## Friday, July 17, 2020

### Arbitrariness, regularity and comparative probabilities

In “Underdetermination of infinitesimal probabilities”, following a referee’s suggestion, I allow that qualitative (i.e., comparative) probabilities may escape arbitrariness problems for infinitesimal probabilities. But I now think this may be wrong.

Consider an infinite line of independent fair coins, numbered with the integers ..., −2, −1, 0, 1, 2, .... Let Rn be the hypothesis that coins n, n + 1, ... are all heads. Let Ln be the hypothesis that coins ..., n − 2, n − 1, n are all heads.

Suppose “being less likely or equally likely” is transitive, reflexive and total. Write A < B for A being less likely than B, A ≈ B for A and B being equally likely, and A ⪅ B for A being less likely or equally like than B.

We have strong regularity provided that if event A is a proper subset of event B, then A is strictly less likely than B.

Given strong regularity, the events Ln are strictly decreasing in probability: ... > L−2 > L−1 > L0 > L1 > ... and the events Rn are strictly increasing in probability: ... < R−2 < R1 < R0 < R1 < ....

Theorem: Given strong regularity, exactly one of the following options is true:

1. For all n and m, Ln < Rm (heads-runs are right-biased)

2. For all n and m, Rm < Ln (heads-runs are left-biased)

3. There is a unique n such that for all m ≤ n we have Rm ⪅ Lm and for all m > n we have Lm < Rm (there is a switch-over point at n).

But if (1) or (2) is true, it is difficult to see what objective reality could possibly ground whether heads-runs are right-biased or left-biased in our infinite sequence of coin tosses. And if (3) is true, it is difficult to see what objective reality could possibly make n be a switch-over point. The choice between left- and right-bias seems completely arbitrary, and the choice of a switch-over point is also arbitrary.

The Theorem follows from the following lemma:

Lemma: Let (S, ⪅) be a totally preordered set, and let Ln and Rn be sequences of members of S as n ranges over the integers. Suppose Ln is strictly decreasing and Rn is strictly increasing. Then exactly one of the following is true:

1. For all n and m, Ln < Rm

2. For all n and m, Rm < Ln

3. There is a unique n such that for all m ≤ n we have Rm ≤ Lm and for all m > n we have Lm < Rm.

Proof of Lemma: Suppose that for all n we have Ln < Rn. I claim that (1) is true. For suppose that (1) is false and hence (by totality) we have Rm ⪅ Ln for some m and n. Then m ≠ n, since Ln < Rn. Either m < n or n < m. If m < n, then Rm ⪅ Ln < Lm, and if m > n, then Rn < Rm ⪅ Ln. In either case we have a violation of the fact that Ln < Rn for all n′.

Now suppose that for all n we have Rn < Ln. I now claim that (2) is true. For if (2) is false, we have Ln ⪅ Rm for some m ≠ n. Suppose m < n. Then Ln ⪅ Rm < Rn, a contradiction. And if n < m, then Lm < Ln ⪅ Rm, also a contradiction.

Let A(n) be the statement that for all m ≤ n we have Rm ⪅ Lm and for all m > n we have Lm < Rm. There is at most one n such that A(n). For suppose that we had A(n) and A(n′) and that n < n′. Then by A(n) we have Ln < Rn (since n′>n), and by A(n′) we have Rn ⪅ Ln, resulting in a contradiction.

Assume (1) and (2) are false. By what we have shown earlier and totality, if (2) is not true, there is an n such that Ln ⪅ Rn. Note that if Ln ⪅ Rn, then Lm < Ln ⪅ Rn < Rm for all m > n. Hence, either Ln ⪅ Rn is true for all n or else there is a smallest n for which it’s true. If it’s true for all n, then for all n we have Ln + 1 < Ln ⪅ Rn < Rn + 1, and hence for all n we have Ln < Rn, which we saw would imply (1), which we assumed to be false.

So suppose that n is the smallest integer for which Ln ⪅ Rn. Thus, Rm < Lm whenever m < n. Moreover, since Li is strictly decreasing and Ri is strictly increasing, we have Lm < Rn whenever m > n. There are now two possibilities. Either Ln < Rn or Ln ≈ Rn. If Ln ≈ Rn, then we have A(n) and the proof is complete. Suppose now that Ln < Rn. Then we have A(n − 1) and the proof is also complete.

IanS said...

One might give up totality and say the Ls and Rs are not comparable.

This is not merely ad hoc. Intuitively, the coin flips could be described like this: the probability of any particular outcome of any finite set of N of the coins is 2^(-N), regardless of any condition on the outcomes of the other coins (even a condition with classical probability zero).

As in the post, it follows easily that the Ls are strictly decreasing in probability with n and the Rs are strictly increasing. (Use the comparison in your Popper Function post, or my modification of it.) But there seems no natural way to compare any L with any R. For simplicity, take a disjoint L and R. What is P(L | L ∪ R)? As far as I can see, it cannot be deduced from the intuitive description of the flips.

Alexander R Pruss said...

Yeah. I wonder if there is a non-arbitrary way to work out when we have lack comparability between two events here.

When the classical probabilities are different, we should have comparability.
When one event is a subset of the other, we should have comparability.
When the two events depend on only finitely many flips each, we should have comparability.

Here's another interesting test case. Let A be the event that the strong law of large numbers holds in the leftward direction (i.e., that the frequencies of heads converge to 1/2 leftward) and let B be the event that it holds in the rightward direction. Note that it doesn't matter where we start when we define A and B. Can we say that A and B are comparable? I don't think so. Let A' and B' be the events that the strong laws of large numbers hold in the leftward and rightward directions when we restrict to the even-numbered coins. Then A and B are, respectively, proper subsets of A' and B', so they should be respectively less probable. But there is no more reason to think that A is equiprobable with B' than that A is equiprobable with B.

This suggests the hypothesis that when A and B depend on disjoint infinite sets of coin tosses, and their classical probabilities are equal, then they should be incomparable. There may be a counterexample to that hypothesis.

Alexander R Pruss said...

Ian:

One more thought. If you deny comparability between L and R, then for the same reason you have to say that if the experiment is repeated, and for the repeat you have events L' and R', then you have to deny comparability between L and L', as well as between R and R' (for by symmetry there is no more reason to say that L and L' are comparable than that L and R' are). But this seems to commit one to the view that in infinitary cases one cannot repeat an experiment with the same chances as before. And the repeatability of experiments with the same chances seems foundational to our concept of chance.

IanS said...

Here are some half-baked thoughts. You may make something of them. I’m still thinking about it.

Yes, the approach I have suggested would say that no L can be compared with any L’. This seems reasonable. For example, L1 implies L0, and, granted the non-classical intuition given earlier, P(L1|L0) = 1/2. So L1 and L0 can be compared. But no such relation applies between any L and any L’. Note also, there is no natural correspondence between the indexing of the flips in the two experiments. You could shift it to make L0 correspond the L’1, or any other L’.

One moral I draw is that no totally ordered probabilities, standard or not, can reflect the structure implied by the intuitive description given above.

That said, classical probabilities seem adequate in practice (if the infinite can ever be practical…). All the Ls and Rs and L’s and R’s have classical probability zero. They are all bad bets at any odds. What more do you need to know in practice? The comparisons that are not captured by classical probabilities are certainly of conceptual interest, but relevant only in cases that happen with classical probability zero.