# Almost Sure Convergence in a Case of Convergence in Distribution

Suppose $$X_n - Y_n \xrightarrow{a.s.} 0$$ and $$X_n \xrightarrow{d} X$$ for some random variable $$X$$ and sequences of random variables $$X_n, Y_n$$.

I want to show that $$Y_n \xrightarrow{d} X$$, but I don't really know how to do this formally.

Obviously I have that $$E(f(X_n)) \rightarrow E(f(X))$$ for all bounded continuous $$f$$, so I can write $$E(f(X_n)) = E[f(X_n)1(\text{lim} X_n = \text{lim} Y_n)]$$, but as almost sure convergence is pointwise the $$n$$ varies with $$\omega \in \Omega$$ and I don't know how to get these things sorted.

Any ideas?

• If you have Slutsky's theorem, then $$Y_n=(Y_n-X_n)+X_n \xrightarrow{d} 0+X=X$$ since the first summand converges to $0$ in probability and the second one converges weakly.
– NCh
Feb 20 '20 at 12:04

This is standard and the statement holds true under the weaker assumption $$X_n - Y_n \xrightarrow{P} 0$$.
Consider $$f$$ a $$K$$-Lipschitz function bounded by some $$M$$ and write $$|E[f(Y_n)]-E[f(X)]|\leq E[|f(Y_n)-f(X_n)|] + |E[f(X_n)]-E[f(X)]|$$
• $$\lim_n|E[f(X_n)]-E[f(X)]| = 0$$ because $$X_n \xrightarrow{d} X$$.
• for any $$\epsilon >0$$, \begin{aligned}[t] E[|f(Y_n)-f(X_n)|] &= E[|f(Y_n)-f(X_n)|1_{|X_n-Y_n|\leq \epsilon}]+E[|f(Y_n)-f(X_n)|1_{|X_n-Y_n|> \epsilon}] \\ &\leq K \epsilon P(|X_n-Y_n|\leq \epsilon) + 2MP(|X_n-Y_n|> \epsilon)\\ &\leq K \epsilon + 2MP(|X_n-Y_n|> \epsilon) \end{aligned} Hence $$\limsup_n E[|f(Y_n)-f(X_n)|] \leq K \epsilon$$ for any $$\epsilon>0$$, hence $$\lim_n E[|f(Y_n)-f(X_n)|] = 0$$