tag:blogger.com,1999:blog-3891434218564545511.post7883785794345737873..comments2021-05-13T00:37:21.978-05:00Comments on Alexander Pruss's Blog: Complete Probabilistic CharacterizationsAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-3891434218564545511.post-33764466019041173952020-08-24T12:05:23.480-05:002020-08-24T12:05:23.480-05:00These are interesting suggestions. It's partic...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. Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-87917937290639221692020-08-24T00:52:21.985-05:002020-08-24T00:52:21.985-05:00More on the lottery example (Again, this may all b...More on the lottery example (Again, this may all be way off-beam.):<br /><br />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.<br /><br />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.<br /><br />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.<br /><br />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.<br /><br />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.)IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-25715662427957737372020-08-23T06:52:25.151-05:002020-08-23T06:52:25.151-05:00Warning: the following may be way off-beam.
Assum...Warning: the following may be way off-beam.<br /><br />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.<br /> <br />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.<br /><br />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.<br /><br />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.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.com