tag:blogger.com,1999:blog-3891434218564545511.post5078652150385454998..comments2018-02-24T10:46:34.919-06:00Comments on Alexander Pruss's Blog: Two kinds of non-measurable eventsAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-3891434218564545511.post-70626934165091214522017-10-05T04:06:37.306-05:002017-10-05T04:06:37.306-05:00If you accept causal independence and unrestricted...If you accept causal independence and unrestricted finite conglomerability, you have to give up countable additivity (and with it, countable conglomerability).<br /><br />To 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.]<br /><br />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.<br /><br />Of course, opinions differ on countable additivity.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-55695049828550090422017-09-13T05:32:44.531-05:002017-09-13T05:32:44.531-05:00That’s interesting. It shows, if I understand corr...That’s interesting. It shows, if I understand correctly, that you cannot make <i>all</i> ‘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.<br /><br />Grant 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.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-49597503638529209502017-09-12T12:02:59.009-05:002017-09-12T12:02:59.009-05:00There may be limits to this sort of reasoning, tho...There may be limits to this sort of reasoning, though: https://mathoverflow.net/questions/280883/extending-the-product-measure-on-2-omega/280982Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-9476526168796454982017-09-12T07:58:59.751-05:002017-09-12T07:58:59.751-05:00Yeah. I wonder if finite conglomerability over non...Yeah. I wonder if finite conglomerability over non-measurable events works.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-59640527101341543132017-09-12T07:56:46.496-05:002017-09-12T07:56:46.496-05:00This comment has been removed by the author.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-51763632000712069962017-09-12T04:51:57.959-05:002017-09-12T04:51:57.959-05:00In the construction given in Footnote 2, ON is non...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.<br /><br />Of course, this reasoning goes beyond classical probability theory (as it must). In effect, it invokes (finite) conglomerability over two classically non-measurable events.<br /><br />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.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-25959184988575714592017-09-11T11:20:49.926-05:002017-09-11T11:20:49.926-05:00Nope, my construction doesn't work. Instead:
h...Nope, my construction doesn't work. Instead:<br />http://alexander-pruss-lx.baylor.edu/alex/blog/footnotes/46-49-10-23-4-114-5-142-1-2.html<br /><br />(For context: http://alexanderpruss.blogspot.com/2014/05/thomson-lamp-and-axiom-of-choice.html )Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-18516864599590846852017-09-11T11:19:35.983-05:002017-09-11T11:19:35.983-05:00This comment has been removed by the author.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-62630942692814315682017-09-11T09:11:15.908-05:002017-09-11T09:11:15.908-05:00Nice. Or avoiding the detour through infinite lott...Nice. 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...Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-32231614654122363122017-09-11T03:28:26.117-05:002017-09-11T03:28:26.117-05:00Here is a similar case. Take a permutation symmetr...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.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.com