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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – 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 $\lim_n E[|f(Y_n)-f(X_n)|] = 0$


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