# 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/… – Jimmy R. 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. – Jose Antonio 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