Non-measurable events are ones to which the probability function in the situation assigns no probability. Philosophically speaking, non-measurable events come in two varieties:
Non-measurable events that should not have any probability assignment.
Non-measurable events that should have a probability assignment.
Type (1) non-measurable events are the kinds of weird events that can be constructed from the Hausdorff and Banach-Tarski paradoxes, as well as perhaps (this is less clear) the Vitali non-measurable sets.
But I think there are also type (2) non-measurable events relative to standard choices of probability functions. For instance, suppose that in each universe of an infinite multiverse a fair coin is tossed countably infinitely often.
How likely is it that in at least one universe all the coin tosses are heads? If the universes form a countable infinity, classical probability theory gives an answer: zero. But if the universes form an uncountable infinity, classical probability theory gives no answer at all—the standard completed product measure makes the event be non-measurable. However, intuitively, there should be an answer in at least some cases. If the number of universes is much larger than the number of possible countable sequences of coin tosses (i.e., is much larger than 2ω), we would expect the probability to be 1 or close to it. We can coherently extend the standard probability function to give that answer. But we can also coherently extend it to give a different answer, including the answer that the probability of an all-heads universe is zero, even if the number of universes is a gigantic infinite cardinality.
We don’t want to just make up an answer here. We want the answer to be derivable in some way resembling the proof of the theorem that if you toss a coin infinitely many times, you’ve got probability 1 of getting heads at least once.
I suppose we could take it to be a metaphysical axiom that if you have K disjoint collections each with M coin tosses, then if K and M are infinite and K > M, then with probability one at least one collection yields all heads. But it would be nice to have more than just intuition here, and in similar problems.
Here is a similar case. Take a permutation symmetric fair infinite lottery (e.g. your construction posted April 29, 2014). Code the outcome as Even or Odd. Repeat the setup a countably infinite number of times. Intuitively, it seems that you should expect at least one Even. But classically, Even and Odd are both saturated non-measurable, so it is not obvious whether or how you could prove anything.
ReplyDeleteNice. Or avoiding the detour through infinite lotteries, say two countable sequences of heads/tails are equivalent iff they differ in an even number of places. Let E contain a member of each equivalence class. You should expect to hit E at least once given infinite repeats...
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ReplyDeleteNope, my construction doesn't work. Instead:
ReplyDeletehttp://alexander-pruss-lx.baylor.edu/alex/blog/footnotes/46-49-10-23-4-114-5-142-1-2.html
(For context: http://alexanderpruss.blogspot.com/2014/05/thomson-lamp-and-axiom-of-choice.html )
In the construction given in Footnote 2, ON is non-measurable in the standard product measure. But note that switching the first coin outcome from Heads to Tails always switches ON and not-ON, whatever the other outcomes. So if you take the first coin as fair and causally, not just mathematically, independent of the others, it seems reasonable to take the probability of ON as 1/2.
ReplyDeleteOf course, this reasoning goes beyond classical probability theory (as it must). In effect, it invokes (finite) conglomerability over two classically non-measurable events.
In the construction I quoted, the lottery outcome does not depend in such a simple way on any single coin flip, so this sort of reasoning cannot be applied.
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ReplyDeleteYeah. I wonder if finite conglomerability over non-measurable events works.
ReplyDeleteThere may be limits to this sort of reasoning, though: https://mathoverflow.net/questions/280883/extending-the-product-measure-on-2-omega/280982
ReplyDeleteThat’s interesting. It shows, if I understand correctly, that you cannot make all ‘switchable’ events measurable. But the reasoning I gave only has to make one such event measurable. This is a straightforward extension. But yes, it is an interesting question how far you can take finite conglomerability over non-measurable events.
ReplyDeleteGrant that the coin flips are fair, objectively random, and causally independent. Then, for what it’s worth, I would be prepared to bet on the Footnote 2 version based on a probability of 1/2. But the version I gave feels ‘non-probabilistically’ random.
If you accept causal independence and unrestricted finite conglomerability, you have to give up countable additivity (and with it, countable conglomerability).
ReplyDeleteTo see this, construct an infinite lottery as follows. Define countably infinite sequences of Heads and Tails as equivalent if they differ in at most a finite number of places. Use Choice to select a reference sequence for each equivalence class. Run a countably infinite sequence of coin flips. Code the differences of the resulting sequence from its reference sequence as 0 for ‘same’, 1 for ‘different’. Read the resulting sequence a binary integer, least significant digit first. [I think you have given an equivalent construction in an earlier post.]
Grant causal independence and unrestricted finite conglomerability. Then this lottery gives definite probability 1/2 to ‘odd’ and to ‘even’, definite probability 1/4 to ‘multiple of 4’, and in general definite probability 2^(-n) to ‘k mod 2^n’. So, in a framework of imprecise probability, each lottery outcome must have definite probability 0 (at least if you give it a probability at all, and if you reject infinitesimals.) This is incompatible with countable additivity.
Of course, opinions differ on countable additivity.