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Let $X_n, Y_n, X$ be real random variables such that $X_n \to X$ weakly and $\mathbb{P}_{Y_n} = N(0, 1/n)$ for all positive integers $n$.

I am trying to prove that $X_n + Y_n \to X$ weakly as well.

I have been trying to prove that $E[f(Y_n)] \to 0$ for all continuous bounded $f$, which would imply the statement. But i wasn't able to prove this, thought that this might be wrong approach.

I would be grateful for your hints or ideas. Thanks.

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1 Answer 1

up vote 2 down vote accepted

For every $n\geqslant1$, $f(Y_n)$ is distributed as $Z_n=f(Y_1/\sqrt{n})$ and $Z_n\to f(0)$ almost surely because $f$ is continuous at $0$. Since $f$ is bounded, the sequence $(Z_n)_{n\geqslant1}$ is dominated hence, by dominated convergence, $\mathbb E(f(Y_n))=\mathbb E(Z_n)\to f(0)$ when $n\to\infty$.

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