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.