## Tuesday, 5 March 2019

### Dutch Books, Money Pumps, and 'By Their Fruits' Reasoning

There is a species of reasoning deployed in some of the central arguments of formal epistemology and decision theory that we might call 'by their fruits' reasoning. It seeks to establish certain norms of rationality that govern our mental states by showing that, if your mental states fail to satisfy those norms, they lead you to make choices that have some undesirable feature. Thus, just as we might know false prophets by their behaviour, and corrupt trees by their evil fruit, so can we know that certain attitudes are irrational by looking not to them directly but to their consequences. For instance, the Dutch Book argument seeks to establish the norm of Probabilism for credences, which says that your credences should satisfy the axioms of the probability calculus. And it does this by showing that, if your credences do not satisfy those axioms, they will lead you to enter into a series of bets that, taken together, lose you money for sure (Ramsey 1931, de Finetti 1937). The Money Pump argument seeks to establish, among other norms, the norm of Transitivity for preferences, which says that if you prefer one option to another and that other to a third, you should prefer the first option to the third. And it does this by showing that, if your preferences are not transitive, they will lead you, again, to make a series of choices that loses you money for sure (Davidson, et al. 1955). Both of these arguments use 'by their fruits' reasoning. In this paper, I will argue that such arguments fail. I will focus particularly on the Dutch Book argument so that I can illustrate the points with examples. But the objections I raise apply equally to Money Pump arguments.

## The Dutch Book argument: an example

Joachim is more confident that Sarah is an astrophysicist and a climate activist (proposition $A\ \&\ B$) than he is that she is an astrophysicist (proposition $A$). He is 60% confident in $A\ \&\ B$ and only 30% confident in $A$. But $A\ \&\ B$ entails $A$. So, intuitively, Joachim's credences are irrational.

How can we establish this? According to the Dutch Book argument, we look to the choices that Joachim's credences will lead him to make. The first premise of that argument posits a connection between credences and betting behaviour. Suppose $X$ is a proposition and $S$ is a number $S$, positive, negative, or zero. Then a £$S$ bet on $X$ is a bet that pays £$S$ if $X$ is true and £$0$ if $X$ is false. £$S$ is the stake of the bet. The first premise of the Dutch Book argument says that, if you have credence $p$ in $X$, you will buy a £$S$ bet on $X$ for anything less than £$pS$. That is, the more confident you are in a proposition, the greater a proportion of the stake you are prepared to pay to buy it. Thus, in particular:
• Bet 1: Joachim will buy a £$100$ bet on $A\ \&\ B$ for £$50$;
• Bet 2: Joachim will sell a £$100$ bet on $A$ for £$40$.
The total net gain of these bets, taken together, is guaranteed to be negative. Thus, his credences will lead him to a perform a pair of actions that, taken together, loses him money for sure. This is the second premise of the Dutch Book argument against Joachim. We say that this pair of actions (buy the first bet for £$40$; sell the second for £$50$) is dominated by the pair of actions in which he refuses to enter into each bets (refuse the first bet; refuse the second). The latter pair is guaranteed to result in greater total value than the former pair; the latter pair results in no loss and no gain, while the former results in a loss for sure. The third premise of the Dutch Book argument contends that, since it is undesirable to choose a pair of dominated options, it is irrational to have credences that lead you to do this. Ye shall know them by their fruits.

Thus, a Dutch Book argument has three premises. The first premise posits a connection between having a particular credence in a proposition and accepting certain bets on that proposition. The second is a mathematical theorem that shows that, if the first premise is true, and if your credences do not satisfy the probability axioms, they will lead you to make a series of choices that is dominated by some alternative series of choices you might have made instead; the third premise says that your credences are irrational if, together with the connection posited in the first premise, they lead you to choose a dominated series of options. My objection is this: there is no account of the connection between credences and betting behaviour that makes both the first and third premise plausible; those accounts strong enough to make the third premise plausible are too strong to make the first premise plausible. Our strategy will be to enumerate the possible putative accounts of that connection and show either that either the first or the third premise is false when we adopt that account.

Let $C(p, X)$ be the proposition that you have credence $p$ in proposition $X$;  and let $B(x, S, X)$ be the proposition that you pay £$x$ for a £$S$ bet on $X$. Then the first premise of the Dutch Book argument has the following form:

For all credences $p$, propositions $X$, prices $x$, and stakes $S$, if $x < pS$
$$O(C(p, X) \rightarrow B(x, S, X))$$where $O$ is a modal operator. But which modal operator? Different answers to this constitute different versions of the connection between credences and betting behaviour that appears in the first and third premise of the Dutch Book argument. We will consider six different candidate operators and argue that none makes the first and third premises both true. The six candidates are: metaphysical necessity; nomological necessity; nomological high probability; deontic necessity; deontic possibility; and the modality of defeasible reasons.

## $O$ is metaphysical necessity

We begin with metaphysical modality. According to this account, the first premise of the Dutch Book argument says that it is metaphysically impossible to have a credence of $p$ in $X$ while refusing to pay £$x$ for a £$S$ bet on $X$ (for $x < pS$). If you were to refuse such a bet, that would simply mean that you do not have that credence. This sort of account would be appealing to a behaviourist, who seeks an operational definition of what it means to have a particular precise credence in a proposition---a definition in terms of betting behaviour might well satisfy them.

If this account were true, the third premise of the Dutch Book argument would be plausible. If having a set of mental states were to guarantee as a matter of metaphysical necessity that you'd make a dominated series of choices when faced with a particular series of decisions, that seems sufficient to show that those credences are irrational. The problem is that, as David Christensen (1996) shows, the account itself cannot be true. Christensen's central point is this: credences are often and perhaps typically connected to betting behaviour and decision-making more generally; but they are often and perhaps typically connected to other things as well, such as emotional states, conative states, and other doxastic states. If I have a high credence that my partner loves me, I'm likely to pay a high proportion of the stake for a bet on it; but I'm also likely to feel joy, plan to spend more time with him, hope that his love continues, and believe that we will still be together in five years' time. What's more, none of these connections is obviously more important than any other in determining that a mental state is a credence. And each might fail while the others hold. Indeed, as Christensen notes, in Dutch Book arguments, we are concerned precisely with those cases in which there is a breakdown of the rationally required connections between credences, namely, the connections described by the probability axioms. Having a credence in one proposition usually leads you to have at least as high a credence in another proposition it entails. But, as we saw in Joachim's case, this connection can break down. So, just as Joachim's case shows that it is metaphysically possible to have a particular credence  that has all the other connections that we typically associate with it except the connection to other credences, so it must be at least metaphysically possible to have a credence has all the other connections that we associate with it but not the connection to betting behaviour posited by the first premise. Such a mental state would still count as the credence in question because of all the other connections; but it wouldn't give rise to the apparently characteristic betting behaviour that is required to run the Dutch Book argument. Moreover, note that we need not assume that the credence has none of the usual connections to betting behaviour. Consider Joachim again. Every Dutch Book to which he is vulnerable involves him buying a bet on $A\ \&\ B$ and selling a bet on $A$. That is, it involves him buying a bet on $A\ \&\ B$ with a positive stake and buying a bet on $A$ with a negative stake. So he would evade the argument if his credence in $A\ \&\ B$ were to lead him to buy the bets with any stake that the first premise says they will, while his credence in $A$ were only to lead him to buy the bets with positive stake that the first premise says they will. In this case, we'd surely say he has the credences we assign to him. But he would not be vulnerable to a Dutch Book argument.

Thus, if $O$ is metaphysical necessity, the third premise might well be true; but the first premise is false.

## $O$ is nomological necessity

Learning from the problems with the previous proposal, we might retreat to a weaker modality. For instance, we might suggest that $O$ is a nomological modality. There are two that it might be. We might say that the connection between credences and betting behaviour posited by the first premise is nomologically necessary---that is, it is entailed by the laws of nature. Or, we might say that it is nomologically highly probable---that is, the objective chance of the consequent given the antecedent is high. Let's take them in turn.

First, $O$ is nomological necessity. The problem with this is the same as the problem with the suggestion from the previous section that $O$ is metaphysical necessity. Above, we imagined a mental state that had all the other features we'd typically expect of a particular credence in a proposition, except some range of connections to betting behaviour that was crucial for the Dutch Book argument. We noted that this would still count as the credence in question. All that needs to be added here is that the example we considered is not only metaphysically possible, but also nomologically possible. That is, this is not akin to an example in which the fine structure constant is different from what it actually is---in that case, it would be metaphysically possible, but nomologically impossible. There is no law of nature that entails that your credence will lead to particular betting behaviour.

Thus, again, the first premise is false.

## $O$ is nomological high probability

Nonetheless, while it is not guaranteed by the laws of nature that an individual with a particular credence in a proposition will engage in the betting behaviour posited by the first premise, it does seem plausible that they are very likely to do so---that is, the objective chance that they will display the behaviour given that they have the credence is high. In other words, while weakening from metaphysical to nomological necessity doesn't make the first premise plausible, weakening further to nomological high probability does. So let's suppose, then, that $O$ is nomological high probability. Unfortunately, this causes two problems for the third premise.

Here's the first. Suppose I have credences in 1,000 mutually exclusive and exhaustive propositions. And suppose each credence is $\frac{1}{1,001}$. So they violate Probabilism. Suppose further that each credence is 99% likely to give rise to the betting behaviour mentioned in the first premise of the Dutch Book argument; and suppose that whether one of the credences does or not is independent of whether any of the others does or not. Then the objective chance that the set of 1,000 credences will give rise to the betting behaviour that will lose me money for sure is $0.99^{1,000} = 0.00004 \approx \frac{1}{23,163}$. And this tells against the third premise. After all, what is so irrational about a set of credences that will lead to a dominated series of choices less than once in every 20,000 times I face the bets described in the Dutch Book argument against me?

Here's the second problem. On the account we are considering, having a particular credence in a proposition makes it highly likely that you'll bet in a particular way. Let's say, then, that you violate Probabilism, and your credences do indeed result in you making a dominated series of choices. The third premise infers from this that your credences are irrational. But why lay the blame at the credences' door? After all, there is another possible culprit, namely, the probabilistic connection between the credence and the betting behaviour. Consider an analogy. Suppose that, as the result of some bizarre causal pathway, when I fall in love, it is very likely that I will feed myself a diet of nothing but mud and leaves for a week. I hate the taste of the mud and the leaves make me very sick, and so I lower my utility considerably by responding in this way. But I do it anyway. In this case, we would not, I think, say that it is irrational to fall in love. Rather, we'd say that what is irrational is my response to falling in love. Similarly, suppose I make a dominated series of choices and thus reveal some irrationality in myself. Then, for all the Dutch Book argument says, it might be that the irrationality lies not in the credences, but rather in my response to having those credences.

Thus, on this account, the first premise is plausible, but the third premise is unmotivated, for it imputes irrationality to my credences when it might instead lie in my response to having those credences.

## $O$ is deontic necessity

A natural response to the argument of the previous section is that the analogy between the credence-betting connection and the love-diet connection fails because the first is a rationally appropriate connection, while the latter is not. This leads us to suggest, along with Christensen (1996), that the connection between credences and betting behaviour at the heart of the Dutch Book argument is not governed by a descriptive modality, such as metaphysical or nomological modality, but rather by a prescriptive modality, such as deontic modality. In particular, it suggests that what the first premise says is not that someone with a particular credence in a proposition will or might or probably will accept certain bets on that proposition; but rather that they should or may or have good but defeasible reason to do so.

Let's begin with deontic necessity. Here, my objection is that, if this is the modality at play in the first and third premise, then the argument is self-defeating. To see why, consider Joachim again. Suppose the modality is deontic necessity, and suppose that the first premise is true. So Joachim is rationally required to make a dominated series of choices---buy the £$100$ bet on $A\ \&\ B$ for £$50$; sell the £$100$ bet on $A$ for £$40$. Now suppose further that the third premise is true as well---it does, after all, seem plausible on this account of the modality involved. Then we conclude that Joachim's credences are irrational. But surely it is not rationally required to choose in line with irrational credences. Surely what is rationally required of Joachim instead is that he should correct his irrational credences so that they are now rational, and he should then choose in line with his new rational credences. Now, whatever other features they have, his new rational credences must obey Probabilism. If not, they will be vulnerable to the Dutch Book argument and thus irrational. But the Converse Dutch Theorem shows that, if they obey Probabilism, they will not rationally require or even permit Joachim to make a dominated series of choices. And, in particular, they neither require nor permit him to accept both of the bets described in the original argument. But from this we can conclude that the first premise is false. Joachim's original credences do not rationally require him to accept both of the bets; instead, rationality requires him to fix up those credences and choose in line with the credences that result. But those new fixed up credences do not require what the first premise says they require. Indeed, they don't even permit what the first premise says they require. So, if the premises of the Dutch Book argument are true, Joachim's credences are irrational, and thus the first premise of the argument is false.

Thus, on this account, the Dutch Book argument is self-defeating: if it succeeds, its first premise is false.

## $O$ is deontic possibility

A similar problem arises if we take the modality to be deontic possibility, rather than necessity. On this account, the first premise says not that Joachim is required to make each of the choices in the dominated series of choices, but rather that he is permitted to do so. The third premise must then judge a person irrational if they are permitted to accept each choice in a dominated series of choices. If we grant that, we can conclude that Joachim's credences are irrational. And again, we note that rationality therefore requires him to fix up those credences first and then to choose in line with the fixed up credences. But just as those fixed up credences don't require him to make each of the choices in the dominated series, so they don't permit him to make them either. So the Dutch Book argument, if successful, undermines its first premise again.

Again, the Dutch Book argument is self-defeating.

## $O$ is the modality of defeasible reasons

The final possibility we will consider: Joachim's credences neither rationally require nor rationally permit him to make each of the choices in the dominated series; but perhaps we might say that each credence gives him a pro tanto or defeasible reason to accept the corresponding bet. That is, we might say that Joachim's credence of 60% in $A\ \&\ B$ gives him a pro tanto or defeasible reason to buy a £$100$ bet on $A\ \&\ B$ for £$50$, while his credence of 30% in $A$ gives him a pro tanto or defeasible reason to sell a £$100$ bet on $A$ for £$40$. As we saw above, those reasons must be defeasible, since they will be defeated by the fact that Joachim's credences, taken together, are irrational. Since they are irrational, he has stronger reason to fix up those credences and choose in line with the fixed up one than he has to choose in line with his original credences. But his original credences nonetheless still provide some reason in favour of accepting the bets.*

Rendered thus, I think the first premise is quite plausible. The problem is that the third premise is not. It must say that it is irrational to have any set of mental states where (i) each state in the set gives pro tanto reason to make a particular choice and (ii) taken together, that series of choices is dominated by another series of choices. But that is surely false. Suppose I believe this car in front of me is two years old and I also believe it's done 200,000 miles. The first belief gives me pro tanto or defeasible reason to pay £$5,000$ for it. The second gives me pro tanto reason to sell it for £$500$ as soon as I own it. Doing both of these things will lose me £$4,500$ for sure. But there is nothing irrational about my two beliefs. The problem arises only if I make decisions in line with the reasons given by just one of the beliefs, rather than taking into account my whole doxastic state. If I were to attend to my whole doxastic state, I'd never pay £$5,000$ for the car in the first place.  And the same might be said of Joachim. If he pays attention only to the reasons given by his credence in $A\ \&\ B$ when he considers the bet on that proposition, and pays attention only to the reasons given by his credence in $A$ when he considers the bet on that proposition, he will choose a dominated series of options. But if he looks to the whole credal state, and if the Dutch Book argument succeeds, he will see that its irrationality defeats those reasons and gives him stronger reason to fix up his credences and act in line with those. In sum, there is nothing irrational about a set of mental states each of which individually gives you pro tanto or defeasible reason to choose an option in a dominated series of options.

On this account, the first premise may be true, but the third is false.

## Conclusion

In conclusion, there is no account of the modality involved in the first and third premises of the Dutch Book argument that can make both premises true. Metaphysical and nomological necessity are too strong to make the first premise true. Nomological high probability is not, but it does not make the third premise true. Deontic necessity and possibility render the argument self-defeating, for if the arguments succeeds, the first premise must be false. Finally, the modality of defeasible reasons, like nomological high probability, renders the first premise plausible. But it is not sufficient to secure to the third premise.

Before we conclude, let's consider briefly how these considerations affect money pump arguments. The first premise of a money pump argument does not posit a connection between credences and betting behaviour, but between preferences and betting behaviour. In particular: if I prefer one option to another, there will be some small amount of money I'll be prepared to pay to receive the first option rather than the second. As with the Dutch Book argument, the question arises what the modal force of this connection is. And indeed the same candidates are available. What's more, the same considerations tell against each of those candidates. Just as credences are typically connected not only to betting behaviour but also to emotional states, intentional states, and other doxastic states, so preferences are typically connected to emotional states, intentional states, and other preferences. If I prefer one option to another, then this might typically lead me to pay a little to receive the first rather than the second; but it will also typically lead me to hope that I will receive the first rather than the second, to fear that I'll receive the second, to intend to choose the first over the second when faced with such a choice, and to have a further preference for the first and a small loss of money over the second. And again the connections to behaviour are no more central to this preference than the connections to the emotional states of hope and fear, the intentions to choose, and the other preferences. So the modal force of the connection posited by the first premise cannot be metaphysical or nomological necessity. And for the same reasons as above, it cannot be nomological high probability, deontic necessity or possibility, or the modality of defeasible reasons. In each case, the same objections hold.

So these two central instances of 'by their fruits' reasoning fail. We cannot give an account of the connection between the mental states and their evil fruit that renders the argument successful.

[* Thanks to Jason Konek for pushing me to consider this account.]

## References

• Christensen, D. (1996). Dutch-Book Arguments Depragmatized: Epistemic Consistency for Partial Believers. The Journal of Philosophy, 93(9), 450–479.
• Davidson, D., McKinsey, J. C. C., & Suppes, P. (1955). Outlines of a Formal Theory of Value, I. Philosophy of Science, 22(2), 140–60.
• de Finetti, B. (1937). Foresight: Its Logical Laws, Its Subjective Sources. In H. E. Kyburg, & H. E. K. Smokler (Eds.) Studies in Subjective Probability. Huntingdon, N. Y.: Robert E. Kreiger Publishing Co.
• Ramsey, F. P. (1931). Truth and Probability. The Foundations of Mathematics and Other Logical Essays, (pp. 156–198).

## Saturday, 2 February 2019

### Credences in vague propositions: supervaluationist semantics and Dempster-Shafer belief functions

Safet is considering the proposition $R$, which says that the handkerchief in his pocket is red. Now, suppose we take red to be a vague concept. And suppose we favour a supervaluationist semantics for propositions that involve vague concepts. According to such a semantics, there is a set of legitimate precisifications of the concept red, and a proposition that involves that concept is true if it is true relative to all legitimate precisifications, false if false relative to all legitimate precisifications, and neither if true relative to some and false relative to others. So London buses are red is true, Daffodils are red is false, and Cherry blossom is red is neither.

Safet is assigning a credence to $R$ and a credence to its negation $\overline{R}$. He assigns 20% to $R$ and 20% to $\overline{R}$. Normally, we'd say that he is irrational, since his credences in mutually exclusive and exhaustive propositions don't sum to 100%. What's more, we'd demonstrate his irrationality using either

(i) a sure loss betting argument, which shows there is a finite series of bets, each of which his credences require him to accept but which, taken together, are guaranteed to lose him money; or

(ii) an accuracy argument, which shows that there are alternative credences in those two propositions that are guaranteed to be closer to the ideal credences.

However, in Safet's case, both arguments fail.

Take the sure loss betting argument first. According to that, my credences require me to sell a £100 bet on $R$ for £30 and sell a £100 bet on $\overline{R}$ for £30. Thus, I will receive £60 from the sale of these two bets. Usually the argument proceeds by noting that, however the world turns out, either $R$ is true or $\overline{R}$ is true. So I will have to pay out £100 regardless. And I'm therefore guaranteed to lose £40 overall. But, in a supervaluationist semantics, this assumption isn't true. If Safet's handkerchief is a sort of pinkish colour, $R$ will be neither true nor false, and $\overline{R}$ will be neither true nor false. So I won't have to pay out either bet, and I'll gain £60 overall.

Next, take the accuracy argument. According to that, my credences are more accurate the closer they lie to the ideal credences; and the ideal credence in a true proposition is 100% while the ideal credence in a proposition that isn't true is 0%. Then, given that the measure of distance between credence functions has a particular property, then we usually show that there are alternative credences in $R$ and $\overline{R}$ that are closer to each set of ideal credences than Safet's are. For instance, if we measure the distance between two credence functions $c$ and $c'$ using the so-called squared Euclidean distance, so that $$SED(c, c') = (c(R) - c'(R))^2 + (c(\overline{R}) - c'(\overline{R}))^2$$ then credences of 50% in both $R$ and $\overline{R}$ are guaranteed to be closer than Safet's to the credences of 100% in $R$ and 0% in $\overline{R}$, which are ideal if $R$ is true, and closer than Safet's to the credences of 0% in $R$ and 100% in $\overline{R}$, which are ideal if $\overline{R}$ is true. Now, if $R$ is a classical proposition, then this covers all the bases--either $R$ is true or $\overline{R}$ is. But since $R$ has a supervaluationist semantics, there is a further possibility. After all, if Safet's handkerchief is a sort of pinkish colour, $R$ will be neither true nor false, and $\overline{R}$ will be neither true nor false. So the ideal credences will be 0% in $R$ and 0% in $\overline{R}$. And 50% in $R$ and 50% in $\overline{R}$ is not closer than Safet's to those credences. Indeed, Safet's are closer.

So our usual arguments that try to demonstrate that Safet is irrational fail. So what happens next? The answer was given by Jeff Paris ('A Note on the Dutch Book Method'). He argued that the correct norm for Safet is not Probabilism, which requires that his credence function is a probability function, and therefore declares him irrational. Instead, it is Dempster-Shaferism, which requires that his credence function is a Dempster-Shafer belief function, and therefore declares him rational. To establish this, Paris showed how to tweak the standard sure loss betting argument for Probabilism, which depends on a background logic that is classical, to give a sure loss betting argument for Dempster-Shaferism, which depends on a background logic that comes from the supervaluationist semantics. To do this, he borrowed an insight from Jean-Yves Jaffray ('Coherent bets under partially resolving uncertainty and belief functions'). Robbie Williams then appealed to Jaffray's theorem to tweak the accuracy argument for Probabilism to give an accuracy argument for Dempster-Shaferism ('Generalized Probabilism: Dutch Books and Accuracy Domination'). However, Jaffray's result doesn't explicitly mention supervaluationist semantics. And neither Paris nor Williams fill in the missing details. So I thought it might be helpful to lay out those details here.

I'll start by sketching the argument. Then I'll go into the mathematical detail. So first, the law of credences that we'll be justifying. We begin with a definition. Throughout we'll consider only credence functions on a finite Boolean algebra $\mathcal{F}$. We'll represent the propositions in $\mathcal{F}$ as subsets of a set of possible worlds.

Definition (belief function) Suppose $c : \mathcal{F} \rightarrow [0, 1]$. Then $c$ is a Dempster-Shafer belief function if
• (DS1a) $c(\bot) = 0$
• (DS1b) $c(\top) = 1$
• (DS2) For any proposition $A$ in $\mathcal{F}$,$$c(A) \geq \sum_{B \subsetneqq A} (-1)^{|A-B|+1}c(B)$$
Then we state the law:

Dempster-Shaferism $c$ should be a D-S belief function.

Now, suppose $Q$ is a set of legitimate precisifications of the concepts that are involved in the propositions in $\mathcal{F}$. Essentially, $Q$ is a set of functions each of which takes a possible world and returns a classically consistent assignment of truth values to the propositions in $\mathcal{F}$. Given a possible world $w$, let $A_w$ be the strongest proposition that is true at $w$ on all legitimate precisifications in $Q$. If $A = A_w$ for some world $w$, we say that $A$ is a state description for $w$.

Definition (belief function$^*$) Suppose $c : \mathcal{F} \rightarrow [0, 1]$. Then $c$ is a Dempster-Shafer belief function$^*$ relative to a set of precisifications if $c$ is a Dempster-Shafer belief function and
• (DS3) For any proposition $A$ in $\mathcal{F}$ that is not a state description for any world, $$c(A) = \sum_{B \subsetneqq A} (-1)^{|A-B|+1}c(B)$$
Dempster-Shaferism$^*$ $c$ should be a Dempster-Shafer belief function$^*$.

It turns out that Dempster-Shaferism$^*$ is the strongest credal norm that we can justify using sure loss betting arguments and accuracy arguments. The sure loss betting argument is based on the following assumption: Let's say that a £$S$ bet on a proposition $A$ pays out £$S$ if $A$ is true and £0 otherwise. Then if your credence in $A$ is $p$, then you are required to pay anything less than £$pS$ for a £$S$ bet on $A$. With that in hand, we can show that you are immune to a sure loss betting arrgument iff your credence function is a Dempster-Shafer belief function$^*$. That is, if your credence function violates Dempster-Shaferism$^*$, then there is a finite set of bets on propositions in $\mathcal{F}$ such that (i) your credences require you to accept each of them, and (ii) together, they lose you money in all possible worlds. If your credence function satisfies Dempster-Shaferism$^*$, there is no such set of bets.

The accuracy argument is based on the following assumption: The ideal credence in a proposition at a world is 1 if that proposition is true at the world, and 0 otherwise; and the distance from one credence function to another is measured by a particular sort of measure called a Bregman divergence. With that in hand, we can show that you are immune to an accuracy dominance argument iff your credence function is a Dempster-Shafer belief function$^*$. That is, if your credence function violates Dempster-Shaferism$^*$, then there is an alternative credence function that is closer to the ideal credence function than yours at every possible world. If your credence function satisfies Dempster-Shaferism$^*$, there is no such alternative.

So much for the sketch of the arguments. Now for some more details. Suppose $c : \mathcal{F} \rightarrow [0, 1]$ is a credence function defined on the set of propositions $\mathcal{F}$. Often, we don't have to assume anything about $\mathcal{F}$, but in the case we're considering here, we must assume that it is a finite Boolean algebra. In both sure loss arguments and accuracy arguments, we need to define a set of functions, one for each possible world. In the sure loss arguments, these specify when certain bets payout; in the accuracy arguments, they specify the ideal credences. In the classical case and in the supervaluationist case that we consider here, they coincide. Given a possible world $w$, we abuse notation and write $w : \mathcal{F} \rightarrow \{0, 1\}$ for the following function:
• $w(A) = 1$ if $X$ is true at $w$---that is, if $A$ is true on all legitimate precisifications at $w$;
• $w(A) = 0$ if $X$ is not true at $w$---that is, if $A$ is false on some (possibly all) legitimate precisifications at $w$.
Then, given our assumptions, we have that a £$S$ bet on $A$ pays out £$Sw(A)$ at $w$; and we have that $w(A)$ is the ideal credence in $A$ at $w$. Now, let $\mathcal{W}$ be the set of these functions. And let $\mathcal{W}^+$ be the convex hull of $\mathcal{W}$. That is, $\mathcal{W}^+$ is the smallest convex set that contains $\mathcal{W}$. In other words, $\mathcal{W}^+$ is the set of convex combinations of the functions in $\mathcal{W}$. There is then a general result that says that $c$ is vulnerable to a sure loss betting argument iff $c$ is not in $\mathcal{W}^+$. And another general result that says that $c$ is accuracy dominated iff $c$ is not in $\mathcal{W}^+$. To complete our argument, therefore, we must show that $\mathcal{W}^+$ is precisely the set of Dempster-Shafer belief functions$^*$. That's the central purpose of this post. And that's what we turn to now.

We start with some definitions that allow us to given an alternative characterization of the Dempster-Shafer belief functions and belief functions$^*$.

Definition (mass function) Suppose $m : \mathcal{F} \rightarrow [0, 1]$. Then $m$ is a mass function if
• (M1) $m(\bot) = 0$
• (M2) $\sum_{A \in \mathcal{F}} m(A) = 1$
Definition (mass function$^*$) Suppose $m : \mathcal{F} \rightarrow [0, 1]$. Then $m$ is a mass function$^*$ relative to a set of precisifications if $m$ is a mass function and
• (M3) For any proposition $A$ in $\mathcal{F}$ that is not the state description of any world, $m(A) = 0$.
Definition ($m$ generates $c$) If $m$ is a mass function and $c$ is a credence function, we say that $m$ generates $c$ if, for all $A$ in $\mathcal{F}$, $$c(A) = \sum_{B \subseteq A} m(B)$$ That is, a mass function generates a credence function iff the credence assigned to a proposition is the sum of the masses assigned to the propositions that entail it.

Theorem 1
• $c$ is a Dempster-Shafer belief function iff there is a mass function $m$ that generates $c$.
• $c$ is a Dempster-Shafer belief function$^*$ iff there is a mass function$^*$ $m$ that generates $c$.
Proof of Theorem 1 Suppose $m$ is a mass function that generates $c$. Then it is straightforward to verify that $c$ is a D-S belief function. Suppose $c$ is a D-S belief function. Then let$$m(A) = c(A) - \sum_{B \subsetneqq A} (-1)^{|A-B|+1}c(B)$$ This is positive, since $c$ is a belief function. It is then straightforward to verify that $m$ is a mass function. And it is straightforward to see that $m(A) = 0$ iff $c(A) = \sum_{B \subsetneqq A} (-1)^{|A-B|+1}c(B)$. That completes the proof.

Theorem 2 $c$ is in $\mathcal{W}^+$ iff $c$ is a Dempster-Shafer belief function$^*$.

Proof of Theorem 2 Suppose $c$ is in $\mathcal{W}^+$. So $c(-) = \sum_{w \in \mathcal{W}} \lambda_w w(-)$. Then:
• if $A$ is the state description for world $w$ (that is, $A = A_w$), then let $m(A) = m(A_w) = \lambda_w$;
• if $A$ is not a state description of any world, then let $m(A) = 0$.
Then $m$ is a mass function$^*$. And $m$ generates $c$. So $c$ is a Dempster-Shafer belief function$^*$.

Suppose $c$ is a Dempster-Shafer belief function$^*$ generated by a mass function$^*$ $m$. Then for world $w$, let $\lambda_w = m(A_w)$. Then $c(-) = \sum_{w \in \mathcal{W}} \lambda_w w(-)$. So $c$ is in $\mathcal{W}^+$.

This completes the proof. And with the proof we have the sure loss betting argument and the accuracy dominance argument for Dempster-Shaferism$^*$ when the propositions about which you have an opinion are governed by a supervaluationist semantics.

## 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.