Does $max_{x\in X}E[{Y|X=x}] = max_{P_X}\sum_{x\in X}P_X(x)E[{Y|X=x}]$ [closed]

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, AretinoNov 11 '17 at 13:17

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• Can you show that LHS$\leqslant$RHS or that LHS$\geqslant$RHS? – Did Oct 25 '17 at 18:55
• 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 $(X,Y)$? – kimchi lover Oct 26 '17 at 0:29
• 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. – Dilip Sarwate Oct 26 '17 at 3:19