Is this proof of convergence is distribution of product of random variables correct?

Question

Given that $X_n \Rightarrow X$ and $Y_n \Rightarrow Y$ prove that $X_nY_n \Rightarrow XY$, if both $\{X_n, Y_n\}; \{X,Y\}$ are independent.

This is my attempt:

I aim to prove \begin{align} \Bbb E_{X_n Y_n}(e^{itX_n Y_n}) = \Bbb E_{XY}(e^{itXY}) \forall t\in \Bbb R \; \; \text{(Because by [Levy's continuity theorem][1] the statement will follow)} \\ \end{align} Now, Taking limit $n \to \infty$ on LHS and using

1. Tower Property of Expectation
2. Dominated Convergence Theorem
3. using Portmanteau lemma (ii) (as Characteristic function is uniformly continuous)
4. Independence

, we have $\lim_{n \to \infty} \Bbb E_{X_n Y_n}(e^{itX_n Y_n}) \overset{(1)} = \lim_{n \to \infty} \Bbb E_{Y_n} \Bbb E_{X_n|Y_n}(e^{itX_nY_n}) \\ \overset{(4)} = \lim_{n \to \infty} \Bbb E_{Y_n} \Bbb E_{X_n}(e^{itX_nY_n}) \overset{(2)} = \lim_{n \to \infty} \Bbb E_{Y_n} \Bbb \lim_{n \to \infty} E_{X_n}(e^{itX_nY_n}) \\ \overset{(3)} = \lim_{n \to \infty} \Bbb E_{Y_n} \Bbb E_{X}(e^{itXY_n}) \overset{(3)} = \Bbb E_{Y} \Bbb E_{X}(e^{itXY}) \overset{(1)} = \Bbb E_{X Y}(e^{itX Y}) $

Is my proof correct? Thank you!

Edited to reflect d.k.o's comments: From Wikipedia, it is stated that the product of two independent random variables $Z = XY$ has the following characteristic function:

\begin{align} \varphi_Z(t) & =\operatorname{E}(e^{itX Y}) \\ & = \operatorname{E}_Y ( \operatorname{E}_{X Y \mid Y} (e^{itX Y} \mid Y)) \\ & = \operatorname{E}_Y ( \operatorname{E}_{X \mid Y} (e^{itX Y} \mid Y)) \\ & = \operatorname{E}_Y ( \varphi_X(tY)) \end{align}

Therefore, we have the characteristic function of $Z_n = X_nY_n$ as $\varphi_{Z_n}(t) = \operatorname{E}_{Y_n} ( \varphi_{X_n}(tY_n)) = \operatorname{E}_{Y_n} ( \operatorname{E}_{X_n} (e^{itX_nY_n})) $.

Answer 1

By independence $(X_n,Y_n)\xrightarrow{d}(X,Y)$ because $$ \mathsf{E}e^{\mathbf{i}(sX_n+tY_n)}=\mathsf{E}e^{\mathbf{i}sX_n}\mathsf{E}e^{\mathbf{i}tY_n}\to\mathsf{E}e^{\mathbf{i}sX}\mathsf{E}e^{\mathbf{i}tY}=\mathsf{E}e^{\mathbf{i}(sX+tY)} $$ as $n\to\infty$. Then use the continuous mapping theorem.

Answer 2

I can provide a proof for this that does not invoke Levy's theorem. First, by independence you have: $$\mathbb{P}[X_n \leq x, Y_n \leq y] = \mathbb{P}[X_n \leq x]\mathbb{P}[Y_n \leq y] \overset{d}{\to} \mathbb{P}[X \leq x]\mathbb{P}[Y \leq y] = \mathbb{P}[X \leq x, Y \leq y]$$ This proves $(X_n, Y_n) \overset{d}{\to} (X,Y)$. Now write $g(X_n, Y_n) = X_nY_n$ and $g(X,Y) = XY$. You can now invoke continuous mapping theorem and write $$(X_n, Y_n) \overset{d}{\to} (X,Y) \implies g(X_n,Y_n) \overset{d}{\to} g(X,Y) \implies X_nY_n \overset{d}{\to} XY$$ establishing the result.