# Convergence in distribution, $X_n \xrightarrow{d} X$ and $|X_n-Y_n| \xrightarrow{P} 0$ implies $Y_n \xrightarrow{d} X$

I find this problem and I'd like to know if my answer is correct. Thank you

Let $(X, \mathscr{A}, P)$ a probability space. Suppose that $X$ is a r.v. and $\{ X_n \}$ is a sequence of r.v.'s such that $X_n \xrightarrow{d} X$ (convergence in distribution) and $\{Y_n\}$ a sequence of r.v.'s such that $|X_n-Y_n| \xrightarrow{P} 0$. Then $Y_n \xrightarrow{d} X$

Proof: Let $f: \mathbb R \to \mathbb R$ be bounded and uniformly continous function. Let $K$ a constant such that $|f|\le K$ and let $\epsilon>0$ given, so there is a $\delta>0$ such that for $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$. Thus

\begin{align*} |E f(X_n)-& Ef(Y_n)|\le E |f(X_n)- f(Y_n)|\\ &=\int_{\{|X_n- Y_n|<\delta\}} |f(X_n)- f(Y_n)| dP+E |f(X_n)- f(Y_n)|+\int_{\{|X_n- Y_n|\ge \delta\}} |f(X_n)- f(Y_n)| dP\\ &\le \epsilon P \{|X_n- Y_n|<\delta\} +2K P\{|X_n- Y_n|\ge\delta\}\\ &\le \epsilon + 2K P\{|X_n- Y_n|\ge\delta\} \end{align*}

Letting $n\to \infty$ we have $|E f(X_n)- Ef(Y_n)|\le \epsilon$ and since $\epsilon$ was arbitrary thus $\{Ef(y_n)\}$ converges at the same value that $\{Ef(X_n)\}$, that is, $\{Ef(y_n)\}\to Ef(X)$ and since this holds for all uniformly continuous and bounded function thus $Y_n \xrightarrow{d} X$

• Check it here: en.wikipedia.org/wiki/… Nov 30 '15 at 21:39
• Thank you @Stef I think the ideas are very similar. But I'm using uniformly continuous functions instead of the strongest condition of Lipschitz. But in the argument is very similar. Nov 30 '15 at 21:49

In fact, such problem may generalize to the case of random elements. More specifically, let $(\Omega,\mathscr{B},p)$ be a probability space. and $S$ be a metric space. A mapping $X: \Omega \to S$ is called random element of $S$ if it is measurable in the sense that $\{ \omega:X(\omega) \in A \} = X^{-}A \in \mathscr{B}$.
If $X_n \xrightarrow{d} X$ and the distance $\rho(X_n,Y_n)\xrightarrow{p} 0$ then we have $Y_n \xrightarrow{d} X$. The proof can be found in Billingsley's Covergence of Probability Measures (first edition) Page 25, Theorem 4.1