
> 
> 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 $\mathbb{P}(X\_i'<Y\_i')$...



How to show $\max_{1\leq i\leq n}(X_i+Y_1)\preceq \max_{1\leq i\leq n}(X_i+Y_i)$?
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 $0\le F\le1$. 


