Sure. So, just to be clear, the situation is: We have real-valued random variable X depending on a single real-valued parameter t. And I claim it is possible (indeed, usual) that the choice of t that maximizes E(log X) is not the same as the choice of t that maximizes E(X).
My X will have two possible values for any given t, both with probability 1⁄2. They are t exp t and exp −2t.
E(log X) = 1⁄2 (log(t exp t) + log(exp −2t)) = 1⁄2 (log t + t − 2t) = 1⁄2 (log t—t). This is maximized at t=1. (It’s also undefined for t<=0; I’ll fix that in a moment.)
E(X) is obviously monotone increasing for large positive t, so it’s “maximized at t=+oo”. (It doesn’t have an actual maximum; I’ll fix that in a moment.)
OK, now let me fix those two parenthetical quibbles. I said X depends on t, but actually it turns out that t = 100.5 + 100 sin u, where u is an angle (i.e., varies mod 2pi). Now E(X) is maximized when sin u = 1, so for u = pi2; and E(log X) is maximized when 100 sin u = −99.5, so for two values of u close to -pi/2. (Two local maxima, with equal values of E(log X).)
Okay, I accept that I’m wrong and you’re right. Now the interesting part is that my mathematical intuition is not that great, but this is a pretty big fail even for it. So in between googling for crow recipes, I think I need to poke around my own mind and figure out which wrong turn did it happily take… I suspect I got confused about the expectation operator, but to confirm I’ll need to drag my math intuition into the interrogation room and start asking it pointed questions.
Sure. So, just to be clear, the situation is: We have real-valued random variable X depending on a single real-valued parameter t. And I claim it is possible (indeed, usual) that the choice of t that maximizes E(log X) is not the same as the choice of t that maximizes E(X).
My X will have two possible values for any given t, both with probability 1⁄2. They are t exp t and exp −2t.
E(log X) = 1⁄2 (log(t exp t) + log(exp −2t)) = 1⁄2 (log t + t − 2t) = 1⁄2 (log t—t). This is maximized at t=1. (It’s also undefined for t<=0; I’ll fix that in a moment.)
E(X) is obviously monotone increasing for large positive t, so it’s “maximized at t=+oo”. (It doesn’t have an actual maximum; I’ll fix that in a moment.)
OK, now let me fix those two parenthetical quibbles. I said X depends on t, but actually it turns out that t = 100.5 + 100 sin u, where u is an angle (i.e., varies mod 2pi). Now E(X) is maximized when sin u = 1, so for u = pi2; and E(log X) is maximized when 100 sin u = −99.5, so for two values of u close to -pi/2. (Two local maxima, with equal values of E(log X).)
Okay, I accept that I’m wrong and you’re right. Now the interesting part is that my mathematical intuition is not that great, but this is a pretty big fail even for it. So in between googling for crow recipes, I think I need to poke around my own mind and figure out which wrong turn did it happily take… I suspect I got confused about the expectation operator, but to confirm I’ll need to drag my math intuition into the interrogation room and start asking it pointed questions.
Upvoted for public admission of error :-).
(In the unlikely event that I can help with the brain-fixing, e.g. by supplying more counterexamples to things, let me know.)