$X_n, Y_n$ are random variables on the same probability space and it's known that $X_n \rightarrow X$ and $Y_n \rightarrow 0 =: Y$ in distribution. Proof that $X_n + Y_n \rightarrow X$ in distribution.
I have a problem because of the case, when $x\ge 0$ (then $F_Y$ = 1).
I need to proof that $|F_{X_n + Y_n} – F_Y| < \varepsilon$ for large n, I've tried to estimate
$F_{X_n + Y_n}(x) = P(X_n + Y_n \le x) \le P(X_n \le x) + P(Y_n \le 0)$, but it doesn't work, because $P(Y_n \le 0) = 1$
Best Answer
Hint: $P(X_n+Y_n \leq z) \leq P(X_n+Y_n \leq z, |Y_n| <\epsilon)+P(|Y_n|\geq \epsilon)$. Also $P(X_n+Y_n \leq z, |Y_n| <\epsilon) \leq P(X_n \leq z+\epsilon)$. Can you show using these that $\lim \sup P(X_n+Y_n \leq z) \leq P(X \leq z)$? A similar argument gives $\lim \inf P(X_n+Y_n \leq z) \geq P(X \leq z)$ provided $(X=z)=0$.