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Suppose we have two countable collections of random variables $\{X_n \vert n \in \mathbb{N} \} $ and $\{Y_n \vert n \in \mathbb{N} \} $ such that $ \forall n \in \mathbb{N}:$ $X_n$ and $Y_n $ have the same distribution, i.e. $\mathbb{P}\{X_n \in A\} = \mathbb{P}\{Y_n \in A\} $ for each Borel subset $A$ of $\mathbb{R}$.

Is it true that $ \sup_{n \in \mathbb{N}}X_n $ and $\sup_{n \in \mathbb{N}}Y_n$ have the same distribution?

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  • $\begingroup$ Did you consider accepting the (full) answer you received below? $\endgroup$ – Did Mar 2 '18 at 16:57
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If the sequences $\left(X_n\right)$ and $\left(Y_n\right)$ have the same distribution (equivalently, that for each $N$, $(X_1,\dots,X_N)$ and $(Y_1,\dots,Y_N)$ have the same distribution), then it is true. Indeed, for any fixed $t$, $$\mathbb P\left\{\sup_{n\in\mathbb N}X_n\leqslant t\right\}= \mathbb P\left(\bigcap_{n\in\mathbb N}\left\{X_n\leqslant t\right\}\right)=\lim_{N\to +\infty}\mathbb P\left(\bigcap_{n=0}^N\left\{X_n\leqslant t\right\}\right)$$ and since for each $N$ and each $t$, $\mathbb P\left(\bigcap_{n=0}^N\left\{X_n\leqslant t\right\}\right)=\mathbb P\left(\bigcap_{n=0}^N\left\{Y_n\leqslant t\right\}\right)$, we can reach the wanted conclusion.

However, if we only assume that for any fixed $n$, the random variables $X_n$ and $Y_n$ have the same distribution, then it is not true. Indeed, let $Z$ be a random variable taking the values $1$ and $-1$ with respective probability $1/2$. Let $X_1=Z$, $X_2=-Z$, $Y_1=Y_2=Z$ and for $n\geqslant 2$, let $X_n=Y_n=-1$. Then $\sup_nX_n=1$ and $\sup_nY_n$ takes the values $-1$ or $1$ with probability $1/2$.

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