Friday, 26 October 2018

Dutch Books and Reflection

I've just signed a contract with Cambridge University Press to write a book on the Dutch Book Argument for their Elements in Decision Theory and Philosophy series. So over the next few months, I'm going to be posting some bits and pieces as I get properly immersed in the literature.

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In yesterday's post, I walked through the Dutch Strategy argument for Conditionalization. Today, I'd like to think about a standard objection to it. As van Fraassen (1984) pointed out, we can give a seemingly analogous Dutch Strategy argument for his Reflection Principle, which says:

Reflection Principle  If $c(C_{c_i}) > 0$, then $c(X | C_{c_i}) = c_i(X)$.

We'll consider the details of the argument below. Now, van Fraassen took this argument to count in favour of his Reflection Principle. Indeed, since the Reflection Principle looks implausible on all accounts of credence except van Fraassen's voluntarism, he appealed to this Dutch Strategy Argument for the Reflection Principle in an argument for voluntarism. But most philosophers have seen a modus tollens where van Fraassen saw a modus ponens. After all, the Reflection Principle demands a level of deference to your future credences that is sometimes simply not rationally permitted, let alone required. For instance, if Sandy knows that between Monday and Tuesday, he will take a drug that makes him enormously and irrationally under-confident in extreme climate scenarios and over-confident in more moderate scenarios, his confidence in Medium on Monday conditional on being 90% confident in Medium on Tuesday, for instance, should not be 90%---it should be less than that (Christensen 1991). Thus, from the denial of the Reflection Principle, many infer that the Dutch Strategy arguments in its favour is invalid, and from that they infer that all such arguments are invalid, and thus they cast doubt on the particular Dutch Strategy argument for conditionalizing.

R. A. Briggs responds to this objection to the Dutch Strategy argument for conditionalizing by arguing that, contrary to appearances, there is a disanalogy between the two Dutch Strategy arguments. This allows us to reject the argument for the Reflection Principle as invalid, while retaining the argument for conditionalizing as valid. To see Briggs' point, let's place the two arguments side by side. First, the Dutch Strategy Argument for Rule Conditionalization. Suppose my credence function at $t$ is $c$, suppose $c(E) > 0$, and suppose that, if I learn $E$ and nothing more between $t$ and $t'$, I will adopt $c'$ at $t'$, where $c'(X) \neq c(X | E)$, for some $X$. Then there are sets of bets $B$, $B'_E$, and $B'_{\overline{E}}$ such that $c$ requires me to accept $B$, $c'$ requires me to accept $B'_E$, and any credence function requires me to accept $B'_{\overline{E}}$, where, taken together, $B$ and $B'_E$ will lose me money in all worlds at which $E$ is true and, taken together, $B$ and $B'_{\overline{E}}$ will lose me money in all worlds at which $E$ is false. Second, the Dutch Strategy Argument for Reflection. Suppose $c(C_{c_i}) > 0$ and suppose that $c(X | C_{c_i}) \neq c_i(X)$, for some $X$. Then there are bets $B$, $B'_{C_{c_i}}$, and $B'_{\overline{C_{c_i}}}$ such that $c$ requires me to accept $B$, $c_i$ requires me to accept $B'_{C_{c_i}}$, and any credence function requires me to accept $B'_{\overline{C_{c_i}}}$, where, taken together, the bets in $B$ and $B_{C_{c_i}}$ will lose me money in all worlds at which my credence function at $t'$ is indeed $c_i$, and, taken together, $B$ and $B'_{\overline{E}}$ will lose me money in all worlds at which $c_i$ is not my credence function at $t'$. So, as Briggs points out, if you will update other than conditionalizing if you learn $E$, then whatever evidence comes your way---whether $E$ or something else---the strategy described will generate bets that, taken together, will lose you money at all worlds at which your evidence is true. That is, they will lose you money at all epistemically possible worlds, which is what is required to establish irrationality. But, if you violate Reflection, then whatever credence function you adopt at $t'$---whether $c_i$ or something else---the strategy described will generate bets that, taken together, will lose you money at all worlds at which you adopt that credence function. However, that is not necessarily all epistemically possible worlds. For you might not know what your credences are at $t'$. In that case, even if I actually adopt $c_i$ at $t'$, there will nonetheless be an epistemically possible world at which I didn't adopt that, and then the bets in $B$ and $B'_{C_{c_i}}$, taken together, might not lose me money. And that blocks the Dutch Strategy argument for Reflection.

However, while Briggs successfully blocks the argument for Reflection in its strong, general form, Anna Mahtani (2012) points out that they do not block a Dutch Strategy argument for a weak, more specific version of Reflection:

Weak Reflection Principle If at $t'$ you will know what your credence function is, then if $c(C_{c_i}) > 0$, then $c(X | C_{c_i}) = c_i(X)$.

After all, if you satisfy the antecedent of the principle, then it cannot be that, after you adopt credence function $c_i$, it is still epistemically possible that you have some different credence function.

Now, the Weak Reflection Principle is hardly more plausible than the stronger version. That is, it is still very implausible. Knowing that his credences will be completely luminous to him after I take the mind-altering drug should not make Sandy any more inclined to defer to the credences he will end up having after taking it. Thus, the objection to the Dutch Strategy argument for Conditionalization remains intact.

How, then, should we respond to this objection? The first thing to note is that, in a sense, the Dutch Strategy argument for Reflection does not actually target Reflection. Or, at least, it doesn't target it directly. One way to see this is to note that, unlike the versions of conditionalization we have been considering, Reflection is a synchronic norm. It says something about how your credences should be at $t$. It says nothing about how your credences at $t$ should relate to your credences at $t'$, only how your credences at $t$ about your credences at $t'$ should relate to your other credences at $t$. But the Dutch Strategy argument involves bets that your credences at $t$ require you to accept, as well as bets that your credences at $t'$ require you to accept. You can violate Reflection, and have probabilistic credences---so the Converse Dutch Book Theorem shows that there is no synchronic Dutch Book argument against your credences; that is, there is no set of bets that $c$ alone requires you to accept that will lose you money at all epistemically possible worlds.

So what's going on? The key fact is this: if you violate Reflection, and you have a deterministic updating rule, then that updating rule cannot possibly be a conditionalizing rule. After all, suppose $c(C_{c_i}) > 0$ and $c(X | C_{C_{c_i}}) \neq c_i(X)$ and you learn $C_{c_i}$ and nothing more. Then, since you learn $C_{c_i}$, it must be true and thus your new credence function must be $c_i$. But your violation of Reflection ensures that $c_i$ is not the result of conditionalizing on your evidence $C_{c_i}$. So the Dutch Strategy argument for Reflection does not target Reflection itself; rather, it targets the updating rule you are forced to adopt because you violate Reflection.

Consider an analogous case---Briggs also considers this analogy, but draws a different moral from it. Suppose you think that it is irrational to have a set of beliefs that can't possibly all be true. Now, suppose you have the following second-order belief: you believe that you believe a contradiction, such as $X\ \&\ \overline{X}$. Then that belief itself might be true. So, by your standards, on its own, it is not irrational. However, suppose we now consider what your attitude to $X\ \&\ \overline{X}$ is. Whatever attitude you have, you are guaranteed to have a false belief: if you do believe the contradiction, your second-order belief is true, but your first-order belief is false; if you don't believe the contradiction, then your second-order belief itself is false. In this case, we might say that the belief itself is not irrational---it might be true, and it might be supported by your evidence. But its presence forces your total doxastic state to be irrational.

The same thing is going on in the case of Reflection. Just as you think that it is irrational to have beliefs that cannot all be true, you also think it is irrational to have credences that require you to enter into bets that lose you money for sure. And just as the single second-order belief that you believe a contradiction is possibly true, so by the Converse Dutch Book Theorem, a credence function that violates Reflection doesn't require you to accept any bets that will lose you money for sure. However, just as the single belief that you believe a contradiction forces you to have an attitude to the contradiction (either believing it or not) that ensures that your total doxastic state (first- and second-order beliefs together) includes a false belief, so your violation of Reflection forces you to adopt an updating rule that is vulnerable to a Dutch Strategy. For this reason, we can allow that you are irrational if you are vulnerable to a Dutch Strategy without rendering violations of Reflection irrational, just as we can allow that you are irrational if you have beliefs that are guaranteed to include some falsehoods without rendering the second-order belief that you believe a contradiction irrational. Both force you to adopt some other sort of doxastic state---a first-order belief or an updating rule---that is irrational. But they are not themselves irrational. This saves the Dutch Strategy argument for Conditionalization.

References

• Briggs, R. A. (2009). Distorted Reflection. Philosophical Review, 118(1), 59–85.
• Christensen, D. (1991). Clever Bookies and Coherent Beliefs. Philosophical Review, 100(2), 229–247.
• Mahtani, A. (2012). Diachronic Dutch Book Arguments. Philosophical Review, 121(3), 443–450.

Thursday, 25 October 2018

Dutch Books and Conditionalization

I've just signed a contract with Cambridge University Press to write a book on the Dutch Book Argument for their Elements in Decision Theory and Philosophy series. So over the next few months, I'm going to be posting some bits and pieces as I get properly immersed in the literature.

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In this post, I'm interested in Dutch Book or sure loss arguments for updating norms, which say how you should change your credences in response to new evidence.

To illustrate these points, let's pursue an example throughout the chapter. Consider Sandy. On Monday, Sandy is 40% confident that the global mean surface temperature will rise by between 0 and 1 degrees Celsius in the next 100 years. That is his unconditional credence in the proposition Medium on Monday. As well as that unconditional credence, he also has various conditional credences on Monday. For instance, let CO$_2$ High be the proposition that current CO$_2$ levels are greater than 420ppm, and let CO$_2$ Low be its negation. On Monday, Sandy has a conditional credence in Medium given CO$_2$ High---it is 60%. And he has a conditional credence in Medium given CO$_2$ Low---it is 20%. By definition, his conditional credence in one proposition $A$ given another $B$ is the proportion of his credence in $B$ he gives also to $A$. That is, it is the ratio of his credence in the conjunction $A\ \&\ B$ to his credence in $B$. Now, on Tuesday, Sandy learns CO$_2$ High. In the light of this, he updates his credences. His new unconditional credence in Medium is 70%. We naturally judge him irrational. If he were rational, his unconditional credences on Tuesday would be the same as his conditional credences on Monday given CO$_2$ High, the proposition he gained as evidence in between times. In the jargon, he would update by conditionalizing on his new evidence, CO$_2$ High. But he doesn't.

So Sandy violates a putative norm that governs his actual updating behaviour:

Actual Conditionalization  If
• $c$ is your credence function at $t$,
• $c'$ is your credence function at a later time $t'$,
• $E$ is the strongest evidence you obtained between $t$ and $t'$, and
• $c(E) > 0$,
then it ought to be that $c'(X) = c(X | E)$, for all $X$.

Is there a sure loss argument against Sandy? As we said at the beginning, there is one sort of sure loss argument against him, but this proves too much; there is also another sort of sure loss argument, but it does not apply to Sandy only on the basis of his actual updating behaviour---if it applies to him at all, it is on the basis of other modal features of him.

Throughout, we'll assume that Sandy's credences are probabilistic on Monday and probabilistic on Tuesday. As a result, the Converse Dutch Book Theorem tells us that there is no Dutch Book that can be made against his credences on Monday and no Dutch Book that can be made against his credences on Tuesday. Perhaps, though, there is some sort of Dutch Book we can make against the combination of his credences at the two different times. That is, perhaps there is a diachronic Dutch Book against Sandy. This would consist of a set of bets offered on Monday and a set of bets offered on Tuesday; his Monday credences would have to require him to accept the Monday bets and his Tuesday credences would have to require him to accept the Tuesday bets; and, taken together, those bets would be guaranteed to lose him money. We might say that you are irrational if you are vulnerable to a diachronic Dutch Book.

In fact, it turns out that there is such a diachronic Dutch Book against Sandy. His credences require him to sell a £100 bet on Medium for £45 on Monday, since he's 40% confident in Medium on Monday. And they require him to buy a £100 bet on Medium for £55 on Tuesday, since he's 70% confident in Medium on Tuesday. Taken together, these two bets will lose Sandy £10 in all epistemically possible worlds. Thus, if vulnerability to a diachronic Dutch Book is sufficient for irrationality, then Sandy is irrational.

The problem with this argument is that Sandy is also required to accept that same pair of bets on Monday and Tuesday if he updates by conditionalizing on the evidence he learns, namely, $E$. In that case, his credences on Monday are the same as before and thus require him to sell a £100 bet on Medium for £45 on Monday. And his credence in Medium on Tuesday is 60%, rather than 70%, and that still requires him to pay £55 for a £100 bet on Medium. So he is sure to lose £10. And indeed, unless he retains exactly the same credences between Monday and Tuesday, there will always be a pair of bets, one offered on Monday that his Monday credences require him to accept, and one offered on Tuesday that his Tuesday credences require him to accept, that, taken together, will lose him money at all epistemically possible worlds. So, if vulnerability to a diachronic Dutch Book is sufficient for irrationality, then Sandy is irrational, but so is anyone who ever changes any of their credences. And that surely can't be right.

So the existence of a diachronic Dutch Book against your actual updating behaviour is not sufficient for irrationality. But why not? One natural answer is this. Come Tuesday, both we and Sandy know that he in fact learned CO$_2$ High (which we'll call $E$) and updated his credence in Medium (which we'll abbreviate $M$) on the basis of that. But, on Monday, it is still open at least from Sandy's own point of view which of $E$ or $\overline{E}$ he will learn. And, were he to learn $\overline{E}$ instead, he might well have updated his credences in a different way.

Let's suppose he would. In fact, contrary to the description of the example so far, let's suppose that, whichever he learns, he'll update by conditionalizing. So, if he learns $E$, he'll become 60% confident in $M$, and if he learns $\overline{E}$, he'll become 20% confident in $M$. Then his Monday credences require him to sell a £100 bet on $M$ for £45, and his Tuesday credences should he learn $E$ require him to buy a £100 bet on $M$ for £55, thereby losing him money whether or not $M$ is true. But his Tuesday credences should he learn $\overline{E}$ do not require him to buy a £100 bet on $M$ for £55. Indeed, they require him to refuse to buy that bet.

This suggests the following sort of argument against someone who will update by something other than conditionalizing in the face of some evidence they might acquire. Suppose $c$ is your credence function at time $t$---it is defined on $\mathcal{F}$. There's some proposition $E$ in $\mathcal{F}$ that you might learn as evidence between an earlier time $t$ and a later time $t'$. And you'll learn $E$ just in case it's true. And suppose $c(E) > 0$. If you learn $E$---that is, if $E$ is true---you'll adopt credence function $c'$. If you don't learn $E$---that is, if $E$ is false---we don't know how you'll respond---perhaps it isn't determined. Then we'll say that you are vulnerable to a moderate Dutch Strategy if there are
• bets $B$ that $c$ requires you to accept,
• bets $B'_E$ that $c'$ requires you to accept, and
• bets $B'_{\overline{E}}$ that any credence function requires you to accept
such that
• the bets in $B$ and $B'_E$, taken together, lose you money in all worlds at which $E$ is true, and
• the bets in $B$ and $B'_{\overline{E}}$, taken together, lose you money in all worlds at which $E$ is false.
And we'll say that you are irrational if you are vulnerable to a moderate Dutch Strategy. Now, if we accept this, we can give an argument for updating by conditionalizing that appeals to sure loss bets. Following R. A. Briggs' presentation of David Lewis' argument, here's how (Lewis 1999, Briggs 2009).

Suppose $c'(X) = r' < r = c(X|E)$ and $c(E) = d > 0$. Then let $0 < \varepsilon < \frac{d(r-r')}{3}$. Then it is easy to see that your credences require you to accept the following bets:

After all, Bets 1 and 2 have positive expected value relative to $c$, and Bet 3 has positive expected value relative to $c'$, which you will adopt at $t'$ if $E$ is true. And it is easy to calculate that, if $E$ is true, then Bets 1, 2, and 3 taken together lose you money; and if $E$ is false, then Bets 1 and 2 taken together lose you money.

Now notice that this argument is directed against someone who will update by something other than conditionalization on certain evidence she might receive. Thus, at least on the face of it, it is not directed against Sandy's actual updating behaviour, but rather against his dispositions to update in different ways depending on the evidence he receives---what we might call his updating rule. That is, the object of criticism against which the Dutch Strategy argument is posed is Sandy's updating rule. One way to see this is to ask what would happen if Sandy instead learned $\overline{E}$ and updated on $\overline{E}$ by conditionalizing on it. Then, even though his actual updating behaviour would have been in line with Actual Conditionalization, he would nonetheless still have been vulnerable to a Dutch Strategy because he would have strayed from conditionalization had he learned $E$ instead. This shows that Dutch Strategy arguments target irreducibly modal features of an agent---that is, they target rules or dispositions, not actual behaviour. We will see this again below. Thus, we might take the Dutch Strategy argument to establish the following norm, at least in the first instance:

Rule Conditionalization  If
• $c$ is your credence function at $t$,
• if $E$ is the strongest evidence you obtain between $t$ and $t'$, then you will adopt $c'$ as your credence function at $t'$,
• $c(E) > 0$,
then it ought to be that $c'(X) = c(X|E)$, for all $X$.

The crucial difference between Actual and Rule Conditionalization lies in the modal status of the second clause. Whereas Actual Conditionalization targets what you actually have done, Rule Conditionalization targets what you will do.

Now, it might seem that we can salvage an argument for Actual Conditionalization from Rule Conditionalization.  Sandy violates Actual Conditionalization because his unconditional credence in $M$ on Tuesday is 70% while his conditional credence in $M$ given $E$ on Monday is 60%. But surely it is then true on Monday that he will adopt a credence of 70% in $M$ on Tuesday if he learns $E$. That is, he violates Rule Conditionalization as well.

But there is a problem with that reasoning. Suppose Sandy's credences don't evolve deterministically. That is, suppose that, while on Tuesday it turns out that he in fact responded to learning $E$ by raising his confidence in $M$ to 70%, he might have responded differently. For instance, suppose that there was some possibility that he responded to the evidence $E$ by dropping his confidence to 50%. Then the Dutch strategy against Sandy described above has a hole. It tells us what to do if he learns $E$ and responds by becoming 70% confident in $M$. And it tells us what to do if he learns $\overline{E}$. But it says nothing about what to do if he learns $E$ and he drops his confidence in $M$ to 50%. And indeed it turns out that it isn't always possible to fill that gap. Thus, what the standard Dutch Strategy argument sketched above shows is that there always a Dutch strategy against someone with a deterministic update rule that would make them stray from conditionalizing in some cases. Now, it turns out that, for certain non-deterministic updating rules, we can create Dutch Strategies against them too. But not all of them. Indeed, there are non-deterministic ways to update your credences that always lead you to not conditionalize, but for which there is no strategy for creating a Dutch Book against you.

I'll illustrate this with an example first, and then I'll state the central fact of which the example is a particular case. Suppose Sandy does not update by a purely deterministic rule. Rather, his credences develop as depicted in Figure 1.

 Figure 1: Sandy's credences on Monday are given by $c$. On Tuesday, he learns $E$ or $\overline{E}$. If $E$, he adopts either $c_1$ or $c_2$, but it is not determined which. If $\overline{E}$, he adopts either $c_3$ or $c_4$, but it is not determined which.

Thus, whatever happens, Sandy will not update by conditionalizing. However, there is no Dutch strategy against him. The reason is that, while Sandy does not update by conditionalizing on his strongest evidence on Tuesday, there is a way of representing him as if he were updating by conditionalization on some evidence, namely, the identity of his credence function on Tuesday. First, notice that Sandy's credence function $c$ on Monday is the average of the possible credence functions he might adopt on Tuesday---that is, for any $X$ in $\mathcal{F}$,
$$c(X) = \frac{1}{4}c_1(X) + \frac{1}{4}c_2(X) + \frac{1}{4}c_3(X) + \frac{1}{4}c_4(X)$$
Now, suppose we expand the set of propositions $\mathcal{F}$ to which Sandy assigns credences by adding, for each possible future credence function $c_i$, the proposition $C_{c_i}$, which says that $c_i$ is Sandy's credence function on Tuesday. And then suppose we extend $c$ to $c^*$, which is defined on this expanded set of propositions as follows: given a possible world $w$, let
$c^*(w\ \&\ C_{c_i}) = \frac{1}{4}c_i(w)$
Then it's easy to verify that $c^*(X) = c(X)$ for any $X$ in $\mathcal{F}$. So $c^*$ really is an extension of $c$. And we can also see that
$c^*(M | C_{c_i}) = \frac{c^*(M\ \&\ C_{c_i})}{c^*(C_{c_i})} = \frac{\frac{1}{4}c_i(M)}{\frac{1}{4}} = c_i(M)$
So, $c_i$ is the result of conditionalizing $c^*$ on the proposition that $c_i$ is Sandy's credence function on Tuesday. As we see in Theorem 1 below, there can be no Dutch Strategy against Sandy because he can be represented as if he is updating on these propositions about his Tuesday credences, even though that is not in fact how his updating proceeds. This shows again that whatever sure loss argument we have for updating, it does not target actual updating, but rather updating rules or dispositions. For it is possible to have an update rule that makes it certain that you will violate Actual Conditionalization, and yet not be vulnerable to a Dutch Strategy.

Before we state our theorem, some terminology: Suppose $c$ is your credence function at $t$. It is defined on a set of propositions $\mathcal{F}$. Suppose $\mathcal{E} = \{E_1, \ldots, E_n\} \subseteq \mathcal{F}$ is a partition that contains the strongest propositions you might learn between $t$ and $t'$. Suppose $\mathcal{C} = \{c_1, \ldots, c_m\}$ is the set of possible credence functions you might adopt at $t'$. They are also defined on $\mathcal{F}$. So:
• for each $E_i$ in $\mathcal{E}$, there is at least one $c_j$ in $\mathcal{C}$ such that $c_j(E_i) = 1$---that is, for every possible piece of evidence you might acquire, there is some possible future credence function that is a response to that evidence;
• for each $c_j$ in $\mathcal{C}$, there is exactly one $E_i$ in $\mathcal{E}$ such that $c_j(E_i) = 1$---that is, every possible future credence function is a response to exactly one of these possible pieces of evidence.
Then we say that there is a strong Dutch Strategy against you iff there are sets of bets $B$ and $B'$ such that
• $c$ requires you to accept the bets in $B$,
• $c_j$ requires you to accept the bets in $B'$, for all $1 \leq j \leq m$, and
• taken together, the bets in $B$ and $B'$ lose you money in all epistemically possible worlds.
And we say that there is a weak Dutch Strategy against you iff there are sets of bets $B, B'_1, \ldots, B'_n$ such that
• $c$ requires you to accept the bets in $B$,
• $c_j$ requires you to accept the bets in $B'_j$, for all $1 \leq j \leq m$, and
• the bets in $B$ and $B_j$, taken together, lose you money at all worlds at which you have credence function $c_j$ at time $t'$.
Note: if you are vulnerable to a strong Dutch Strategy, you're certainly vulnerable to a weak Dutch Strategy; if you are not vulnerable to a weak Dutch Strategy, then you cannot be vulnerable to a strong Dutch Strategy.

We say that you are representable as a conditionalizer iff there is an extension of $c, c_1, \ldots, c_m$ to credence functions $c^*, c^*_1, \ldots, c^*_m$ defined on $(\mathcal{F} \cup \{C_{c_1}, \ldots, C_{c_m}\})^*$ such that
• $c^*_i(C_{c_i}) = 1$, for $1 \leq i \leq m$;
• $c^*_i(X) = c^*(X | C_{c_i})$ for $X$ in $(\mathcal{F} \cup \{C_{c_1}, \ldots, C_{c_m}\})^*$
where $(\mathcal{F} \cup \{C_{c_1}, \ldots, C_{c_m}\})^*$ is the smallest algebra to contain $\mathcal{F}$ and each $C_{c_i}$.

Theorem 1
• If you are not representable as a conditionalizer, there is a strong Dutch Strategy against you;
• If you are representable as a conditionalizer, there is no weak (or strong) Dutch Strategy against you.
I won't give the full proof here, but it runs roughly as follows: you are representable as a conditionalizer iff $c$ is in the convex hull of $\{c_1, \ldots, c_n\}$ and if $c$ is not in the convex hull of $\{c_1, \ldots, c_n\}$, then there is a set of bets that $c$ requires you to buy for one price, while each $c_i$ requires you to sell them for a lower price.

Now, suppose your updating rule is deterministic. Then, for each $E_i$, there is exactly one $c_j$ in $\mathcal{C}$ such that $c_j(E_i) = 1$. Thus, in this case, your updating rule is vulnerable to a strong Dutch Strategy if it is not a conditionalizing rule, and not vulnerable even to a weak Dutch Strategy if it is. Thus, we have an extra argument for Rule Conditionalization. In some ways it strengthens the standard argument presented above, for it shows that the same set of bets can be offered at $t'$ regardless of what credence function you end up having. But in some ways it weakens that argument, for it relies on the assumption that there is a finite set of possible future credence functions you might adopt at $t'$.

Now, of course, you might object that updating other than by a deterministic rule is irrational: your evidence, together with your prior credences, should determine your new credences; there should not be many possible ways you might respond to the same piece of evidence. This may be true, and if we supplement the Dutch Strategy argument with this assumption, we obtain a Dutch Strategy for conditionalizing. But note that the argument is no longer purely pragmatic. It is now partly evidentialist, because it incorporates this evidentialist principle that we have not and cannot justify on pragmatic grounds---we cannot specify how you will go wrong in your decisions if you update using a non-deterministic rule.

References

• Briggs, R. A. (2009). Distorted Reflection. Philosophical Review, 118(1), 59–85.
• Lewis, D. (1999). Why Conditionalize? In Papers in Metaphysics and Epistemology. Cambridge, UK: Cambridge University Press.

Wednesday, 26 September 2018

Assistant professorship in formal philosophy (Gdansk)

A tenure-track job in formal philosophy in Gdansk is available. Polish language skills not required. The application deadline is November 23, 2018. Details here

The exploitability-implies-irrationality argumentative strategy

In decision theory, we often wish to impose normative constraints either on an agent's preference ordering or directly on the utility function that partly determines it. We might demand, for instance, that your preferences should not be cyclical, or that your utility function should discount the future exponentially. And in Bayesian epistemology, we often wish to impose normative constraints on credences. We might demand, for instance, that your credence in one proposition should be no greater than your credence in another proposition that it entails. In both cases, we often use a particular argumentative strategy to establish these norms: we'll call it the exploitability-implies-irrationality strategy (or EII, for short). I want to start by arguing that this is a bad argumentative strategy; and then I want to describe a way to replace it with a good argumentative strategy that is inspired by the problem we have identified with EII. I want to finish by sketching a version of the good argumentative strategy that would replace the EII strategy in the case of credal norms; that is, in the case of the Dutch Book argument. I leave it open here whether a similar strategy can be made to work in the case of preferences or utility functions. (I think this alternative argument strategy is new---it essentially combines an old result by Mark Schervish (1989) with a more recent result by Joel Predd and his co-authors at Princeton (2009); so it wouldn't surprise me at all if something similar has been proposed before---I'd welcome any information about this.)

The EII strategy runs as follows:

(I) Mental state-action link. It begins by claiming that, for anyone with a particular mental state---a preference ordering, a utility function, a credence function, or some combination of these---it is rationally required of them to choose in a particular way when faced with a decision problem.

Some examples:
(i) someone with preference ordering $a \prec b$ is rationally required to pay some amount of money to receive $b$ rather than $a$;
(ii) someone with credence $p$ in proposition $X$ should pay £$(p-\epsilon)$ for a bet that pays out £1 if $X$ is true and £0 if $X$ is false---call this a £1 bet on $X$.

(II) Mathematical theorem. It proceeds to show that, for anyone with a mental state that violates the norm in question, there are decision problems the agent might face such that, if she does, then there are choices she might make in response to them that dominate the choices that premise (I) says are rationally required of her as a result of her mental state. That is, the first set of choices is guaranteed to leave her better off than the second set of choices.

Some examples:
(i) if $c \prec a \prec b \prec c$, then rationality requires you to pay to get $a$ rather than $c$, pay again to get $b$ rather than $a$, and pay again to get $c$ rather than $a$. If, instead, you'd just chosen $c$ at the beginning, and refused to pay anything to swap, you'd be better off for sure now.
(ii) if you have credence $p$ in $XY$ and a credence $q < p$ in $X$, then you will sell a £1 bet on $X$ for £$(q + \varepsilon)$, and you'll buy a £1 bet on $XY$ for £$(p-\varepsilon)$. Providing $3\varepsilon < p - q$, it is easy to see that, taken together, these bets lose you money for sure, and thus refusing both bets is guaranteed to leave you better off.

(III) Action-rationality link. The final premise says that, if there is some series of decision problems such that the choices your mental states rationally require you to make are dominated by some other set of choices you might have made instead, then your mental states are irrational.

Some examples:
(i) By (I-III)(i), we conclude that preferences $c \prec a \prec b \prec c$ are irrational.
(ii) By (I-III)(ii), we conclude that having a higher credence in a conjunction than in one of the conjuncts is irrational.

Now, there are often problems with the instance of (I) that is used in such EII arguments. For instance, there are many reasons to think rationality does not require someone with credence $p$ in $X$ to pay £$(p - \varepsilon)$ for a £1 bet on $X$. But my focus here is on (III).

The Problem with the Action-Rationality Link

The problem with (III) is this: It is clear that it is irrational to make a series of decisions when there is an alternative series that is guaranteed to do better---it is irrational because, when you act, you are attempting to maximise your utility and doing what you have done is guaranteed to be suboptimal as a means to that end; there is an alternative you can know a priori would serve that end better. But it is much less clear why it is irrational to have mental states that require you to make a dominated series of decisions when faced with a particular decision problem. When you choose a dominated option, you are irrational because there's something else you could have done that is guaranteed to serve your ends better. But when you have mental states that require you to choose a dominated option, that alone doesn't tell us that there is anything else you could have done---any alternative mental states you could have had---that are guaranteed to serve your ends better.

Of course, there is often something else you could have done that would not have required you to make the dominated choice. Let's focus on the case of credences. The Dutch Book Theorem shows that, if your credences are not probabilistic, then there's a series of decision problems and a dominated series of options from them that those credences require you to choose. The Converse Dutch Book Theorem shows that, if your credences are instead probabilistic, then there is no such series of decision problems and options. So it's true that there's something else you could do that's guaranteed not to require you to make a dominated choice. But making a dominated choice is not an eventuality so dreadful and awful that, if your credences require you to do it in the face of one particular sort of decision problem, they are automatically irrational, regardless of what they lead you to do in the face of any other decision problem and regardless of how likely it is that you face a decision problem in which they require it of you.

After all, for all the Dutch Book or Converse Dutch Book Theorem tell you, it might be that your non-probabilistic credences lead you to choose badly when faced with the very particular Dutch Book decision problem, but lead you to choose extremely profitably when faced with many other decision problems. Any indeed, even in the case of the Dutch Book decision problem, it might be that your non-probabilistic credences require you to choose in a way that leaves you a little poorer for sure, while all the alternative probabilistic credences require you to choose in a way that leaves you with the possibility of great gain, but also the risk of great loss. In this case, it is not obvious that the probabilistic credences are to be preferred. Furthermore, you might have reason to think that it is extremely unlikely you will ever face the Dutch Book decision problem itself. Or at least much more probable that you'll face other decision problems where your credences don't lead you to choose a dominated series of options. For all these reasons, the mere possibility of a series of decision problems from which your credences require you to choose a dominated series of options is not sufficient to show that your credences are irrational. To do this, we need to show that there are some alternative credences that are in some sense sure to serve you better as you face the decision problems that make up your life. Without these alternative that do better, pointing out a flaw in some mental state does not show that it is irrational, even if there are other mental states without the flaw---for those alternative mental states might have other strikes against them that the mental state in question does not have.

A new Dutch Book argument

So our question is now: Is there any sense in which, when you have non-probabilistic credences, there are some alternative credences that are guaranteed to serve you better as a guide in your decision-making? Borrowing from work by Mark Schervish ('A General Method for Comparing Probability Assessors', 1989) and Ben Levinstein ('A Pragmatist's Guide to Epistemic Utility', 2017), I want to argue that there is.

The pragmatic utility of an individual credence

Our first order of business is to create a utility function that measures how good individual credences are as a guide to decision-making. Then we'll take the utility of a whole credence function to be the sum of the utilities of the credences that comprise it. (In fact, I think there's a way to do all this without that additivity assumption, but I'm still ironing out the creases in that.)

Suppose you assign credence $p$ to proposition $X$. Our job is to say how good this credence is as a guide to action. The idea is this:
• an act is a function from states of the world to utilities---let $\mathcal{A}$ be the set of all acts;
• an $X$-act is an act that assigns the same utility to all the worlds at which $X$ is true, and assigns the same utility to all worlds at which $X$ is false---let $\mathcal{A}_X$ be the set of all $X$-acts;
• a decision problem is a set of acts; that is, a subset of $\mathcal{A}$---let $\mathcal{D}$ be the set of all decision problems;
• an $X$-decision problem is a set of $X$-acts; that is, a subset of $\mathcal{A}_X$---let $\mathcal{D}_X$ be the set of all $X$-decision problems.
We suppose that there is a probability function $P$ that says how likely it is that the agent will face different $X$-decision problems---since the set of $X$-decision problems is infinite, we actually take $P$ to be a probability density function. The idea here is that $P$ is something like an objective chance function. With that in hand, we take the pragmatic utility of credence $p$ in proposition $X$ to be the expected utility of the choices that credence $p$ in $X$ will lead you to make when faced with the decision problems you will encounter. That is, it is the integral, relative to measure $P$, over the possible $X$-decision problems $D$ in $\mathcal{D}_X$ you might face, of the utility of the act you'd choose from $D$ using $p$, discounted by the probability that you'd face $D$. Given $D$ in $\mathcal{D}_X$, let $D^p$ be the act you'd choose from $D$ using $p$---that is, $D^p$ is one of the acts in $D$ that maximises expected utility by the lights of $p$. Thus, for any $D$ in $\mathcal{D}_X$, and any act $a$ in $D$,$$\mathrm{Exp}_p(u(a)) \leq \mathrm{Exp}_p(u(D^p))$$ Then we define the pragmatic utility of credence $p$ in $X$ when $X$ is true as follows:
$$g_X(1, p) = \int_{\mathcal{D}_X}u(D^p, X) dP$$ And we define the pragmatic utility of credence $p$ in $X$ when $X$ is false as follows:
$$g_X(0, p) = \int_{\mathcal{D}_X}u(D^p, \overline{X}) dP$$ These are slight modifications of Schervish's and Levinstein's definitions.

$g$ is a strictly proper scoring rule

Our next order of business is to show that this utility function for $g$ is a strictly proper scoring rule. That is, $\mathrm{Exp}_p(g_X(q)) = pg_X(1, q) + (1-p)g_X(0, q)$ is uniquely maximised, as a function of $q$, at $p = q$. We show this now:
\begin{eqnarray*}
\mathrm{Exp}_p(g_X(q)) & = & pg_X(1, q) + (1-p)g_X(0, q)\\
& = & p \int_{\mathcal{D}_X}u(D^q, X) dP + (1-p) \int_{\mathcal{D}_X}u(D^q, \overline{X}) dP \\
& = & \int_{\mathcal{D}_X}p u(D^q, X) + (1-p) u(D^q, \overline{X}) dP\\
& = & \int_{\mathcal{D}_X} \mathrm{Exp}_p(u(D^q)) dP
\end{eqnarray*}
But, by the definition of $D^q$, if $q \neq p$, then, for all $D$ in $\mathcal{D}_X$,
$$\mathrm{Exp}_p(u(D^q)) \leq \mathrm{Exp}_p(u(D^p))$$
and, for some $D$ in $\mathcal{D}_X$,
$$\mathrm{Exp}_p(u(D^q)) < \mathrm{Exp}_p(u(D^p))$$
Now, for two credences $p$ and $q$ in $X$, we say that a set of decision problems separates $p$ and $q$ if (i) each decision problem in the set contains only two available acts, (ii) for each decision problem in the set, $p$ expects one act to have higher expected value and $q$ expects the other to have higher expected value. Then, as long as there is some set of decision problems such that (i) that set separates $p$ and $q$ and (ii) $P$ assigns positive probability to this set, then
$$\mathrm{Exp}_p(g(q)) < \mathrm{Exp}_p(g(p))$$ And so the scoring rule $g$ is strictly proper.

The pragmatic utility of a whole credence function

The scoring rule $g_X$ we have just defined assigns pragmatic utilities to individual credences in $X$. In the next step, we define $G$, a pragmatic utility function that assigns pragmatic utilities to whole credence functions. We take the utility of a credence function to be the sum of the utilities of the individual credences it assigns. Suppose $c : \mathcal{F} \rightarrow [0, 1]$ is a credence function defined on the set of propositions $\mathcal{F}$. Then: $$G(c, w) = \sum_{X \in \mathcal{F}} g_X(w(X), c(X))$$ where $w(X) = 1$ if $X$ is true at $w$ and $w(X) = 0$ if $X$ is false at $w$. In this situation, we say that $G$ is generated from the scoring rules $g_X$ for $X$ in $\mathcal{F}$.

Predd, et al.'s Dominance Result

Finally, we appeal to a theorem due to Predd, et al. ('Probabilistic Coherence and Proper Scoring Rules', 2009):

Theorem (Predd, et al 2015) Suppose $G$ is generated from strictly proper scoring rules $g_X$ for $X$ in $\mathcal{F}$. Then,
(I) if $c$ is not a probability function, then there is a probability function $c^*$ such that, $G(c, w) < G(c^*, w)$ for all worlds $w$;
(II) if $c$ is a probability function, then there is no credence function $c^* \neq c$ such that $G(c, w) \leq G(c^*, w)$ for all worlds $w$.

This furnishes us with a new pragmatic argument for probabilism. And indeed, now that we have a pragmatic utility function that is generated from strictly proper scoring rules, we can take advantage of all of the epistemic utility arguments that make that same assumption, such as Greaves and Wallace's argument for Conditionalization, my arguments for the Principal Principle, the Principle of Indifference, linear pooling in judgment aggregation cases, and so on.

In this argument, we see that non-probabilistic credences are irrational not because there is some series of decision problems such that, when faced with them, the credences require you to make a dominated series of choices. Rather, they are irrational because there are alternative credences that are guaranteed to serve you better on average as a guide to action---however the world turns out, the expected or average utility you'll gain from making decisions using those alternative credences is greater than the expected or average utility you'll gain from making decisions using the original credences.

Monday, 6 August 2018

Postdoc in formal epistemology & law

A postdoc position (3 years, fixed term) in the Chair of Logic, Philosophy of Science and Epistemology is available at the Department of Philosophy, Sociology, and Journalism, University of Gdansk, Poland. The application deadline is September 15, 2018. More details here.

Monday, 30 July 2018

The Dutch Book Argument for Regularity

I've just signed a contract with Cambridge University Press to write a book on the Dutch Book Argument for their Elements in Decision Theory and Philosophy series. So over the next few months, I'm going to be posting some bits and pieces as I get properly immersed in the literature.

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We say that a probabilistic credence function $c : \mathcal{F} \rightarrow [0, 1]$ is regular if $c(A) > 0$ for all propositions $A$ in $\mathcal{F}$ such that there is some world at which $A$ is true.

The Principle of Regularity (standard version) If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, rationality requires that $c$ is regular.

I won't specify which worlds are in the scope of the quantifier over worlds that occurs in the antecedent of this norm. It might be all the logically possible worlds, or the metaphysically possible worlds, or the conceptually possible worlds; it might be the epistemically possible worlds. Different answers will give different norms. But we needn't decide the issue here. We'll just specify that it's the same set of worlds that we quantify over in the Dutch Book argument for Probabilism when we say that, if your credences aren't probabilistic, then there's a series of bets they'll lead you to enter into that will lose you money at all possible worlds.

In this post, I want to consider the almost-Dutch Book Argument for the norm of Regularity. Here's how it goes: Suppose you have a credence $c(A) = 0$ in a proposition $A$, and suppose that $A$ is true at world $w$. Then, recall, the first premise of the standard Dutch Book argument for Probabilism:

Ramsey's Thesis If your credence in a proposition $X$ is $c(X) = p$, then you're permitted to pay £$pS$ for a bet that returns £$S$ if $X$ is true and £$0$ if $X$ is false, for any $S$, positive or negative or zero.

So, since $c(A) = 0$, your credences will permit you to sell the following bet for £0: if $A$, you must pay out £1; if $\overline{A}$, you will pay out £0. But selling this bet for this price is weakly dominated by refusing the bet. Selling the bet at that price loses you money in all $A$-worlds, and gains you nothing in $\overline{A}$-worlds. Whereas refusing the bet neither loses nor gains you anything in any world. Thus, your credences permit you to choose a weakly dominated act. So they are irrational. Or so the argument goes. I call this the almost-Dutch Book argument for Regularity since it doesn't punish you with a sure loss, but rather with a possible loss with no compensating possible gain.

If this argument works, it establishes the standard version of Regularity stated above. But consider the following case. $A$ and $B$ are two logically independent propositions -- It will be rainy tomorrow and It will be hot tomorrow, for instance. You have only credences in $A$ and in the conjunction $AB$. You don't have credences in $\overline{A}$, $A \vee B$, $A\overline{B}$, and so on. What's more, your credences in $A$ and $AB$ are equal, i.e., $c(A) = c(AB)$. That is, you are exactly as confident in $A$ as you are in its conjunction with $B$. Then, in some sense, you violate Regularity, though you don't violate the standard version we stated above. After all, since your credence in $A$ is the same as your credence in $AB$, you must give no credence whatsoever to the worlds in which $A$ is true and $B$ is false. If you did, then you would set $c(AB) < c(A)$. But you don't have a credence in $A\overline{B}$. So there is no proposition true at some worlds to which you assign a credence of 0. Thus, the almost-Dutch Book argument sketched above will not work. We need a different Dutch Book argument for the following version of Regularity:

The Principle of Regularity (full version) If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, then rationality requires that there is an extension $c^*$ of $c$ to a full algebra $\mathcal{F}^*$ that contains $\mathcal{F}$ such that $c^* : \mathcal{F}^* \rightarrow [0, 1]$ is regular.

It is this principle that you violate if $c(A) = c(AB)$ when $A$ and $B$ are logically independent. For any probabilistic extension $c^*$ of $c$ that assigns a credence to $A\overline{B}$ must assign it credence 0 even though there is a world at which it is true.

How are we to give an almost-Dutch Book argument for this version of Regularity? There are two possible approaches.

On the first, we strengthen the first premise of the standard Dutch Book argument. Ramsey's Thesis says: if you have credence $c(X) = p$ in $X$, then you are permitted to pay £$pS$ for a bet that pays £$S$ if $X$ and £$0$ if $\overline{X}$. The stronger version says:

Strong Ramsey's Thesis If every extension $c^*$ of $c$ to a full algebra $\mathcal{F}^*$ that contains $\mathcal{F}$ is such that $c^*(X) = p$, then you are permitted to pay £$pS$ for a bet that pays £$S$ if $X$ and £$0$ if $\overline{X}$.

The idea is that, if every extension assigns the same credence $p$ to $X$, then you are in some sense committed to assigning credence $p$ to $X$. And thus, you are permitted to enter into which ever bets you'd be permitted to enter into if you actually had credence $p$.

On the second approach to giving an almost-Dutch Book argument for the full version of the Regularity principle, we actually provide an almost-Dutch Book using just the credences that you do in fact assign. Suppose, for instance, you have credence $c(A) = c(AB) = 0.5$. Then you will sell for £5 a bet that pays out £10 if $A$ and £0 if $\overline{A}$, while you will buy for £5 a bet that pays £10 if $AB$ and £0 if $\overline{AB}$. Then, if $A$ is true and $B$ is true, you will have a net gain of £0, and similarly if $A$ is false. But if $A$ is true and $B$ is false, you will lose £10. Thus, you face the possibility of loss with no possibility of gain. Now, the question is: can we always construct such almost-Dutch Books? And the answer is that we can, as the following theorem shows:

Theorem 1 (Almost-Dutch Book Theorem for Full Regularity) Suppose $\mathcal{F} = \{X_1, \ldots, X_n\}$ is a set of propositions. Suppose $c : \mathcal{F} \rightarrow [0, 1]$ is a credence function that cannot be extended to a regular probability function on a full algebra $\mathcal{F}^*$ that contains $\mathcal{F}$. Then there is a sequence of stakes $S = (S_1, \ldots, S_n)$, such that if, for each $1 \leq i \leq n$, you pay £$(c(X_i) \times S_i)$ for a bet that pays out £$S_i$ if $X_i$ and £0 if $\overline{X_i}$, then the total price you'll pay is at least the pay off of these bets at all worlds, and more than the payoff at some.

That is,
(i) for all worlds $w$,
$$S\cdot (w - c) = S \cdot w - S \cdot c = \sum^n_{i=1} S_iw(X_i) + \sum^n_{i=1} S_ic(X_i) \leq 0$$
(ii) for some worlds $w$,
$$S\cdot (w - c) = S \cdot w - S \cdot c = \sum^n_{i=1} S_iw(X_i) + \sum^n_{i=1} S_ic(X_i) \leq 0$$
where $w(X_i) = 1$ if $X_i$ is true at $w$ and $w(X_i) = 0$ if $X_i$ is false at $w$. We call $w(-)$ the indicator function of $w$.

Proof sketch. First, recall de Finetti's observation that your credence function $c : \mathcal{F} \rightarrow [0, 1]$ is a probability function iff it is in the convex hull of the indicator functions of the possible worlds -- that is, iff $c$ is in $\{w(-) : w \mbox{ is a possible world}\}^+$. Second, note that, if your credence function can't be extended to a regular credence function, it sits on the boundary of this convex hull. In particular, if $W_c = \{w' : c = \sum_w \lambda_w w \Rightarrow \lambda_{w'} > 0\}$, then $c$ lies on the boundary surface created by the convex hull of $W_c$. Third, by the Supporting Hyperplane Theorem, there is a vector $S$ such that $S$ is orthogonal to this boundary surface and thus:
(i) $S \cdot (w-c) = S \cdot w - S \cdot c = 0$ for all $w$ in $W_c$; and
(ii) $S \cdot (w-c) = S \cdot w - S \cdot c < 0$ for all $w$ not in $W_c$.
Fourth, recall that $S \cdot w$ is the total payout of the bets at world $w$ and $S \cdot c$ is the price you'll pay for it. $\Box$

Thursday, 26 July 2018

Dutch Strategy Theorems for Conditionalization and Superconditionalization

I've just signed a contract with Cambridge University Press to write a book on the Dutch Book Argument for their Elements in Decision Theory and Philosophy series. So over the next few months, I'm going to be posting some bits and pieces as I get properly immersed in the literature.

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Many Bayesians formulate the update norm of Bayesian epistemology as follows:

Bayesian Conditionalization  If
(i) your credence function at $t$ is $c : \mathcal{F} \rightarrow [0, 1]$,
(ii) your credence function at a later time $t'$ is $c' : \mathcal{F} \rightarrow [0, 1]$,
(iii) $E$ is the strongest evidence you acquire between $t$ and $t'$,
(iv) $E$ is in $\mathcal{F}$,
then rationality requires that, if $c(E) > 0$, then for all $X$ in $\mathcal{F}$, $$c'(X) = c(X|E) = \frac{c(XE)}{c(E)}$$

I don't. One reason you might fail to conditionalize between $t$ and $t'$ is that you re-evaluate the options between those times. You might disavow the prior that you had at the earlier time, perhaps decide it was too biased in one way or another, or not biased enough; perhaps you come to think that it doesn't give enough consideration to the explanatory power one hypothesis would have were it true, or gives too much consideration to the adhocness of another hypothesis; and so on. Now, it isn't irrational to change your mind. So surely it can't be irrational to fail to conditionalize as a result of changing your mind in this way. On this, I agree with van Fraassen.

Instead, I prefer to formulate the update norm as follows -- I borrow the name from Kenny Easwaran:

Plan Conditionalization If
(i) your credence function at $t$ is $c: \mathcal{F} \rightarrow [0, 1]$,
(ii) between $t$ and $t'$ you will receive evidence from the partition $\{E_1, \ldots, E_n\}$,
(iii) each $E_i$ is in $\mathcal{F}$
(iv) at $t$, your updating plan is $c'$, so that $c'_i : \mathcal{F} \rightarrow [0, 1]$ is the credence function you will adopt if $E_i$,
then rationality requires that, if $c(E_i) > 0$, then for all $X$ in $\mathcal{F}$, $$c'_i(X) = c(X | E_i)$$

I want to do two things in this post. First, I'll offer what I think is a new proof of the Dutch Strategy or Diachronic Dutch Book Theorem that justifies Plan Conditionalization (I haven't come across it elsewhere, though Ray Briggs and I used the trick at the heart of it for our accuracy dominance theorem in this paper). Second, I'll explore how that might help us justify other norms of updating that concern situations in which you don't come to learn any proposition with certainty. We will see that we can use the proof I give to justify the following standard constraint on updating rules: Suppose the evidence I receive between $t$ and $t'$ is not captured by any of the propositions to which I assign a credence -- that is, there is no proposition $e$ to which I assign a credence that is true at all and only the worlds at which I receive the evidence I actually receive between $t$ and $t'$. As a result, there is no proposition $e$ that I learn with certainty as a result of receiving that evidence. Nonetheless, I should update my credence function from $c$ to $c'$ in such a way that it is possible to extend my earlier credence function $c$ to a credence function $c^*$ so that: (i) $c^*$ does assign a credence to $e$, and (ii) my later credence $c'(X)$ in a proposition $X$ is the credence that this extended credence function $c^*$ assigns to $X$ conditional on me receiving evidence $e$ -- that is, $c'(X) = c^*(X | e)$. That is, I should update as if I had assigned a credence to $e$ at the earlier time and then updated by conditionalizing on it.

Here's the Dutch Strategy or Diachronic Dutch Book Theorem for Plan Conditionalization:

Definition (Conditionalizing pair) Suppose $c$ is a credence function and $c'$ is an updating rule defined on $\{E_1, \ldots, E_n\}$. We say that $(c, c')$ is a conditionalizing pair if, whenever $c(E_i) > 0$, then for all $X$, $c'_i(X) = c(X | E_i)$.

Dutch Strategy Theorem Suppose $(c, c')$ is not a conditionalizing pair. Then
(i) there are two acts $A$ and $B$ such that $c$ prefers $A$ to $B$, and
(ii) for each $E_i$, there are two acts $A_i$ and $B_i$ such that $c'_i$ prefers $A_i$ to $B_i$,
and, for each $E_i$, $A + A_i$ has greater utility than $B + B_i$ at all worlds at which $E_i$ is true.

We'll now give the proof of this.

First, we describe a way of representing pairs $(c, c')$. Both $c$ and each $c'_i$ are defined on the same set $\mathcal{F} = \{X_1, \ldots, X_m\}$. So we can represent $c$ by the vector $(c(X_1), \ldots, c(X_m))$ in $[0, 1]^m$, and we can represent each $c'_i$ by the vector $(c'_i(X_1), \ldots, c'_i(X_m))$ in $[0, 1]^m$. And we can represent $(c, c')$ by concatenating all of these representations to give:
$$(c, c') = c \frown c'_1 \frown c'_2 \frown \ldots \frown c'_n$$
which is a vector in $[0, 1]^{m(n+1)}$.

Second, we use this representation to give an alternative characterization of conditionalizing pairs. First, three pieces of notation:
• Let $W$ be the set of all possible worlds.
• For any $w$ in $W$, abuse notation and write $w$ also for the credence function on $\mathcal{F}$ such that $w(X) = 1$ if $X$ is true at $w$, and $w(X) = 0$ if $X$ is false at $w$.
• For any $w$ in $W$, let $$(c, c')_w = w \frown c'_1 \frown \ldots \frown c'_{i-1} \frown w \frown c'_{i+1} \frown \ldots \frown c'_n$$ where $E_i$ is the element of the partition that is true at $w$.
Lemma 1 If $(c, c')$ is not a conditionalizing pair, then $(c, c')$ is not in the convex hull of $\{(c, c')_w : w \in W\}$, which we write $\{(c, c')_w : w \in W\}^+$.

Proof of Lemma 1. If $(c, c')$ is in $\{(c, c')_w : w \in W\}^+$, then there are $\lambda_w \geq 0$ such that

(1) $\sum_{w \in W} \lambda_w = 1$,
(2) $c(X) = \sum_{w \in W} \lambda_w w(X)$
(3) $c'_i(X) = \sum_{w \in E_i} \lambda_w w(X) + \sum_{w \not \in E_i} \lambda_w c'_i(X)$.

By (2), we have $\lambda_w = c(w)$. So by (3), we have $$c'_i(X) = c(XE_i) + (1-c(E_i))c'_i(X)$$ So, if $c(E_i) > 0$, then $c'_i(X) = c(X | E_i)$.

Third, we use this alternative characterization of conditionalizing pairs to specify the acts in question. Suppose $(c, c')$ is not a conditionalizing pair. Then $(c, c')$ is outside $\{(c, c')_w : w \in W\}^+$. Now, let $(p, p')$ be the orthogonal projection of $(c, c')$ into $\{(c, c')_w : w \in W\}^+$. Then let $(S, S') = (c, c') - (p, p')$. That is, $S = c - p$ and $S'_i = c'_i - p'_i$. Now pick $w$ in $W$. Then the angle between $(S, S')$ and $(c, c')_w - (c, c')$ is obtuse and thus
$$(S, S') \cdot ((c, c')_w - (c, c')) = -\varepsilon_w < 0$$

Thus, define the acts $A$, $B$, $A'_i$ and $B'_i$ as follows:
• The utility of $A$ at $w$ is $S \cdot (w - c) + \frac{1}{3}\varepsilon_w$:
• The utility of $B$ at $w$ is 0;
• The utility of $A'_i$ at $w$ is $S'_i \cdot (w - c'_i) + \frac{1}{3}\varepsilon_w$;
• The utility of $B'_i$ at $w$ is 0.
Then the expected utility of $A$ by the lights of $c$ is $\sum^w c(w)\frac{1}{3}\varepsilon_w > 0$, while the expected utility of $B$ is 0, so $c$ prefers $A$ to $B$. And the expected utility of $A'_i$ by the lights of $c'_i$ is $\sum_w c'_i(w)\frac{1}{3}\varepsilon_w > 0$, while the expected utility of $B'_i$ is 0, so $c'_i$ prefers $A'_i$ to $B'_i$. But the utility of $A + A'_i$ at $w$ is
$$S \cdot (w - c) + S'_i \cdot (w - c'_i) + \frac{2}{3}\varepsilon_w = (S, S') \cdot ((c, c')_w - (c, c')) + \frac{2}{3}\varepsilon_w = - \frac{1}{3}\varepsilon_w < 0$$
where $E_i$ is true at $w$. While the utility of $B + B'_i$ at $w$ is 0.

This completes our proof. $\Box$

You might be forgiven for wondering why we are bothering to give an alternative proof for a theorem that is already well-known. David Lewis proved the Dutch Strategy Theorem in a handout for a seminar at Princeton in 1972, Paul Teller then reproduced it (with full permission and acknowledgment) in a paper in 1973, and Lewis finally published his handout in 1997 in his collected works. Why offer a new proof?

It turns out that this style of proof is actually a little more powerful. To see why, it's worth comparing it to an alternative proof of the Dutch Book Theorem for Probabilism, which I described in this post (it's not original to me, though I'm afraid I can't remember where I first saw it!). In the standard Dutch Book Theorem for Probabilism, we work through each of the axioms of the probability calculus, and say how you would Dutch Book an agent who violates it. The axioms are: Normalization, which says that $c(\top) = 1$ and $c(\bot) = 0$; and Additivity, which says that $c(A \vee B) = c(A) + c(B) - c(AB)$. But consider an agent with credences only in the propositions $\top$, $A$, and $A\ \&\ B$.  Her credences are: $c(\top) = 1$, $c(A) = 0.4$, $c(A\ \&\ B) = 0.7$. Then there is no axiom of the probability calculus that she violates. And thus the standard proof of the Dutch Book Theorem is no help in identifying any Dutch Book against her. Yet she is Dutch Bookable. And she violates a more expansive formulation of Probabilism that says, not only are you irrational if your credence function is not a probability function, but also if your credence function cannot be extended to a probability function. So the standard proof of the Dutch Book Theorem can't establish this more expansive version. But the alternative proof I mentioned above can.

Now, something similar is true of the alternative proof of the Dutch Strategy Theorem that I offered above (I happened upon this while discussing Superconditionalizing with Jason Konek, who uses similar techniques in his argument for J-Kon, the alternative to Jeffrey's Probability Kinematics that he proposes in his paper, 'The Art of Learning', which was runner-up for last year's Sander's Prize in Epistemology). In Lewis' proof of that theorem: First, if you violate Plan Conditionalization, there must be $E_i$ and $X$ such that $c(E_i) > 0$ and $c'_i(X) \neq c(X|E_i)$. Then you place bets on $XE_i$, $\overline{E_i}$ at the earlier time $t$, and a bet on $X$ at $t'$. These bets then together lose you money in any world at which $E_i$ is true. Now, it might seem that you must have the required credences to make those bets just in virtue of violating Plan Conditionalization. But imagine the following is true of you: between $t$ and $t'$, you'll obtain evidence from the partition $\{E_1, \ldots, E_n\}$. And, at $t'$, you'll update on this evidence using the rule $c'$. That is, if $E_i$, then you'll adopt the new credence function $c'_i$ at time $t'$. Now, you don't assign credences to the propositions in $\{E_1, \ldots, E_n\}$. Perhaps this is because you don't have the conceptual resources to formulate these propositions. So while you will update using the rule $c'$, this is not a rule you consciously or explicitly adopt, since to state it would require you to use the propositions in $\{E_1, \ldots, E_n\}$. So it's more like you have a disposition to update in this way. Now, how might we state Plan Conditionalization for such an agent? We can't demand that $c'_i(X) = c(X|E_i)$, since $c(X | E_i)$ is not defined. Rather, we demand that there is some extension $c^*$ of $c$ to a set of propositions that does include each $E_i$ such that $c'_i(X) = c^*(X | E_i)$. Thus, we have:

Plan Superconditionalization If
(i) your credence function at $t$ is $c : \mathcal{F} \rightarrow [0, 1]$,
(ii) between $t$ and $t'$ you will receive evidence from the partition $\{E_1, \ldots, E_n\}$,
(iii) at $t$, your updating plan is $c'$, so that $c'_i : \mathcal{F} \rightarrow [0, 1]$ is the credence function you plan to adopt if $E_i$,
then rationality requires that there is some extension $c^*$ of $c$ for which, if $c^*(E_i) > 0$, then for all $X$, $$c'_i(X) = c^*(X | E_i)$$

And it turns out that we can adapt the proof above for this purpose. Say that $(c, c')$ is a superconditionalizing pair if there is an extension $c^*$ of $c$ such that, if $c^*(E_i) > 0$, then for all $X$, $c'_i(X) = c^*(X | E_i)$. Then we can prove that if $(c, c')$ is not a superconditionalizing pair, then $(c, c')$ is not in $\{(c, c')_w : w \in W\}^+$. Here's the proof from above adapted to our case: If $(c, c')$ is in $\{(c, c')_w : w \in W\}^+$, then there are $\lambda_w \geq 0$ such that

(1) $\sum_{w \in W} \lambda_w = 1$,
(2) $c(X) = \sum_{w \in W} \lambda_w w(X)$
(3) $c'_i(X) = \sum_{w \in E_i} \lambda_w w(X) + \sum_{w \not \in E_i} \lambda_w c'_i(X)$.

Define the following extension $c^*$ of $c$: $c^*(w) = \lambda_w$. Then, by (3), we have $$c'_i(X) = c^*(XE_i) + (1-c^*(E_i))c'_i(X)$$ So, if $c^*(E_i) > 0$, then $c'_i(X) = c^*(X | E_i)$, as required. $\Box$

Now, this is a reasonably powerful version of conditionalization. For instance, as Skyrms showed here, if we make one or two further assumptions on the extension of $c$ to $c^*$, we can derive Richard Jeffrey's Probability Kinematics from Plan Superconditionalization. That is, if the evidence $E_i$ will lead you to set your new credences across the partition $\{B_1, \ldots, B_k\}$ to $q_1, \ldots, q_k$, respectively, so that $c'_i(B_j) = q_j$, then your new credence $c'_i(X)$ must be $\sum^k_{j=1} c(X | B_j)q_j$, as Probability Kinematics demands. Thus, Plan Superconditionalization places a powerful constraint on updating rules for situations in which the proposition stating your evidence is not one to which you assign a credence. Other cases of this sort include the Judy Benjamin problem and the many cases in which MaxEnt is applied.

Wednesday, 25 July 2018

Deadline for PhD position in formal epistemology & law extended

This position is still available. Deadline extended to September 7, 2018.

On the Expected Utility Objection to the Dutch Book Argument for Probabilism

I've just signed a contract with Cambridge University Press to write a book on the Dutch Book Argument for their Elements in Decision Theory and Philosophy series. So over the next few months, I'm going to be posting some bits and pieces as I get properly immersed in the literature. The following came up while thinking about Brian Hedden's paper 'Incoherence without Exploitability'.

What is Probabilism?

Probabilism says that your credences should obey the axioms of the probability calculus. Suppose $\mathcal{F}$ is the algebra of propositions to which you assign a credence. Then we let $0$ represent the lowest possible credence you can assign, and we let $1$ represent the highest possible credence you can assign. We then represent your credences by your credence function $c : \mathcal{F} \rightarrow [0, 1]$, where, for each $A$ in $\mathcal{F}$, $c(A)$ is your credence in $A$.

Probabilism
If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, then rationality requires that:
(P1a) $c(\bot) = 0$, where $\bot$ is a necessarily false proposition;
(P1b) $c(\top) = 1$, where $\top$ is a necessarily true proposition;
(P2) $c(A \vee B) = c(A) + c(B)$, for any mutually exclusive propositions $A$ and $B$ in $\mathcal{F}$.

This is equivalent to:

Partition Probabilism
If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, then rationality requires that, for any two partitions $\mathcal{X} = \{X_1, \ldots, X_m\}$ and $\mathcal{Y} = \{Y_1, \ldots, Y_n\}$,$$\sum^m_{i=1} c(X_i) = 1= \sum^n_{j=1} c(Y_j)$$

The Dutch Book Argument for Probabilism

The Dutch Book Argument for Probabilism has three premises. The first, which I will call Ramsey's Thesis and abbreviate RT, posits a connection between your credence in a proposition and the prices you are rationally permitted or rationally required to pay for a bet on that proposition. The second, known as the Dutch Book Theorem, establishes that, if you violate Probabilism, there is a set of bets you might face, each with a price attached, such that (i) by Ramsey's Thesis, for each bet, you are rationally required to pay the attached price for it, but (ii) the sum of the prices of the bets exceeds the highest possible payout of the bets, so that, having paid each of those prices, you are guaranteed to lose money. The third premise, which we might call the Domination Thesis, says that credences are irrational if they mandate you to make a series of decisions (i.e, paying certain prices for the bets) that is guaranteed to leave you worse off than another series of decisions (i.e., refusing to pay those prices for the bets)---in the language of decision theory, paying the attached price for each of the bets is dominated by refusing each of the bets, and credences that mandate you to choose dominated options are irrational. The conclusion of the Dutch Book Argument is then Probabilism. Thus, the argument runs:

The Dutch Book Argument for Probabilism
(DBA1) Ramsey's Thesis
(DBA2) Dutch Book Theorem
(DBA3) Domination Thesis
Therefore,
(DBAC) Probabilism

The argument is valid. The second premise is a mathematical theorem. Thus, if the argument fails, it must be because the first or third premise is false, or both. In this paper, we focus on the first premise, and the expected utility objection to it. So, let's set out that premise in a little more detail.

In what follows, we assume that (i) you are risk-neutral, and (ii) that there is some quantity such that your utility is linear in that quantity---indeed, we will speak as if your utility is linear in money, but that is just for ease of notation and familiarity; any quantity would do. Neither (i) nor (ii) is realistic, and indeed these idealisations are the source of other objections to Ramsey's Thesis. But they are not our concern here, so we will grant them.

Ramsey's Thesis (RT) Suppose your credence in $X$ in $c(X)$. Consider a bet that pays you £$S$ if $X$ is true and £0 if $X$ is false, where $S$ is a real number, either positive, negative, or zero---$S$ is called the stake of the bet. You are offered this bet for the price £$x$, where again $x$ is a real number, either positive, negative, or zero. Then:
(i) If $x < c(X) \times S$, you are rationally required to pay £$x$ to enter into this bet;
(ii) If $x = c(X) \times S$, you are rationally permitted to pay £$x$ and rationally permitted to refuse;
(iii) If $x > c(X) \times S$, you are rationally required to refuse.

Roughly speaking, Ramsey's Thesis says that, the more confident you are in a proposition, the more you should be prepared to pay for a bet on it. More precisely, it says: (a) if you have minimal confidence in that proposition (i.e. 0), then you should be prepared to pay nothing for it; (b) if you have maximal confidence in it (i.e. 1), then you should be prepared to pay the full stake for it; (c) for levels of confidence in between, the amount you should be prepared to pay increases linearly with your credence.

The Expected Utility Objection

We turn now to the objection to Ramsey's Thesis (RT) we wish to treat here. Hedden (2013) begins by pointing out that we have a general theory of how credences and utilities should guide action:

Given a set of options available to you, expected utility theory says that your credences license you to choose the option with the highest expected utility, defined as:
$$\mathrm{EU}(A) = \sum_i P(O_i|A) \times U(O_i)$$
On this view, we should evaluate which bets your credences license you to accept by looking at the expected utilities of those bets. (Hedden, 2013, 485)

He considers the objection that this only applies when credences satisfy Probabilism, but rejects it:

In general, we should judge actions by taking the sum of the values of each possible outcome of that action, weighted by one's credence that the action will result in that outcome. This is a very intuitive proposal for how to evaluate actions that applies even in the context of incoherent credences. (Hedden, 2013, 486)

Thus, Hedden contends that we should always choose by maximising expected utility relative to our credences, whether or not those credences are coherent. Let's call this principle Maximise Subjective Expected Utility and abbreviate it MSEU. He then observes that MSEU conflicts with RT. Consider, for instance, Cináed, who is 60% confident it will rain and 20% confident it won't. According to RT, he is rationally required to sell for £65 a bet in which he pays out £100 if it rains and £0 if is doesn't. But the expected utility of this bet for him is$$0.6 \times (-100 + 65) + 0.2 \times (-0 + 65) = -8$$That is, it has lower expected utility than refusing to sell the bet, since his expected utility for doing that is$$0.6 \times 0 + 0.2 \times 0 = 0$$So, while RT says you must sell that bet for that price, MSEU says you must not. So RT and MSEU are incompatible, and Hedden claims that we should favour MSEU. There are two ways to respond to this. On the first, we try to retain RT in some form in spite of Hedden's objection---I call this the permissive response below. On the second, we try to give a pragmatic argument for Probabilism using MSEU instead of RT---I call this the bookless response below. In the following sections, I will consider these in turn.

The Permissive Response

While Hedden is right to say that maximising expected utility in line with Maximise Subjective Expected Utility (MSEU) is intuitively rational even when your credences are incoherent, so is Ramsey's Thesis (RT). It is certainly intuitively correct that, to quote Hedden, ''we should judge actions by taking the sum of the values of each possible outcome of that action, weighted by one's credence that the action will result in that outcome.'' But it is also intuitively correct that, to quote from our gloss of Ramsey's Thesis above, ''(a) if you have minimal confidence in that proposition (i.e. 0), then you should be prepared to pay nothing for it; (b) if you have maximal confidence in it (i.e. 1), then you should be prepared to pay the full stake for it; (c) for levels of confidence in between, the amount you should be prepared to pay increases linearly with your credence.'' What are we to do in the face of this conflict between our intuitions?

One natural response is to say that choosing in line with RT is rationally permissible and choosing in line with MSEU is also rationally permissible. When your credences are coherent, the dictates of MSEU and RT are the same. But when you are incoherent, they are sometimes different, and in that situation you are allowed to follow either. In particular, faced with a bet and proposed price, you are permitted to pay that price if it is permitted by RT and you are permitted to pay it if it is permitted by MSEU.

If this is right, then we can resurrect the Dutch Book Argument with a permissive version of RT as the first premise:

Permissive Ramsey's Thesis Suppose your credence in $X$ in $c(X)$. Consider a bet that pays you £$S$ if $X$ is true and £0 if $X$ is false. You are offered this bet for the price £$x$. Then:
(i) If $x \leq c(X) \times S$, you are rationally permitted to pay £$x$ to enter into this bet.

And we could then amend the third premise---the Domination Thesis (DBA3)---to ensure we could still derive our conclusion. Instead of saying that credences are irrational if they mandate you to make a series of decisions that is guaranteed to leave you worse off than another series of decisions, we might say that credences are irrational if they permit you to make a series of decisions that is guaranteed to leave you worse off than another series of decisions. In the language of decision theory, instead of saying only that credences that mandate you to choose dominated options are irrational, we say also that credences that permit you to choose dominated options are irrational. We might call this the Permissive Domination Thesis.

Now, by weakening the first premise in this way, we respond to Hedden's objection and make the premise more plausible. But we strengthen the third premise to compensate and perhaps thereby make it less plausible. However, I imagine that anyone who accepts one of the versions of the third premise---either the Domination Thesis or the Permissive Domination Thesis---will also accept the other. Having credences that mandate dominated choices may be worse than having credences that permit such choices, but both seem sufficient for irrationality. Perhaps the former makes you more irrational than the latter, but it seems clear that the ideally rational agent will have credences that do neither. And if that's the case, then we can replace the standard Dutch Book Argument with a slight modification:

The Permissive Dutch Book Argument for Probabilism
(PDBA1) Permissive Ramsey's Thesis
(PDBA2) Dutch Book Theorem
(PDBA3) Permissive Domination Thesis
Therefore,
(PDBAC) Probabilism

The Bookless Response

Suppose you refuse even the permissive version of RT, and insist that coherent and incoherent agents alike should choose in line with MSEU. Then what becomes of the Dutch Book Argument? As we noted above, Hedden shows that it fails---MSEU is not sufficient to establish the conclusion. In particular, Hedden gives an example of an incoherent credence function that is not Dutch Bookable via MSEU. That is, there are no sets of bets with accompanying prices such that (a) MSEU will demand that you pay each of those prices, and (b) the sum of those prices is guaranteed to exceed the sum of the payouts of that set of bets. However, as we will see, accepting individual members of such a set of bets is just one way to make bad decisions based on your credences.

Consider Hedden's example. In it, you assign credences to propositions in the algebra built up from three possible worlds, $w_1$, $w_2$, and $w_3$. Here are some of your credences:
• $c(w_1 \vee w_2) = 0.8$ and $c(w_3) = 0$
• $c(w_1) = 0.7$ and $c(w_2 \vee w_3) = 0$
Now, consider the following two options, $A$ and $B$, whose utilities in each state of the world are set out in the following table:

Then notice first that $A$ dominates $B$---that is, the utility of $A$ is higher than $B$ in every possible state of the world. But, using your incoherent credences, you assign a higher expected utility to $B$ than to $A$. Your expected utility for $A$---which must be calculated relative to your credences in $w_1$ and $w_2 \vee w_3$, since  the utility of $A$ given $w_1 \vee w_2$ is undefined---is $0.7 \times 78 + 0 \times 77 = 54.6$. And your expected utility for $B$---which must be calculated relative to your credences in $w_1 \vee w_2$ and $w_3$, since the utility of $B$ given $w_2 \vee w_3$ is undefined---is $0.8 \times 74 + 0 \times 75 = 59.2$. So, while Hedden might be right that MSEU won't leave you vulnerable to a Dutch Book, it will leave you vulnerable to choosing a dominated option. And since what is bad about entering a Dutch Book is that it is a dominated option---it is dominated by the option of refusing the bets---the invulnerability to Dutch Books should be no comfort to you.

Now, this raises the question: For which incoherence credences is it guaranteed that MSEU won't lead you to choose a dominated option? Is it all incoherent credences, in which case we would have a new Dutch Book Argument for Probabilism from MSEU rather than RT? Or is it some subset? Below, we prove a theorem that answers that. First, a weakened version of Probabilism:

Bounded Probabilism If $c : \mathcal{F}\rightarrow [0, 1]$ is your credence function, then rationality requires that:
(BP1a) $c(\bot) = 0$, where $\bot$ is a necessarily false proposition;
(BP1b) There is $0 < M \leq 1$ such that $c(\top) = M$, where $\top$ is a necessarily true proposition;
(BP2) $c(A \vee B) = c(A) + c(B)$, if $A$ and $B$ are mutually exclusive.

Bounded Probabilism says that you should have lowest possible credence in necessary falsehoods, some positive credence---not necessarily 1---in necessary truths, and your credence in a disjunction of two incompatible propositions should be the sum of your credences in the disjuncts.

Theorem 1 The following are equivalent:
(i) $c$ satisfies Bounded Probabilism
(ii) For all options $A$, $B$, if $A$ dominates $B$, then $\mathrm{EU}_c(A) > \mathrm{EU}_c(B)$.

The proof is in the Appendix below. Thus, even without Ramsey's Thesis or the permissive version described above, you can still give a pragmatic argument for a norm that lies very close to Probabilism, namely, Bounded Probabilism. On its own, this argument cannot say what is wrong with someone who gives less than the highest possible credence to necessary truths, but it does establish the other requirements that Probabilism imposes. To see just how close to Probabilism lies Bounded Probabilism, consider the following two norms, which are equivalent to it:

Scaled Probabilism  If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, then rationality requires that there is $0 < M \leq 1$ and a probability function $p : \mathcal{F} \rightarrow [0, 1]$ such that $c(-) = M \times p(-)$.

Bounded Partition Probabilism  If $c : \mathcal{F} \rightarrow [0, 1]$ is your credence function, then rationality requires that, for any two partitions $\mathcal{X} = \{X_1, \ldots, X_m\}$ and $\mathcal{Y} = \{Y_1, \ldots, Y_n\}$,$$\sum^m_{i=1} c(X_i) = \sum^n_{j=1} c(Y_j)$$Then

Lemma 2 The following are equivalent:
(i) Bounded Probabilism
(ii) Scaled Probabilism
(iii) Bounded Partition Probabilism

As before, the proof is in the Appendix.

So, on its own, MSEU can deliver us very close to Probabilism. But it cannot establish (P1b), namely, $c(\top) = 1$. However, I think we can also appeal to a highly restricted version of the Permissive Ramsey's Thesis to secure (P1b) and push us all the way to Probabilism.

Consider Dima and Esther. They both have minimal confidence---i.e. 0---that it won't rain tomorrow. But Dima has credence 0.01 that it will rain, while Esther has credence 0.99 that it will. If we permit only actions that maximise expected utility, then Dima and Esther are required to pay exactly the same prices for bets on rain---that is, Dima will be required to pay a price exactly when Esther is. After all, if £$S$ is the payoff when it rains, £0 is the payoff when it doesn't, and $x$ is a proposed price, then $0.01\times (S- x) + 0 \times (0-x) \geq 0$ iff $0.99 \times (S-x) + 0 \times (0-x) \geq 0$ iff $S \geq x$. So, according to MSEU, Dima and Esther are rationally required to pay anything up to the stake of the bet for such a bet. But this is surely wrong. It is surely at least permissible for Dima to refuse to pay a price that Esther accepts. It is surely permissible for Esther to pay £99 for a bet on rain that pays £100 if it rains and £0 if it doesn't, while Dima refuses to pay anything more than £1 for such a bet, in line with Ramsey's Thesis. Suppose Dima were offered such a bet for the price of £99, and suppose she then defended her refusal to pay that price saying, 'Well, I only think it's 1% likely to rain, so I don't want to risk such a great loss with so little possible gain when I think the gain is so unlikely'. Then surely we would accept that as a rational defence.

In response to this, defenders of MSEU might concede that RT is sometimes the correct norm of action when you are incoherent, but only in very specific cases, namely, those in which you have a positive credence in a proposition, minimal credence (i.e. 0) in its negation, and you are considering the price you might pay for a bet on that proposition. In all other cases---that is, in any case in which your credences in the proposition and its negation are both positive, or in which you are considering an action other than a bet on a proposition---you should use MSEU. I have some sympathy with this. But, fortunately, this restricted version is all we need. After all, it is precisely by applying Ramsey's Thesis to such a case that we can produce a Dutch Book against someone with $c(\bot) = 0$ and $c(\top) < 1$---we simply offer to pay them £$c(\top) \times 100$ for a bet in which they will pay out £100 if $\top$ is true and £0 if it is false; this is then guaranteed to lose them £$100 \times (1-c(X))$, which is positive. Thus, we end up with a disjunctive pragmatic argument for Probabilism: if $c(\bot) = 0$ and $c(\top) < 1$, then RT applies and we can produce a Dutch Book against you; if you violate Probabilism in any other way, then you violate Bounded Probabilism and we can then produce two options $A$ and $B$ such that $A$ dominates $B$, but your credences, via MSEU, dictate that you should choose $B$ over $A$. This, then, is our bookless pragmatic argument for Probabilism:

Bookless Pragmatic Argument for Probabilism
(BPA1) If $c$ violates Probabilism, then either (i) $c(\bot) = 0$ and $c(\top) < 1$, or (ii) $c$ violates Bounded Probabilism.
(BPA2) If $c(\bot) = 0$ and $c(\top) < 1$, then RT applies, and there is a bet on $\top$ such that you are required by RT to pay a higher price for that bet than its guaranteed payoff. Thus, there are options $A$ and $B$ (namely, refuse the bets and pay the price), such that $A$ dominates $B$, but RT demands that you choose $B$ over $A$.
(BPA3) If $c$ violates Bounded Probabilism, then by Theorem 1, there are options $A$ and $B$ such that $A$ dominates $B$, but RT demands that you choose $B$ over $A$. Therefore, by (BPA1), (BPA2), and (BPA3),
(BPA4) If $c$ violates Probabilism, then there are options $A$ and $B$ such that $A$ dominates $B$, but rationality requires you to choose $B$ over $A$.
(BPA5) Dominance Thesis
Therefore,
(BPAC) Probabilism

Conclusion

The Dutch Book Argument for Probabilism assumes Ramsey's Thesis, which determines the prices an agent is rationally required to pay for a bet. Hedden argues that Ramsey's Thesis is wrong. He claims that Maximise Subjective Expected Utility  determines those prices, and it often disagrees with RT. In our Permissive Dutch Book Argument, I suggested that, in the face of that disagreement, we might be permissive: agents are permitted to pay any price that is required or permitted by RT and they are permitted to pay any price that is required or permitted by MSEU. In our Bookless Pragmatic Argument, I then explored what we might do if we reject this permissive response and insist that only prices permitted or required by MSEU are permissible. I showed that, in that case, we can give a pragmatic argument for Bounded Probabilism, which comes close to Probabilism, but doesn't quite reach; and I showed that, if we allow RT in the very particular cases in which it agrees better with intuition than MSEU does, we can give a pragmatic argument for Probabilism.

Appendix: Proof of Theorem 1

Theorem 1 The following are equivalent:
(i) $c$ satisfies Bounded Probabilism
(ii) For all options $A$, $B$, if $A$ dominates $B$, then $\mathrm{EU}_c(A) > \mathrm{EU}_c(B)$.

($\Rightarrow$) Suppose $c$ satisfies Bounded Probabilism. Then, by Lemma 2, there is $0 < M \leq 1$ and a probability function $p$ such that $c(-) = M \times p(-)$. Now suppose $A$ and $B$ are actions. Then
• $\mathrm{EU}_c(A) = \mathrm{EU}_{M \times p}(A) = M \times \mathrm{EU}_p(A)$
• $\mathrm{EU}_c(B) = \mathrm{EU}_{M \times p}(B) = M \times \mathrm{EU}_p(B)$
Thus, $\mathrm{EU}_c(A) > \mathrm{EU}_c(B)$ iff $\mathrm{EU}_p(A) > \mathrm{EU}_p(B)$. And we know that, if $A$ dominates $B$ and $p$ is a probability function, then $\mathrm{EU}_p(A) > \mathrm{EU}_p(B)$.

($\Leftarrow$) Suppose $c$ violates Bounded Probabilism. Then there are partitions $\mathcal{X} = \{X_1, \ldots, X_m\}$ and $\mathcal{Y} = \{Y_1, \ldots, Y_n\}$ such that $$\sum^m_{i=1} c(X_i) = x < y = \sum^n_{j=1} c(Y_j)$$We will now define two acts $A$ and $B$ such that $A$ dominates $B$, but $\mathrm{EU}_c(A) < \mathrm{EU}_c(B)$.
• For any $X_i$ in $\mathcal{X}$, $$U(A, X_i) = y - i\frac{y-x}{2(m + 1)}$$
• For any $Y_j$ in $\mathcal{Y}$,$$U(B, Y_j) = x + j\frac{y-x}{2(n + 1)}$$
Then the crucial facts are:
• For any two $X_i \neq X_j$ in $\mathcal{X}$,$$U(A, X_i) \neq U(A, X_j)$$
• For any two $Y_i \neq Y_j$ in $\mathcal{Y}$,$$U(B, Y_i) \neq U(B, Y_j)$$
• For any $X_i$ in $\mathcal{X}$ and $Y_j$ in $\mathcal{Y}$, $$x < U(B, Y_j) < \frac{x+y}{2} < U(A, X_i) < y$$
So $A$ dominates $B$, but$$\mathrm{EU}_c(A) = \sum^m_{i=1} c(X_i) U(A, X_i) < \sum^m_{i=1} c(X_i) \times y = xy$$
while$$\mathrm{EU}_c(B) = \sum^n_{j=1} c(Y_i) U(B, Y_j) > \sum^n_{j=1} c(Y_j) \times x = yx$$So $\mathrm{EU}_c(B) > \mathrm{EU}_c(A)$, as required.