I know we don't have an equality $\sup_{y \in \mathbb{R}} \mathbb{E} f(X,y) = \mathbb{E} \sup_{y \in \mathbb{R}}f(X,y)$ in general. But how about a simpler case, namely, is equality
$$\sup_{y \in \mathbb{R}} \mathbb{E} \sum_{k=1}^nX_kf(ky) = \mathbb{E}\sup_{y \in \mathbb{R}} \sum_{k=1}^nX_kf(ky)$$
true, where $f$ is some nice, bounded and continuous function, and $X_k$ are Bernoulli r.v.?