Convergence weakly of distributions

Question

Let $X_1,X_2,...$ and $Y_1,Y_2,...$ be two sequences of random variables such that $\mathcal{L}(X_n)\xrightarrow{w}\mathcal{L}(X)$ and $|Y_n-X_n|\xrightarrow{w}0$ as $n\rightarrow\infty$ Show that $$\mathcal{L}(Y_n)\xrightarrow{w}\mathcal{L}(X)$$ as $n\rightarrow\infty$.

I have the next attempt:

I know that convergence in probability implies convergence weakly or in distribution, then given $\epsilon>0$ I can say that:

$$|Y_n-X_n|<\epsilon$$

Also since $\mathcal{L}(X_n)\xrightarrow{w}\mathcal{L}(X)$, I can say that $Ef(X_n)\rightarrow Ef(X)$ for a bounded Lipschitz function using the Portamnteau theorem.

I don't know how to use these facts to conclude.

Answer 1

The Portmanteau theorem works here. Let $f$ be bounded and Lipschitz, so there exists $B>0$ such that $|f(x)|\le B$ for all $x$, and there exists $K>0$ such that $|f(y)-f(x)|\le K|y-x|$ for all $y$ and $x$.

To show that $Ef(Y_n)\to Ef(X)$, start with the triangle inequality: $$ |Ef(Y_n)-Ef(X)|\le E|f(Y_n)-f(X_n)|+|Ef(X_n)-Ef(X)|\tag1 $$ The second term on the RHS of (1) tends to zero since $X_n$ converges weakly to $X$. To show the first term tends to $0$, pick $\epsilon>0$ and break up the expectation $E|f(Y_n)-f(X_n)|$ according to the cases $|Y_n-X_n|>\epsilon$ and $|Y_n-X_n|\le\epsilon$. In the first case use the fact that $f$ is bounded to see $$ E\left[|f(Y_n)-f(X_n)|I(|Y_n-X_n|>\epsilon)\right]\le 2B\,P(|Y_n-X_n|>\epsilon),\tag2 $$ which tends to zero because $Y_n-X_n$ converges in probability to zero. In the second case use the fact that $f$ is Lipschitz: $$E\left[|f(Y_n)-f(X_n)|I(|Y_n-X_n|\le\epsilon)\right]\le E K|Y_n-X_n|I(|Y_n-X_n|\le\epsilon)\le K\epsilon.\tag3 $$ But $\epsilon$ is arbitrary positive, so the result follows.