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March 27[edit]

Function on events in a probability space satisfying certain conditions[edit]

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.

$P(Q|A)>P(Q|B)\implies f_{Q}(A)>f_{Q}(B)$ .

$f_{Q}(A\cap B)\geq f_{Q}(A)$ 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)[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 $E$ 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)[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

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)[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

f_{Q}(Q)\geq f_{Q}(\Omega )

. --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

\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 )\leq 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)[reply]

Thanks for your help.--49.184.160.10 (talk) 00:08, 29 March 2018 (UTC)[reply]