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Let $X_1,\dots,X_n$ and $Y_1,\dots,Y_n$ be random variables for which $X_i \geq Y_i$ a.s. and the $(Y_i)$ are identically distributed. Is it true that \begin{align*} E(\max X_i - \max Y_j) \geq E(X_i-Y_i) \end{align*} for any $i$?

I tried reducing the problem to a question about the tail probabilities using \begin{align*} E(\max X_i - \max Y_j) = \int_0^\infty P(\max X_i - \max Y_i > t) dt, \end{align*} since $\max X_i - \max Y_j \geq 0$, but this doesn't appear to directly give the inequality. Can anyone provide a proof or counterexample?

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  • $\begingroup$ The tail integral only holds if all the random variables are a.s. positive. $\endgroup$ Nov 30, 2019 at 22:00
  • $\begingroup$ $X_i \geq Y_i$ for all $i$ should imply that $\max X_i - \max Y_j \geq 0$, no? $\endgroup$ Nov 30, 2019 at 22:01
  • $\begingroup$ Yes, it does. ${}{}{}$ $\endgroup$
    – Clement C.
    Nov 30, 2019 at 22:06

1 Answer 1

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Note that $X_i - Y_i\geq 0$ for all $i$. You do have $$ \max_{1\leq i\leq n}\mathbb{E}[X_i - Y_i] \leq \mathbb{E}[\max_{1\leq i\leq n}(X_i - Y_i)] \tag{1} $$ but you don't have, in general, $$ \max_{1\leq i\leq n}\mathbb{E}[X_i - Y_i] \leq \mathbb{E}[\max_{1\leq i\leq n}X_i - \max_{1\leq i\leq n}Y_i]\tag{2} $$ To see why (2) can fail to hold, consider $Y_1,\dots,Y_n$ i.i.d. Normal $N(0,1)$ r.v.'s, and $X_i=\max_{1\leq j\leq n}Y_j$ for all $i$. Then the RHS of (2) is $0$, but the LHS is $$ \max_{1\leq i\leq n}\mathbb{E}[\max_{1\leq j\leq n} Y_j - Y_i]= \max_{1\leq i\leq n}(\mathbb{E}[\max_{1\leq j\leq n} Y_j] - \mathbb{E}[Y_i]) = \max_{1\leq i\leq n}(\mathbb{E}[\max_{1\leq j\leq n} Y_j] - 0) = \mathbb{E}[\max_{1\leq j\leq n} Y_j] \operatorname*{\sim}_{n\to\infty} \sqrt{2\log n} $$ which is strictly positive for any not-too-small $n$.

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