Wednesday, January 17, 2018

Arbitrariness, probability and infinitesimals

A well-known objection to replacing the zero probability of some events—such as getting heads infinitely many times in a row—with an infinitesimal is arbitrariness. Infinitesimals are usually taken to be hyperreals and there are infinitely many hyperreal extensions of the reals.

This version of the arbitrariness has an objection. There are extensions of the reals that one can unambiguously define. Three examples: (1) the surreals, (2) formal Laurent series and (3) the Kanovei-Shelah model.

But it turns out that there is still an arbitrariness objection in these contexts. Instead of saying that the choice of extension of the reals is arbitrary, we can say say that the choice of particular infinitesimals within the system to be assigned to events is arbitrary.

Here is a fun fact. Let R be the reals and let R* be any extension of R that is a totally ordered vector space over the reals, with the order agreeing with that on R. (This is a weaker assumption than taking R* to be an ordered field extension of the reals.) Say that an infinitesimal is an x in R* such that −y < x < y for any real y > 0.

Theorem: Suppose that P is an R*-valued finitely additive probability on some algebra of sets, and suppose that P assigns a non-real number to some set. Then there are uncountably different many R*-valued finitely additive probability assignments Q on the same algebra of sets such that:

  1. If P(A) is real if and only if Q(A) is real, and then P(A)=Q(A).

  2. All corresponding linear combinations of P and Q are ordinally equivalent to each other, i.e., for any sets A1, ..., An, B1, ..., Bm in the algebra and any real a1, ..., an, b1, ..., bm, we have ∑aiP(Ai)<∑biP(Bi) if and only if ∑aiQ(Ai)<∑biQ(Bi).

  3. P(A)−Q(A) differ by a non-zero infinitesimal whenever P(A) is non-real.

Condition (ii) has some important consequences. First, it follows that ordinal comparisons of probabilities will be equally preserved by P and by Q. Second, it follows that both probabilities will assign the same results to decision problems with real-number utilities. Third, it follows that P(A)=P(B) if and only if Q(A)=Q(B), so any symmetries preserved by P will be preserved by Q. These remarks show that it is difficult indeed to hold that the choice of P over Q (or any of the other uncountably many options) is non-arbitrary, since it seems epistemic, decision-theoretic and symmetry constraints satisfied by P will be satisfied by Q.

Sketch of proof: For any finite member x of R* (x is finite if and only if there is a real y such that −y < x < y), let s(x) be the unique real number such that x − s(x) is infinitesimal. Let i(x)=x − s(x). Then for any real number r > 0, let Qr(A)=s(P(A)) + ri(P(A)). Note that s and i are linear transformations, from which it follows that Qr is a finitely additive probability assignment. It is not difficult to show that (i) and (ii) hold, and that (iii) holds if r ≠ 1.

Remark 1: I remember seeing the s + ri construction, but I can’t remember where. Maybe it was in my own work, maybe in something by someone else (Adam Elga?).

Remark 2: What if we want to preserve facts about conditional probabilities? This is a bit trickier. We’ll need to assume that R* is a totally ordered field rather than a totally ordered vector space. I haven’t yet checked what properties will be preserved by the construction above then.


Alexander R Pruss said...

Independence may not be preserved.

Alexander R Pruss said...

A paper based on the result in this post has just been accepted by Synthese.