If $X_t = Y_t$ in distribution, for any $t \in T$ (compact), is it true that $\mathbb E \sup_{t \in T} X_t = \mathbb E\sup_{t \in T} Y_t$?

Let $T$ be a compact topological space and for any $t \in T$, let $X_t$ and $Y_t$ be random variables which have the same distribution. We may assume that $X_t$ and $Y_t$ depend on $t$ in continuous / smooth manner.

Question. Under what minimal additional conditions can we conclude that $\mathbb E \sup_{t \in T} X_t = \mathbb E \sup_{t \in T} Y_t$ ?

My rough guess is that the answer to the above question is No!

The scenario I have in mind is the following. Let $T$ be the unit-ball in $\mathbb R^n$. Let $V$ be uniformly distributed over the grassmannian of $k$-dimensional subspaces of $\mathbb R^n$, and let $(v_j)$ be an orthonomal basis for $V$. Let $P_V:\mathbb R^n \to V$ be the orthogonal projector for $V$. Finally, let $P_k$ be the random $k \times n$ matrix by obtained selecting uniformly selecting (without replacement) $n-k+1$ rows of the $n \times n$ identity matrix and deleting them. Let $(u_j)_j$ be an orthonormal basis for the row space of $P_k$. Fix a nonnegative scalars $\lambda_1,\ldots,\lambda_k$.

Now, by rotation-invariance, it is clear that $X_t := \sum_{j=1}^k\lambda_j \langle v_j,P_Vt\rangle^2$ and $Y_t := \sum_{j=1}^k\lambda_j \langle u_j,P_k t\rangle^2$ are equal in distribution, for any $t \in T$. The question then is

Question 2. Is $\mathbb E \sup_{t \in T}X_t = \mathbb E\sup_{t \in T} Y_t$ ?

Solution 1:

Let $T= {1, 2}$ and $X_t, Y_t$ be symmetric Bernoulli random variables. Let $X_1$ and $X_2$ satisfy $P(X_1X_2=1)=1$, and $Y_1$ and $Y_2$ satisfy $P(Y_1Y_2=-1)=1$.

Then we have $P(\max_{t \in T}X_t=1)=P(\max_{t \in T}X_t=-1)=1/2$ and $P(\max_{t\in T}Y_t = 1)=1$. So, this gives a counterexample to your question.