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March 27

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Function on events in a probability space satisfying certain conditions

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I would like to find a function on events in a probability space satisfying two conditions. The function is with respect to an event Q.

.

for any A and B. So shrinking a set increases f.

I had originally thought was an example, but alas I discovered that it doesn't satisfy condition 2.

The motivation is this: Think of Q as a "goal", and E as your "belief state". Obviously, you want to act so that be high. But it's possible that you will gain new evidence that reduces , but you want a utility function that seeks out that information anyway (and doesn't throw it away, because that would be irrational). Condition 2 ensures that gaining new evidence never harms you, even if it reduces P(Q|E). So maximize instead. --49.184.160.10 (talk) 10:19, 27 March 2018 (UTC)[reply]

  • That is not possible except if Q is realized on the whole probability space or nowhere at all (in such a trivial case, any function would do).
Denote by the whole probability space. Condition 1 with yields (except in the trivial cases) . Condition 2 with yields . That's a contradiction. TigraanClick here to contact me 14:42, 27 March 2018 (UTC)[reply]
Well, actually, this might not be exactly correct as written; the trivial cases are not Q is realized on the whole probability space or nowhere at all, but which is a bit different. So technically, events that happen almost surely or almost never are not covered by the demo, but it should be easy enough to carve a proper subset of E such that the demonstration holds. TigraanClick here to contact me 14:53, 27 March 2018 (UTC)[reply]
No, I don't think that's right. I think you have the inequality the wrong way around in condition 2. Shrinking a set increases f. So condition 2 implies . --49.184.160.10 (talk) 19:48, 27 March 2018 (UTC)[reply]
Huh, yes, sorry. I tried to "simplify" the proof I had written on paper (which used three sets), and made it wrong. The correct sets to consider are and .
Compared to the incorrect proof above, the inequality of condition 2 is kept in the "same" direction because we still shrink the set ( but condition 1 is in "reverse" because we lose on event probability (rather than gaining) (. (The same nitpick about events of zero or one probability applies.) TigraanClick here to contact me 12:51, 28 March 2018 (UTC)[reply]
Thanks for your help.--49.184.160.10 (talk) 00:08, 29 March 2018 (UTC)[reply]