in which we have opinions about a proposition and its negation -- the point generalizes to richer sets of propositions. We'll represent the credence functions as pairs $(c(X), c(\overline{X}))$. And let's measure epistemic utility using the Brier score. So, when $X$ is true, the epistemic utility of $(x, y)$ is $-(1-x)^2 - y^2$, and when $X$ is false, it is $-x^2 - (1-y)^2$. Then, for $r > -0.25$, there is no credence function that guarantees you at least epistemic value $-0.25$ -- if you have at least that epistemic value at one world, you have less than that epistemic value at a different world. For $r = 0.25$, there is exactly one credence function that guarantees you at least epistemic value $-0.25$ -- it is the uniform credence function $(0.5, 0.5)$. And for $r < -0.25$, there are both probabilistic and non-probabilistic credence functions that guarantee you at least epistemic utility $r$. So, Horowitz concludes, a certain level of guaranteed epistemic utility can't be what separates the rational from the irrational for the moderate permissivist, since for any level, either no credence function guarantees it, exactly one does, or there are both credence functions the moderate considers rational and credence functions they consider irrational that guarantee it.<br /></p><p>She identifies a similar problem if we think not about guaranteed accuracy but about expected accuracy. Suppose, as the moderate permissivist urges, that some but not all probability functions are rationally permissible. Then for many rational credence functions, there will be irrational ones that they expect to be better than they expect some rational credence functions to be. Horowitz gives the example of a case in which the rational credence in $X$ is between 0.6 and 0.8 inclusive. Then someone with credence 0.8 will expect the irrational credence 0.81 to be better than it expects the rational credence 0.7 to be -- at least according to many many strictly proper measures of epistemic utility. So, Horowitz concludes, whatever separates the rational from the irrational, it cannot be considerations of expected epistemic utility.</p><p>I'd like to argue that, in fact, Horowitz should be happy with appeals to guaranteed or expected epistemic utility. Let's take guaranteed utility first. All that the moderate permissivist needs to say to answer the value question is that there are two valuable things that you obtain by being rational: immodesty <i>and</i> a guaranteed level of epistemic utility. Immodesty rules out all non-probabilistic credence functions, while the guaranteed level of epistemic utility narrows further -- how narrow depends on how much epistemic utility you wish to guarantee. So, for instance, suppose we say that the rational credence functions are exactly those $(x, 1-x)$ with $0.4 \leq x \leq 0.6$. Then each is immodest. And each has a guaranteed epistemic utility of at least $-(1-0.4)^2 - 0.6^2 = -0.72$. If Horowitz is satisfied with the immodesty answer to the value question when the extreme permissivist gives it, I think she should also be satisfied with it when the moderate permissivist combines it with a requirement not to risk certain low epistemic utilities (in this case, utilities below $-0.72$). And this combination of principles rules in all of the credence functions that the moderate counts as rational and rules out all they count as irrational.<br /></p><p>Next, let's think about expected epistemic utility. Suppose that the set of credence functions that the moderate permissivist counts as rational is a closed convex set. For instance, perhaps the set of rational credence function is $$R = \{c : \{X, \overline{X}\} \rightarrow [0, 1] : 0.6 \leq c(X) \leq 0.8\ \&\ c(\overline{X}) = 1- c(X)\}$$ Then we can prove the following: if a credence function is not in $R$, then there is $c^*$ in $R$ such that each $p$ in $R$ expects $c^*$ to be better than it expects $c$ to be (for the proof strategy, see Section 3.2 <a href="https://philpapers.org/rec/PETAEW" target="_blank">here</a>, but replace the possible chance functions with the rational credence functions). Thus, just as the extreme impermissivist answers the value question by saying that, if you're irrational, there's a credence function <i>the unique</i> <i>rational credence function</i> prefers to yours, while if you're rational, there isn't, the moderate permissivist can say that, if you're irrational, there is a credence function that <i>all the rational credence functions </i>prefer to yours, while if you're rational, there isn't. </p><p>Of course, you might think that it is still a problem for moderate permissivists that there are rational credence functions that expect some irrational credence functions to be better than some alternative rational ones. But I don't think Horowitz will have this worry. After all, the same problem affects extreme permissivism, and she doesn't take issue with this -- at least, not in the paper we're considering. For any two probabilistic credence functions $p_1$ and $p_2$, there will be some non-probabilistic credence function $p'_1$ that $p_1$ will expect to be better than it expects $p_2$ to be -- $p'_1$ is just a very slight perturbation of $p_1$ that makes it incoherent; a perturbation small enough to ensure it lies closer to $p_1$ than $p_2$ does.<br /></p><p>A different worry about the account of the value of rationality that I have just offered on behalf of the moderate permissivist is that it seems to do no more than push the problem back a step. It says that all irrational credence functions have a flaw that all rational credence functions lack. The flaw is this: there is an alternative preferred by all rational credence functions. But to assume that this is indeed a flaw seems to presuppose that we should care how rational credence functions evaluate themselves and other credence functions. But isn't the reason for caring what they say exactly what we have been asking for? Isn't the person who posed the value question in the first place simply going to respond: OK, but what's so great about all the rational credence functions expecting something else to be better, when the question on the table is exactly why rational credence functions are so good?</p><p>This is a powerful objection, but note that it applies equally well to Horowitz's response to the value question on behalf of the impermissivist. There, she claims that what is good about being rational is that you thereby maximise expected accuracy from the point of view of the unique rational credence function. But without an account of what's so good about being rational, I think we equally lack an account of what's so good about maximizing expected accuracy from the point of view of the rational credence functions.</p><p>So, in the end, I think Horowitz's answer to the value question on behalf of the impermissivist and my proposed expected epistemic utility answer on behalf of the moderate permissivist are ultimately unsatisfying.</p><p>What's more, Horowitz's answer on behalf of the extreme permissivist is also a little unsatisfying. The answer turns on the claim that immodesty is a virtue, together with the fact that precisely those credence functions identified as rational by subjective Bayesianism have that virtue. But is it a virtue? Just as arrogance in a person might seem excusable if they genuinely are very competent, but not if they are incompetent, so immodesty in a credence function only seems virtuous if the credence function itself is good. If the credence function is bad, then evaluating itself as uniquely the best seems just another vice to add to its collection. </p><p>So I think Horowitz's answer to the value question on behalf of the extreme permissivist is a little unsatisfactory. But it lies very close to an answer I find compelling. That answer appeals not to immodesty, but to non-dominance. Having a credence function that is dominated is bad. It leaves free epistemic utility on the table in just the same way that a dominated action in practical decision theory leaves free pragmatic utility on the table. For the extreme permissivist, what is valuable about rationality is that it ensures that you don't suffer from this flaw. </p><p>One noteworthy feature of this answer is the conception of rationality to which it appeals. On this conception, the value of rationality does not derive fundamentally from the possession of a positive feature, but from the lack of a negative feature. Ultimately, the primary notion here is irrationality. A credence function is irrational if it exhibits certain flaws, which are spelled out in terms of its success in the pursuit of epistemic utility. You are rational if you are free of these flaws. Thus, for the extreme permissivist, there is just one such flaw -- being dominated. So the rational credences are simply those that lack that flaw -- and the maths tells us that those are precisely the probabilistic credence functions.</p><p>We can retain this conception of rationality, motivate moderate permissivism, and answer the value question for it. In fact, there are at least two ways to do this. We have met something very close to one of these ways when we tried to rehabilitate the moderate permissivist's appeal to guaranteed epistemic utility above. There, we said that what makes rationality good is that it ensures that you are immodest and also ensures a certain guaranteed level of accuracy. But, a few paragraphs back, we argued that immodesty is no virtue. So that answer can't be quite right. But we can replace the appeal to immodesty with an appeal to non-dominance, and then the answer will be more satisfying. Thus, the moderate permissivist who says that the rational credence functions are exactly those $(x, 1-x)$ with $0.4 \leq x \leq 0.6$ can say that being rational is valuable for two reasons: (i) if you're rational, you aren't dominated; (ii) if you're rational you are guaranteed to have epistemic utility at least $-0.72$; (iii) only if you are rational will (i) and (ii) both hold. This answers the value question by appealing to how well credence functions promote epistemic utility, and it separates out the rational from the irrational precisely. <br /></p><p>To explain the second way we might do this, we invoke William James. Famously, in <i>The Will to Believe</i>, James said that we have two goals when we believe: to believe truth, and to avoid error. But these pull in different directions. If we pursue the first by believing something, we open ourselves up to the possibility of error. If we pursue the second by suspending judgment on something, we foreclose the possibility of believing the truth about it. Thus, to govern our epistemic life, we must balance these two goals. James held that how we do this is a subjective matter of personal judgment, and a number of different ways of weighing them are permissible. <a href="https://philpapers.org/rec/KELECB" target="_blank">Thomas Kelly has argued</a> that this can motivate permissivism in the case of full beliefs. Suppose the epistemic utility you assign to getting things right -- that is, believing truths and disbelieving falsehoods -- is $R > 0$. And suppose you assign epistemic utility $-W < 0$ to getting things wrong -- that is, disbelieving truths and believing falsehoods. And suppose you assign $0$ to suspending judgment. And suppose $W > R$. Then, as <a href="https://philpapers.org/rec/EASDTO" target="_blank">Kenny Easwaran</a> and <a href="https://philpapers.org/rec/DORLME-2" target="_blank">Kevin Dorst</a> have independently pointed out, if $r$ is the evidential probability of $X$, believing $X$ maximises expected epistemic utility from its point of view iff $\frac{W}{R + W} \leq r$, while suspending on $X$ maximises expected epistemic utility iff $\frac{R}{W+R} \leq r \leq \frac{W}{R+W}$. If William James is right, different values for $R$ and $W$ are permissible. The more you value believing truths, the greater will be $R$. The more you value avoiding falsehoods, the greater will be $W$ (and the lower will be $-W$). Thus, there will be a possible evidential probability $r$ for $X$, as well as permissible values $R$, $R'$ for getting things right and permissible values $W$, $W'$ for getting things wrong such that $$\frac{W}{R+W} < r < \frac{W'}{R'+W'}$$So, for someone with epistemic utilities characterised by $R$, $W$, it is rational to suspend judgment on $X$, while for someone with $W'$, $R'$, it is rational to believe $X$. Hence, permissivism about full beliefs.</p><p>As <a href="https://philpapers.org/rec/HOREVA" target="_blank">Horowitz points out</a>, however, the same trick won't work for credences. After all, as we've seen, all legitimate measures of epistemic utility for credences are strictly proper measures. And thus, if $r$ is the evidential probability of $X$, then credence $r$ in $X$ uniquely maximises expected epistemic utility relative to any one of those measures. So, a Jamesian permissivism about measures of epistemic value gives permissivism about doxastic states in the case of full belief, but not in the case of credence.</p><p>Nonetheless, I think we can derive permissivism about credences from James' insight. The key is to encode our attitudes towards James' two great goals for belief not in our epistemic utilities but in the rule we adopt when we use those epistemic utilities to pick our credences. Here's one suggestion, which I pursued at greater length in <a href="https://philpapers.org/rec/PETJEF-2" target="_blank">this paper</a> a few years ago, and that I generalised in <a href="https://m-phi.blogspot.com/2020/07/taking-risks-and-picking-priors.html" target="_blank">some blog posts</a> over the summer -- I won't actually present the generalization here, since it's not required to make the basic point. James recognised that, by giving yourself the opportunity to be right about something, you thereby run the risk of being wrong. In the credal case, by giving yourself the opportunity to be very accurate about something, you thereby run the risk of being very inaccurate. In the full belief case, to avoid that risk completely, you must never commit on anything. It was precisely this terror of being wrong that he lamented in Clifford. By ensuring he could never be wrong, there were true beliefs to which Clifford closed himself off. James believed that the extent to which you are prepared to take these epistemic risks is a passional matter -- that is, a matter of subjective preference. We might formalize it using a decision rule called <i>the Hurwicz criterion</i>. This rule was developed by Leonid Hurwicz for situations in which no probabilities are not available to guide our decisions, so it is ideally suited for the situation in which we must pick our prior credences. </p><p>Maximin is the rule that says you should pay attention only to the worst-case scenario and choose a credence function that does best there -- you should maximise your minimum possible utility. Maximax is the rule that says you should pay attention only to the best-case scenario and choose a credence function that does best there -- you should maximise your maximum possible utility. The former is maximally risk averse, the latter maximally risk seeking. As I showed <a href="https://philpapers.org/rec/PETARA-4" target="_blank">here</a>, if you measure epistemic utility in a standard way, maximin demands that you adopt the uniform credence function -- its worst case is best. And almost however you measure epistemic utility, maximax demands that you pick a possible world and assign maximal credence to all propositions that are true there and minimal credence to all propositions that are false there -- its best case, which obviously occurs at the world you picked, is best, because it is perfect there. </p><p>The Hurwicz criterion is a continuum of decision rules with maximin at one end and maximax at the other. You pick a weighting $0 \leq \lambda \leq 1$ that measures how risk-seeking you are and you define the <i>Hurwicz score</i> of an option $a$, with utility $a(w)$ at world $w$, to be$$H^\lambda(a) = \lambda \max \{a(w) : w \in W\} + (1-\lambda) \min \{a(w) : w \in W\}$$And you pick an option with the highest Hurwicz score.</p><p>Let's see how this works out in the simplest case, namely, that in which you have credences only in $X$ and $\overline{X}$. As before, we write credence functions defined on these two propositions as $(c(X), c(\overline{X})$. Then, if $\lambda \leq \frac{1}{2}$ --- that is, if you give at least as much weight to the worst case as to the best case --- then the uniform distribution $(\frac{1}{2}, \frac{1}{2})$ maximises the Hurwicz score relative to any strictly proper measure. And if $\lambda > \frac{1}{2}$ --- that is, if you are risk seeking and give more weight to the best case than the worst --- then $(\lambda, 1 - \lambda)$ and $(1-\lambda, \lambda)$ both maximise the Hurwicz score.</p><p>Now, if any $0 \leq \lambda \leq 1$ is permissible, then so is any credence function $(x, 1-x)$, and we get extreme permissivism. But I think we're inclined to say that there are extreme attitudes to risk that are not rationally permissible, just as there are preferences relating the scratching of one's finger and the destruction of the world that are not rationally permissible. I think we're inclined to think there is some range from $a$ to $b$ with $0 \leq a < b \leq 1$ such that the only rational attitudes to risk are precisely those encoded by the Hurwicz weights that lie between $a$ and $b$. If that's the case, we obtain moderate permissivism.</p><p>To be a bit more precise, this gives us both moderate interpersonal and intrapersonal permissivism. It gives us moderate interpersonal permissivism if $\frac{1}{2} < b < 1$ -- that is, if we are permitted to give more than half our weight to the best case epistemic utility. For then, since $a < b$, there is $b'$ such that $\frac{1}{2} < b' < b$, and then both $(b, 1-b)$ and $(b', 1-b')$ are both rationally permissible. But there is also $b < b'' < 1$, and for any such $b''$, $(b'', 1-b'')$ is not rationally permissible. It also gives us moderate intrapersonal permissivism under the same condition. For if $\frac{1}{2} < b$ and $b$ is your Hurwicz weight, then for you, both $(b, 1-b)$ and $(1-b, b)$ are different, but both are rationally permissible.<br /></p><p>How does this motivation for moderate permissivism fare with respect to the value question? I think it fares as well as the non-dominance-based answer I sketched above for the extreme permissivist. There, I appealed to a single flaw that a credence function might have: it might be dominated by another. Here, I introduced another flaw. It might be rationalised only by Jamesian attitudes to epistemic risk that are too extreme or otherwise beyond the pale. Like being dominated, this is a flaw that relates to the pursuit of epistemic utility. If you exhibit it, you are irrational. And to be rational is to be free of such flaws. The moderate permissivist can thereby answer the value question that Horowitz poses.<br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com1tag:blogger.com,1999:blog-4987609114415205593.post-87600337975819463592020-12-15T16:22:00.067+00:002022-07-27T19:11:54.889+01:00Deferring to rationality -- does it preclude permissivism?<p>Permissivism about epistemic rationality is the view that there are bodies of evidence in response to which rationality permits a number of different doxastic attitudes. I'll be thinking here about the case of credences. Credal permissivism says: there are bodies of evidence in response to which rationality permits a number of different credence functions.</p><p><span></span></p><a name='more'></a>Over the past year, I've watched friends on social media adopt remarkably different credence functions based on the same information about aspects of the COVID-19 pandemic, the outcome of the US election, and the withdrawal of the UK from European Union. And while I watch them scream at each other, cajole each other, and sometimes simply ignore each other, I can't shake the feeling that they are all taking rational stances. While they disagree dramatically, and while some will end up closer to the truth than others when it is finally revealed, it seems to me that all are responding rationally to their shared evidence, their opponents' protestations to the contrary. So permissivism is a very timely epistemic puzzle for 2020. What's more, <a href="https://cambridgereview.cargo.site/Dr-Rachel-Fraser" target="_blank">this wonderful piece</a> by Rachel Fraser made me see how my own William James-inspired approach to epistemology connects with a central motivation for believing in conspiracy theories, another major theme of this unloveable year. <br /><p></p><p>One type of argument against credal permissivism turns on the claim that rationality is worthy of deference. The argument begins with a precise version of this claim, stated as a norm that governs credences. It proceeds by showing that, if epistemic rationality is permissive, then it is sometimes impossible to meet the demands of this norm. Taking this to be a reductio, the argument concludes that rationality cannot be permissive. I know of two versions of the argument, one due to <a href="https://www.pdcnet.org/jphil/content/jphil_2016_0113_0008_0365_0395" target="_blank">Daniel Greco and Brian Hedden</a>, and one due to <a href="https://onlinelibrary.wiley.com/doi/abs/10.1111/phpr.12225" target="_blank">Ben Levinstein</a>. I'll mainly consider Levinstein's, since it fixes some problems with Greco and Hedden's. I'll consider <a href="https://academic.oup.com/mind/article-abstract/128/511/907/5133308" target="_blank">David Thorstad's response</a> to Greco and Hedden's argument, which would also work against Levinstein's argument were it to work at all. But I'll conclude that, while it provides a crucial insight, it doesn't quite work, and I'll offer my own alternative response.<br /></p><p>Roughly speaking, you defer to someone on an issue if, upon learning their attitude to that issue, you adopt it as your own. So, for instance, if you ask me what I'd like to eat for dinner tonight, and I say that I defer to you on that issue, I'm saying that I will want to eat whatever I learn you would like to eat. That's a case of deferring to someone else's preferences---it's a case where we defer conatively to them. Here, we are interested in cases in which we defer to someone else's beliefs---that is, where we defer doxastically to them. Thus, I defer doxastically to my radiographer on the issue of whether I've got a broken finger if I commit to adopting whatever credence they announce in that diagnosis. By analogy, we sometimes say that we defer doxastically to a feature of the world if we commit to setting our credence in some way that is determined by that feature of the world. Thus, I might defer doxastically to a particular computer simulation model of sea level change on the issue of sea level rise by 2030 if I commit to setting my credence in a rise of 10cm to whatever probability that model reports when I run it repeatedly while perturbing its parameters and initial conditions slightly around my best estimate of their true values.<br /><br />In philosophy, there are a handful of well-known theses that turn on the claim that we are required to defer doxastically to this individual or that feature of the world---and we're required to do it on all matters. For instance, van Fraassen's Reflection Principle says that you should defer doxastically to your future self on all matters. That is, for any proposition $X$, conditional on your future self having credence $r$ in $X$, you should have credence $r$ in $X$. In symbols:$$c(X\, |\, \text{my credence in $X$ at future time $t$ is $r$}) = r$$And the Principal Principle says that you should defer to the objective chances on all doxastic matters by setting your credences to match the probabilities that they report. That is, for any proposition $X$, conditional on the objective chance of $X$ being $r$, you should have credence $r$ in $X$. In symbols:$$c(X\, |\, \text{the objective chance of $X$ now is $r$}) = r$$Notice that, in both cases, there is a single expert value to which you defer on the matter in question. At time $t$, you have exactly one credence in $X$, and the Reflection Principle says that, upon learning that single value, you should set your credence in $X$ to it. And there is exactly one objective chance of $X$ now, and the Principal Principle says that, upon learning it, you should set your credence in $X$ equal to it. You might be uncertain about what that single value is, but it is fixed and unique. So this account of deference does not cover cases in which there is more than one expert. For instance, it doesn't obviously apply if I defer not to a specific climate model, but to a group of them. In those cases, there is usually no fixed, unique value that is the credence they all assign to a proposition. So principles of the same form as the Reflection or Principal Principle do not say what to do if you learn one of those values, or some of them, or all of them. This problem lies at the heart of the deference argument against permissivism. Those who make the argument think that deference to groups should work in one way; those who defend permissivism against it think it should work in some different way. <br /><br />As I mentioned above, the deference argument begins with a specific, precise norm that is said to govern the deference we should show to rationality. The argument continues by claiming that, if rationality is permissive, then it is not possible to satisfy this norm. Here is the norm as Levinstein states it, where $c \in R_E$ means that $c$ is in the set $R_E$ of rational responses to evidence $E$: </p><p><b>Deference to Rationality</b> Suppose:</p><ol style="text-align: left;"><li>$c$ is your credence function;</li><li>$E$ is your total evidence;</li><li>$c(c \in R_E) = 0$;</li><li>$c'$ is a probabilistic credence function;</li><li>$c(c' \in R_E) > 0$;</li></ol><p>then rationality requires$$c(-|c' \in R_E) = c'(-|c' \in R_E)$$That is, if you are certain that your credence function is not a rational response to your total evidence, then, conditional on some alternative probabilistic credence function being a rational response to that evidence, you should set your credences in line with that alternative once you've brought it up to speed with your new evidence that it is a rational response to your original total evidence.<br /><br />Notice, first, that Levinstein's principle is quite weak. It does not say of just anyone that they should defer to rationality. It says only that, if you are in the dire situation of being certain that you are yourself irrational, then you should defer to rationality. If you are sure you're irrational, then your conditional credences should be such that, were you to learn of a credence function that it's a rational response to your evidence, you should fall in line with the credences that it assigns conditional on that same assumption that it is rational. Restricting its scope in this way makes it more palatable to permissivists who will typically not think that someone who is already pretty sure that they are rational must switch credences when they learn that there are alternative rational responses out there. <br /><br />Notice also that you need only show such deference to rational credence functions that satisfy the probability axioms. This restriction is essential, for otherwise (DtR) will force you to violate the probability axioms yourself. After all, if $c(-)$ is probabilistic, then so is $c(-|X)$ for any $X$ with $c(X) > 0$. Thus, if $c'(-|c' \in R_E)$ is not probabilistic, and $c$ defers to $c'$ in the way Levinstein describes, then $c(-|c' \in R_E)$ is not probabilistic, and thus neither is $c$.<br /><br />Now, suppose:</p><ul style="text-align: left;"><li>$c$ is your credence function;</li><li>$E$ is your total evidence;</li><li>$c'$ and $c''$ are probabilistic credence functions with$$c'(-|c' \in R_E\ \&\ c'' \in R_E) \neq c''(-|c' \in R_E\ \&\ c'' \in R_E)$$That is, $c'$ and $c''$ are distinct and remain distinct even once they become aware that both are rational responses to $E$;</li><li>$c(c' \in R_E\ \&\ c'' \in R_E) > 0$. That is, you give some credence to both of them being rational responses to $E$;</li><li>$c(c \in R_E) = 0$. That is, you are certain that your own credence function is not a rational response to $E$.</li></ul><p>Then, by (DtR),</p><ul style="text-align: left;"><li>$c(-|c' \in R_E) = c'(-|c' \in R_E)$</li><li>$c(-|c'' \in R_E) = c''(-|c'' \in R_E)$ </li></ul><p>Thus, conditioning both sides of the first identity on $c'' \in R_E$ and both sides of the second identity on $c' \in R_E$, we obtain</p><ul style="text-align: left;"><li>$c(-|c' \in R_E\ \&\ c'' \in R_E) = c'(-|c' \in R_E\ \&\ c'' \in R_E)$ </li><li>$c(-|c'' \in R_E\ \&\ c' \in R_E) = c''(-|c' \in R_E\ \&\ c'' \in R_E)$</li></ul><p>But, by assumption, $c'(-| c' \in R_E\ \&\ c'' \in R_E) \neq c''(-|c' \in R_E\ \&\ c'' \in R_E)$. So (DtR) cannot be satisfied.<br /><br />One thing to note about this argument: if it works, it establishes not only that there can be no two different rational responses to the same evidence, but that it is irrational to be anything less than certain of this. After all, what is required to derive the contradiction from DtR is not that there are two probabilistic credence functions $c'$ and $c''$ such that $c'(-|c' \in R_E\ \&\ c'' \in R_E) \neq c''(-|c' \in R_E\ \&\ c'' \in R_E)$<i> that are both rational responses to $E$</i>. Rather, what is required is only that there are two probabilistic credence functions $c'$ and $c''$ with $c'(-|c' \in R_E\ \&\ c'' \in R_E) \neq c''(-|c' \in R_E\ \&\ c'' \in R_E)$ <i>that you think might both be rational responses to $E$</i>---that is, $c(c' \in R_E\ \&\ c'' \in R_E) > 0$. The conclusion that it is irrational to even entertain permissivism strikes me as too strong, but perhaps those who reject permissivism will be happy to accept it.<br /><br />Let's turn, then, to a more substantial worry, given compelling voice by David Thorstad: (DtR) is too strong because the deontic modality that features in it is too strong. As I hinted above, the point is that the form of the deference principles that Greco & Hedden and Levinstein use is borrowed from cases---such as the Reflection Principle and the Principal Principle---in which there is just one expert value, though it might be unknown to you. In those cases, it is appropriate to say that, upon learning the single value and nothing more, you are <i>required</i> to set your credence in line with it. But, unless we simply beg the question against permissivism and assume there is a single rational response to every body of evidence, this isn't our situation. Rather, it's more like the case where you defer to a group of experts, such as a group of climate models. And in this case, Thorstad says, it is inappropriate to <i>demand</i> that you set your credence in line with an expert's credence when you learn what it is. Rather, it is at most appropriate to <i>permit</i> you to do that. That is, Levinstein's principle should not say that rationality <i>requires</i> your credence function to assign the conditional credences stated in its consequent; it should say instead that rationality <i>allows</i> it. <br /><br />Thorstad motivates his claim by drawing an analogy with a moral case that he describes. Suppose you see two people drowning. They're called John and James, and you know that you will be able to save at most one. So the actions available to you are: save John, save James, save neither. And the moral actions are: save John, save James. But now consider a deference principle governing this situation that is analogous to (DtR): it demands that, upon learning that it is moral to save James, you must do that; and upon learning that it is moral to save John, you must do that. From this, we can derive a contradiction in a manner somewhat analogous to that in which we derived the contradiction from (DtR) above: if you learn both that it is moral to save John and moral to save James, you should do both; but that isn't an available action; so moral permissivism must be false. But I take it no moral theory will tolerate that in this case. So, Thorstad argues, there must be something wrong with the moral deference principle; and, by analogy, there must be something wrong with the analogous doxastic principle (DtR).<br /><br />Thorstad's diagnosis is this: the correct deference principle in the moral case should say: upon learning that it is moral to save James, you may do that; upon learning that it is moral to save John, you may do that. You thereby avoid the contradiction, and moral permissivism is safe. Similarly, the correct doxastic deference principle is this: upon learning that a credence function is rational, it is permissible to defer to it. In Levinstein's framework, the following is rationally permissible, not rationally mandated:$$c(-|c' \in R_E) = c'(-|c' \in R_E)$$</p><p>I think Thorstad's example is extremely illuminating, but for reasons rather different from his. Recall that a crucial feature of Levinstein's version of the deference argument against permissivism is that it applies only to people who are certain that their current credences are irrational. If we add the analogous assumption to Thorstad's case, his verdict is less compelling. Suppose, for instance, you are currently committed to saving neither John nor James from drowning; that's what you plan to do; it's the action you have formed an intention to perform. What's more, you're certain that this action is not moral. But you're uncertain whether either of the other two available actions are moral. And let's add a further twist to drive home the point. Suppose, furthermore, that you are certain that you are just about learn, of exactly one of them, that it is permissible. And add to that the fact that, immediately after you learn, of exactly one of them, that it is moral, you must act---failing to do so will leave both John and James to drown. In this case, I think, it's quite reasonable to say that, upon learning that saving James is permissible, you are not only morally permitted to drop your intention to save neither and replace it with the intention to save James, but you are also morally required to do so; and the same should you learn that it is permissible to save John. It would, I think, be impermissible to save neither, since you're certain that's immoral and you know of an alternative that is moral; and it would be impermissible to save John, since you are still uncertain about the moral status of that action, while you are certain that saving James is moral; and it would be morally required to save James, since you are certain of that action alone that it is moral. Now, Levinstein's principle might seem to holds for individuals in an analogous situation. Suppose you're certain that your current credences are irrational. And suppose you will learn of only one credence function that it is rationally permissible. At least in this situation, it might seem that it is rationally required that you adopt the credence function you learn is rationally permissible, just as you are morally required to perform the single act you learn is moral. So, is Levinstein's argument rehabilitated?<br /><br />I think not. Thorstad's example is useful, but not because the case of rationality and morality are analogous; rather, precisely because it draws attention to the fact that they are disanalogous. After all, all moral actions are better than all immoral ones. So, if you are committed to an action you know is immoral, and you learn of another that it is moral, and you know you'll learn nothing more about morality, you must commit to perform the action you've learned is moral. Doing so is the only way you know how to improve the action you'll perform for sure. But this is not the case for rational attitudes. It is not the case that all rational attitudes are better than all irrational attitudes. Let's see a few examples.<br /><br />Suppose my preferences over a set of acts $a_1, \ldots, a_N$ are as follows, where $N$ is some very large number:$$a_1 \prec a_2 \prec a_3 \prec \ldots \prec a_{N-3} \prec a_{N-2} \prec a_{N-1} \prec a_N \prec a_{N-2}$$This is irrational, because, if the ordering is irreflexive, then it is not transitive: $a_{N-2} \prec a_{N-1} \prec a_N \prec a_{N-2}$, but $a_{N-2} \not \prec a_{N-2}$. And suppose I learn that the following preferences are rational:$$a_1 \succ a_2 \succ a_3 \succ \ldots \succ a_{N-3} \succ a_{N-2} \succ a_{N-1} \succ a_N$$Then surely it is not rationally required of me to adopt these alternative preferences. (Indeed, it seems to me that rationality might even prohibit me from transitioning from the first irrational set to the second rational set, but I don't need that stronger claim.) In the end, my original preferences are irrational because of a small, localised flaw. But they nonetheless express coherent opinions about a lot of comparisons. And, concerning all of those comparisons, the alternative preferences take exactly the opposite view. Moving to the latter in order to avoid having preferences that are flawed in the way that the original set are flawed does not seem rationally required, and indeed might seem irrational. <br /><br />Something similar happens in the credal case, at least according to the accuracy-first epistemologist. Suppose I have credence $0.1$ in $X$ and $1$ in $\overline{X}$. And suppose the single legitimate measure of inaccuracy is the Brier score. I don't know this, but I do know a few things: first, I know that accuracy is the only fundamental epistemic value, and I know that a credence function's accuracy scores at different possible worlds determine its rationality at this world; furthermore, I know that my credences are accuracy dominated and therefore irrational, but I don't know what dominates them. Now suppose I learn that the following credences are rational: $0.95$ in $X$ and $0.05$ in $\overline{X}$. It seems that I am not required to adopt these credences (and, again, it seems that I am not even rationally permitted to do so, though again this latter claim is stronger than I need). While my old credences are irrational, they do nonetheless encode something like a point of view. And, from that point of view, the alternative credences look much much worse than staying put. While I know that mine are irrational and accuracy dominated, though I don't know what by, I also know that, from my current, slightly incoherent point of view, the rational ones look a lot less accurate than mine. And indeed they will be much less accurate than mine if $X$ turns out to be false.<br /><br />So, even in the situation in which Levinstein's principle is most compelling, namely, when you are certain you're irrational and you will learn of only one credence function that it is rational, still it doesn't hold. It is possible to be sure that your credence function is an irrational response to your evidence, sure that an alternative is a rational response, and yet not be required to adopt the alternative because learning that the alternative is rational does not teach you that it's better than your current irrational credence function for sure---it might be much worse. This is different from the moral case. So, as stated, Levinstein's principle is false.<br /><br />However, to make the deference argument work, Levinstein's principle need only hold in a single case. Levinstein describes a family of cases---those in which you're certain you're irrational---and claims that it holds in all of those. Thorstad's objection shows that it doesn't. Responding on Levinstein's behalf, I narrowed the family of cases to avoid Thorstad's objection---perhaps Levinstein's principle holds when you're certain you're irrational<i> and know you'll only learn of one credence function that it's rational</i>. After all, the analogous moral principle holds in those cases. But we've just seen that the doxastic version doesn't always hold there, because learning that an alternative credence function is rational does not teach you that it is better than your irrational credence function in the way that learning an act is moral teaches you that it's better than the immoral act you intend to perform. But perhaps we can narrow the range of cases yet further to find one in which the principle does hold.<br /><br />Suppose, for instance, you are certain you're irrational, you know you'll learn of just one credence function that it's rational, and moreover you know you'll learn that it is better than yours. Thus, in the accuracy-first framework, suppose you'll learn that it accuracy dominates you. Then surely Levinstein's principle holds here? And this would be sufficient for Levinstein's argument, since each non-probabilistic credence function is accuracy dominated by many different probabilistic credence functions; so we could find the distinct $c'$ and $c''$ we need for the reductio.<br /><br />Not so fast, I think. How you should respond when you learn that $c'$ is rational depends on what else you think about what determines the rationality of a credence function. Suppose, for instance, you think that a credence function is rational just in case it is not accuracy dominated, but you don't know which are the legitimate measures of accuracy. Perhaps you think there is only one legitimate measure of accuracy, and you know it's either the Brier score---$\mathfrak{B}(c, i) = \sum_{X \in \mathcal{F}} |w_i(X) - c(X)|^2$---or the absolute value score---$\mathfrak{A}(c, i) = \sum_{X \in \mathcal{F}} |w_i(X) - c(X)|^2$---but you don't know which. And suppose your credence function is $c(X) = 0.1$ and $c(\overline{X}) = 1$, as above. Now you learn that $c'(X) = 0.05$ and $c'(\overline{X}) = 0.95$ is rational and an accuracy dominator. So you learn that $c'$ is more accurate than $c$ at all worlds, and, since $c'$ is rational, there is nothing that is more accurate than $c'$ at all worlds. Then you thereby learn that the Brier score is the only legitimate measure of accuracy. After all, according to the absolute value score, $c'$ does not accuracy dominate $c$; in fact, $c$ and $c'$ have exactly the same absolute value score at both worlds. You thereby learn that the credence functions that accuracy dominate you without themselves being accuracy dominated are those for which $c(X)$ lies strictly between the solution of $(1-x)^2 + (1-x)^2 = (1-0.05)^2 + (0-1)^2$ that lies in $[0, 1]$ and the solution of $(0-x)^2 + (1-(1-x))^2 = (0-0.05)^2 + (1-1)^2$ that lies in $[0, 1]$, and $c(\overline{X}) = 1 - c(X)$. You are then permitted to pick any one of them---they are all guaranteed to be better than yours. You are not obliged to pick $c'$ itself. <br /><br />The crucial point is this: learning that $c'$ is rational teaches you something about the features of a credence function that determine whether it is rational---it teaches you that they render $c'$ rational! And that teaches you a bit about the set of rational credence functions---you learn it contains $c'$, of course, but you also learn other normative facts, such as the correct measure of inaccuracy, perhaps, or the correct decision principle to apply with the correct measure of inaccuracy to identify the rational credence functions. And learning those things may well shift your current credences, but you are not compelled to adopt $c'$.<br /><br />Indeed, you might be compelled to adopt something other than $c'$. An example: suppose that, instead of learning that $c'$ is rational and accuracy dominates $c$, you learn that $c''$ is rational and accuracy dominates $c$, where $c''$ is a probability function that Brier dominates $c$, and $c'' \neq c'$. Then, as before, you learn that the Brier score and not the absolute value score is the correct measure of inaccuracy, and thereby learn the set of credence functions that accuracy dominates yours. Perhaps rationality then requires you to fix up your credence function so that it is rational, but in a way that minimizes the amount by which you change your current credences. How to measure this? Well, perhaps you're required to pick an undominated dominator $c^*$ such that the expected inaccuracy of $c$ from the point of view of $c^*$ is minimal. That is, you pick the credence function that dominates you and isn't itself dominated <i>and which thinks most highly of your original credence function</i>. Measuring accuracy using the Brier score, this turns out to be the credence function $c'$ described above. Thus, given this reasonable account of how to respond when you learn what the rational credence functions are, upon learning that $c''$ is rational, rationality then requires you to adopt $c'$. <br /><br />In sum: For someone certain their credence function $c$ is irrational, learning only that $c'$ is rational is not enough to compel them to move to $c'$, nor indeed to change their credences at all, since they've no guarantee that doing so will improve their situation. To compel them to change their credences, you must teach them how to improve their epistemic situation. But when you teach them that doing a particular thing will improve their epistemic situation, that usually teaches them normative facts of which they were uncertain before---how to measure epistemic value, or the principles for choosing credences once you've fixed how to measure epistemic value---and doing that will typically teach them other ways to improve their epistemic situation besides the one you've explicitly taught them. Sometimes there will be nothing to tell between all the ways they've learned to improve their epistemic situation, and so all will be permissible, as Thorstad imagines; and sometimes there will be reason to pick just one of those ways, and so that will be mandated, even if epistemic rationality is permissive. In either case, Levinstein's argument does not go through. The deference principle on which it is based is not true.<br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com3tag:blogger.com,1999:blog-4987609114415205593.post-51839273472087298062020-09-03T12:19:00.001+01:002022-07-27T19:12:24.536+01:00Accuracy and Explanation in a Social Setting: thoughts on Douven and Wenmackers<p>For a PDF version of this post, see <a href="https://drive.google.com/file/d/1LDmS7qRgkkkTOSL-rOY4LEANa6GGTE2Y/view?usp=sharing" target="_blank">here</a>. <br /></p><p>In this post, I want to <a href="https://m-phi.blogspot.com/2020/08/accuracy-and-explanation-thoughts-on.html" target="_blank">continue my discussion</a> of the part of van Fraassen's argument against inference to the best explanation (IBE) that turns on its alleged clash with Bayesian Conditionalization (BC). In the previous post, I looked at <a href="https://onlinelibrary.wiley.com/doi/abs/10.1111/1467-9213.12032" target="_blank">Igor Douven's argument</a> that there are at least some ways of valuing accuracy on which updating by IBE comes out better than BC. I concluded that Douven's arguments don't save IBE; BC is still the only rational way to update. <br /><br />The setting for Douven's arguments was individualist epistemology. That is, he considered only the single agent collecting evidence directly from the world and updating in the light of it. But of course we often receive evidence not directly from the world, but indirectly through the opinions of others. I learn how many positive SARS-CoV-2 tests there have been in my area in the past week not my inspecting the test results myself but by listening to the local health authority. In their 2017 paper, <a href="https://doi.org/10.1093/bjps/axv025" target="_blank">'Inference to the Best Explanation versus Bayes’s Rule in a Social Setting'</a>, Douven joined with Sylvia Wenmackers to ask how IBE and BC fare in a context in which some of my evidence comes from the world and some from learning the opinions of others, where those others are also receiving some of their evidence from the world and some from others, and where one of those others from whom they're learning might be me. Like Douven's study of IBE vs BC in the individual setting, Douven and Wenmackers conclude in favour of IBE. Indeed, their conclusion in this case is considerably stronger than in the individual case:</p><p></p><blockquote>The upshot will be that if agents not only update their degrees of belief on the basis of evidence, but also take into account the degrees of belief of their epistemic neighbours, then the noted advantage of Bayesian updating [from Douven's earlier paper] evaporates and IBE does better than Bayes’s rule on every reasonable understanding of inaccuracy minimization. (536-7)</blockquote><p></p><p>As in the previous post, I want to stick up for BC. As in the individualist setting, I think this is the update rule we should use in the social setting.<br /><br /><span></span></p><a name='more'></a>Following van Fraassen's original discussion and the strategy pursued in Douven's solo piece, Douven and Wenmackers take the general and ill-specified question whether IBE is better than BC and make it precise by asking it in a very specific case. We imagine a group of individuals. Each has a coin. All coins have the same bias. No individual knows what this shared bias is, but they do know that it is the same bias for each coin, and they know that the options are given by the following bias hypotheses:<p></p><p>$B_0$: coin has 0% chance of landing heads</p><p>$B_1$: coin has 10% chance of landing heads</p><p>$\ldots$ <br /></p><p>$B_9$: coin has 90% chance of landing heads</p><p>$B_{10}$: coin has 100% chance of landing heads</p><p>Though they don't say so, I think Douven and Wenmackers assume that all individuals have the same prior over $B_0, \ldots, B_{10}$, namely, the uniform prior; and each satisfies the Principal Principle, and so their credences in everything else follows from their credences in $B_0, \ldots, B_{10}$. As we'll see, we needn't assume that they all have the uniform prior over the bias hypotheses. In any case, they assume that things proceed as follows:<br /></p><p><i>Step (i)</i> Each member tosses their coin some fixed number of times. This produces their worldly evidence for this round.</p><p><i>Step (ii)</i> Each then updates their credence function on this worldly evidence they've obtained. To do this, each member uses the same updating rule, either BC or a version of IBE. We'll specify these in more detail below. </p><p><i>Step (iii) </i>Each then learns the updated credence functions of the others in the group. This produces their social evidence for this round.<br /><br /><i>Step (iv) </i>They then update their own credence function by taking the average of their credence function and the other credence functions in the group that lie within a certain distance of theirs. The set of credence functions that lie within a certain distance of one's own, Douven and Wenmackers call one's bounded confidence interval.</p><p>They then repeat this cycle a number of times, each time an individual begins with the credence function they reached at the end of the previous cycle.<br /><br />Douven and Wenmackers use simulation techniques to see how this group of individuals perform for different updating rules used in step (ii) and different specifications of how close a credence function must lie to yours in order to be included in the average in step (iv). Here's the class of updating rules that they consider: if $P$ is your prior and $E$ is your evidence then your updated credence function should be$$P^c_E(B_i) = \frac{P(B_i)P(E|B_i) + f_c(B_i, E)}{\sum^{10}_{k=0} \left (P(B_k)P(E|B_k) + f_c(B_k, E) \right )}$$where$$f_c(B_i, E) = \left \{ \begin{array}{ll} c & \mbox{if } P(E | B_i) > P(E | B_j) \mbox{ for all } j \neq i \\ \frac{1}{2}c & \mbox{if } P(E | B_i) = P(E|B_j) > P(E | B_k) \mbox{ for all } k \neq j, i \\ 0 & \mbox{otherwise} \end{array} \right. $$That is, for $c = 0$, this update rule is just BC, while for $c > 0$, it gives a little boost to whichever hypothesis best explains the evidence $E$, where providing the best explanation for a series of coin tosses amounts to making it most likely, and if two bias hypotheses make the evidence most likely, they split the boost between them. Douven and Wenmackers consider $c = 0, 0.1, \ldots, 0.9, 1$. For each rule, specified by $c$, they also consider different sizes of bounded confidence intervals. These are specified by the parameter $\varepsilon$. Your bounded confidence interval for $\varepsilon$ includes each credence function for which the average difference between the credences it assigns and the credences you assign is at most $\varepsilon$. Thus, $\varepsilon = 0$ is the most exclusive, and includes only your own credence function, while $\varepsilon = 1$ is the most inclusive, and includes all credence functions in the group. Again, Douven and Wenmackers consider $\varepsilon = 0, 0.1, \ldots, 0.9, 1$. Here are two of their main results:<br /></p><ol style="text-align: left;"><li>For each bias other than $p = 0.1$ or $0.9$, there is an explanationist rule (i.e. $c > 0$ and some specific $\varepsilon$) that gives rise to a lower average inaccuracy at the end of the process than all BC rules (i.e. $c = 0$ and any $\varepsilon$).</li><li>There is an averaging explanationist rule (i.e. $c > 0$ and $\varepsilon > 0$) such that, for each bias other than $p = 0, 0.1, 0.9, 1$, it gives rise to lower average inaccuracy than all BC rules (i.e. $c = 0$ and any $\varepsilon$).</li></ol><p>Inaccuracy is measured by the Brier score throughout. <br /><br />Now, you can ask whether these results are enough to tell so strongly in favour of IBE. But that isn't my concern here. Rather, I want to focus on a more fundamental problem: Douven and Wenmackers' argument doesn't really compare BC with IBE. They're comparing BC-for-worldly-data-plus-Averaging-for-social-data with IBE-for-worldly-data-plus-Averaging-for-social-data. So their simulation results don't really impugn BC, because the average inaccuracies that they attribute to BC don't really arise from it. They arise from using BC in step (ii), but something quite different in step (iv). Douven and Wenmackers ask the Bayesian to respond to the social evidence they receive using a non-Bayesian rule, namely, Averaging. And we can see just how far Averaging lies from BC by considering the following version of the example we have been using throughout.<br /><br />Consider the biased coin case, and suppose there are just three members of the group. And suppose they all start with the uniform prior over the bias hypotheses. At step (i), they each toss their coin twice. The first individual's coin lands $HT$, the second's $HH$, and the third's $TH$. So, at step (ii), if they all use BC (i.e. $c = 0$), they update on this worldly evidence as follows, where $P$ is the shared prior:<br />$$\begin{array}{r|ccccccccccc}<br />& B_0 & B_1& B_2& B_3& B_4& B_5& B_6& B_7& B_8& B_9& B_{10} \\<br />\hline<br />&&&&&&&&&& \\<br />P & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} & \frac{1}{11} \\<br />&&&&&&&&&& \\<br />P(-|HT) & 0 & \frac{9}{165} & \frac{16}{165}& \frac{21}{165}& \frac{24}{165} & \frac{25}{165}& \frac{24}{165}& \frac{21}{165}& \frac{16}{165}& \frac{9}{165}& 0\\<br />&&&&&&&&&& \\<br />P(-|HH) & 0 & \frac{1}{385} & \frac{4}{385}& \frac{9}{385}& \frac{16}{385}& \frac{25}{385}& \frac{36}{385}& \frac{49}{385}& \frac{64}{385}& \frac{81}{385}& \frac{100}{385}\\<br />&&&&&&&&&& \\<br />P(-|TH) & 0 & \frac{9}{165} & \frac{16}{165}& \frac{21}{165}& \frac{24}{165} & \frac{25}{165}& \frac{24}{165}& \frac{21}{165}& \frac{16}{165}& \frac{9}{165}& 0\\<br />\end{array}$$<br />Now, at step (iii), they each learn the other's distribution. And they average on that. Let's suppose I'm the first individual. Then I have two choices for my BCI. It either includes my own credence function $P(-|HT)$ and the third individual's $P(-|TH)$, which are identical, or it includes all three, $P(-|HT), P(-|HH), P(-|TH)$. Let's suppose it includes all three. Here is the outcome of averaging:$$\begin{array}{r|ccccccccccc}<br />& B_0 & B_1& B_2& B_3& B_4& B_5& B_6& B_7& B_8& B_9& B_{10} \\<br />\hline<br />&&&&&&&&&& \\<br />\mbox{Av} & 0 & \frac{129}{3465} & \frac{236}{3465}& \frac{321}{3465}& \frac{384}{3465}& \frac{425}{3465}& \frac{444}{3465}& \frac{441}{3465}& \frac{416}{3465}& \frac{369}{3465}& \frac{243}{3465}<br />\end{array}$$<br />And now compare that with what they would do if they updated at step (iv) using BC rather than Averaging. I learn the distributions of the second and third individuals. Now, since I know how many times they tossed their coin, and I know that they updated by BC at step (ii), I thereby learn something about how their coin landed. I know that it landed in such a way that would lead them to update to $P(-|HH)$ and $P(-|TH)$, respectively. Now what exactly does this tell me? In the case of the second individual, it tells me that their coin landed $HH$, since that's the only evidence that would lead them to update to $P(-|HH)$. In the case of the third individual, my evidence is not quite so specific. I learn that their coin either landed $HT$ or $TH$, since either of those, and only one of those, would lead them to update to $P(-|TH)$. In general, learning an individual's posteriors when you know their prior and the number of times they've tossed the coin will teach you how many heads they saw and how many tails, though it won't tell you the order in which they saw them. But that's fine. We can still update on that information using BC, and indeed BC will tell us to adopt the same credence as we would if we were to learn the more specific evidence of the order in which the coin tosses landed. If we do so in this case, we get:<br />$$\begin{array}{r|ccccccccccc}<br />& B_0 & B_1& B_2& B_3& B_4& B_5& B_6& B_7& B_8& B_9& B_{10} \\<br />\hline&&&&&&&&&& \\<br />\mbox{Bayes} & 0 & \frac{81}{95205} & \frac{1024}{95205} & \frac{3969}{95205} & \frac{9216}{95205} & \frac{15625}{95205} & \frac{20736}{95205} & \frac{21609}{95205} & \frac{16384}{95205} & \frac{6561}{95205} &0 \\<br />\end{array}<br />$$And this is pretty far from what I got by Averaging at step (iv).<br /><br />So updating using BC is very different from averaging. Why, then, do Douven and Wenmackers use Averaging rather than BC for step (iv)? Here is their motivation:</p><p></p><blockquote>[T]aking a convex combination of the probability functions of the individual agents in a group is the best studied method of forming social probability functions. Authors concerned with social probability functions have mostly considered assigning different weights to the probability functions of the various agents, typically in order to reflect agents’ opinions about other agents’ expertise or past performance. The averaging part of our update rule is in some regards simpler and in others less simple than those procedures. It is simpler in that we form probability functions from individual probability functions by taking only straight averages of individual probability functions, and it is less simple in that we do not take a straight average of the probability functions of all given agents, but only of those whose probability function is close enough to that of the agent whose probability is being updated. (552)</blockquote><p></p><p>In some sense, they're right. Averaging or linear pooling or taking a convex combination of individual credence functions is indeed the best studied method of forming social credence functions. And there are good justifications for it: <a href="https://www.math.utk.edu/~wagner/papers/arithmetic.pdf" target="_blank">János Aczél and Carl Wagner</a> and, independently, <a href="https://www.jstor.org/stable/2287843" target="_blank">Kevin J. McConway</a>, give a neat axiomatic characterization; and <a href="https://philpapers.org/rec/PETOTA-3" target="_blank">I've argued</a> that there are accuracy-based reasons to use it in particular cases. The problem is that our situation in step (iv) is not the sort of situation in which you should use Averaging. Arguments for Averaging concern those situations in which you have a group of individuals, possibly experts, and each has a credence function over the same set of propositions, and you want to produce a single credence function that could be called the group's collective credence function. Thus, for instance, if I wish to give the SAGE group's collective credence that there will be a safe and effective SARS-CoV-2 vaccine by March 2021, I might take the average of their individual credences. But this is quite a different task from the one that faces me as the first individual when I reach step (iv) of Douven and Wenmackers' process. There, I already have credences in the propositions in question. What's more, I know how the other individuals update and the sort of evidence they will have received, even if I don't know which particular evidence of that sort they have. And that allows me to infer from their credences after the update at step (ii) a lot about the evidence they receive. And I have opinions about the propositions in question conditional on the different evidence my fellow group members received. And so, in this situation, I'm not trying to summarise our individual opinions as a single opinion. Rather, I'm trying to use their opinions as evidence to inform my own. And, in that case, BC is better than Averaging. So, in order to show that IBE is superior to BC in some respect, it doesn't help to compare BC at step (ii) + Averaging at step (iv) with IBE at (ii) + Averaging at (iv). It would be better to compare BC at (ii) and (iv) with IBE at (ii) and (iv). <br /><br />So how do things look if we do that? Well, it turns out that we don't need simulations to answer the question. We can simply appeal to the mathematical results we mentioned in the previous post: first, <a href="https://philpapers.org/rec/GREJCC" target="_blank">Hilary Greaves and David Wallace's expected accuracy argument</a>; and second, the accuracy dominance argument that <a href="https://onlinelibrary.wiley.com/doi/full/10.1111/nous.12258" target="_blank">Ray Briggs and I</a> gave. Or, more precisely, we use the slight extensions of those results to multiple learning experiences that I sketched in the previous post. For both of those results, the background framework is the same. We begin with a prior, which we hold at $t_0$, before we begin gathering evidence. And we then look forward to a series of times $t_1, \ldots, t_n$ at each of which we will learn some evidence. And, for each time, we know the possible pieces of evidence we might receive, and we plan, for each time, which credence function we would adopt in response to each of the pieces of evidence we might learn at that time. Thus, formally, for each $t_i$ there is a partition from which our evidence at $t_i$ will come. For each $t_{i+1}$, the partition is a fine-graining of the partition at $t_i$. That is, our evidence gets more specific as we proceed. In the case we've been considering, at $t_1$, we'll learn the outcome of our own coin tosses; at $t_2$, we'll add to that our fellow group members' credence functions at $t_1$, from which we can derive a lot about the outcome of their first run of coin tosses; at $t_3$, we'll add to that the outcome of our next run of our own coin tosses; at $t_4$, we'll add our outcomes of the other group members' coin tosses by learning their credences at $t_3$; and so on. The results are then as follows: </p><p><b>Theorem (Extended Greaves and Wallace)</b> <i>For any strictly proper inaccuracy measure, the updating rule that minimizes expected inaccuracy from the point of view of the prior is BC</i>.</p><p><b>Theorem (Extended Briggs and Pettigrew)</b> <i>For any continuous and strictly proper inaccuracy measure, if your updating rule is not BC, then there is an alternative prior and alternative updating rule that accuracy dominates your prior and your updating rule</i>.</p><p>Now, these results immediately settle one question: if you are an individual in the group, and you know which update rules the others have chosen to use, then you should certainly choose BC for yourself. After all, if you have picked your prior, then it expects picking BC to minimize your inaccuracy, and thus expects picking BC to minimize the total inaccuracy of the group that includes you; and if you have not picked your prior, then if you consider a prior together with something other than BC as your updating rule, there's some other combination you could chose instead that is guaranteed to do better, and thus some other combination you could choose that is guaranteed to improve the total accuracy of the group. But Douven and Wenmackers don't set up the problem like this. Rather, they assume that all members of the group use the same updating rule. So the question is whether everyone picking BC is better than everyone picking something else. Fortunately, at least in the case of the coin tosses, this does follow. As we'll see, things could get more complicated with other sorts of evidence.<br /><br />If you know the updating rules that others will use, then you pick your updating rule simply on the basis of its ability to get you the best accuracy possible; the others have made their choices and you can't affect that. But if you are picking an updating rule for everyone to use, you must consider not only its properties as an updating rule for the individual, but also its properties as a means of signalling to the other members what evidence you have. Thus, prior to considering the details of this, you might think that there could be an updating rule that is very good at producing accurate responses to evidence, but poor at producing a signal to others of the evidence you've received---there might be a wide range of different pieces of evidence you could receive that would lead you to update to the same posterior using this rule, and in that case, learning your posterior would give little information about your evidence. If that were so, we might prefer an updating rule that does not produce such accurate updates, but does signal very clearly what evidence is received. For, in that situation, each individual would produce a less accurate update at step (ii), but would then receive a lot more evidence at step (iv), because the update at step (ii) would signal the evidence that the other members of the group received much more clearly. However, in the coin toss set up that Douven and Wenmackers consider, this isn't an issue. In the coin toss case, learning someone's posterior when you know their prior and how many coin tosses they have observed allows you to learn exactly how many heads and how many tails they observed. It doesn't tell you the order in which you learned them, but knowing that further information wouldn't affect how you would update anyway, either on the BC rule or on the IBE rule---learning $HT \vee TH$ leads to the same update as learning $HT$ for both Bayesian and IBEist. So when we are comparing them, we can consider the information learned at step (ii) and step (iv) both to be worldly information. Both give us information about the tosses of the coin that our peers witnessed. So when we are comparing them, we needn't take into account how good they are at signalling the evidence you have. They are both equally good and both very good. So comparing them when choosing a single rule that each member of the group must use, we need only compare the accuracy of using them as update rules. And the theorems above indicate that BC wins out on that measure.</p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com76tag:blogger.com,1999:blog-4987609114415205593.post-9501253408257158292020-08-24T07:55:00.003+01:002022-07-27T19:12:48.586+01:00Accuracy and explanation: thoughts on Douven<p>For a PDF of this post, see <a href="https://drive.google.com/file/d/1DUmhOSySmEIn2TCKiaUtUcs2D2zDlVzi/view?usp=sharing" target="_blank">here</a>. <br /></p><p>Igor has eleven coins in his pocket. The first has 0% chance of landing heads, the second 10% chance, the third 20%, and so on up to the tenth, which has 90% chance, and the eleventh, which has 100% chance. He picks one out without letting me know which, and he starts to toss it. After the first 10 tosses, it has landed tails 5 times. How confident should I be that the coin is fair? That is, how confident should I be that it is the sixth coin from Igor's pocket; the one with 50% chance of landing heads? According to the Bayesian, the answer is calculated as follows:$$P_E(H_5) = P(H_5 | E) = \frac{P(H_5)P(E | H_5)}{\sum^{10}_{i=0} P(H_i) P(E|H_i)}$$where</p><ul style="text-align: left;"><li>$E$ is my evidence, which says that 5 out of 10 of the tosses landed heads,</li><li>$P_E$ is my new posterior updating credence upon learning the evidence $E$,</li><li>$P$ is my prior,</li><li>$H_i$ is the hypothesis that the coin has $\frac{i}{10}$ chance of landing heads,</li><li>$P(H_0) = \ldots = P(H_{10}) = \frac{1}{11}$, since I know nothing about which coin Igor pulled from his pocket, and</li><li>$P(E | H_i) = \left ( \frac{i}{10} \right )^5 \left (\frac{10-i}{10} \right )^5$, by the Principal Principle, and since each coin toss is independent of each other one.</li></ul><p>So, upon learning that the coin landed heads five times out of ten, my posterior should be:$$P_E(H_5) = P(H_5 | E) = \frac{P(H_5)P(E | H_5)}{\sum^{10}_{i=0} P(H_i) P(E|H_i)} = \frac{\frac{1}{11} \left ( \frac{5}{10} \right )^5\left ( \frac{5}{10} \right )^5}{\sum^{10}_{i=1}\frac{1}{11} \left ( \frac{i}{10} \right )^5 \left (\frac{10-i}{10} \right )^5 } \approx 0.2707$$But some philosophers have suggested that this is too low. The Bayesian calculation takes into account how likely the hypothesis in question makes the evidence, as well as how likely I thought the hypothesis in the first place, but it doesn't take into account that the hypothesis explains the evidence. We'll call these philosophers explanationists. Upon learning that the coin landed heads five times out of ten, the explanationist says, we should be most confident in $H_5$, the hypothesis that the coin is fair, and the Bayesian calculation does indeed give this. But we should be most confident in part because $H_5$ best explains the evidence, and the Bayesian calculation takes no account of this.<br /><br /><span></span></p><a name='more'></a>To accommodate the explanationist's demand, <a href="https://onlinelibrary.wiley.com/doi/abs/10.1111/1467-9213.12032" target="_blank">Igor Douven</a> proposes the following alternative updating rule:$$P_E(H_k) = P(H_k | E) = \frac{P(H_k)P(E | H_k) + f(H_k, E)}{\sum^{10}_{i=0} (P(H_i) P(E|H_i) + f(H_i, E))}$$where $f$ gives a little boost to $H_k$ if it is the best explanation of $E$ and not if it isn't. Perhaps, for instance,<p></p><ul style="text-align: left;"><li>$f(H_k, E) = 0.1$, if the frequency of heads among the coin tosses that $E$ reports is uniquely closest to the chance of heads according to $H_k$, namely, $\frac{k}{10}$,</li><li>$f(H_k, E) = 0.05$, if the frequency of heads among the coin tosses that $E$ reports is equally closest to the chance of heads according to $H_k$ and another hypothesis,</li><li>$f(H_k, E) = 0$, otherwise.</li></ul><p>Thus, according to this:$$P_E(H_5) = \frac{P(H_5)P(E | H_5) + 0.1}{\left (\sum^{10}_{i=0} P(H_i) P(E|H_i) \right ) + 0.1} = \frac{\frac{1}{11} \left ( \frac{5}{10} \right )^5\left ( \frac{5}{10} \right )^5 + 0.1}{\sum^{10}_{i=1}\frac{1}{11} \left ( \frac{i}{10} \right )^5 \left (\frac{10-i}{10} \right )^5 + 0.1 } \approx 0.9746$$So, as required, $H_5$ certainly gets a boost in posterior probability because it best explains the run of heads and tails we observe.<br /><br />Before we move on, it's worth noting a distinctive feature of this case. In many cases where we wish to apply something like abduction or inference to the best explanation, we might think that we can record our enthusiasm for good explanations in the priors. For instance, suppose I have two scientific theories, $T_1$ and $T_2$, both of which predict the evidence I've collected. So, they both make the evidence equally likely. But I want to assign higher probability to $T_1$ upon receipt of that evidence because it provides a better explanation for the evidence. Then I should simply encode this in my prior. That is, I should assign $P(T_1) > P(T_2)$. But that sort of move isn't open to us in Douven's example. The reason is that none of the chance hypotheses are better explanations in themselves: none is simpler or more general or what have you. But rather, for each, there is evidence we might obtain such that it is a better explanation of that evidence. But before we obtain the evidence, we don't know which will prove the better explanation of it, and so can't accommodate our explanationist instincts by giving that hypothesis a boost in our prior.<br /><br />Now let's return to the example. There are well known objections to updating in the explanationist way Douven suggests. Most famously, van Fraassen pointed out that we have good reasons to comply with the Bayesian method of updating, and the explanationist method deviates quite dramatically from that (<i>Laws and Symmetry</i>, chapter 6) . When he was writing, the most compelling argument was <a href="https://philpapers.org/rec/LEWWC" target="_blank">David Lewis' diachronic Dutch Book argument</a>. If you plan to update as Douven suggests, by giving an extra-Bayesian boost to the hypothesis that best explains the evidence, then there is a series of bets you'll accept before you receive the evidence and another set you'll accept afterwards that, taken together, will lose you money for sure. Douven is unfazed. He first suggests that vulnerability to a Dutch Book does not impugn your epistemic rationality, but only your practical rationality. He notes <a href="http://fitelson.org/probability/skyrms_coherence.pdf" target="_blank">Skyrms's claim</a> that, in the case of synchronic Dutch Books, such vulnerability reveals an inconsistency in your assessment of the same bet presented in different ways, and therefore perhaps some epistemic failure, but notes that this cannot be extended to the diachronic case. In any case, he says, avoiding the machinations of malevolent bookies is only one practical concern that we have, and, let's be honest, not a very pressing one. What's more, he points out that, while updating in the Bayesian fashion serves one practical end, namely, making us immune to these sorts of diachronic sure losses, there are other practical ends it might not serve as well. For instance, he uses computer simulations to show that, if we update in his explanationist way, we'll tend to assign credence greater than 0.99 in the true hypothesis much more quickly than if we update in the Bayesian way. He admits that we'll also tend to assign credence greater than 0.99 in a false hypothesis much more quickly than if we use Bayesian updating. But he responds, again with the results of a computer simulation result: suppose we keep tossing the coin until one of the rules assigns more than 0.99 to a hypothesis; then award points to that rule if the hypothesis it becomes very confident in is true, and deduct them if it is false; then the explanationist updating rule will perform better on average than the Bayesian rule. So, if there is some practical decision that you will make only when your credence in a hypothesis exceeds 0.99 -- perhaps the choice is to administer a particular medical treatment, and you need to be very certain in your diagnosis before doing so -- then you will be better off on average updating as Douven suggests, rather than as the Bayesian requires.<br /><br />So much for the practical implications of updating in one way or another. I am more interested in the epistemic implications, and so is Douven. He notes that, since van Fraassen gave his argument, there is a new way of justifying the Bayesian demand to update by conditioning on your evidence. These are the accuracy arguments. While Douven largely works with the argument for conditioning that <a href="https://philpapers.org/rec/LEIAOJ" target="_blank">Hannes Leitgeb and I</a> gave, I think the better version of that argument is due to <a href="https://philpapers.org/rec/GREJCC" target="_blank">Hilary Greaves and David Wallace</a>. The idea is that, as usual, we measure the inaccuracy of a credence function using a strictly proper inaccuracy measure $\mathfrak{I}$. That is, if $P$ is a probabilistic credence function and $w$ is a possible world, then $\mathfrak{I}(P, w)$ gives the inaccuracy of $P$ at $w$. And, if $P$ is a probabilistic credence function, $P$ expects itself to be least inaccurate. That is, $\sum_w P(w) \mathfrak{I}(P, w) < \sum_w P(w) \mathfrak{I}(Q, w)$, for any credence function $Q \neq P$. Then Greaves and Wallace ask us to consider how you might plan to update your credence function in response to different pieces of evidence you might receive. Thus, suppose you know that the evidence you'll receive will be one of the following propositions, $E_1, \ldots, E_m$, which form a partition. This is the situation you're in if you know that you're about to witness 10 tosses of a coin, for instance, as in Douven's example: $E_1$ might be $HHHHHHHHHH$, $E_2$ might be $HHHHHHHHHT$, and so on. Then suppose you plan how you'll respond to each. If you learn $E_i$, you'll adopt $P_i$. Then we'll call this updating plan $\mathcal{R}$ and write it $(P_1, \ldots, P_m)$. Then we can calculate the expected inaccuracy of a given updating plan. Its inaccuracy at a world is the inaccuracy of the credence function it recommends in response to learning the element of the partition that is true at that world. That is, for world $w$ at which $E_i$ is true,$$\mathfrak{I}(\mathcal{R}, w) = \mathfrak{I}(P_i, w)$$And Greaves and Wallace show that the updating rule your prior expects to be best is the Bayesian one. That is, if there is $E_i$ and $P(E_i) > 0$ and $P_i(-) \neq P(X|E_i)$, then there is an alternative updating rule $\mathcal{R}^\star = (P^\star_1, \ldots, P^\star_m)$ such that$$\sum_w P(w) \mathfrak{I}(\mathcal{R}^\star, w) < \sum_w P(w) \mathfrak{I}(\mathcal{R}, w)$$So, in particular, your prior expects the Bayesian rule to be more accurate than Douven's rule. <br /><br />In response to this, Douven points out that there are many ways in which we might value the accuracy of our updating plans. For instance, the Greaves and Wallace argument considers only your accuracy at a single later point in time, after you've received a single piece of evidence and updated only on it. But, Douven argues, we might be interested not in the one-off inaccuracy of a single application of an updating rule, but rather in its inaccuracy in the long run. And we might be interested in different features of the long-run total inaccuracy of using that rule: we might be interested in just adding up all of the inaccuracies of the various credence functions you obtain from multiple applications of the rule; or we might be less interested in the inaccuracies of the interim credence functions and more interested in the inaccuracy of the final credence function you obtain after multiple updates. And, Douven claims, the accuracy arguments do not tell us anything about which performs better out of the Bayesian and explanationist approaches when viewed in these different ways.<br /><br />However, that's not quite right. It turns out that we can, in fact, adapt the Greaves and Wallace argument to cover these cases. To see how, it's probably best to illustrate it with the simplest possible case, but it should be obvious how to scale up the idea. So suppose: </p><ul style="text-align: left;"><li>my credences are defined over four worlds, $XY$, $X\overline{Y}$, $\overline{X}Y$, and $\overline{X}\overline{Y}$;</li><li>my prior at $t_0$ is $P$;</li><li>at $t_1$, I'll learn either $X$ or its negation $\overline{X}$, and I'll respond with $P_X$ or $P_{\overline{X}}$, respectively;</li><li>at $t_2$, I'll learn $XY$, $X\overline{Y}$, $\overline{X}Y$, or $\overline{X} \overline{Y}$, and I'll respond with $P_{XY}$, $P_{X\overline{Y}}$, $P_{\overline{X}Y}$, or $P_{\overline{X}\overline{Y}}$, respectively.</li></ul><p>For instance, I might know that a coin is going to be tossed twice, once just before $t_1$ and once just before $t_2$. So $X$ is the proposition that it lands heads on the first toss, i.e., $X = \{HH, HT\}$, while $\overline{X}$ is the proposition it lands tails on the first toss $\overline{X} = \{TH, TT\}$. And then $Y$ is the proposition it lands heads on the second toss. So $XY = \{HH\}$, $X\overline{Y} = \{HT\}$, and so on. <br /><br />Now, taken together, $P_X$, $P_{\overline{X}}$, $P_{XY}$, $P_{X\overline{Y}}$, $P_{\overline{X}Y}$, and $P_{\overline{X}\overline{Y}}$ constitute my updating plan---let's denote that $\mathcal{R}$. Now, how might be measure the inaccuracy of this plan $\mathcal{R}$? Well, we want to assign a weight to the inaccuracy of the credence function it demands after the first update -- let's call that $\alpha_1$; and we want a weight for the result of the second update -- let's call that $\alpha_2$. So, for instance, if I'm interested in the total inaccuracy obtained by following this rule, and each time is just as important as each other time, I just set $\alpha_1 = \alpha_2$; but if I care much more about my final inaccuracy, then I let $\alpha_1 \ll \alpha_2$. Then the inaccuracy of my updating rule is$$\begin{eqnarray*}<br />\mathfrak{I}(\mathcal{R}, XY) & = & \alpha_1 \mathfrak{I}(P_X, XY) + \alpha_2\mathfrak{I}(P_{XY}, XY) \\<br />\mathfrak{I}(\mathcal{R}, X\overline{Y}) & = & \alpha_1 \mathfrak{I}(P_X, X\overline{Y}) + \alpha_2\mathfrak{I}(P_{\overline{X}Y}, X\overline{Y}) \\<br />\mathfrak{I}(\mathcal{R}, \overline{X}Y) & = & \alpha_1 \mathfrak{I}(P_{\overline{X}}, \overline{X}Y) + \alpha_2\mathfrak{I}(P_{\overline{X}Y}, \overline{X}Y) \\<br />\mathfrak{I}(\mathcal{R}, \overline{X}\overline{Y}) & = & \alpha_1 \mathfrak{I}(P_{\overline{X}}, \overline{X}\overline{Y}) + \alpha_2\mathfrak{I}(P_{\overline{X}\overline{Y}}, \overline{X}\overline{Y})<br />\end{eqnarray*}$$Thus, the expected inaccuracy of $\mathcal{R}$ from the point of view of my prior $P$ is:<br /><br />$P(XY)\mathfrak{I}(\mathcal{R}, XY) + P(X\overline{Y})\mathfrak{I}(\mathcal{R}, X\overline{Y}) + P(\overline{X}Y)\mathfrak{I}(\mathcal{R}, \overline{X}Y) + P(\overline{X} \overline{Y})\mathfrak{I}(\mathcal{R}, \overline{X}\overline{Y}) = $<br /><br />$P(XY)[\alpha_1 \mathfrak{I}(P_X, XY) + \alpha_2\mathfrak{I}(P_{XY}, XY)] + $<br /><br />$P(X\overline{Y})[\alpha_1 \mathfrak{I}(P_X, X\overline{Y}) + \alpha_2\mathfrak{I}(P_{X\overline{Y}}, X\overline{Y})] + $<br /><br />$P(\overline{X}Y)[\alpha_1 \mathfrak{I}(P_{\overline{X}}, \overline{X}Y) + \alpha_2\mathfrak{I}(P_{\overline{X}Y}, \overline{X}Y)] + $<br /><br />$P(\overline{X}\overline{Y})[\alpha_1 \mathfrak{I}(P_{\overline{X}}, \overline{X}\overline{Y}) + \alpha_2\mathfrak{I}(P_{\overline{X}\overline{Y}}, \overline{X}\overline{Y})]$<br /><br />But it's easy to see that this is equal to:<br /><br />$\alpha_1[P(XY)\mathfrak{I}(P_X, XY) + P(X\overline{Y})\mathfrak{I}(P_X, X\overline{Y}) + $<br /><br />$P(\overline{X}Y)\mathfrak{I}(P_{\overline{X}}, \overline{X}Y) + P(\overline{X}\overline{Y})\mathfrak{I}(P_{\overline{X}}, \overline{X}\overline{Y})] + $<br /><br />$\alpha_2[\mathfrak{I}(P_{XY}, XY) + P(X\overline{Y})\mathfrak{I}(P_{X\overline{Y}}, X\overline{Y}) + $<br /><br />$P(\overline{X}Y)\mathfrak{I}(P_{\overline{X}Y}, \overline{X}Y) + P(\overline{X}\overline{Y})\mathfrak{I}(P_{\overline{X}\overline{Y}}, \overline{X}\overline{Y})]$<br /><br />Now, this is the weighted sum of the expected inaccuracies of the two parts of my updating plan taken separately; the part that kicks in at $t_1$, and the part that kicks in at $t_2$. And, thanks to Greaves and Wallace's result, we know that each of those expected inaccuracies is minimized by the rule that demands you condition on your evidence. Now, we also know that conditioning $P$ on $XY$ is the same as conditioning $P(-|X)$ on $XY$, and so on. So a rule that tells you, at $t_2$, to update your $t_0$ credence function on your total evidence at $t_2$ is also one that tells you, at $t_2$, to update your $t_1$ credence function on your total evidence at $t_2$. So, of the updating rules that cover the two times $t_1$ and $t_2$, the one that minimizes expected inaccuracy is the one that results from conditioning at each time. That is, if the part of $\mathcal{R}$ that kicks in at $t_1$ doesn't demand I condition my prior on my evidence at $t_1$, or if the part of $\mathcal{R}$ that kicks in at $t_2$ doesn't demand I condition my credence function at $t_1$ on my evidence at $t_2$, then there is an alternative rule $\mathcal{R}^\star$, that $P$ expects to be more accurate: that is,$$\sum_w P(w)\mathfrak{I}(\mathcal{R}^\star, w) < \sum_w P(w)\mathfrak{I}(\mathcal{R}, w)$$And, as I mentioned above, it's clear how to generalize this to cover not just updating plans that cover two different times at which you receive evidence, but any finite number.<br /><br />However, I think Douven would not be entirely moved by this. After all, while he is certainly interested in the long-run effects on inaccuracy of using one updating rule or another, he thinks that looking only to expected inaccuracy is a mistake. He thinks that we care about other features of updating rules. Indeed, he provides us with one, and uses computer simulations to show that, in the toy coin tossing case that we've been using, the explanationist account has that desirable feature to a greater degree than the Bayesian account.</p><blockquote>For each possible bias value, we ran 1000 simulations of a sequence of 1000 tosses. As previously, the explanationist and the Bayesian updated their degrees of belief after each toss. We registered in how many of those 1000 simulations the explanationist incurred a lower penalty than the Bayesian at various reference points [100 tosses, 250, 500, 750, 1000], at which we calculated both Brier penalties and log score penalties. The outcomes [...] show that, on either measure of inaccuracy, IBE is most often the winner—it incurs the lowest penalty -- at each reference point. Hence, at least in the present kind of context, IBE seems a better choice than Bayes' rule. (page 439)</blockquote>How can we square this with the Greaves and Wallace result? Well, as Douven goes on to explain: "[the explanationist rule] in general achieves greater accuracy than [the Bayesian], even if typically not much greater accuracy"; but "[the Bayesian rule] is less likely than [explanationist rule] to ever make one vastly inaccurate, even though the former typically makes one somewhat more inaccurate than the latter." So the explanationist is most often more accurate, but when it is more accurate, it's only a little more, while when it is less accurate, it's a lot less. So, in expectation, the Bayesian rule wins. Douven then argues that you might be more interested in being more likely to be more accurate, rather than being expectedly more accurate. <br /><br />Perhaps. But in any case there's another accuracy argument for the Bayesian way of updating that doesn't assume that expected inaccuracy is the thing you want to minimize. This is an argument that <a href="https://philpapers.org/rec/BRIAAA-11" target="_blank">Ray Briggs and I</a> gave a couple of years ago. I'll illustrate it in the same setting we used above, where we have prior $P$, at $t_1$ we'll learn $X$ or $\overline{X}$, and at $t_2$ we'll learn $XY$, $X\overline{Y}$, $\overline{X}Y$, or $\overline{X} \overline{Y}$. And we measure the inaccuracy of an updating rule $\mathcal{R} = (P_X, P_{\overline{X}}, P_{XY}, P_{X\overline{Y}}, P_{\overline{X}Y}, P_{\overline{X}\overline{Y}})$ for this as follows: <br />$$\begin{eqnarray*}<br />\mathfrak{I}(\mathcal{R}, XY) & = & \alpha_1 \mathfrak{I}(P_X, XY) + \alpha_2\mathfrak{I}(P_{XY}, XY) \\<br />\mathfrak{I}(\mathcal{R}, X\overline{Y}) & = & \alpha_1 \mathfrak{I}(P_X, \overline{X}Y) + \alpha_2\mathfrak{I}(P_{\overline{X}Y}, X\overline{Y}) \\<br />\mathfrak{I}(\mathcal{R}, \overline{X}Y) & = & \alpha_1 \mathfrak{I}(P_{\overline{X}}, \overline{X}Y) + \alpha_2\mathfrak{I}(P_{X\overline{Y}}, \overline{X}Y) \\<br />\mathfrak{I}(\mathcal{R}, \overline{X}\overline{Y}) & = & \alpha_1 \mathfrak{I}(P_{\overline{X}}, \overline{X}\overline{Y}) + \alpha_2\mathfrak{I}(P_{\overline{X}\overline{Y}}, \overline{X}\overline{Y})<br />\end{eqnarray*}$$Then the following is true: if the part of my plan that kicks in at $t_1$ doesn't demand I condition my prior on my evidence at $t_1$, or if the part of my plan that kicks in at $t_2$ doesn't demand I condition my $t_1$ credence function on my evidence at $t_2$, then, for any $0 < \beta < 1$, there is an alternative prior $P^\star$ and its associated Bayesian updating rule $\mathcal{R}^\star$, such that, for all worlds $w$,$$\beta\mathfrak{I}(P^\star, w) + (1-\beta)\mathfrak{I}(\mathcal{R}^\star, w) < \beta \mathfrak{I}(P, w) + (1-\beta)\mathfrak{I}(\mathcal{R}, w)$$And, again, this result generalizes to cases that include any number of times at which we receive new evidence, and in which, at each time, the set of propositions we might receive as evidence forms a partition. So it certainly covers the case of the coin of unknown bias that we've been using throughout. So, if you plan to update in some way other than by Bayesian conditionalization starting with your prior, there is an alternative prior and plan that, taken together, is guaranteed to have greater accuracy than yours; that is, they will have greater total accuracy than yours however the world turns out.<br /><br />How do we square this with Douven's simulation results? The key is that this dominance result includes the prior in it. It does not say that, if $\mathcal{R}$ requires you not to condition $P$ on your evidence at any point, then a rule that does require that is guaranteed to be better. It says that if $\mathcal{R}$ requires you not to condition $P$ on your evidence at any point, then there is an alternative prior $P^\star$ such that it, together with a rule that requires you to condition it on your evidence, are better than $P$ and $\mathcal{R}$ for sure. Douven's results compare the performance of conditioning on $P$ and performing the explanationist update on it. This shows that while conditioning might not always give a better result than the explanationist, there is an alternative prior such that conditioning on it is guaranteed to be better than retaining the original prior and performing the explanationist rule. And that, I think, is the reason we should prefer conditioning on our evidence to giving the little explanationist boosts that Douven suggests. If we update by conditioning, our prior and update rule, taken together, are never accuracy dominated; it we update using Douven's explanationist rule, our prior and update rule, taken together, are accuracy dominated.<br /><br />Before wrapping up, it's worth mentioning that there's a little wrinkle to iron out. It might be that, while the original prior and the posteriors it generates at the various times all satisfy the Principal Principle, the dominating prior and updating rule don't. While being dominated is clearly bad, you might think that being dominated by something that is itself irrational -- because it violates the Principal Principle, or for other reasons -- isn't so bad. But in fact we can tweak things to avoid this situation. The following is true: if the part of my plan that kicks in at $t_1$ doesn't demand I condition my prior on my evidence at $t_1$, or if the part of my plan that kicks in at $t_2$ doesn't demand I condition my $t_1$ credence function on my evidence at $t_2$, then, for any $0 < \beta < 1$, there is an alternative prior $P^\star$ and its associated Bayesian updating rule $\mathcal{R}^\star$, such that, $P^\star$ obeys the Principal Principle and, for all possible objective chance functions $ch$,<br /><br />$\beta\sum_{w} ch(w) \mathfrak{I}(P^\star, w) + (1-\beta)\sum_{w} ch(w) \mathfrak{I}(\mathcal{R}^\star, w) < $<br /><br />$\beta \sum_{w} ch(w) \mathfrak{I}(P, w) + (1-\beta)\sum_{w} ch(w) \mathfrak{I}(\mathcal{R}, w)$ <br /><br />So I'm inclined to think that Douven's critique of the Dutch Book argument against the explanationist updating rule hits the mark; and I can see why he thinks the expected accuracy argument against it is also less than watertight; but I think the accuracy dominance argument against it is stronger. We shouldn't use that updating rule, with its extra boost for explanatory hypotheses, because if we do so, there will be an alternative prior such that applying the Bayesian updating rule to that prior is guaranteed to be more accurate than applying the explanationist rule to our actual prior.<br /><p></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com14tag:blogger.com,1999:blog-4987609114415205593.post-55270550802447985852020-08-11T08:42:00.024+01:002022-07-27T19:13:41.439+01:00The only symmetric inaccuracy measure is the Brier score<p>If you'd like a PDF of this post, see here. </p><p><span></span></p><a name='more'></a>[UPDATE 1: I should have made this clear in the original post. The Normality condition makes the proof go through more easily, but it isn't really necessary. Suppose we simply assume instead that $$\mathfrak{I}(w^i, j)= \left \{ \begin{array}{ll} b & \mbox{if } i \neq j \\ a & \mbox{if } i = j \end{array} \right.$$Then we can show that, if $\mathfrak{I}$ is symmetric then, for any probabilistic credence function $p$ and any world $w_i$,$$\mathfrak{I}(p, i) = (b-a)\frac{1}{2} \left (1 - 2p_i + \sum_j p^2_j \right ) + a$$End Update 1.]<p></p><p>[UPDATE 2: There's something puzzling about the result below. Suppose $\mathcal{W} = \{w_1, \ldots, w_n\}$ is the set of possible worlds. And suppose $\mathcal{F}$ is the full algebra of propositions built out of those worlds. That is, $\mathcal{F}$ is the set of subsets of $\mathcal{W}$. Then there are two versions of the Brier score over a probabilistic credence function $p$ defined on $\mathcal{F}$. The first considers only the credences that $p$ assigns to the possible worlds. Thus,$$\mathfrak{B}(p, i) = \sum^n_{j=1} (w^i_j - p_j)^2 = 1 - 2p_i + \sum_j p^2_j$$But there is another that considers also the credences that $p$ assigns to the other propositions in $\mathcal{F}$. Thus,$$\mathfrak{B}^\star(p, i) = \sum_{X \in \mathcal{F}} (w_i(X) - p(X))^2$$Now, at first sight, these look related, but not very closely. However, notice that both are symmetric. Thus, by the extension of Selten's theorem below (plus update 1 above), if $\mathfrak{I}(w^i, j) = b$ for $i \neq j$ and 0 for $i = j$, then $\mathfrak{I}(p, i) = \frac{1}{2}b\mathfrak{B}(p, i)$. Now, $\mathfrak{B}(w^i, j) = 2$ for $i \neq j$, and $\mathfrak{B}(w^i, j) = 0$ for $i = j$, and so this checks out. But what about $\mathfrak{B}^\star$? Well, according to our extension of Selten's theorem, since $\mathfrak{B}^\star$ is symmetric, we can see that it is just a multiple of $\mathfrak{B}$, the factor determined by $\mathfrak{B}^\star(w^i, j)$. So what is this number? Well, it turns out that, if $i \neq j$, then$$\mathfrak{B}^\star(w^i, j) = 2\sum^{n-2}_{k=0} {n-2 \choose k}$$Thus, it follows that$$\mathfrak{B}^\star(p, i) = \sum^{n-2}_{k=0} {n-2 \choose k}\mathfrak{B}(p, i)$$And you can verify this by other means as well. This is quite a nice result independently of all this stuff about symmetry. After all, there doesn't seem any particular reason to favour $\mathfrak{B}$ over $\mathfrak{B}^\star$ or vice versa. This result shows that using one for the sorts of purposes we have in accuracy-first epistemology won't give different results from using the other. End update 2.]</p><p>So, as is probably obvious, I've been trying recently to find out what things look like in accuracy-first epistemology if you drop the assumption that the inaccuracy of a whole credal state is the sum of the inaccuracies of the individual credences that it comprises --- this assumption is sometimes called Additivity or Separability. In this post, I want to think about a result concerning additive inaccuracy measures that intrigued me in the past and on the basis of which I tried to mount an argument in favour of the Brier score. The result dates back to <a href="https://link.springer.com/article/10.1023/A:1009957816843" target="_blank">Reinhard Selten</a>, the German economist who shared the 1994 Nobel prize with John Harsanyi and John Nash for his contributions to game theory. In this post, I'll show that the result goes through even if we don't assume additivity.<br /></p><p>Suppose $\mathfrak{I}$ is an inaccuracy measure. Thus, if $c$ is a credence function defined on the full algebra built over the possible worlds $w_1, \ldots, w_n$, then $\mathfrak{I}(c, i)$ measures the inaccuracy of $c$ at world $w_i$. Then define the following function on pairs of probabilistic credence functions:$$\mathfrak{D}_\mathfrak{I}(p, q) = \sum_i p_i \mathfrak{I}(q, i) - \sum_i p_i\mathfrak{I}(p, i)$$$\mathfrak{D}_\mathfrak{I}$ measures how much more inaccurate $p$ expects $q$ to be than it expects itself to be; equivalently, how much more accurate $p$ expects itself to be than it expects $q$ to be. Now, if $\mathfrak{I}$ is strictly proper, $\mathfrak{D}_\mathfrak{I}$ is positive whenever $p$ and $q$ are different, and zero when they are the same, so in that case $\mathfrak{D}_\mathfrak{I}$ is a divergence. But we won't be assuming that here -- rather remarkably, we don't need to.</p><p>Now, it's not hard to see that $\mathfrak{D}_\mathfrak{I}$ is not necessarily symmetric. For instance, consider the log score$$\mathfrak{L}(p, i) = -\log p_i$$Then$$\mathfrak{D}_\mathfrak{L}(p, q) = p_i \log \frac{p_i}{q_i}$$This is the so-called Kullback-Leibler divergence and it is not symmetric. Nonetheless, it's equally easy to see that it is at least possible for $\mathfrak{D}_\mathfrak{I}$ to be symmetric. For instance, consider the Brier score$$\mathfrak{B}(p, i) = 1-2p_i + \sum_j p^2_j$$Then$$\mathfrak{D}_\mathfrak{B}(p, q) = \sum_i (p_i - q_i)^2$$So the natural question arises: how many inaccuracy measures are symmetric in this way? That is, how many generate symmetric divergences in the way that the Brier score does? It turns out: none, except the Brier score.</p><p>First, a quick bit of notation: Given a possible world $w_i$, we write $w^i$ for the probabilistic credence function that assigns credence 1 to world $w_i$ and 0 to any world $w_j$ with $j \neq i$. </p><p>And two definitions:</p><p><b>Definition </b><b>(Normal inaccuracy measure) </b><i>An inaccuracy measure $\mathfrak{I}$ is </i>normal<i> if $$\mathfrak{I}(w^i, j) = \left \{ \begin{array}{ll} 1 & \mbox{if } i \neq j \\ 0 & \mbox{if } i = j \end{array} \right.$$</i></p><p><b>Definition (Symmetric inaccuracy measure)</b><i> An inaccuracy measure is </i>symmetric<i> if </i><i>$$\mathfrak{D}_\mathfrak{I}(p, q) = \mathfrak{D}_\mathfrak{I}(q, p)$$for all probabilistic credence functions $p$ and $q$.</i></p><p>Thus, $\mathfrak{I}$ is symmetric if, for any probability functions $p$ and $q$, the loss of accuracy that $p$ expects to suffer by moving to $q$ is the same as the loss of accuracy that $q$ expects to suffer by moving to $p$. </p><p><b>Theorem</b> <i>The only normal and symmetric inaccuracy measure agrees with the Brier score for probabilistic credence functions.<br /></i></p><p><i>Proof</i>. (This just adapts Selten's proof in exactly the way you'd expect.) Suppose $\mathfrak{D}_\mathfrak{I}(p, q) = \mathfrak{D}_\mathfrak{I}(q, p)$ for all probabilistic $p$, $q$. Then, in particular, for any world $w_i$ and any probabilistic $p$,$$\sum_j w^i_j \mathfrak{I}(p, j) - \sum_j w^i_j \mathfrak{I}(w^i, j) = \sum_j p_j \mathfrak{I}(w^i, j) -\sum_j p_j \mathfrak{I}(p, j)$$So,$$\mathfrak{I}(p, i) = (1-p_i) - \sum_j p_j \mathfrak{I}(p, j)$$So,$$\sum_j p_j \mathfrak{I}(p, j) = 1 - \sum_j p^2_j- \sum_j p_j \mathfrak{I}(p, j)$$So,$$\sum_j p_j \mathfrak{I}(p, j) = \frac{1}{2}[1 - \sum_j p^2_j]$$So,$$\mathfrak{I}(p, i) = 1-p_i -\frac{1}{2}[1 - \sum_j p^2_j] = \frac{1}{2} \left (1 - 2p_i + \sum_j p^2_j \right )$$as required. $\Box$</p><p>There are a number of notable features of this result:</p><p>First, the theorem does not assume that the inaccuracy measure is strictly proper, but since the Brier score is strictly proper, it follows that symmetry entails strict propriety.</p><p>Second, the theorem does not assume additivity, but since the Brier score is additive, it follows that symmetry entails additivity. </p><p><br /></p><p><br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com8tag:blogger.com,1999:blog-4987609114415205593.post-17636646622945989472020-08-10T18:44:00.002+01:002022-07-27T19:13:59.005+01:00The Accuracy Dominance Argument for Conditionalization without the Additivity assumption<p> For a PDF of this post, see <a href="https://drive.google.com/file/d/1WpVipvzgrX6zaS1qeEbHxTWP03bZma9z/view?usp=sharing" target="_blank">here</a>. <br /></p><p><a href="https://m-phi.blogspot.com/2020/08/accuracy-without-additivity.html" target="_blank">Last week</a>, I explained how you can give an accuracy dominance argument for Probabilism without assuming that your inaccuracy measures are additive -- that is, without assuming that the inaccuracy of a whole credence function is obtained by adding up the inaccuracy of all the individual credences that it assigns. The mathematical result behind that also allows us to give <a href="https://drive.google.com/file/d/1z7Z4rE0BlEtCeD_YPktCEaj1Vm24RnSG/view?usp=sharing" target="_blank">my chance dominance argument for the Principal Principle</a> without assuming additivity, and ditto for <a href="https://drive.google.com/file/d/1_NeXKXQ84xL2plPDWz3ioA12Q3IDcIYf/view?usp=sharing" target="_blank">my accuracy-based argument for linear pooling</a>. In this post, I turn to another Bayesian norm, namely, Conditionalization. The first accuracy argument for this was given by <a href="https://philpapers.org/rec/GREJCC" target="_blank">Hilary Greaves and David Wallace</a>, building on ideas developed by <a href="https://philpapers.org/rec/ODDCCA" target="_blank">Graham Oddie</a>. It was an expected accuracy argument, and it didn't assume additivity. More recently, <a href="https://drive.google.com/open?id=1kjY_wQ0nlXIGfnla_MhWF1uQmcJtcTUB" target="_blank">Ray Briggs and I</a> offered an accuracy dominance argument for the norm, and we did assume additivity. It's this latter argument I'd like to consider here. I'd like to show how it goes through even without assuming additivity. And indeed I'd like to generalise it at the same time. The generalisation is inspired by a recent paper by <a href="https://philpapers.org/rec/RESAID" target="_blank">Michael Rescorla</a>. In it, Rescorla notes that all the existing arguments for Conditionalization assume that, when your evidence comes in the form of a proposition learned with certainty, that proposition must be true. He then offers a Dutch Book argument for Conditionalization that doesn't make this assumption, and he issues a challenge for other sorts of arguments to do the same. Here, I take up that challenge. To do so, I will offer an argument for what I call the Weak Reflection Principle.</p><p><b>Weak Reflection Principle (WRP)</b> <i>Your current credence function should be a linear combination of the possible future credence functions that you endorse.</i></p><p><span></span></p><a name='more'></a>A lot might happen between now and tomorrow. I might see new sights, think new thoughts; I might forget things I know today, take mind-altering drugs that enhance or impair my thinking; and so on. So perhaps there is a set of credence functions I think I might have tomorrow. Some of those I'll endorse -- perhaps those that I'd get if I saw certain new things, or enhanced my cognition in various ways. And some of them I'll disavow -- perhaps those that I'd get if I forget certain things, or impaired my cognition. WRP asks you to separate out the wheat from the chaff, and once you've identified the ones you endorse, it tells you that your current credence function should lie within the span of those future ones; it should be in their convex hull; it should be a weighted sum or convex combination of them.<p></p><p>One nice thing about WRP is that it gives back Conditionalization in certain cases. Suppose $c^0$ is my current credence function. Suppose I know that between now and tomorrow I'll learn exactly one member of the partition $E_1, \ldots, E_m$ with certainty --- this is the situation that Greaves and Wallace envisage. And suppose I endorse credence function $c^1$ as a response to learning $E_1$, $c^2$ as a response to learning $E_2$, and so on. Then, if I satisfy WRP, and if $c^k(E_k) = 1$, since I did after all learn it with certainty, then it follows that, whenever $c^0(E_k) > 0$, $c^k(X) = c^0(X | E_k)$, which is exactly what Conditionalization asks of you. And notice that, at no point did we assume that if I learn $E_k$, then $E_k$ is true. So we've answered Rescorla's challenge if we can establish WRP.</p><p>To do that, we need Theorem 1 below. And to get there, we need to go via Lemmas 1 and 2. Just to remind ourselves of the framework:</p><ul style="text-align: left;"><li>$w_1, \ldots, w_n$ are the possible worlds;</li><li>credence functions are defined on the full algebra built on top of these possible worlds;</li><li>given a credence function $c$, we write $c_i$ for the credence that $c$ assigns to $w_i$. <b> <br /></b></li></ul><p><b>Lemma 1</b> If $c^0$ is not in the convex combination of $c^1, \ldots, c^m$, then $(c^0, c^1, \ldots, c^m)$ is not in the convex hull of $\mathcal{X}$, where$$\mathcal{X} := \{(w^i, c^1, \ldots, c^{k-1}, w^i, c^{k+1}, \ldots, c^m) : 1 \leq i \leq n\ \&\ 1 \leq k \leq m\}$$</p><p><b>Definition 1</b> Suppose $\mathfrak{I}$ is a continuous strictly proper inaccuracy measure. Then let$$\mathfrak{D}_\mathfrak{I}((p^0, p^1, \ldots, p^m), (c^0, c^1, \ldots, c^m)) = \sum^m_{k=0} \left ( \sum^n_{i=1} p^k_i \mathfrak{I}(c^k, i) - \sum^n_{i=1} p^k_i \mathfrak{I}(p^k, i) \right )$$<br /></p><p><b>Lemma 2</b> Suppose $\mathfrak{I}$ is a continuous strictly proper inaccuracy measure. Suppose $\mathcal{X}$ is a closed convex set of $(n+1)$-tuples of probabilistic credence functions. And suppose $(c^0, c^1, \ldots, c^n)$ is not in $\mathcal{X}$. Then there is $(q^0, q^1, \ldots, q^m)$ in $\mathcal{X}$ such that </p><p>(i) for all $(p^0, p^1, \ldots, p^m) \neq (q^0, q^1, \ldots, q^m)$ in $\mathcal{X}$,</p><p>$\mathfrak{D}_\mathfrak{I}((q^0, q^1, \ldots, q^m), (c^0, c^1, \ldots, c^m)) <$</p><p>$\mathfrak{D}_\mathfrak{I}((p^0, p^1, \ldots, p^m), (c^0, c^1, \ldots, c^m))$;</p><p>(ii) for all $(p^0, p^1, \ldots, p^n)$ in $\mathcal{X}$,</p><p>$\mathfrak{D}_\mathfrak{I}((p^0, p^1, \ldots, p^m), (c^0, c^1, \ldots, c^m)) \geq$</p><p>$\mathfrak{D}_\mathfrak{I}((p^0, p^1, \ldots, p^m), (q^0, q^1, \ldots, q^m)) +$</p><p>$\mathfrak{D}_\mathfrak{I}((q^0, q^1, \ldots, q^m), (c^0, c^1, \ldots, c^m))$<b>.<br /></b></p><p><b>Theorem 1</b> Suppose each $c^0, c^1, \ldots, c^n$ is a probabilistic credence function. If $c^0$ is not in the convex hull of $c^1, \ldots, c^m$, then there are probabilistic credence functions $q^0, q^1, \ldots, q^m$ such that for all worlds $w_i$ and $1 \leq k \leq m$,$$\mathfrak{I}(q^0, i) + \mathfrak{I}(q^k, i) < \mathfrak{I}(c^0, i) + \mathfrak{I}(c^k, i)$$ </p><p>Let's keep the proofs on ice for a moment. What does this show exactly? It says that, if you don't do as WRP demands, there is some alternative current credence function and, for each of the possible future credence functions in the set you endorse, there is an alternative such that having your current credence function now and then one of your endorsed future credence functions later is guaranteed to make you less accurate overall than having the alternative to your current credence function now and the alternative to that endorsed future credence function later. This, I claim, establishes WRP.</p><p>Now for the proofs.<br /></p><p><i>Proof of Lemma 1</i>. We prove the contrapositive. Suppose $(c^0, c^1, \ldots, c^m)$ is in $\mathcal{X}$. Then there are $0 \leq \lambda_{i, k} \leq 1$ such that $\sum^n_{i=1}\sum^m_{k=1} \lambda_{i, k} = 1$ and$$(c^0, c^1, \ldots, c^m) = \sum^n_{i=1} \sum^m_{k=1} \lambda_{i, k} (w^i, c^1, \ldots, c^{k-1}, w^i, c^{k+1}, \ldots, c^m)$$Thus,$$c^0 = \sum^n_{i=1}\sum^m_{k=1} \lambda_{i,k} w^i$$<br />and$$c^k = \sum^n_{i=1} \lambda_{i, k} w^i + \sum^n_{i=1} \sum_{l \neq k} \lambda_{i, l} c^k$$So$$(\sum^n_{i=1} \lambda_{i, k}) c^k = \sum^n_{i=1} \lambda_{i, k} w^i$$So let $\lambda_k = \sum^n_{i=1} \lambda_{i, k}$. Then, for $1 \leq k \leq m$,$$\lambda_k c_k = \sum^n_{i=1} \lambda_{i, k} w^i$$And thus$$\sum^m_{k=1} \lambda^k c^k = \sum^m_{k=1} \sum^n_{i=1} \lambda_{i, k} w^i = c^0$$as required. $\Box$ <br /></p><p><i>Proof of Lemma 2</i>. This proceeds exactly like the corresponding theorem from the previous blogpost. $\Box$</p><p><i>Proof of Theorem 1</i>. So, if $c^0$ is not in the convex hull of $c^1, \ldots, c^m$, there is $(q^0, q^1, \ldots, q^m)$ such that, for all $(p^0, p^1, \ldots, p^m)$ in $\mathcal{X}$,$$\mathfrak{D}((p^0, p^1, \ldots, p^m), (q^0, q^1, \ldots, q^m)) < \mathfrak{D}((p^0, p^1, \ldots, p^m), (c^0, c^1, \ldots, c^m))$$In particular, for any world $w_i$ and $1 \leq k \leq m$,</p><p>$\mathfrak{D}((w^i, c^1, \ldots, c^{k-1}, w^i, c^{k+1}, \ldots, c^m), (q^0, q^1, \ldots, q^m)) <$</p><p>$\mathfrak{D}((w^i, c^1, \ldots, c^{k-1}, w^i, c^{k+1}, \ldots, c^m), (c^0, c^1, \ldots, c^m))$</p><p>But$$\begin{eqnarray*}<br />& & \mathfrak{I}(q^0, i) + \mathfrak{I}(q^k, i) \\<br />& = & \mathfrak{D}(w^i, q^0) + \mathfrak{D}(w^i, q^k) \\<br />& \leq & \mathfrak{D}((w^i, c^1, \ldots, c^{k-1}, w^i, c^{k+1}, \ldots, c^m), (q^0, q^1, \ldots, q^m)) \\<br />& < & \mathfrak{D}((w^i, c^1, \ldots, c^{k-1}, w^i, c^{k+1}, \ldots, c^m), (c^0, c^1, \ldots, c^m)) \\<br />& = & \mathfrak{D}(w^i, c^0) + \mathfrak{D}(w^i, c^k) \\<br />& = & \mathfrak{I}(c^0, i) + \mathfrak{I}(c^k, i) <br />\end{eqnarray*}$$as required.<br /><br /> <br /></p>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com5tag:blogger.com,1999:blog-4987609114415205593.post-54675969383751584272020-08-07T09:27:00.003+01:002022-07-27T19:14:41.324+01:00The Accuracy Dominance Argument for Probabilism without the Additivity assumption<div>For a PDF of this post, see <a href="https://drive.google.com/file/d/1EidttBl-pYE8rjqbE3iL3tsXBvxCF639/view?usp=sharing" target="_blank">here</a>.<br /></div><div><br /></div><div>One of the central arguments in accuracy-first epistemology -- the one that gets the project off the ground, I think -- is the accuracy-dominance argument for Probabilism. This started life in a more pragmatic guise in <a href="https://onlinelibrary.wiley.com/doi/book/10.1002/9781119286387" target="_blank">de Finetti's proof</a> that, if your credences are not probabilistic, there are alternatives that would lose less than yours would if they were penalised using the Brier score, which levies a price of $(1-x)^2$ on every credence $x$ in a truth and $x^2$ on every credence $x$ in a falsehood. This was then adapted to an accuracy-based argument by <a href="https://philpapers.org/rec/ROSFAA-3">Roger Rosencrantz</a>, where he interpreted the Brier score as a measure of inaccuracy, not a penalty score. Interpreted thus, de Finetti's result says that any non-probabilistic credences are accuracy-dominated by some probabilistic credences. <a href="https://philpapers.org/rec/JOYANV">Jim Joyce</a> then noted that this argument only establishes Probabilism if you have a further argument that inaccuracy should be measured by the Brier score. He thought there was no particular reason to think that's right, so he greatly generalized de Finetti's result to show that, relative to a much wider range of inaccuracy measures, all non-probabilistic credences are accuracy dominated. One problem with this, which <a href="https://philpapers.org/rec/HJEAFA-2" target="_blank">Al Hájek</a> pointed out, was that he didn't give a converse argument -- that is, he didn't show that, for each of his inaccuracy measures, each probabilistic credence function is not accuracy dominated. <a href="https://philpapers.org/rec/PREPCA" target="_blank">Joel Predd and his Princeton collaborators</a> then addressed this concern and proved a very general result, namely, that for any additive, continuous, and strictly proper inaccuracy measure, any non-probabilistic credences are accuracy-dominated, while no probabilistic credences are.</div><div><br /></div><div><span><a name='more'></a></span>That brings us to this blogpost. Additivity is a controversial claim. It says that the inaccuracy of a credence function is the (possibly weighted) sum of the inaccuracies of the credences it assigns. So the question arises: can we do without additivity? In this post, I'll give a quick proof of the accuracy-dominance argument that doesn't assume anything about the inaccuracy measures other than that they are continuous and strictly proper. Anyone familiar with the Predd, et al. paper will see that the proof strategy draws very heavily on theirs. But it bypasses out the construction of the Bregman divergence that corresponds to the strictly proper inaccuracy measure. For that, you'll have to wait for Jason Konek's forthcoming work...</div><div><br />Suppose:<br /><ul style="text-align: left;"><li>$\mathcal{F}$ is a set of propositions;</li><li>$\mathcal{W} = \{w_1, \ldots, w_n\}$ be the set of possible worlds relative to $\mathcal{F}$;</li><li>$\mathcal{C}$ be the set of credence functions on $\mathcal{F}$;</li><li>$\mathcal{P}$ be the set of probability functions on $\mathcal{F}$. So, by de Finetti's theorem, $\mathcal{P} = \{v_w : w \in \mathcal{W}\}^+$. If $p$ is in $\mathcal{P}$, we write $p_i$ for $p(w_i)$.</li></ul><b>Theorem</b> Suppose $\mathfrak{I}$ is a strictly proper inaccuracy measure on the credence functions in $\mathcal{F}$. Then if $c$ is not in $\mathcal{P}$, there is $c^\star$ in $\mathcal{P}$ such that, for all $w_i$ in $\mathcal{W}$, <br />$$<br />\mathfrak{I}(c^\star, w_i) < \mathfrak{I}(c, w_i)<br />$$<br /><br /><i>Proof</i>. We begin by defining a divergence $\mathfrak{D} : \mathcal{P} \times \mathcal{C} \rightarrow [0, \infty]$ that takes a probability function $p$ and a credence function $c$ and measures the divergence from the former to the latter:<br />$$<br />\mathfrak{D}(p, c) = \sum_i p_i \mathfrak{I}(c, w_i) - \sum_i p_i \mathfrak{I}(p, w_i)<br />$$<br />Three quick points about $\mathfrak{D}$.<br /><br />(1) $\mathfrak{D}$ is a divergence. Since $\mathfrak{I}$ is strictly proper, $\mathfrak{D}(p, c) \geq 0$ with equality iff $c = p$.<br /><br />(2) $\mathfrak{D}(v_{w_i}, c) = \mathfrak{I}(c, w_i)$, for all $w_i$ in $\mathcal{W}$.<br /><br />(3) $\mathfrak{D}$ is strictly convex in its first argument. Suppose $p$ and $q$ are in $\mathcal{P}$, and suppose $0 < \lambda < 1$. Then let $r = \lambda p + \lambda q$. Then, since $\sum_i p_i\mathfrak{I}(c, w_i)$ is uniquely minimized, as a function of $c$, at $c = p$, and $\sum_i q_i\mathfrak{I}(c, w_i)$ is uniquely minimized, as a function of $c$, at $c = q$, we have$$\begin{eqnarray*}<br />\sum_i p_i \mathfrak{I}(c, w_i) & < & \sum_i p_i \mathfrak{I}(r, w_i) \\<br />\sum_i q_i \mathfrak{I}(c, w_i) & < & \sum_i q_i \mathfrak{I}(r, w_i)<br />\end{eqnarray*}$$Thus<br /><br /> $\lambda [-\sum_i p_i \mathfrak{I}(p, w_i)] + (1-\lambda) [-\sum_i q_i \mathfrak{I}(q, w_i)] >$<br /><br />$ \lambda [-\sum_i p_i \mathfrak{I}(r, w_i)] + (1-\lambda) [-\sum_i q_i \mathfrak{I}(r, w_i)] = $<br /><br />$-\sum_i r_i \mathfrak{I}(r, w_i)$<br /><br />Now, adding</div><div><br /></div><div>$\lambda \sum_i p_i \mathfrak{I}(c, w_i) + (1-\lambda)\sum_i q_i\mathfrak{I}(c, w_i) =$</div><div><br /></div><div>$\sum_i (\lambda p_i + (1-\lambda)q_i) \mathfrak{I}(c, w_i) = \sum_i r_i \mathfrak{I}(c, w_i)$<br /></div><div><br /></div><div>to both sides gives<br /><br />$\lambda [\sum_i p_i \mathfrak{I}(c, w_i)-\sum_i p_i \mathfrak{I}(p, w_i)]+ $<br /><br />$(1-\lambda) [\sum_i q_i\mathfrak{I}(c, w_i)-\sum_i q_i \mathfrak{I}(q, w_i)] > $<br /><br /> $\sum_i r_i \mathfrak{I}(c, w_i)-\sum_i r_i \mathfrak{I}(r, w_i)$<br /><br />That is,$$\lambda \mathfrak{D}(p, c) + (1-\lambda) \mathfrak{D}(q, c) > \mathfrak{D}(\lambda p + (1-\lambda)q, c)$$as required.<br /><br />Now, suppose $c$ is not in $\mathcal{P}$. Then, since $\mathcal{P}$ is a closed convex set, there is a unique $c^\star$ in $\mathcal{P}$ that minimizes $\mathfrak{D}(x, c)$ as a function of $x$. Now, suppose $p$ is in $\mathcal{P}$. We wish to show that$$\mathfrak{D}(p, c) \geq \mathfrak{D}(p, c^\star) + \mathfrak{D}(c^\star, c)$$We can see that this holds iff$$\sum_i (p_i - c^\star_i) (\mathfrak{I}(c, w_i) - \mathfrak{I}(c^\star, w_i)) \geq 0$$After all,<br />$$\begin{eqnarray*}<br />& & \mathfrak{D}(p, c) - \mathfrak{D}(p, c^\star) - \mathfrak{D}(c^\star, c) \\<br />& = & [\sum_i p_i \mathfrak{I}(c, w_i) - \sum_i p_i \mathfrak{I}(p, w_i)] - \\<br />&& [\sum_i p_i \mathfrak{I}(c^\star, w_i) - \sum_i p_i \mathfrak{I}(p, w_i)] - \\<br />&& [\sum_i c^\star_i \mathfrak{I}(c, w_i) - \sum_i c^\star_i \mathfrak{I}(c^\star, w_i)] \\<br />& = & \sum_i (p_i - c^\star_i)(\mathfrak{I}(c, w_i) - \mathfrak{I}(c^\star, w_i))<br />\end{eqnarray*}$$<br />Now we prove this inequality. We begin by observing that, since $p$, $c^\star$ are in $\mathcal{P}$, since $\mathcal{P}$ is convex, and since $\mathfrak{D}(x, c)$ is minimized uniquely at $x = c^\star$, if $0 < \varepsilon < 1$, then$$\frac{1}{\varepsilon}[\mathfrak{D}(\varepsilon p + (1-\varepsilon) c^\star, c) - \mathfrak{D}(c^\star, c)] > 0$$Expanding that, we get<br /><br />$\frac{1}{\varepsilon}[\sum_i (\varepsilon p_i + (1- \varepsilon) c^\star_i)\mathfrak{I}(c, w_i) -$<br /><br />$\sum_i (\varepsilon p_i + (1-\varepsilon)c^\star_i)\mathfrak{I}(\varepsilon p + (1-\varepsilon) c^\star, w_i) - $<br /><br />$\sum_i c^\star_i\mathfrak{I}(c, w_i) + \sum_i c^\star_i \mathfrak{I}(c^\star, i)] > 0$\medskip<br /><br /> So<br /><br />$\frac{1}{\varepsilon}[\sum_i ( c^\star_i + \varepsilon(p_i - c^\star_i))\mathfrak{I}(c, w_i) -$<br /><br />$\sum_i ( c^\star_i + \varepsilon(p_i-c^\star_i))\mathfrak{I}(\varepsilon p + (1-\varepsilon) c^\star, w_i) - $<br /><br />$\sum_i c^\star_i\mathfrak{I}(c, w_i) + \sum_i c^\star_i \mathfrak{I}(c^\star, w_i)] > 0 $\medskip<br /><br /> So\medskip<br /><br />$\sum_i (p_i - c^\star_i)(\mathfrak{I}(c, w_i) - \mathfrak{I}(\varepsilon p+ (1-\varepsilon) c^\star), w_i) +$<br /><br />$ \frac{1}{\varepsilon}[\sum_i c^\star_i \mathfrak{I}(c^\star, w_i) - \sum_ic^\star_i \mathfrak{I}(\varepsilon p + (1-\varepsilon) c^\star, w_i)] > 0$\medskip<br /><br />Now, since $\mathfrak{I}$ is strictly proper,<br />$$\frac{1}{\varepsilon}[\sum_i c^\star_i \mathfrak{I}(c^\star, w_i) - \sum_ic^\star_i \mathfrak{I}(\varepsilon p + (1-\varepsilon) c^\star, w_i)] < 0$$<br />So, for all $\varepsilon > 0$,$$\sum_i (p_i - c^\star_i)(\mathfrak{I}(c, w_i) - \mathfrak{I}(\varepsilon p+ (1-\varepsilon) c^\star, w_i) > 0$$<br />So, since $\mathfrak{I}$ is continuous$$\sum_i (p_i - c^\star_i)(\mathfrak{I}(c, w_i) - \mathfrak{I}(c^\star, w_i)) \geq 0$$which is what we wanted to show. So, by above,$$\mathfrak{D}(p,c) \geq \mathfrak{D}(p, c^\star) + \mathfrak{D}(c^\star, c) $$In particular, since each $w_i$ is in $\mathcal{P}$,$$\mathfrak{D}(v_{w_i}, c) \geq \mathfrak{D}(v_{w_i}, c^\star) + \mathfrak{D}(c^\star, c)$$But, since $c^\star$ is in $\mathcal{P}$ and $c$ is not, and since $\mathfrak{D}$ is a divergence, $\mathfrak{D}(c^\star, c) > 0$. So$$\mathfrak{I}(c, w_i) = \mathfrak{D}(v_{w_i}, c) > \mathfrak{D}(v_{w_i}, c^\star) = \mathfrak{I}(c^\star, w_i)$$as required. $\Box$<br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com8tag:blogger.com,1999:blog-4987609114415205593.post-25380882643116155392020-07-26T16:16:00.006+01:002022-07-27T19:15:09.174+01:00Decomposing Bregman divergencesFor a PDF of this post, see <a href="https://drive.google.com/file/d/1Ek85k8ri-B0EB7_g-CDcC25QgaPO00Sx/view?usp=sharing" target="_blank">here</a>. <br /><br /><div>Here are a couple of neat little results about Bregman divergences that I just happened upon. They might help to prove some more decomposition theorems along the lines of <a href="https://www.jstor.org/stable/2987588" target="_blank">this classic result</a> by Morris DeGroot and Stephen Fienberg and, more recently, <a href="https://link.springer.com/chapter/10.1007/978-3-319-23528-8_5" target="_blank">this paper</a> by my colleagues in the computer science department at Bristol. I should say that a lot is known about Bregman divergences because of their role in information geometry, so these results are almost certainly known already, but I don't know where.</div><span><a name='more'></a></span><div><br /></div><div><h2 style="text-align: left;">Refresher on Bregman divergences</h2></div><br />First up, what's a divergence? It's essentially generalization of the notion of a measure of distance from one point to another. The points live in some closed convex subset $\mathcal{X} \subseteq \mathbf{R}^n$. A divergence is a function $D : \mathcal{X} \times \mathcal{X} \rightarrow [0, \infty]$ such that<br /><ul><li>$D(x, y) \geq 0$, for all $x$, $y$ in $\mathcal{X}$, and</li><li>$D(x, y) = 0$ iff $x = y$.</li></ul>Note: We do not assume that a divergence is <i>symmetric</i>. So the distance from $x$ to $y$ need not be the same as the distance from $y$ to $x$. That is, we do not assume $D(x, y) = D(y, x)$ for all $x$, $y$ in $\mathcal{X}$. Indeed, among the family of divergences that we will consider -- the Bregman divergences -- only one is symmetric -- the squared Euclidean distance. And we do not assume <i>the triangle inequality</i>. That is, we don't assume that the divergence from $x$ to $z$ is at most the sum of the divergence from $x$ to $y$ and the divergence from $y$ to $z$. That is, we do not assume $D(x, z) \leq D(x, y) + D(y, z)$. Indeed, the conditions under which $D(x, z) = D(x, y) + D(y, z)$ for a Bregman divergence $D$ will be our concern here. <br /><br />So, what's a Bregman divergence? $D : \mathcal{X} \times \mathcal{X} \rightarrow [0, \infty]$ is a Bregman divergence if there is a strictly convex function $\Phi : \mathcal{X} \rightarrow \mathbb{R}$ that is differentiable on the interior of $\mathcal{X}$ such that$$D(x, y) = \Phi(x) - \Phi(y) - \nabla \Phi(y) (x-y)$$In other words, to find the divergence from $x$ to $y$, you go to $y$, find the tangent to $\Phi$ at $y$. Then hop over to $x$ and subtract the value at $x$ of the tangent you just drew at $y$ from the value at $x$ of $\Phi$. That is, you subtract $\nabla \Phi(y) (x-y) + \Phi(y)$ from $\Phi(x)$. Because $\Phi$ is convex, it is always curving away from the tangent, and so $\nabla \Phi(y) (x-y) + \Phi(y)$, the value at $x$ of the tangent you drew at $y$, is always less than $\Phi(x)$, the value at $x$ of $\Phi$.<br /><br />The two most famous Bregman divergences are:<br /><ul><li><i>Squared Euclidean distance</i>. Let $\Phi(x) = ||x||^2 = \sum_i x_i^2$, in which case$$D(x, y) = ||x-y||^2 = \sum_i (x_i - y_i)^2$$</li><li><i>Generalized Kullback-Leibler divergence</i>. Let $\Phi(x) = \sum_i x_i \log x_i$, in which case$$D(x, y) = \sum_i x_i\log\frac{x_i}{y_i} - x_i + y_i$$</li></ul>Bregman divergences are convex in the first argument. Thus, we can define, for $z$ in $\mathcal{X}$ and for a closed convex subset $C \subseteq \mathcal{X}$, the $D$-projection of $z$ into $C$ is the point $\pi_{z, C}$ in $C$ such that $D(y, z)$ is minimized, as a function of $y$, at $y = \pi_{z, C}$. Now, we have the following theorem about Bregman divergences, due to Imre Csiszár:<br /><br /><div><b>Theorem (Generalized Pythagorean Theorem)</b> If $\mathcal{C} \subseteq \mathcal{X}$ is closed and convex, then$$D(x, \pi_{z, C}) + D(\pi_{z, C}, z) \leq D(x, z)$$</div><div><br /></div><div><h2 style="text-align: left;">Decomposing Bregman divergences</h2></div><br />This invites the question: when does equality hold? The following result gives a particular class of cases, and in doing so provides us with a recipe for creating decompositions of Bregman divergences into their component parts. Essentially, it says that the above inequality is an equality if $C$ is a plane in $\mathbb{R}^n$.<br /><br /><b>Theorem 1 </b> Suppose $r$ is in $\mathbb{R}$ and $0 \leq \alpha_1, \ldots, \alpha_n \leq 1$ with $\sum_i \alpha_i = 1$. Then let $C := \{(x_1, \ldots, x_n) : \sum_i \alpha_ix_i = r\}$. Then if $z$ in $\mathcal{X}$ and $x$ is in $C$,$$D_\Phi(x, z) = D_\Phi(x, \pi_{z, C}) + D_\Phi(\pi_{z, C}, z)$$<br /><br /><i>Proof of Theorem 1. </i> We begin by showing:<br /><br /><b>Lemma 1</b> For any $x$, $y$, $z$ in $\mathcal{X}$,$$D_\Phi(x, z) = D_\Phi(x, y) + D_\Phi(y, z) \Leftrightarrow (\nabla \Phi(y) - \nabla \Phi(z))(x-y) = 0$$<br /><br /><i>Proof of Lemma 1</i>. $$D_\Phi(x, z) = D_\Phi(x, y) + D_\Phi(y, z)$$iff<br /><br />$\Phi(x) - \Phi(z) - \nabla(z)(x-z)$<br /><br />$= \Phi(x) - \Phi(y) - \nabla(y)(x-y) + \Phi(y) - \Phi(z) - \nabla(z)(y-z)$<br /><br />iff$$(\nabla \Phi(y) - \nabla \Phi(z))(x-y) = 0$$as required.<br /><br /><i>Return to Proof of Theorem 1</i>. Now we show that if $x$ is in $C$, then$$(\nabla \Phi(\pi_{z, C}) - \Phi(z))(x-\pi_{z, C}) = 0$$We know that $D(y, z)$ is minimized on $C$, as a function of $y$, at $y = \pi_{z, C}$. Thus, let $y = \pi_{z, C}$. And let $h(x) := \sum_i \alpha_ix^i - r$. Then $\frac{\partial}{\partial x_i} h(x) = \alpha_i$. So, by the KKT conditions, there is $\lambda$ such that,$$\nabla \Phi(y) - \nabla \Phi(z) + (\lambda \alpha_1, \ldots, \lambda \alpha_n) = (0, \ldots, 0)$$Thus,$$\frac{\partial}{\partial y_i} \Phi(y) - \frac{\partial}{\partial z_i} \Phi(z) = -\lambda \alpha_i$$for all $i = 1, \ldots, n$. <br /><br />Thus, finally, <br />\begin{eqnarray*}<br />& &(\nabla \Phi(y) - \nabla \Phi(z))(x-y) \\<br />& = & \sum_i \left (\frac{\partial}{\partial y_i} \Phi(y) - \frac{\partial}{\partial z_i} \Phi(z)\right )(x_i-y_i) \\<br />& = & \sum_i (-\lambda \alpha_i) (x_i - y_i) \\<br />& = & -\lambda \left (\sum_i \alpha_i x_i - \sum_i \alpha_i y_i\right ) \\<br />& = & -\lambda (r-r) \\<br />& = & 0<br />\end{eqnarray*}<br />as required. $\Box$<br /><br /><b>Theorem 2 </b> Suppose $1 \leq k \leq n$. Let $C := \{(x_1, \ldots, x_n) : x_1 = x_2 = \ldots = x_k\}$. Then if $z$ in $\mathcal{X}$ and $x$ is in $C$,$$D_\Phi(x, z) = D_\Phi(x, \pi_{z, C}) + D_\Phi(\pi_{z, C}, z)$$<br /><br /><i>Proof of Theorem 2</i>. We know that $D(y, z)$ is minimized on $C$, as a function of $y$, at $y = \pi_{z, C}$. Thus, let $y = \pi_{z, C}$. And let $h_i(x) := x_{i+1} - x_i$, for $i = 1, \ldots, k-1$. Then$$\frac{\partial}{\partial x_j} h_i(x) = \left \{ \begin{array}{ll} 1 & \mbox{if } i+1 = j \\ -1 & \mbox{if } i = j \\ 0 & \mbox{otherwise}\end{array} \right.$$ So, by the KKT conditions, there are $\lambda_1, \ldots, \lambda_k$ such that,<br /><br />$\nabla \Phi(y) - \nabla \Phi(z)$<br /><br />$+ (-\lambda_1, \lambda_1, 0, \ldots, 0) + (0, -\lambda_2, \lambda_2, 0, \ldots, 0) + \ldots$<br /><br />$+ (0, \ldots, 0, -\lambda_k, \lambda_k, 0, \ldots, 0) = (0, \ldots, 0)$<br /><br />Thus,$$\begin{eqnarray*}\frac{\partial}{\partial y_1} \Phi(y) - \frac{\partial}{\partial z_1} \Phi(z) & = & - \lambda_1 \\ \frac{\partial}{\partial y_2} \Phi(y) - \frac{\partial}{\partial z_2} \Phi(z) & = & \lambda_1 - \lambda_2 \\ \vdots & \vdots & \vdots \\ \frac{\partial}{\partial y_{k-1}} \Phi(y) - \frac{\partial}{\partial z_{k-1}} \Phi(z) & = & \lambda_{k-2}- \lambda_{k-1} \\ \frac{\partial}{\partial y_k} \Phi(y) - \frac{\partial}{\partial z_k} \Phi(z) & = & \lambda_{k-1} \\ \frac{\partial}{\partial y_{k+1}} \Phi(y) - \frac{\partial}{\partial z_{k+1}} \Phi(z) & = & 0 \\ \vdots & \vdots & \vdots \\ \frac{\partial}{\partial y_n} \Phi(y) - \frac{\partial}{\partial z_n} \Phi(z) & = & 0 \end{eqnarray*}$$<br /><br />Thus, finally, <br />\begin{eqnarray*}<br />& &(\nabla \Phi(y) - \nabla \Phi(z))(x-y) \\<br />& = & \sum_i \left (\frac{\partial}{\partial y_i} \Phi(y) - \frac{\partial}{\partial z_i} \Phi(z)\right )(x_i-y_i) \\<br />& = & -\lambda_1(x_1-y_1) + (\lambda_1 - \lambda_2)(x_2-y_2) + \ldots \\<br />&& + (\lambda_{k-2} - \lambda_{k-1})(x_{k-1}-y_{k-1}) + \lambda_{k-1}(x_k-y_k) \\<br />&& + 0(x_{k+1} - y_{k+1}) + \ldots + 0 (x_n - y_n) \\<br />& = & \sum^{k-1}_{i=1} \lambda_i (x_{i+1} - x_i) + \sum^{k-1}_{i=1} \lambda_i (y_i - y_{i+1})\\<br />& = & 0<br />\end{eqnarray*}<br /><div>as required. $\Box$</div><div><br /></div><div><h2 style="text-align: left;">DeGroot and Fienberg's calibration and refinement decomposition</h2></div><br />To obtain these two decomposition results, we needed to assume nothing more than that $D$ is a Bregman divergence. The classic result by DeGroot and Fienberg requires a little more. We can see this by considering a very special case of it. Suppose $(X_1, \ldots, X_n)$ is a sequence of propositions that forms a partition. And suppose $w$ is a possible world. Then we can represent $w$ as the vector $w = (0, \ldots, 0, 1, 0, \ldots, 0)$, which takes value 1 at the proposition that is true in $w$ and 0 everywhere else. Now suppose $c = (c, \ldots, c)$ is an assignment of the same credence to each proposition. Then one very particular case of DeGroot and Fienberg's result says that, if $(0, \ldots, 0, 1, 0, \ldots, 0)$ is the world at which $X_i$ is true, then<br /><br />$D((0, \ldots, 0, 1, 0, \ldots, 0), (c, \ldots, c))$<br /><br />$= D((0, \ldots, 0, 1, 0, \ldots, 0), (\frac{1}{n}, \ldots, \frac{1}{n})) + D((\frac{1}{n}, \ldots, \frac{1}{n}), (c, \ldots, c))$<br /><br />Now, we know from Lemma 1 that this is true iff$$(\nabla \Phi(\frac{1}{n}, \ldots, \frac{1}{n}) - \nabla \Phi(c, \ldots, c))((0, \ldots, 0, 1, 0, \ldots, 0) - (\frac{1}{n}, \ldots, \frac{1}{n})) = 0$$which is true iff<br /><br />$\left ( \frac{\partial}{\partial x_i} \Phi(\frac{1}{n}, \ldots, \frac{1}{n}) - \frac{\partial}{\partial x_i} \Phi(c, \ldots, c) \right )$<br /><br />$= \frac{1}{n} \sum^n_{j=1} \left ( \frac{\partial}{\partial x_j} \Phi(\frac{1}{n}, \ldots, \frac{1}{n}) - \frac{\partial}{\partial x_j} \Phi(c, \ldots, c) \right )$<br /><br />and that is true iff<br /><br />$\frac{\partial}{\partial x_i} \Phi(\frac{1}{n}, \ldots, \frac{1}{n}) - \frac{\partial}{\partial x_i} \Phi(c, \ldots, c)$<br /><br />$= \frac{\partial}{\partial x_j} \Phi(\frac{1}{n}, \ldots, \frac{1}{n}) - \frac{\partial}{\partial x_j} \Phi(c, \ldots, c)$<br /><br />for all $1 \leq i, j, \leq n$, which is true iff, for any $x$, $1 \leq i, j \leq n$,$$\frac{\partial}{\partial x_i} \Phi(x, \ldots, x) = \frac{\partial}{\partial x_j} \Phi(x, \ldots, x)$$Now, this is true if $\Phi(x_1, \ldots, x_n) = \sum^n_{i=1} \varphi(x_i)$ for some $\varphi$. That is, it is true if $D$ is an additive Bregman divergence. But it is also true for certain non-additive Bregman divergences, such as the one generated from the log-sum-exp function:<br /><br /><b>Definition (log-sum-exp)</b> Suppose $0 \leq \alpha_1, \ldots, \alpha_n \leq 1$ with $\sum^n_{i=1} \alpha_i = 1$. Then let $$\Phi^A(x_1, \ldots, x_n) = \log(1 + \alpha_1e^{x_1} + \ldots \alpha_ne^{x_n})$$Then <br />$$D(x, y) = \log (1 + \sum_i \alpha_ie^{x_i}) - \log(1 + \sum_i \alpha_ie^{y_i}) - \sum_k \frac{\alpha_k(x_k - y_k)e^{y_k}}{1 + \sum_i \alpha_ie^{y_i}}$$<br /><br />Now$$\frac{\partial}{\partial x_i} \Phi^A(x_1, \ldots, x_n) = \frac{\alpha_i e^{x_i}}{1 + \alpha_1 e^{x_1} + \ldots + \alpha_ne^{x_n}}$$So, if $\alpha_i = \alpha_j$ for all $1 \leq i, j \leq n$, then$$\frac{\partial}{\partial x_i} \Phi^A(x, \ldots, x) = \frac{\alpha e^x}{1 + e^x} = \frac{\partial}{\partial x_j} \Phi^A(x, \ldots, x)$$But if $\alpha_i \neq \alpha_j$ for some $1 \leq i, j \leq n$, then$$\frac{\partial}{\partial x_i} \Phi^A(x, \ldots, x) = \frac{\alpha_ie^x}{1 + e^x} \neq \frac{\alpha_je^x}{1 + e^x} = \frac{\partial}{\partial x_j} \Phi^A(x, \ldots, x)$$<br /><br /><div>And indeed, the result even fails if we have a semi-additive Bregman divergence. That is, there are different $\phi_1, \ldots, \phi_n$ such that $\Phi(x) = \sum^n_{i=1} \phi_i(x_i)$. For instance, suppose $\phi_1(x) = x^2$ and $\phi_2(x) = x\log x$ and $\Phi(x, y) = \phi_1(x) + \phi_2(y) = x^2 + y\log y$. Then$$\frac{\partial}{\partial x_1} \Phi(x, x) = 2x \neq 1 + \log x = \frac{\partial}{\partial x_2} \Phi(x, x)$$</div><div><br /></div><div><h2 style="text-align: left;">Proving the Generalized Pythagorean Theorem</h2></div><div><br /></div><div>In this section, I really just spell out in more detail the proof that <a href="https://ieeexplore.ieee.org/document/5238758" target="_blank">Predd, et al.</a> give of the Generalized Pythagorean Theorem, which is their Proposition 3. But that proof contains some important general facts that might be helpful for people working with Bregman divergences. I collect these together here into one lemma.</div><div><br /></div><div><b>Lemma 2</b> Suppose $D$ is a Bregman divergence generated from $\Phi$. And suppose $x, y, z \in \mathcal{X}$. Then$$\begin{eqnarray*} & & D(x, z) - [D(x, y) + D(y, z)] \\ & = & (\nabla \Phi(y) - \nabla \Phi(z))(x - y) \\ & = & \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} [D(y + \varepsilon (x - y), z) - D(y, z)] \\ & = & \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} [D(\varepsilon x + (1-\varepsilon)y, z) - D(y, z)] \end{eqnarray*}$$</div><div><br /></div><div>We can then prove the Generalized Pythagorean Theorem easily. After all, if $x$ is in a closed convex set $C$ and $y$ is the point in $C$ that minimizes $D(y, z)$ as a function of $y$. Then, for all $0 \leq \varepsilon \leq 1$, $\varepsilon x + (1-\varepsilon)y$ is in $C$. And since $y$ minimizes,$$D(\varepsilon x + (1-\varepsilon)y, z) \geq D(y, z)$$. So $D(\varepsilon x + (1-\varepsilon)y, z) - D(y, z) \geq 0$. So $$\lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon}D(\varepsilon x + (1-\varepsilon)y, z) - D(y, z) \geq 0$$So, by Lemma 2,$$D(x, z) \geq D(x, y) + D(y, z)$$</div><div><br /></div><div><i>Proof of Lemma 2. </i>$$\begin{eqnarray*} && \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} [D(\varepsilon x + (1-\varepsilon)y, z) - D(y, z)] \\ & = & \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} [(\Phi(\varepsilon x + (1-\varepsilon)y) - \Phi(z) - \nabla \Phi(z)(\varepsilon x + (1-\varepsilon)y - z)) - \\ & & (\Phi(y) - \Phi(z) - \nabla\Phi(z)(y-z))]\\ & = & \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} [(\Phi(\varepsilon x + (1-\varepsilon)y) - \Phi(y) - \varepsilon\nabla \Phi(z)(x -y)] \\ & = & \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} [(\Phi(\varepsilon x + (1-\varepsilon)y) - \Phi(y)] - \nabla \Phi(z)(x -y) \\ & = & \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} [(\Phi(y + \varepsilon (x -y)) - \Phi(y)] - \nabla \Phi(z)(x -y) \\ & = & \nabla \Phi(y)(x -y) - \nabla \Phi(z)(x -y) \\ & = & (\nabla \Phi(y) - \nabla \Phi(z))(x -y)\end{eqnarray*}$$<br /></div>Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com4tag:blogger.com,1999:blog-4987609114415205593.post-46220199378213929972020-07-23T10:28:00.003+01:002020-07-23T10:29:28.555+01:00Epistemic risk and permissive rationality (part I): an overviewI got interested in epistemic risk again, after a hiatus of four or five years, by thinking about the debate in epistemology between permissivists and impermissivists about epistemic rationality. Roughly speaking, according to the impermissivist, every body of evidence you might obtain mandates a unique rational set of attitudes in response --- this is sometimes called the uniqueness thesis. According to the permissivist, there is evidence you might obtain that doesn't mandate a unique rational set of attitudes in response --- there are, instead, multiple rational responses.<br /><br />I want to argue for permissivism. And I want to do it by appealing to the sorts of claims about how to set your priors and posteriors that I've been developing over this series of blogposts (<a href="https://m-phi.blogspot.com/2020/07/taking-risks-and-picking-priors.html" target="_blank">here</a> and <a href="https://m-phi.blogspot.com/2020/07/taking-risks-and-picking-posteriors.html" target="_blank">here</a>). In the first of those blogposts, I argued that we should pick priors using a decision rule called the generalized Hurwicz criterion (GHC). That is, we should see our choice of priors as a decision we must make; and we should make that decision using a particular decision rule -- namely, GHC -- where we take the available acts to be the different possible credence functions and the utility of an act at a world to be a measure of its accuracy at that world.<br /><br />Now, GHC is, in fact, not a single decision rule, but a family of rules, each specified by some parameters that I call the Hurwicz weights. These encode different attitudes to risk -- they specify the weight you assign to the best-case scenario, the weight you assign to the second-best, and so on down to the weight you assign to the second-worst scenario, and the weight you assign to the worst. And, what's more, many different attitudes to risk are permissible; and therefore many different Hurwicz weights are permissible; and so many versions of GHC are legitimate decision rules to adopt when picking priors. So different permissible attitudes to risk determine different Hurwicz weights; and different Hurwicz weights mandate different rational priors; and different rational priors mandate different rational posteriors given the same evidence. Epistemic rationality, therefore, is permissive. That's the argument in brief.<br /><br />With this post, I'd like to start a series of posts in which I explore how this view plays out in the permissivism debate. If there are many different rationally permissible responses to the same piece of evidence because there are many different rationally permissible attitudes to risk, then how does that allow us to answer the various objections to permissivism that have been raised.<br /><br />In this post, I want to do four things: first, run through a taxonomy of varieties of permissivism that slightly expands on one due to <a href="https://doi.org/10.1017/epi.2019.19" target="_blank">Elizabeth Jackson</a>; second, explain my motivation for offering this argument for permissivism; third, discuss an earlier risk-based argument for the position due to <a href="https://philpapers.org/rec/KELECB" target="_blank">Thomas Kelly</a>; finally, situate within Jackson's taxonomy the version of permissivism that follows from my own risk-based approach to setting priors and posteriors.<br /><br /><h2>Varieties of permissivism</h2><br />Let's start with the taxonomy of permissivisms. I suspect it's not complete; there are likely other dimensions along which permissivists will differ. But it's quite useful for our purposes.<br /><br />First, there are different versions of permissivism for different sorts of doxastic attitudes we might have in response to evidence. So there are versions for credences, beliefs, imprecise probabilities, comparative confidences, ranking functions, and so on. For instance, on the credal version of permissivism, there is evidence that doesn't determine a unique set of credences that rationality requires us to have in response to that evidence. For many different sorts of doxastic attitude, you can be permissive with respect to one but not the other: permissive about beliefs but not about credences, for instance, or vice versa. <br /><br />Second, permissivism comes in interpersonal and intrapersonal versions. According to interpersonal permissivism, it is possible for different individuals to have the same evidence, but different attitudes in response, and yet both be rational. According to the intrapersonal version, there is evidence a single individual might have, and different sets of attitudes such that whichever they have, they'll still be rational. Most people who hold to intrapersonal permissivism for a certain sort of doxastic attitude also hold to the interpersonal version, but there are many who think intrapersonal permissivism is mistaken but interpersonal permissivism is correct.<br /><br />Third, it comes in wide and narrow versions. This is determined by how many different attitudes are permitted in response to a piece of evidence, and how much variation there is between them. On narrow versions, there are not so many different rational responses and they do not vary too widely; on wide versions, there are many and they vary greatly.<br /><br />Fourth, it comes in common and rare versions. On the first, most evidence is permissive; on the latter, permissive evidence is rare. <br /><br />I'll end up defending two versions of permissivism: (i) a wide common version of interpersonal permissivism about credences; and (ii) a narrow common version of intrapersonal permissivism about credences.<br /><br /><h2>Why argue for permissivism?</h2><br />Well, because it's true, mainly. But there's another motivation for adding to the already crowded marketplace of arguments for the position. Many philosophers defend permissivism for negative reasons. They look at two very different sorts of evidence and give reasons to be pessimistic about the prospects of identifying a unique rational credal response to them. They are: very sparse evidence and very complex evidence. In the first, they say, our evidence constrains us too little. There are too many credal states that respect it. If there is a single credal response that rationality mandates to this sparse evidence, there must be some way to whittle down the vast set of states that respect it leave us with only one. For instance, some philosophers claim that, among this vast set of states, we should pick the one that has lowest informational content, since any other will go beyond what is warranted by the evidence. But it has proven extremely difficult to identify that credal state in many cases, such as von Mises' water-wine example, Bertrand's paradox, and van Fraassen's cube factory. Despairing of finding a way to pick a single credal state from this vast range, many philosophers have become permissivist. In the second sort of case, at the other extreme, where our evidence is very complex not very sparse, our evidence points in too many directions at once. In such cases, you might hope to identify a unique way in which to weigh the different sources of evidence and the direction in which they point to give the unique credal state that rationality mandates. And yet again, it has proven difficult to find a principled way of assigning these weights. Despairing, philosophers have become permissivist in these cases too.<br /><br />I'd like to give a positive motivation for permissivism---one that doesn't motivate it by pointing to the difficulty of establishing its negation. My account will be based within accuracy-first epistemology, and it will depend crucially on the notion of epistemic risk. Rationality permits a variety of attitudes to risk in the practical sphere. Faced with the same risky choice, you might be willing to gamble because you are risk-seeking, and I might be unwilling because I am risk-averse, but we are both rational and neither more rational than the other. On my account, rationality also permits different attitudes to risk in the epistemic sphere. And different attitudes to epistemic risk warrant different credal attitudes in response to a body of evidence. Therefore, permissivism.<br /><br /><h2>Epistemic risk encoded in epistemic utility</h2><br />It is worth noting that this is not the first time that the notion of epistemic risk has entered the permissivism debate. In an early paper on the topic, <a href="https://philpapers.org/rec/KELECB" target="_blank">Thomas Kelly</a> appeals to <a href="https://www.gutenberg.org/files/26659/26659-h/26659-h.htm" target="_blank">William James'</a> distinction between the two goals that we have when we have beliefs---believing truths and avoiding errors. When we have a belief, it gives us a chance of being right, but it also runs the risk of being wrong. In constrast, when we withhold judgment on a proposition, we run no risk of being wrong, but we give ourselves no chance of being right. Kelly then notes that whether you should believe on the basis of some evidence depends on how strongly you want to believe truths and how strongly you don't want to believe falsehoods. Using an epistemic utility framework introduced independently by <a href="https://doi.org/10.1111/nous.12099" target="_blank">Kenny Easwaran</a> and <a href="https://doi.org/10.1093/mind/fzx028" target="_blank">Kevin Dorst</a>, we can make this precise. Suppose:<br /><ol><li>I assign a positive epistemic utility of $R > 0$ to believing a truth or disbelieving a falsehood;</li><li>I assign a negative epistemic utility (or positive epistemic disutility) of $-W < 0$ to believing a falsehood or disbelieving a truth; and</li><li>I assign a neutral epistemic utility of 0 to withholding judgment.</li></ol>And suppose $W > R$. And suppose further that there is some way to measure, for each proposition, how likely or probable my evidence makes that proposition---that is, we assume there is a unique evidential probability function of the sort that J. M. Keynes, E. T. Jaynes, and Timothy Williamson envisaged. Then, if $r$ is how likely my evidence makes the proposition $X$, then:<br /><ol><li>the expected value of believing $X$ is $rR + (1-r)(-W)$,</li><li>the expected value of disbelieving $X$ is $r(-W) + (1-r)R$, and</li><li>the expected value of withholding judgment is $0$.</li></ol>A quick calculation shows that believing uniquely maximises expected utility when $r > \frac{W}{R+W}$, disbelieving uniquely maximises when $r < \frac{R}{R+W}$, and withholding uniquely maximises if $\frac{R}{R +W} < r < \frac{W}{R+W}$. What follows is that the more you disvalue being wrong, the stronger the evidence will have to be in order to make it rational to believe. Now, Kelly assumes that various values of $R$ and $W$ are rationally permissible---it is permissible to disvalue believing falsehoods a lot more than you value believing truths, and it is permissible to disvalue that just a little more. And, if that is the case, different individuals might have the same evidence while rationality requires of them different doxastic attitudes---a belief for one of them, who disvalues being wrong only a little more than they value being right, and no belief for the other, where the difference between their disvalue for false belief and value for true belief is much greater. Kelly identifies the values you pick for $R$ and $W$ with your attitudes to epistemic risk. So different doxastic attitudes are permissible in the face of the same evidence because different attitudes to epistemic risk are permissible.<br /><br />Now, there are a number of things worth noting here before I pass to my own alternative approach to epistemic risk.<br /><br />First, note that Kelly manages to show that epistemic rationality might be permissive even if there is a unique evidential probability measure. So even those who think you can solve the problem of what probability is demanded by the very sparse evidence and the very complex evidence we described above, still they should countenance a form of epistemic permissivism if they agree that there are different permissible values for $R$ and $W$.<br /><br />Second, it might seem at first that Kelly's argument gives interpersonal permissivism at most. After all, for fixed $R$ and $W$, and a unique evidential probability $r$ for $X$ given your evidence, it might seem that there is always a single attitude---belief in $X$, disbelief in $X$, or judgment withheld about $X$---that maximises expected epistemic value. But this isn't always true. After all, if $r = \frac{R}{R + W}$, then it turns out that disbelieving and withholding have the same expected epistemic value, and if $r = \frac{W}{R+W}$, then believing and withholding have the same expected epistemic value. And in those cases, it would be rationally permissible for an individual to pick either.<br /><br />Third, and relatedly, it might seem that Kelly's argument gives only narrow permissivism, since it allows for cases in which believing and withholding are both rational, and it allows for cases in which disbelieving and withholding are both rational, but it doesn't allow for cases in which all three are rational. But that again is a mistake. If you value believing truths exactly as much as you value believing falsehoods, so that $R = W$, and if the objective evidential probability of $X$ given your evidence is $r = \frac{1}{2}$, then believing, disbelieving, and withholding judgment are all permissible. Having said that, there is some reason to say that it is not rationally permissible to set $R = W$. After all, if you do, and if $r = \frac{1}{2}$, then it is permissible to both believe $X$ and believe $\overline{X}$ at the same time, and that seems wrong.<br /><br />Fourth, and most importantly for my purposes, Kelly's argument works for beliefs, but not for credences. The problem, briefly stated, is this: suppose $r$ is how likely my evidence makes proposition $X$. And suppose $\mathfrak{s}(1, x)$ is the accuracy of credence $x$ in a truth, while $\mathfrak{s}(0, x)$ is the accuracy of credence $x$ in a falsehood. Then the expected accuracy of credence $x$ in $X$ is<br />\begin{equation}\label{eeu}<br />r\mathfrak{s}(1, x) + (1-r)\mathfrak{s}(0, x)\tag{*}<br />\end{equation}<br />But nearly all advocates of epistemic utility theory for credences agree that rationality requires that $\mathfrak{s}$ is a strictly proper scoring rule. And that means that (*) is maximized, as a function of $x$, at $x = r$. So differences in how you value epistemic utility don't give rise to differences in what credences you should have. Your credences should always match the objective evidential probability of $X$ given your evidence. Epistemic permissivism about credences would therefore be false.<br /><br />I think Kelly's observation, supplemented with Easwaran's precise formulation of epistemic value, furnishes a strong argument for permissivism about beliefs. But I think we can appeal to epistemic risk to give something more, namely, two versions of permissivism about credences: first, an wide common interpersonal version, and second a narrow common intrapersonal version.<br /><br /><h2>Epistemic risk encoded in decision rules</h2><br />To take the first step towards these versions of permissivism for credences, let's begin with the observation that there are two ways in which risk enters into the rational evaluation of a set of options. First, risk might be encoded in the utility function, which measures the value of each option at each possible world; or, second, it might be encoded in the choice rule, which takes in various features of the options, including their utilities at different worlds, and spits out the set of options that are rationally permissible.<br /><br />Before we move to the epistemic case, let's look at how this plays out in the practical case. I am about to flip a fair coin. I make you an offer: pay me £30, and I will pay you £100 if the coin lands heads and nothing if it lands tails. You reject my offer. There are two ways to rationalise your decision. On the first, you choose using expected utility theory, which is a risk-neutral decision rule. However, because the utility you assign to an outcome is a sufficiently concave function of the money you get in that outcome, and your current wealth is sufficiently small, the expected utility of accepting my offer is less than the expected utility of rejecting it. For instance, perhaps your utility for an outcome in which your total wealth is £$n$ is $\log n$. And perhaps your current wealth is £$40$. Then your expected utility for accepting my offer is $\frac{1}{2}\log 110 + \frac{1}{2} \log 10 \approx 3.502$ while your expected utility for rejecting it is $\log 40 \approx 3.689$. So you are rationally required to reject. On this way of understanding your choice, your risk-aversion is encoded in your utility function, while your decision rule is risk-neutral. On the second way of understanding your choice, it is the other way around. Instead of expected utility theory, you choose using a risk-sensitive decision rule, such as Wald's <a href="https://en.wikipedia.org/wiki/Wald's_maximin_model" target="_blank">Maximin</a>, the <a href="https://m-phi.blogspot.com/2020/07/hurwiczs-criterion-of-realism-and.html" target="_blank">Hurwicz criterion</a>, the <a href="https://m-phi.blogspot.com/2020/07/a-generalised-hurwicz-criterion.html" target="_blank">generalized Hurwicz criterion</a>, Quiggin's <a href="https://en.wikipedia.org/wiki/Rank-dependent_expected_utility" target="_blank">rank-dependent utility theory</a>, or Buchak's <a href="https://www.oxfordscholarship.com/view/10.1093/acprof:oso/9780199672165.001.0001/acprof-9780199672165" target="_blank">risk-weighted expected utility theory</a>. According to Maximin, for instance, you are required to choose an option whose worst-case outcome is best. The worst case if you accept the offer is the one in which the coin lands tails and I pay you back nothing, in which case you end up £$30$ down, whereas the worst case if you refuse my offer is the status quo in which you end up with exactly as much as you had before. So, providing you prefer more money to less, the worst-case outcome of accepting the offer is worse than the worse-case outcome of refusing it, so Maximin will lead you to refuse the offer. And it will lead you to do that even if, for instance, you value money linearly. Thus, there is no need to reflect your attitude to risk in your utility function at all, because it is encoded in your decision rule.<br /><br />I take the lesson of the <a href="https://www.oxfordreference.com/view/10.1093/oi/authority.20110803095403122" target="_blank">Allais paradox</a> to be that there is rational risk-sensitive behaviour that we cannot capture entirely using the first method here. That is, there are rational preferences that we cannot recover within expected utility theory by making the utility function concave in money, or applying some other tweak. We must instead permit risk-sensitive choice rules. Now, there are two sorts of such rules: those that require credences among their inputs and those that don't. In the first camp, perhaps the most sophisticated is Lara Buchak's <a href="https://www.oxfordscholarship.com/view/10.1093/acprof:oso/9780199672165.001.0001/acprof-9780199672165" target="_blank">risk-weighted expected utility theory</a>. In the second, we've already met the most famous example, namely, Maximin, which is maximally risk-averse. But there is also Maximax, which is maximally risk-seeking. And there is the Hurwicz criterion, which strikes a balance between the two. And there's my generalization of the Hurwicz criterion, which I'll abbreviate GHC. As I've discussed over the last few blogposts, I favour the latter in the case of picking priors. (For an alternative approach to epistemic risk, see Boris Babic's recent paper <a href="https://www.journals.uchicago.edu/doi/pdfplus/10.1086/703552" target="_blank">here</a>.)<br /><br />To see what happens when you use GHC to pick priors, let's give a quick example in a situation in which there are just three possible states of the world to which you assign credences, $w_1$, $w_2$, $w_3$, and we write $(p_1, p_2, p_3)$ for a credence function $p$ that assigns $p_i$ to world $w_i$. Suppose your Hurwicz weights are these: $\alpha_1$ for the best case, $\alpha_2$ for the second-best (and second-worst) case, and $\alpha_3$ for the worst case. And your accuracy measure is $\mathfrak{I}$. Then we're looking for those that minimize your Hurwicz score, which is$$H^A(p) = \alpha_1\mathfrak{I}(p, w_{i_1}) + \alpha_2\mathfrak{I}(p, w_{i_2}) + \alpha_3\mathfrak{I}(p, w_{i_3})$$when$$\mathfrak{I}(p, w_{i_1}) \geq \mathfrak{I}(p, w_{i_2}) \geq\mathfrak{I}(p, w_{i_3})$$Now suppose for our example that $\alpha_1 \geq \alpha_2 \geq \alpha_3$. Then the credence functions that minimize $H^A_{\mathfrak{I}}$ are$$\begin{array}{ccc} (\alpha_1, \alpha_2, \alpha_3) & (\alpha_1, \alpha_3, \alpha_2) & (\alpha_2, \alpha_1, \alpha_3) \\ (\alpha_2, \alpha_3, \alpha_1) & (\alpha_3, \alpha_1, \alpha_2) & (\alpha_3, \alpha_2, \alpha_1) \end{array}$$<br /><br />With that example in hand, and a little insight into how GHC works when you use it to select priors, let's work through Elizabeth Jackson's taxonomy of permissivism from above.<br /><br />First, since the attitudes we are considering are credences, it's a credal version of permissivism that follows from this risk-based approach in accuracy-first epistemology.<br /><br />Second, we obtain both an interpersonal and an intrapersonal permissivism. A particular person will have risk attitudes represented by specific Hurwicz weights. And yet, even once those are fixed, there will usually be a number of different permissible priors. That is, rationally will permit a number of different credal states in the absence of evidence. For instance, if my Hurwicz weights are $\alpha_1 = 0.5$, $\alpha_2 = 0.3$, $\alpha_3 = 0.2$, then rationality allows me to assign 0.5 to world $w_1$, 0.3 to $w_2$ and 0.2 to $w_3$, but it also permits me to assign $0.3$ to $w_1$, $0.2$ to $w_2$, and $0.5$ to $w_3$.<br /><br />So there is intrapersonal credal permissivism, but it is reasonably narrow---there are only six rationally permissible credence functions for someone with the Hurwicz scores just specified, for instance. On the other hand, the interpersonal permissivism we obtain is very wide. Indeed, it is as wide as range of permissible attitudes to risk. As we noted in a previous post, for any probabilistic credence function over a space of possible worlds, there are Hurwicz weights that will render those credences permissible. So providing those weights are rationally permissible, so are the credences. <br /><br />Finally, is the permissivism we get from this risk-based approach common or rare? So far, we've just considered it in the case of priors. That is, we've only established permissivism in the case in which you have no evidence. But of course, once it's established there, it's also established for many other bodies of evidence, since we obtain the rational credences given a body of evidence by looking to what we obtain by updating rational priors by conditioning on that evidence. And, providing a body of evidence isn't fully informative, if there are multiple rational priors, they will give rise to multiple rational posteriors when we condition them on that evidence. So the wide interpersonal credal permissivism we obtain is common, and so is the narrow intrapersonal credal permissivism.Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com9tag:blogger.com,1999:blog-4987609114415205593.post-33893996719156809942020-07-22T07:23:00.001+01:002020-07-22T14:12:06.290+01:00Taking risks and picking posteriorsFor a PDF of this blog, see <a href="https://drive.google.com/file/d/17FgjpCNg-qwj9zPK1D3PoPG2sh4ecGj2/view?usp=sharing" target="_blank">here</a>.<br /><br />When are my credences rational? In Bayesian epistemology, there's a standard approach to this question. We begin by asking what credences would be rational were you to have no evidence at all; then we ask what ways of updating your credences are rational when you receive new evidence; and finally we say that your current credences are rational if they are the result of updating rational priors in a rational way on the basis of your current total evidence. This account can be read in one of two ways: on the doxastic reading, you're rational if, <i>in fact</i>, when you had no evidence you had priors that were rational and if, <i>in fact</i>, when you received evidence you updated in a rational way; on the propositional reading, you're rational if <i>there exists</i> some rational prior and <i>there exists</i> some rational way of updating such that applying the updating rule to the prior based on your current evidence issues in your current credences.<br /><br />In <a href="https://m-phi.blogspot.com/2020/07/taking-risks-and-picking-priors.html" target="_blank">this previous post</a>, I asked how we might use accuracy-first epistemology and decision-making rules for situations of massive uncertainty to identify the rational priors. I suggested that we should turn to the early days of decision theory, when there was still significant interest in how we might make a decision in situations in which it is not possible to assign probabilities to the different possible states of the world. In particular, I noted Hurwicz's generalization of Wald's Maximin rule, which is now called the Hurwicz Criterion, and I offered a further generalization of my own, which I then applied to the problem of picking priors. Here's my generalization:<br /><br /><b>Generalized Hurwicz Criterion (GHC)</b> Suppose the set of possible states of the world is $W = \{w_1, \ldots, w_n\}$. Pick $0 \leq \alpha_1, \ldots, \alpha_n \leq 1$ with $\alpha_1 + \ldots + \alpha_n = 1$, and denote this sequence of weights $A$. Suppose $a$ is an option defined on $W$, and write $a(w_i)$ for the utility of $a$ at world $w_i$. Then if$$a(w_{i_1}) \geq a(w_{i_2}) \geq \ldots \geq a(w_{i_n})$$then let$$H^A(a) = \alpha_1a(w_{i_1}) + \ldots + \alpha_na(w_{i_n})$$Pick an option that maximises $H^A$.<br /><br />Thus, whereas Wald's Maximin puts all of the weight onto the worst case, and Hurwicz's Criterion distributes all the weight between the best and worst cases, the generalized Hurwicz Criterion allows you to distribute the weight between best, second-best, and so on down to second-worst, and worst. I said that you should pick your priors by applying GHC with a measure of accuracy for credences. Then I described the norms for priors that it imposes.<br /><br />In this post, I'm interested in the second component of the Bayesian approach I described above, namely, the rational ways of updating. How does rationality demand we update our prior when new evidence arrives? Again, I'll be asking this within the accuracy-first framework.<br /><br />As in the previous post, we'll consider the simplest possible case. We'll assume there are just three possible states of the world that you entertain and to which you will assign credences. They are $w_1, w_2, w_3$. If $p$ is a credence function, then we'll write $p_i$ for $p(w_i)$, and we'll denote the whole credence function $(p_1, p_2, p_3)$. At the beginning of your epistemic life you have no evidence, and you must pick your prior. We'll assume that you do as I proposed in the previous post and set your prior using GHC with some weights that you have picked, $\alpha_1$ for the best-case scenario, $\alpha_2$ for the second-best, and $\alpha_3$ for the worst. Later on, let's suppose, you learn evidence $E = \{w_1, w_2\}$. How should you update? As we will see, the problem is that there are many seemingly plausible approaches to this question, most of which disagree and most of which give implausible answers.<br /><br />A natural first proposal is to use the same decision rule to select our posterior as we used to select our prior. To illustrate, let's suppose that our Hurwicz weight for the best-case scenario is $\alpha_1 = 0.5$, for the second-best $\alpha_2 = 0.3$, and for the worst $\alpha_3 = 0.2$. Applying GHC with these weights and any additive and continuous strictly proper (acsp) accuracy measure gives the following as the permissible priors:$$\begin{array}{ccc}(0.5, 0.3, 0.2) & (0.5, 0.2, 0.3) & (0.3, 0.5, 0.2) \\ (0.3, 0.2, 0.5) & (0.2, 0.5, 0.3) &(0.2, 0.3, 0.5)\end{array}$$<br />Let's suppose we pick $(0.5, 0.3, 0.2)$. And now suppose we learn $E = \{w_1, w_2\}$. If we simply apply GHC again, we get the same set of credence functions as permissible posteriors. But none of these even respects the evidence we've obtained -- that is, all of them assign positive credence to world $w_3$, which our evidence has ruled out. So that can't be quite right.<br /><br />Perhaps, then, we should first limit the permissible posteriors to those that respect our evidence -- by assigning credence $0$ to world $w_3$ -- and then find the credence function that maximizes GHC <i>among them</i>. It turns out that the success of this move depends on the measure of accuracy that you use. Suppose, for instance, you use the Brier score $\mathfrak{B}$, whose accuracy for a credence function $p = (p_1, p_2, p_3)$ at world $w_i$ is $$\mathfrak{B}(p, w_i) = 2p_i - (p_1^2 + p_2^2 + p_3^2)$$That is, you find the credence function of the form $q = (q_1, 1-q_1, 0)$ that minimizes $H^A_\mathfrak{B}$. But it turns out that this is $q = (0.6, 0.4, 0)$, which is not the result of conditioning $(0.5, 0.3, 0.2)$ on $E = \{w_1, w_2\}$ -- that would be $q = (0.625, 0.375, 0)$.<br /><br />However, as I explained in <a href="https://m-phi.blogspot.com/2020/07/updating-by-minimizing-expected.html" target="_blank">another previous post</a>, there is a unique additive and continuous strictly proper accuracy measure that will give conditionalization in this way. I called it the enhanced log score $\mathfrak{L}^\star$, and it is also found in Juergen Landes' paper <a href="https://doi.org/10.1016/j.ijar.2015.05.007" target="_blank">here</a> (Proposition 9.1) and Schervish, Seidenfeld, and Kadane's paper <a href="https://doi.org/10.1287/deca.1090.0153" target="_blank">here</a> (Example 6). Its accuracy for a credence function $p = (p_1, p_2, p_3)$ at world $w_i$ is $$\mathfrak{L}^\star(p, w_i) = \log p_i - (p_1 + p_2 + p_3)$$If we apply GHC with that accuracy measure and with the restriction to credence functions that satisfy the evidence, we get $(0.625, 0.375, 0)$ or $(0.375, 0.625, 0)$, as required. So while GHC doesn't mandate conditioning on your evidence, it does at least permit it. However, while this goes smoothly if we pick $(0.5, 0.3, 0.2)$ as our prior, it does not work so well if we pick $(0.2, 0.3, 0.5)$, which, if you recall, is also permitted by the Hurwicz weights we are using. After all, the two permissible posteriors remain the same, but neither is the result of conditioning that prior on $E$. This proposal, then, is a non-starter.<br /><br />There is, in any case, something strange about the approach just mooted. After all, GHC assigns a weight to the accuracy of a candidate posterior in each of the three worlds, even though in world $w_3$ you wouldn't receive evidence $E$ and would thus not adopt this posterior. Let's suppose that you'd receive evidence $\overline{E} = \{w_3\}$ instead at world $w_3$; and let's suppose you'd adopt the only credence function that respects this evidence, namely, $(0, 0, 1)$. If that's the case, we might try applying GHC not to potential posteriors but to potential rules for picking posteriors. I'll call these <i>posterior rules</i>. In the past, I've called them updating rules, but this is a bit misleading. An updating rule would take as inputs both prior and evidence and give the result of updating the former on the latter. But these rules really just take evidence as an input and say which posterior you'll adopt if you receive it. Thus, for our situation, in which you might learn either $E$ or $\overline{E}$, the posterior rule would have the following form:$$p' = \left \{ \begin{array}{rcl}E & \mapsto & p'_E \\ \overline{E} & \mapsto & p'_{\overline{E}}\end{array}\right.$$for some suitable specification of $p'_E$ and $p'_\overline{E}$. Then the accuracy of a rule $p'$ at a world is just the accuracy of the output of that rule at that world. Thus, in this case:$$\begin{array}{rcl}\mathfrak{I}(p', w_1) & = & \mathfrak{I}(p'_E, w_1) \\ \mathfrak{I}(p', w_2) & = & \mathfrak{I}(p'_E, w_2) \\\mathfrak{I}(p', w_3) & = & \mathfrak{I}(p'_\overline{E}, w_3)\end{array}$$The problem is that this move doesn't help. Part of the reason is that whatever was the best-case scenario for the prior, the best case for the posterior is sure to be world $w_3$, since $p'_\overline{E} = (0, 0, 1)$ is perfectly accurate at that world. Thus, suppose you pick $(0.5, 0.3, 0.2)$ as your prior. It turns out that the rules that minimize $H^A_{\mathfrak{L}^\star}$ will give $p'_E = (0.4, 0.6, 0)$ or $p'_E = (0.6, 0.4, 0)$, whereas conditioning your prior on $E$ gives $p'_E = (0.625, 0.375, 0)$ or $p'_E = (0.375, 0.625, 0)$.<br /><br />Throughout our discussion so far, we have dismissed various possible approaches because they are not consistent with conditionalization. But why should that be a restriction? Perhaps the approach we are taking will tell us that the Bayesian fixation with conditionalization is misguided. Perhaps. But there are strong arguments for conditionalization within accuracy-first epistemology, so we'd have to see why they go wrong before we start rewriting Bayesian textbooks. I'll consider three such arguments here. The first isn't as strong as it seems; the second isn't obviously available to someone who used GHC to pick priors; the third is promising but it leads us initially down a tempting road into an inhospitable morass.<br /><br />The first is closely related to a proposal I explored <a href="https://m-phi.blogspot.com/2020/07/updating-by-minimizing-expected.html" target="_blank">in a previous blogpost</a>. So I'll briefly outline the approach here and refer to the issues raised in that post. The idea is this: Your prior is $(p_1, p_2, p_3)$. You learn $E$. You must now adopt a posterior that respects your new evidence, namely, $(q_1, 1-q_1, 0)$. You should choose the posterior of that form that maximises expected accuracy from the point of view of your prior, that is, you're looking for $(x, 1-x, 0)$ that maximizes$$p_1 \mathfrak{I}((x, 1-x, 0), w_1) + p_2 \mathfrak{I}((x, 1-x, 0), w_2) + p_3 \mathfrak{I}((x, 1-x, 0), w_3)$$This approach is taken in a number of places: at the very least, <a href="https://statweb.stanford.edu/~cgates/PERSI/papers/zabell82.pdf" target="_blank">here</a>, <a href="https://drive.google.com/file/d/1YwFOV5abKhsjftoyIEtWI-owO4HjJSEm/view?usp=sharing" target="_blank">here</a>, and <a href="https://philpapers.org/rec/WROBUM" target="_blank">here</a>. Now, it turns out that there is only one additive and continuous strictly proper accuracy measure that is guaranteed always to give conditionalization on this approach. That is, there is only one measure such that, for any prior, the posterior it expects to be best among those that respect the evidence is the one that results from conditioning the prior on the evidence. Indeed, that accuracy measure is one we've already met above, namely, the enhanced log score $\mathfrak{L}^\star$ (see <a href="https://m-phi.blogspot.com/2020/07/updating-by-minimizing-expected.html" target="_blank">here</a>). However, it turns out that it only works if we assume our credences are defined only over the set of possible states of the world, and not over more coarse-grained propositions (see <a href="https://m-phi.blogspot.com/2020/07/update-on-updating-or-fall-from-favour.html" target="_blank">here</a>). So I think this approach is a non-starter.<br /><br />More promising at first sight is the argument by <a href="https://doi.org/10.1093/mind/fzl607" target="_blank">Hilary Greaves and David Wallace</a> from 2006. Here, just as we considered earlier, we look not just at the posterior we will adopt having learned $E$, but also the posterior we would adopt were we to learn $\overline{E}$. Thus, if your prior is $(p_1, p_2, p_3)$, then you are looking for $(x, 1-x, 0)$ that maximizes$$p_1 \mathfrak{I}((x, 1-x, 0), w_1) + p_2 \mathfrak{I}((x, 1-x, 0), w_2) + p_3 \mathfrak{I}((0, 0, 1), w_3)$$And it turns out that this will always be$$x = \frac{p_1}{p_1 + p_2}\ \ 1-x = \frac{p_2}{p_1+p_2}$$providing $\mathfrak{I}$ is strictly proper.<br /><br />Does this help us? Does it show that, if we set our priors using GHC, we should then set our posteriors using conditionalization? One worry might be this: What justifies you in choosing your posteriors using one decision rule -- namely, maximise subjective expected utility -- when you picked your priors using a different one -- namely, GHC? But there seems to be a natural answer. As I emphasised above, GHC is specifically designed for situations in which probabilities, either subjective or objective, are not available. It allows us to make decisions in their absence. But of course when it comes to choosing the posterior, we are no longer in such a situation. At that point, we can simply resort to what became more orthodox decision theory, namely, Savage's subjective expected utility theory.<br /><br />But there's a problem with this. GHC is not a neutral norm for picking priors. When you pick your Hurwicz weights for the best case, the second-best case, and so on down to the second-worst case and the worst case, you reflect an attitude to risk. Give more weight to the worst cases and you're risk averse, choosing options that make those worst cases better; give more weight to the best cases and you're risk seeking; spread the weights equally across all cases and you are risk neutral. But the problem is that subjective expected utility theory is a risk neutral theory. (One way to see this is to note that it is the special case of <a href="https://global.oup.com/academic/product/risk-and-rationality-9780199672165" target="_blank">Lara Buchak's risk-weighted expected utility theory</a> that results from using the neutral risk function $r(x) = x$.) Thus, for those who have picked their prior using a risk-sensitive instance of GHC when they lacked probabilities, the natural decision rule when they have access to probabilities is not going to be straightforward expected utility theory. It's going to be a risk-sensitive rule that can accommodate subjective probabilities. The natural place to look would be Lara Buchak's theory, for instance. And it's straightforward to show that Greaves and Wallace's result does not hold when you use such a rule. (In <a href="https://philpapers.org/rec/CAMAUF" target="_blank">forthcoming work</a>, Catrin Campbell-Moore and Bernhard Salow have been working on what does follow and how we might change our accuracy measures to fit with such a theory and what follows from an argument like Greaves and Wallace's when you do that.) In sum, I think arguments for conditionalization based on maximizing expected accuracy won't help us here.<br /><br />Fortunately, however, there is another argument, and it doesn't run into this problem. As we will see, though, it does face other challenges. In Greaves and Wallace's argument, we took the view from from the prior that we picked using GHC, and we used it to evaluate our way of picking posteriors. In this argument, due to <a href="https://drive.google.com/open?id=1kjY_wQ0nlXIGfnla_MhWF1uQmcJtcTUB" target="_blank">me and Ray Briggs</a>, we take the view from nowhere, and we use it to evaluate the prior and the posterior rule together. Thus, suppose $p$ is your prior and $p'$ is your posterior rule. Then we evaluate them together by taking their joint accuracy to be the sum of their individual accuracies. Thus,$$\mathfrak{I}((p, p'), w) = \mathfrak{I}(p, w) + \mathfrak{I}(p', w)$$Then we have the following fact, where $p'$ is a conditioning rule for $p$ over some partition $\mathcal{E}$ iff, for all $E$ in $\mathcal{E}$, if $p(E) > 0$, then $p'_E(-) = p(-|E)$:<br /><br /><b>Theorem</b> Suppose $\mathfrak{I}$ is an additive and continuous strictly proper scoring rule. Then, if $p'$ is not a conditioning rule for $p$ over $\mathcal{E}$, there are $q$ and $q'$ such that$$\mathfrak{I}((p, p'), w) < \mathfrak{I}((q, q'), w)$$for all worlds $w$.<br /><br />That is, if $p'$ is not a conditioning rule for $p$, then, taken together, they are accuracy-dominated. There is an alternative pair, $q$ and $q'$, that, taken together, are guaranteed to be more accurate than $p$ and $p'$ are, taken together.<br /><br />Notice that this argument establishes a slightly different norm from the one that the expected accuracy argument secures. The latter is a narrow scope norm: if $p$ is your prior, then your posterior rule should be to condition on $p$ with whatever evidence you learn. The former is a wide scope norm: you should not have prior $p$ and a posterior rule that does not condition on the evidence you learn. This suggests that, if you're sitting at the beginning of your epistemic life and you're picking priors and posterior rules together, as a package, you should pick them so that the posterior rule involves conditioning on the prior with the evidence received. Does it also tell you anything about what to do if you're sitting with your prior already fixed and new evidence comes in? I'm not sure. Here's a reason to think it might. You might think that it's only rational to do at a later time what it was rational to plan to do at an earlier time. If that's right, then we can obtain the narrow scope norm from the wide scope one.<br /><br />Let's park those questions for the moment. For the approach taken in this argument suggests something else. In the previous post, we asked how to pick your priors, and we hit upon GHC. Now that we have a way of evaluating priors and posterior rules together, perhaps we should just apply GHC to those? Let's see what happens if we do that. As before, assume the best case receives weight $\alpha_1 = 0.5$, the second-best $\alpha_2 = 0.3$, and the third best $\alpha_3 = 0.2$. Then we know that the priors that GHC permits when we consider them on their own without the posteriors plans appended to them are just$$\begin{array}{ccc}(0.5, 0.3, 0.2) & (0.5, 0.2, 0.3) & (0.3, 0.5, 0.2) \\ (0.3, 0.2, 0.5) & (0.2, 0.5, 0.3) &(0.2, 0.3, 0.5)\end{array}$$Now let's consider what happens when we add in the posterior rules for learning $E = \{w_1, w_2\}$ or $\overline{E} = \{w_3\}$. Then it turns out that the minimizers are the priors$$(0.3, 0.2, 0.5)\ \ (0.2, 0.3, 0.5)$$combined with the corresponding conditionalizing posterior rules. Now, since those two priors are among the ones that GHC permits when applied to the priors alone, this might seem consistent with the original approach. The problem is that these priors are specific to the case in which you'll learn either $E$ or $\overline{E}$. If, on the other hand, you'll learn $F = \{w_1\}$ or $\overline{F} = \{w_2, w_3\}$, the permissible priors are$$(0.5, 0.2, 0.3)\ \ (0.5, 0.3, 0.2)$$And, at the beginning of your epistemic life, you don't know which, if either, is correct.<br /><br />In fact, there's what seems to me a deeper problem. In the previous paragraph we considered a situation in which you might learn either $E$ or $\overline{E}$ or you might learn either $F$ or $\overline{F}$, and you don't know which. But the two options determine different permissible priors. The same thing happens if there are four possible states of the world $\{w_1, w_2, w_3, w_4\}$ and you might learn either $E_1 = \{w_1, w_2\}$ or $E_2 = \{w_3, w_4\}$ or you might learn either $F_1 = \{w_1, w_2\}$ or $F_2 = \{w_3\}$ or $F_3 = \{w_4\}$. Now, suppose you assign the following Hurwicz weights: to the best case, you assign $\alpha_1 = 0.4$, to the second best $\alpha_2 = 0.3$, to the second worst $\alpha_3 = 0.2$ and to the worst $\alpha_1 = 0.1$. Then if you'll learn $E_1 = \{w_1, w_2\}$ or $E_2 = \{w_3, w_4\}$, then the permissible priors are<br /> $$\begin{array}{cccc}(0.1, 0.4, 0.2, 0.3) & (0.4, 0.1, 0.2, 0.3) & (0.1, 0.4, 0.3, 0.2) & (0.4, 0.1, 0.2, 0.3) \\ (0.2, 0.3, 0.1, 0.4) & (0.2, 0.3, 0.4, 0.1) & (0.3, 0.2, 0.1, 0.4) & (0.2, 0.3, 0.4, 0.1) \end{array}$$But if you'll learn $F_1 = \{w_1, w_2\}$ or $F_2 = \{w_3\}$ or $F_3 = \{w_4\}$, then your permissible priors are<br /> $$\begin{array}{cccc}(0.1, 0.2, 0.3, 0.4) & (0.1, 0.2, 0.4, 0.3) & (0.2, 0.1, 0.3, 0.4) & (0.2, 0.1, 0.4, 0.3) \end{array}$$That is, there is no overlap between the two. It seems to me that the reason this is such a problem is that it's always been a bit of an oddity that the two accuracy-first arguments for conditionalization seem to depend on this assumption that there is some partition from which your evidence will come. It seems strange that when you learn $E$, in order to determine how to update, you need to know what alternative propositions you might have learned instead. The reason this assumption hasn't proved so problematic so far is that the update rule is in fact not sensitive to the partition. For instance, if I will learn $E_1 = F_1 = \{w_1, w_2\}$, both the Greaves and Wallace argument and the Briggs and Pettigrew argument for conditionalization say that you should update on that in the same way whether or not you might have learned $E_2 = \{w_3, w_4\}$ instead or whether you might have learned $F_2 = \{w_3\}$ or $F_3 = \{w_4\}$ instead. But here the assumption does seem problematic, because the permissible priors are sensitive to what the partition is from which you'll receive your future evidence.<br /><br />What to conclude from all this? It seems to me that the correct approach is this: choose priors using GHC; choose posterior rules to go with them using the dominance argument that Ray and I gave--that is, update by conditioning.Richard Pettigrewhttp://www.blogger.com/profile/07828399117450825734noreply@blogger.com17