Friday, August 21, 2020

Complete Probabilistic Characterizations

Consider the concept of a complete probabilistic characterization (CPC) of an experiment. It’s a bit of a fuzzy concept, but we can get some idea about it. For instance, if I have a coin loaded in favor of heads, then saying that heads is more likely than tails is not a CPC. Minimally, the CPC will give exact numbers where the probabilities have exact numbers. But the CPC may go beyond giving numerical probabilities. For instance, if you toss infinitely main fair coins, the numerical probability that they are all heads is zero as is the probability that all the even numbered ones are heads. But intuitively it is more likely that the even numbered ones are heads than that all of them are heads. If there is something to this intuition, the CPC will include the relevant information: it may do that by assigning different infinitesimal probabilities to the two events, or by giving conditional probabilities conditioned on various zero-probability events.

A deep question that has sometimes been discussed by philosophers of probability is what CPCs are like. Here are three prominent candidates:

  1. classical real-valued probabilities

  2. hyperreal probabilities assigning non-zero (but perhaps infinitesimal) probability to every possible event

  3. primitive conditional probabilities allowing conditioning on every possible event.

The argument against (1) and for (2) and (3) is that (1) doesn’t distinguish things that should be distinguished—like the heads case above. I want to offer an argument against (2) and (3), however.

Here is a plausible principle:

  1. If X and Y are measurements of two causally independent experiments, then the CPC of the pair (X, Y) is determined by the CPCs of X and Y together with the fact of independence.

If (4) is true, then a challenge for a defender of a particular candidate for CPC is to explain how the CPC of the pair is determined by the individual CPCs of the independent experiments.

In the case of (1), the challenge is easily met: the pair (X, Y) has as its probability measure the product of the probability measures for X and Y.

In the cases of (2) and (3), the challenge has yet to be met, and there is some reason to think it cannot be met. In this post, I will argue for this in the case of (2): the case of (3) follows from the details of the argument in the case of (2) plus the correspondence between Popper functions and hyperreal probabilities.

Consider the case where X and Y are uniformly distributed over the interval [0, 1]. By independence, we want the pair (X, Y) to have a hyperreal finitely additive probability measure P such that P(X ∈ A, Y ∈ B)=P(X ∈ A)P(Y ∈ B) for all events A and B. But it turns out that this requirement on P highly underdetermines P. In particular, it seems to be that for any positive real number r, we can find a hyperreal measure P such that P(X ∈ A, Y ∈ B)=P(X ∈ A)P(Y ∈ B) for all A and B, and such that P(X = Y)=rP(Y = 0). Hence, independence highly underdetermines what value P assigns to the diagonal X = Y as compared to the value it assigns to Y = 0.

Maybe some other conditions can be added that would determine the CPC of the pair. But I think we don’t know what these would be. As it stands, we don’t know how to determine the CPC of the pair in light of the CPC of the members of the pair, if CPCs are of type (2).


IanS said...

Warning: the following may be way off-beam.

Assume your conjecture is true – it sure seems plausible. There could be a different way of looking at it. It could be that the CPC of the pair is well determined, but that it makes the diagonal non-measurable. On this view, the setup really does not determine the probability of the diagonal.

Non-measurability cannot always be wished away. Even classical probabilities on [0, 1] leave some sets non-measurable (as they must, granted Choice, if we require translation symmetry and countable additivity). In setups with symmetry that permits Banach-Tarski style results, the B-T sets cannot be measurable, even by hyperreal probabilities, if we want to respect the symmetry.

As a simple example, think of a ‘label-independent’ countable fair infinite lottery as discussed in your post ‘Label Independence and Lotteries’. Sets that are both infinite and ‘co-infinite’ cannot be given numerical probabilities, even hyperreal ones, if label independence is to be respected (because all such sets can be permuted onto each other, and two can be permuted onto one.) Only the finite and cofinite sets can be given type (2) probabilities.

Note the non-measurable sets in this example are not weird or non-constructible (except perhaps that Countable Choice may be required to number the ‘label independent’ tickets.) Maybe something of this sort is happening in your example.

IanS said...

More on the lottery example (Again, this may all be way off-beam.):

To assign type (2) probabilities to the finite and cofinite sets, it seems natural to take an infinitesimal ε and say that any finite set of size N has probability Nε and its complement has probability 1 – Nε. This, of the face of it, seems to be a CPC.

Now take two independent lotteries. We can explicitly describe the product algebra and probabilities (I think!): sets of the form H ‘horizontal lines’ and V ‘vertical lines’, omitting P points from these lines and adding Q points not on the lines - and their complements. Such a set would have probability (H + V) ε + (Q – P – HV) ε^2 and its complement 1 minus that. This seems like a CPC for the joint lotteries.

What about the ‘diagonal’ set? There is no natural correspondence between the lotteries, so there is no natural diagonal set. But you could set up an arbitrary correspondence. The implied diagonal set would not be in the algebra described above, so it would indeed be non-measurable.

Here are some bounds. The diagonal has an infinite number of points, so Nε^2 for any finite N is a lower bound. By suitable permutations, you can make any finite number of disjoint images of the diagonal, so 1/N for any finite N is an upper bound. This is consistent with rε, as you suggest.

All the above relates to probabilities that apply to single sets. You might be able to do better with comparative or conditional probabilities, which apply to pairs of sets (because permutation symmetry respects set inclusion, even between pairs of sets that are both infinite and co-infinite.)

Alexander R Pruss said...

These are interesting suggestions. It's particularly interesting that there may need to be a tradeoff: classical probabilities are not precise enough to distinguish certain cases from impossibility, while non-classical probabilities instead make these cases be non-measurable.