Skip to main content
13 events
when toggle format what by license comment
Sep 25, 2012 at 1:41 vote accept martin
Sep 24, 2012 at 21:15 answer added Dan timeline score: 14
Sep 24, 2012 at 9:41 comment added pgassiat If the supremum on the lhs is attained at $t^*$, then the inequality is strict unless almost surely $G(t^*,X) = \sup_t G(t,X)$. So the obvious condition that the maximum is always attained at the same $t$ is also almost necessary.
Sep 24, 2012 at 2:21 comment added martin @Yemon Choi: I would like the result to hold for any distribution of $X$, but if you want to impose some conditions on $X$, that's fine. I'm still clueless on how to approach the problem.
Sep 23, 2012 at 18:17 comment added Yemon Choi Is there anything more you're prepared to specify about $X$?
Sep 23, 2012 at 16:00 comment added martin Yes, $I$ and $X$ are fixed. I've also added an assumption that $I=[0,1]$.
Sep 23, 2012 at 14:17 history edited martin CC BY-SA 3.0
added 6 characters in body
Sep 23, 2012 at 14:11 history edited martin CC BY-SA 3.0
deleted 12 characters in body; added 8 characters in body
Sep 23, 2012 at 14:05 history edited martin CC BY-SA 3.0
deleted 12 characters in body
Sep 23, 2012 at 12:55 comment added Gerald Edgar I talked with a mathematician once who expressed frustration with some modern theoretical physics writing, exactly because of this point. They would use this without justification, or even without noticing. If the random variable $X$ is constant a.s. it would have been OK (in that setting, at least); and in thermodynamics it often turns out that they are constant; but some physicists would forge ahead, maximizing the r.v. by maximizing instead the expectation.
Sep 23, 2012 at 11:56 answer added Stanislav timeline score: 3
Sep 23, 2012 at 9:25 comment added Davide Giraudo Are $I$ and $X$ fixed?
Sep 23, 2012 at 6:47 history asked martin CC BY-SA 3.0