I got two sequences of random variables $(X_n)_{n \in \mathbb N}$ and $(Y_n)_{n \in \mathbb N}$ and know that
1) $X_n \to 0, \, n \to \infty,$ almost surely
2) For all $n \in \mathbb N$ the sequences are equal in distribution: $X_n \stackrel{d}{=} Y_n$
Now I'm wondering, does this imply that also $Y_n \to 0, n \to \infty,$ almost surely?
Best Answer
I don't have a simple counterexample at the moment but you can argue using Skorohod's Theorem: let $(Y_n)$ tend to $0$ is probability but not almost surely. Since $Y_n \to 0$ in distribution Skorohod's Theorem tells us that there exits random variable $X_1,X_2,...$ such that $X_n \to 0$ almost surely and $X_n$ has same distribution as $Y_n$ for each $n$.
Reference: https://eventuallyalmosteverywhere.wordpress.com/2014/10/13/skorohod-representation-theorem/
A better example: consider $[0,1)$ with Lebesgue measure. Arrange the intervals $[\frac {i-1} {2^{n}},\frac i {2^{n}})$, $1 \leq i \leq 2^{n}$, $n \geq 1$ in as sequence using the 'natural' ordering. Let $Y_1,Y_2,..$ be the indicator functions of these intervals. Now form $X_1,X_2,...$ by replacing $[\frac {i-1} {2^{n}},\frac i {2^{n}})$ by $[0,\frac 1 {2^{n}})$ for each $i$. Then $X_n \to 0$ almost surely, $Y_n$ does not tend to $0$ almost surely and $X_n$ has the same distribution as $Y_n$ for each $n$.