# Convergence in probability to a sequence converging in distribution

Consider two sequences of real-valued random variables, $\{X_n\}_n$ and $\{Y_n\}_n$, and a real-valued random variable $Y$. Suppose that $X_n\overset{p}{\rightarrow}Y_n$ and $Y_n\overset{d}{\rightarrow}Y$ as $n\rightarrow \infty$. Does this imply $X_n\overset{d}{\rightarrow}Y$?

$\overset{p}{\rightarrow}$ means convergence in probability, $\overset{d}{\rightarrow}$ means convergence in distribution.

• Question edited, thanks – user299158 Dec 21 '15 at 20:02
• Sorry, I have edited again, I hope now is clear: $X_n\rightarrow_p Y_n$ and $Y_n\rightarrow_d Y$ – user299158 Dec 21 '15 at 20:07

Start with the identity $P(U\le a) \le P(V\le a+\epsilon)+P(|U-V|>\epsilon)$ for $\epsilon>0$. Let $a$ be a continuity point of $F_Y$. Then \begin{align} P(X_n\le a) & \le P(Y_n\le a+\epsilon)+P(|X_n-Y_n|>\epsilon) \\ & \le P(Y\le a+\epsilon+\epsilon')+P(|Y_n-Y|>\epsilon')+P(|X_n-Y_n|>\epsilon) \end{align} So $\lim\limits_{n\to\infty} F_{X_n}(a)\le F_Y(a+\epsilon+\epsilon')$, and since the $\epsilon,\epsilon'$ were arbitrary and $a$ is a continuity point, $\lim\limits_{n\to\infty} F_{X_n}(a)\le F_Y(a)$. The reverse inequality is proven analogously.