Is $\liminf\Bbb P(X_nY_n\leq b)\geq\Bbb P(Xy\leq b) $ if $X_n\to X$ in distribution and $Y_n\to y\in\Bbb R$ in probability?

Question

Take $b\in\mathbb{R}$. Suppose $X_n\to X$ in distribution and suppose $Y_n\to y$ in probability, where $y$ is a constant. How to prove $$\Bbb P(Xy\leq b) \leq \liminf_{n\to\infty}\Bbb P(X_nY_n\leq b)?$$

I've asked the question before but I didn't get any answer. So I tried myself and hopefully someone can follow my reasoning.

Note: I know that this follows from Slutsky's theorem, but I'm trying to use this question to prove Slutsky's theorem.

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My try: Since $X_n\to X$ in distribution, then given $\epsilon>0$ there an $n\in\mathbb{N}$ such that for all $n\geq N$ $$\Bbb P(Xy\leq b)\leq\epsilon+\Bbb P(X_ny\leq b),\;\;\text{ and so }\;\;\; \Bbb P(Xy\leq b)\leq\epsilon+\inf_{n\geq N}\Bbb P(X_ny\leq b).$$ Now we need to use the fact that $Y_n\to y$ in probability. We can write the set $$\{X_ny\leq b\}=\{X_ny\leq b,Y_n\in (y-\epsilon,y+\epsilon) \}\cup \{X_ny\leq b,Y_n\not\in (y-\epsilon,y+\epsilon) \},$$ and so $$ \{X_ny\leq b\}\subseteq\underbrace{\{X_nY_n\leq b+\epsilon}_{:=A_n} \}\cup \underbrace{\{Y_n\not\in (y-\epsilon,y+\epsilon) \}}_{:=B_n},$$ Now we know that $\liminf_{n\to\infty}\Bbb P(B_n)=0$. Now let's look at $A_n$. For the sake of simplicity let's assume $X_n>0$. Then we have that
$$A_n=\{X_nY_n\leq b\}\cup \underbrace{\{b<X_nY_n\leq b+\epsilon\}}_{C_n}.$$ In total, we have that $$\Bbb P(X_ny\leq b)\leq \Bbb P(B_n) + \Bbb P(X_nY_n\leq b)+\mathbb{P}(C_n) $$ and $$ \Bbb P(Xy\leq b)\leq \Bbb P(X_nY_n\leq b)+\epsilon + \Bbb P(B_n) +\mathbb{P}(C_n) $$ and so finally we have made the set $\{X_nY_n\leq b\}$ appear. Taking $\liminf$ we have that $$ \Bbb P(Xy\leq b)\leq \liminf_{n\to\infty}\Bbb P(X_nY_n\leq b)+\epsilon +\liminf_{n\to\infty}\mathbb{P}(C_n) $$ But then is it possible to bound the probability of $C_n=\{b<X_nY_n\leq b+\epsilon\}$ as a function of $\epsilon$?

Answer 1

As I said in the comments , what you claim is true if you assume $b$ is a continuity point(which is what you want to prove when showing convergence of the cdf's ).

I'll show in two steps.

We show that for any compatcly supported continuous $f$ , we have $E(f(X_{n}Y_{n}))\to E(f(Xy))$ .

Fix $\epsilon$ positive.

Then by uniform continuity, there exists $\delta(\epsilon)>0$ such that $|f(x)-f(y)|<\epsilon$ when $|x-y|<\delta$ .

Now $\bigg|E(f(X_{n}Y_{n})-f(Xy))\bigg|=\bigg|E(f(X_{n}Y_{n})-E(f(X_{n}y)+E(f(X_{n}y)-Ef(Xy)\bigg|$ .

Now as $Y_{n}\xrightarrow{P} y$ . Hence $P(|Y_{n}-y|\geq \delta)\to 0$ .

Hence we have $|E(f(X_{n}Y_{n})-E(f(X_{n}y)|\leq \\\bigg|\bigg(E(f(X_{n}Y_{n})-E(f(X_{n}y)\bigg)\mathbf{1}_{|Y_{n}-y|< \delta}\bigg| +\bigg|\bigg(E((f(X_{n}Y_{n})-E(f(X_{n}y))\bigg)\mathbf{1}_{|Y_{n}-y|\geq \delta}\bigg| $

Which can be bounded by

$$\epsilon + \sup_{x}|f|\cdot P(|Y_{n}-y|\geq \delta)$$ which can be made less than $2\epsilon$ by choosing $n$ large enough.

And the last term $|E(f(X_{n}y))-E(f(Xy))|\to 0$ as $X_{n}y\xrightarrow{d}Xy$ . (If you don't know this already, this is an equivalent criteria for convergence in distribution. You can directly use Skorohod's Representation theorem(to make it easier, it is not necessary though) and DCT to prove this.

Now let $g=\mathbf{1}_{(-\infty,b)}$ . Take $f_{n}$ to be a sequence of uniformly continuous functions such that $f_{n}$ decrease to $g$. You can certainly do this by just considering $\mathbf{1}_{(-\infty,b-\frac{1}{n}]}$ and then just linearly joining to the $x$ axis.

$P(X_{n}Y_{n}<b)=E(g(X_{n}Y_{n}))\geq E(f_{m}(X_{n}Y_{n}))\to E(f_{m}(Xy))=P(Xy\leq b-\frac{1}{m}) $

Now by monotone convergence theorem $E(f_{m}(Xy))\to E(g(Xy))$ and hence $P(Xy\leq b-\frac{1}{m})\to P(Xy <b)$

Thus $P(X_{n}Y_{n}<b)\geq P(Xy<b)$ for all $n$ large enough. And thus

Thus $$\lim\inf_{n}P(X_{n}Y_{n}\leq b)\geq \lim\inf_{n}P(X_{n}Y_{n}<b)\geq P(Xy<b)$$.

And as $b$ is a continuity point, you have $$\lim\inf_{n}P(X_{n}Y_{n}\leq b)\geq P(Xy<b)=P(Xy\leq b)$$