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Let two collections of random variables $\{X_i\}$ and $\{Y_i\}$ be independent and let $\{Y_i\}$ be i.i.d. Then $$\max_{1\leq i\leq n}(X_i+Y_1)\preceq \max_{1\leq i\leq n}(X_i+Y_i).$$ where $\preceq$ is stochastic domination.

I think it needs to find a coupling such that $X_i'=\max_{1\leq i\leq n}(X_i+Y_1)$ and $Y_i'=\max_{1\leq i\leq n}(X_i+Y_i)$ with $\mathbb{P}(X_i'<Y_i')$...

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  • $\begingroup$ What is your question? $\endgroup$
    – Iosif Pinelis
    Commented Jun 5, 2020 at 17:24
  • $\begingroup$ @IosifPinelis How to prove this statement? $\endgroup$
    – Hermi
    Commented Jun 5, 2020 at 17:45
  • $\begingroup$ @DieterKadelka Your $Y_i$ are not i.i.d. $\endgroup$
    – Yuval Peres
    Commented Jun 5, 2020 at 19:41

1 Answer 1

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Conditioning on the $X_i$'s and using the independence of the $Y_i$'s from the $X_i$'s, we reduce the consideration to the case when $X_i=x_i$ for some real $x_i$'s and all $i$. So, the statement of interest reduces to this: $$P(x_i+Y_1\le y\quad \forall i)\ge P(x_i+Y_i\le y\quad \forall i)$$ for all real $y$, which can be rewritten as $$F(y-\max_i x_i)\ge\prod_i F(y-x_i)$$ where $F$ is the cdf of each $Y_i$. But the latter inequality obviously holds, since $\max_i x_i=x_j$ for some $j$ whereas $0\le F\le1$.

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  • $\begingroup$ @BobO. : Everything seems correct as written in the answer. The expression $P(\max_i x_i+Y_1\le y\quad \forall i)$ in your comment does not seem to make much sense. However, $P(x_i+Y_1\le y\quad \forall i)=P(\max_ix_i+Y_1\le y)=P(Y_1\le y-\max_ix_i)=F(y-\max_ix_i)$. $\endgroup$
    – Iosif Pinelis
    Commented Jun 8, 2020 at 13:48
  • $\begingroup$ Thanks so much. $\endgroup$
    – Hermi
    Commented Jun 8, 2020 at 18:08

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