supremum of expectation = expectation of supremum? In supremum of expectation $\le$ expectation of supremum?, can we have the reverse inequality up to a constant? Like $$ \underset{y\in \mathcal Y} \sup \mathbb E\big[f(X,y)\big] \ge C\mathbb E\big[\underset{y\in \mathcal Y} \sup f(X,y)\big] $$ for some $C>0?$
Or, having some conditions on $f$, can we have $$ \underset{y\in \mathcal Y} \sup \mathbb E\big[f(X,y)\big] = \mathbb E\big[\underset{y\in \mathcal Y} \sup f(X,y)\big]? $$ Let us say $f(X,y)=X^Ty.$ Do we have an equality for this?
Supremum of expectation equals expectation of supremum?: this post seems to be addressing this, but not sure if it's correct!
 A: For every concrete $y$, we have $f(X, y) \le \sup_{y \in \mathcal{Y}} f(X, y)$ as an inequality of random variables, so monotonicity of expectation shows that
$$\sup_{y \in \mathcal{Y}} \mathbb{E}[f(X, y)] \le \mathbb{E}[\sup_{y \in \mathcal{Y}} f(X, y)]$$
Now, consider $X$ to be uniformly distributed on the unit circle in $\mathbb{R}^2$, $y \in \mathcal{Y}$ also running over the unit circle, and $f(X, y) = X^T y$. Then clearly $\mathbb{E}[f(X, y)] = 0$ for every $y$, while  the random variable $\sup_y f(X, y)$ is constantly $1$; consequently, $\mathbb{E}[\sup_{y \in \mathcal{Y}} f(X, y)] = 1$. This is a mild situation in which no statement such as the ones you are proposing can be made.
(One should also keep in mind the justification for $\sup_{y \in Y} f(X, y)$ to be a random variable at all. Everything works out when, for example, $\mathcal{Y}$ can be restricted to a countable subset without affecting the supremum---which happens quite frequently, e.g. when $\mathcal{Y}$ is a separable Hilbert space and $F$ is a suitable bilinear form---because in this case $\sup_{y \in Y} f(X, y)$ equals a supremum of countably many random variables.)
