Does this expression hold? $$max_{x\in X}E[{Y|X=x}] = max_{P_X}\sum_{x\in X}P_X(x)E[{Y|X=x}]$$ Left side says: max over all available $x$ for the expected value of $Y$ given $X=x$.

Right side says: maximize over probabilities distribution of $X$

I can't seem to figure it out. I think that ifyou wan't to max over probablity distribution on the right hand side you would eventually maximize over $E[{Y|X=x}]$ and say you get that maximization is for $x=x'$ then you would simply give $P_X(x')=1$ and then the expressions are equal. Am I right? How can you prove this? Thanks


closed as off-topic by Dilip Sarwate, Claude Leibovici, José Carlos Santos, I am Back, Aretino Nov 11 '17 at 13:17

  • This question does not appear to be about math within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Can you show that LHS$\leqslant$RHS or that LHS$\geqslant$RHS? $\endgroup$ – Did Oct 25 '17 at 18:55
  • 1
    $\begingroup$ So you are assuming that conditional distribution of $Y$ given $X$ is fixed, but the marginal distribution of $X$ is variable, as is the joint distribution of $(X,Y)$? $\endgroup$ – kimchi lover Oct 26 '17 at 0:29
  • 5
    $\begingroup$ I'm voting to close this question as off-topic because it has also been asked on stats.SE where the OP has already accepted an answer. $\endgroup$ – Dilip Sarwate Oct 26 '17 at 3:19