# Almost sure convergence of sequences that are equal in distribution

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?

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 $$n$$.
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 $$n$$.
• Thanks. So if I get this right, this means the statement does not hold on the original probability space that $(X_n)_n$ and $(Y_n)_n$ are defined on, but using Skorohods theorem I can define those sequences on some other space in which the almost surve convergence of $(Y_n)$ holds? – Bazzan Jul 19 at 10:14