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Suppose $\{X_n\}$ converges in distribution to $X$ and $\{Y_n\}$ converges in probability to $0$. How would I show that $\{X_n + Y_n\}$ converges in distribution to $X$?

Thanks for the help.

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How would I use Slutsky's theorem here? –  MathMan Oct 18 '11 at 2:01

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up vote 3 down vote accepted

You can do it like this: you have to demonstrate that for every continuous function with compact support $f\colon \mathbb{R}\to \mathbb{R}$ you have $\mathbb{E}[f(X_n+Y_n)]\to \mathbb{E}[f(X)]$.

Now, $\mathbb{E}[f(X_n+Y_n)]=\mathbb{E}[f(X_n+Y_n)-f(X_n)]+\mathbb{E}[f(X_n)]$ and since $X_n\to X$ in distribution, it is enough to show that $\mathbb{E}[f(X_n+Y_n)-f(X_n)]\to 0$. For this, take $\delta >0$. Then you have $$ \begin{multline*} \mathbb{E}[|f(X_n+Y_n)-f(X_n)|] =\\= \mathbb{E}[|f(X_n+Y_n)-f(X_n)|I_{\{|Y_n|>\delta\}}] + \mathbb{E}[|f(X_n+Y_n)-f(X_n)|I_{\{|Y_n|\leq \delta\}}] \end{multline*} $$

But $$ \mathbb{E}[|f(X_n+Y_n)-f(X_n)|I_{\{|Y_n|>\delta\}}] < 2M \mathbb{P}[|Y_n|>\delta]\to 0, $$ since $Y_n\to 0$ in probability ($M=\sup|f|$). And $$ \mathbb{E}[|f(X_n+Y_n)-f(X_n)|I_{\{|Y_n|\leq \delta\}}] < \epsilon $$ for every $\epsilon >0$ if $\delta$ is small enough, since $f$ is absolutely continuous.

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