Proof verification : $X_n \to X$ in distribution, $Y_n \to 0$ in probability $\implies$ $X_nY_n \to 0$ in probability

Question

I have to show : $X_n \to X$ in distribution, $Y_n \to 0$ in probability $\implies$ $X_nY_n \to 0$ in probability.

Let $\alpha>0, \epsilon>0$. Then $\exists \delta>0$ such that $-\epsilon/\delta$, $\epsilon/\delta$ are continuity points of distribution of $X$ and $$P(|X|>\epsilon/\delta) \leq \alpha$$

Since $X_n \to X$ in distribution, $P(X_n \leq x) \to P(X \leq x)$ for all continuity points $($and in particular $-\epsilon/\delta$ and $\epsilon/\delta)$. There exists $N_1, N_2 \in \mathbb{N}$ such that $$n \geq N_1 \implies |P(X_n \leq -\epsilon/\delta) - P(X \leq -\epsilon/\delta)|<\alpha$$ $$n \geq N_2 \implies |P(X_n \leq \epsilon/\delta) - P(X \leq \epsilon/\delta)|<\alpha$$ Let $N=\max\{N_1,N_2\}$. Then for $n \geq N$,

$$P(|X_n|>\epsilon/\delta)=1-P(|X_n| \leq \epsilon/\delta)=1-P(-\epsilon/\delta \leq X_n \leq \epsilon/\delta) = 1-P(X_n \leq \epsilon/\delta)+P(X_n < -\epsilon/\delta) = 1-P(X_n \leq \epsilon/\delta)+P(X_n \leq -\epsilon/\delta \,\,[\text{by continuity}] \leq 1-P(X \leq \epsilon/\delta)+P(X \leq -\epsilon/\delta)+2\alpha = P(|X|>\epsilon/\delta)+2\alpha \leq 3\alpha$$

Since $Y_n \to 0$ in probability, $\exists$ $N_3 \in \mathbb{N}$ such that $$n \geq N_3 \implies P(|Y_n|>\delta) \leq \alpha$$

Choose $N^{*}=\max\{N,N_3\}$. Note that $$|X_nY_n|>\epsilon \implies |X_n|>\epsilon/\delta \,\,\text{or}\,\, |Y_n|>\delta$$ Hence, $$P(|X_nY_n|>\epsilon) \leq P(|X_n|>\epsilon/\delta \,\,\text{or}\,\, |Y_n|>\delta) \leq P(|X_n|>\epsilon/\delta)+P(|Y_n|>\delta)$$ Thus, $$n \geq N^{*} \implies P(|X_nY_n|>\epsilon) \leq 4\alpha$$ Since, $\alpha>0$ is arbitrary, $X_nY_n \to 0$ in probability.

Is the proof okay? I have a feeling that I have sort of over-killed it. Is it possible to write a shorter proof of the result? Thank you.

Answer 1

I think your proof is essentially correct. Here could be a shorter way. Let $\alpha , \varepsilon\gt 0$ be fixed. Let $R>0$ such that $-R$ and $R$ are continuity points of the cumulative distribution function of $X$ and such that $\mathbb P\left(\left\lvert X\right\rvert\gt R\right)\lt \alpha$. Then \begin{align} \mathbb P\left(\left\lvert X_nY_n\right\rvert\gt \varepsilon\right)&= \mathbb P\left(\left\lvert X_nY_n\right\rvert\gt \varepsilon;\left\lvert X_n\right\rvert\gt R\right)+\mathbb P\left(\left\lvert X_nY_n\right\rvert\gt \varepsilon;\left\lvert X_n\right\rvert\leqslant R\right)\\ &\leqslant \mathbb P\left(\left\lvert X_n\right\rvert\gt R\right)+ \mathbb P\left(R\left\lvert Y_n\right\rvert\gt \varepsilon\right); \end{align} take the $\limsup_{n\to +\infty}$ gives (using the fact that $Y_n\to 0$ in probability), $$ \limsup_{n\to +\infty}\mathbb P\left(\left\lvert X_nY_n\right\rvert\gt \varepsilon\right)\leqslant \limsup_{n\to +\infty}\mathbb P\left(\left\lvert X_n\right\rvert\gt R\right) $$ and from the convergence in distribution, we get $$ \limsup_{n\to +\infty}\mathbb P\left(\left\lvert X_nY_n\right\rvert\gt \varepsilon\right)\leqslant \alpha. $$ Since $\alpha$ is arbitrary, we get the convergence in probability to $0$.