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Let $X_n$ converge in distribution to $X$ and let $Y_n$ converge in probability to a constant $c$. Show that $X_nY_n\overset{\mathcal D}\rightarrow Xc$ and $\frac{X_n}{Y_n}\overset{\mathcal D}\rightarrow\frac{X}{c}$ (where, $\overset{\mathcal D}\rightarrow$ means convergence in distribution)

For a contiuous, bounded and Lipschitz function $f$ with $k$ as Lipschitz constant and $\alpha:=\sup|f(x)|$, we have to show

$|\mathbf E(f(X_nY_n)-f(X_nc))|\to 0$, but

$|\mathbf E(f(X_nY_n)-f(X_nc))|\le\mathbf E(|f(X_nY_n)-f(X_nc)|\mathbf1_{|X_nY_n-X_nc|\le\epsilon})+\mathbf E(|f(X_nY_n)-f(X_nc)|\mathbf1_{|X_nY_n-X_nc|>\epsilon})\le k\epsilon+2\alpha\Pr(|X_nY_n-X_nc|>\epsilon)\le k\epsilon+\Pr(|Y_n-c|>\frac{\epsilon}{|X_n|})$

If $|X_n|\neq0$, otherwise it is automatically satisfied.

Can you verify my steps ?

Thanks.

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I think this is Slutsky's theorem. Proof:

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Sources:

https://en.wikipedia.org/wiki/Convergence_of_random_variables

https://en.wikipedia.org/wiki/Proofs_of_convergence_of_random_variables

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