Wikipedia:Reference desk/Archives/Mathematics/2018 March 27

= March 27 =

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

$$P(Q|A) > P(Q|B) \implies f_Q(A) > f_Q(B)$$.

$$f_Q(A \cap B) \ge f_Q(A)$$ for any A and B. So shrinking a set increases f.

I had originally thought $$f_Q(E) = P(Q|E)^{P(E)}$$ 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 $$P(Q|E)$$ be high. But it's possible that you will gain new evidence that reduces $$P(Q|E)$$, 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 $$f_Q$$ instead. --49.184.160.10 (talk) 10:19, 27 March 2018 (UTC)
 * 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 $$E$$ the whole probability space. Condition 1 with $$A=Q, B=E$$ yields (except in the trivial cases) $$f_Q(Q)>f_Q(E)$$. Condition 2 with $$A=E, B=Q (A \cap B = Q)$$ yields $$f_Q(Q)\leq f_Q(E)$$. That's a contradiction. Tigraan Click here to contact me 14:42, 27 March 2018 (UTC)
 * 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 $$P(Q|E) \in \{0,1\}$$ 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. Tigraan Click here to contact me 14:53, 27 March 2018 (UTC)
 * 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 $$f_Q(Q) \ge f_Q(\Omega)$$. --49.184.160.10 (talk) 19:48, 27 March 2018 (UTC)
 * 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 $$\Omega$$ and $$\bar{Q}=\Omega - Q$$.
 * Compared to the incorrect proof above, the inequality of condition 2 is kept in the "same" direction because we still shrink the set ($$f_Q(\Omega)\le f_Q(\bar{Q})$$ but condition 1 is in "reverse" because we lose on event probability (rather than gaining) ($$f_Q(\Omega)> f_Q(\bar{Q})$$. (The same nitpick about events of zero or one probability applies.) Tigraan Click here to contact me 12:51, 28 March 2018 (UTC)
 * Thanks for your help.--49.184.160.10 (talk) 00:08, 29 March 2018 (UTC)