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Does this expression hold? $$\max_{x\in X}E[Y\mid X=x] = \max_{P_X}\sum_{x\in X} P_X(x)E[Y\mid X=x]$$ 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 $E[Y\mid X=x]$ 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

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    $\begingroup$ This is a bizarre usage of notation, because you're treating $X$ as both a random variable and a state space. $\endgroup$ – Alex R. Oct 25 '17 at 17:23
  • $\begingroup$ say $x \in {0,1}$ is random variable, for x=0 you have a value for $E{[Y|x=0]}$ and for x=1 you have a value for $E{[Y|x=1]}$. now say you can choose the probability $P_x(0)$ and $P_x(1)$. $\endgroup$ – user3921 Oct 25 '17 at 17:24
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Let me call the set in which $X$ takes values $\mathcal X$ and define $h(x) = E(Y\mid X = x)$ for cleaner notation [I take it as given that such a $h$ exists].

Your question statement seems to imply that $h$ attains its supremum on $\mathcal X$: $h(x^\star) = \sup_{x\in \mathcal X}h(x)$ for some $x^\star \in \mathcal X$.

Now for any variable $X$ with distribution $P_X$ on $\mathcal X$:

$$ E(h(X)) = \int_{\mathcal X} h(x)P_X(dx) \leq \int_{\mathcal X}h(x^\star)P_X(dx) = h(x^\star).$$

But if $X$ takes the value $x^\star$ with probability one, then the inequality is an equality.

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If state space is discrete (as your notation implies) the $\max$ exists (e.g. because the state space is finite) then the expression holds almost trivially- your argument says the RHS attains the LHS. Moreover, it cannot exceed the left because it is less than $$ \max_{x\in\mathcal{S}} \mathbb{E}[Y | X = x] \max_{p \in \mathcal{P}(\mathcal{S}) }\sum_{x\in\mathcal{S}} p(x), $$ which is the LHS. (As @Alex R. said your original notation is bizarre and I took the liberty of denoting by $\mathcal{P}(\mathcal{S})$ a suitable space of probability measures on the state space $\mathcal{S}$.

If you are actually talking about $\sup$ (still assuming the space is discrete) then you can take a sequence of $x^\star_n$ such that $$ \mathbb{E}[Y | X = x^\star_n] \to \sup_{x\in\mathcal{S}} \mathbb{E}[Y | X = x] $$ and the argument still works.

Moreover, if your space is continuous and your space $\mathcal{P}{(\mathcal{S})}$ includes discrete measures, again the same argument works but you need integrals instead of the sums etc. If you only want continuous distributions then you need to work a little harder to approximate $\delta$-measures.

So basically, your inequality works with $\max$ replaced by $\sup$.

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