The Accuracy Dominance Argument for Probabilism without the Additivity assumption
For a PDF of this post, see here.
One of the central arguments in accuracy-first epistemology -- the one that gets the project off the ground, I think -- is the accuracy-dominance argument for Probabilism. This started life in a more pragmatic guise in de Finetti's proof that, if your credences are not probabilistic, there are alternatives that would lose less than yours would if they were penalised using the Brier score, which levies a price of on every credence in a truth and on every credence in a falsehood. This was then adapted to an accuracy-based argument by Roger Rosencrantz, where he interpreted the Brier score as a measure of inaccuracy, not a penalty score. Interpreted thus, de Finetti's result says that any non-probabilistic credences are accuracy-dominated by some probabilistic credences. Jim Joyce then noted that this argument only establishes Probabilism if you have a further argument that inaccuracy should be measured by the Brier score. He thought there was no particular reason to think that's right, so he greatly generalized de Finetti's result to show that, relative to a much wider range of inaccuracy measures, all non-probabilistic credences are accuracy dominated. One problem with this, which Al Hájek pointed out, was that he didn't give a converse argument -- that is, he didn't show that, for each of his inaccuracy measures, each probabilistic credence function is not accuracy dominated. Joel Predd and his Princeton collaborators then addressed this concern and proved a very general result, namely, that for any additive, continuous, and strictly proper inaccuracy measure, any non-probabilistic credences are accuracy-dominated, while no probabilistic credences are.
That brings us to this blogpost. Additivity is a controversial claim. It says that the inaccuracy of a credence function is the (possibly weighted) sum of the inaccuracies of the credences it assigns. So the question arises: can we do without additivity? In this post, I'll give a quick proof of the accuracy-dominance argument that doesn't assume anything about the inaccuracy measures other than that they are continuous and strictly proper. Anyone familiar with the Predd, et al. paper will see that the proof strategy draws very heavily on theirs. But it bypasses out the construction of the Bregman divergence that corresponds to the strictly proper inaccuracy measure. For that, you'll have to wait for Jason Konek's forthcoming work...
Suppose:
is a set of propositions; be the set of possible worlds relative to ; be the set of credence functions on ; be the set of probability functions on . So, by de Finetti's theorem, . If is in , we write for .
Proof. We begin by defining a divergence
Three quick points about
(1)
(2)
(3)
Now, adding
to both sides gives
That is, as required.
Now, suppose is not in . Then, since is a closed convex set, there is a unique in that minimizes as a function of . Now, suppose is in . We wish to show that We can see that this holds iff After all,
Now we prove this inequality. We begin by observing that, since , are in , since is convex, and since is minimized uniquely at , if , then Expanding that, we get
\medskip
So
\medskip
So\medskip
\medskip
Now, since is strictly proper,
So, for all ,
So, since is continuous which is what we wanted to show. So, by above, In particular, since each is in , But, since is in and is not, and since is a divergence, . So as required.
That is,
Now, suppose
Now we prove this inequality. We begin by observing that, since
So
So\medskip
Now, since
So, for all
So, since
What do you mean when you say that the inaccuracy measure is continuous, and how do you know this to be true? Is it an additional assumption, or does it follow from strict propriety somehow?
ReplyDeleteSorry, you’re right — I should have included this explicitly. The result covers all continuous strictly proper inaccuracy measures.
DeleteIn the proof of the Theorem, in part (3), where you claim that D is strictly convex in its first argument, I don't see how the first pair of offset strict inequalities could hold. In the argument, c is supposed to be fixed and arbitrary. So for all you've said, we could have c=r there, and the inequalities would not be strict.
ReplyDeleteI think the proof in (3) that is strictly convex in its first argument might be incorrect, and I don't see why this result should hold in general. In the part of the proof where you say "Now adding...to both sides gives...," the quantities you're adding could be infinite, in which case the strict inequality will not be preserved. In general, in order for to be strictly convex in its first argument, there can be no such that for every (otherwise is a constant function ( ) in its first argument, and therefore not strictly convex). But I don't see that continuity and strict propriety rule out the possibility that there is such a .
ReplyDeleteAlso, even if were strictly convex in its first argument, that wouldn't imply (as you claim it does in your paper) that it attains a minimum on a closed convex set. For example, the real function on defined by if and is strictly convex, but it does not attain a minimum due to the discontinuity on the boundary. In order to show that attains a minimum, you need to appeal to some kind of continuity property for .
ReplyDelete