# Convergence in probability: $X_n \overset{P}{\rightarrow} X,\ Y_n \overset{P}{\rightarrow} Y \implies X_n Y_n \overset{P}{\rightarrow} XY$

According to the Wikipedia article on convergence of random variables, the above property holds. How can we prove it?

I guess one way would be to use a second property stated in the article, namely, $$X_n \overset{P}{\rightarrow} X,\ Y_n \overset{P}{\rightarrow} Y \implies (X_n, Y_n) \overset{P}{\rightarrow} (X, Y)$$ and subsequently apply the continuous mapping theorem with $$g: (x, y) \mapsto xy$$. However, the proof for this second property requires definitions of convergence in two dimensions and additional properties of the involved distance metrics.

Can we prove $$X_n \overset{P}{\rightarrow} X,\ Y_n \overset{P}{\rightarrow} Y \implies X_n Y_n \overset{P}{\rightarrow} XY$$ only using the one-dimensional random variables $$X_n, Y_n, X, Y$$?

## 1 Answer

One way I like, is by using subsequential arguments.

$$X_{n}\xrightarrow{P}X$$ if and only if for each subsequence $$X_{n_{k}}$$ there exists a further subsequence $$X_{n_{k_{l}}}\xrightarrow{P}X$$.

(The above criterion is true for convergence in distribution,i.e. weak convergence as well and is often used in functional analysis context. But note that this subsequential criteria is NOT sufficient for almost sure convergence. For pointwise convergence though, I'm sure you have studied it in real analysis and indeed, the proof of the above fact is essentially the same as the proof for real sequences).

We can use this in the following way.

Let $$X_{n_{k}}Y_{n_{k}}$$ be an arbitrary subsequence of $$X_{n}Y_n$$ . Now $$X_{n_{k}}\xrightarrow{P} X$$ and $$Y_{n_{k}}\xrightarrow{P}$$ and hence there exists subsequences of $$X_{n_{k}}$$ and $$Y_{n_{k}}$$ which converge almost surely to $$X$$ and $$Y$$. So chose a common subsequence $$n_{k_{m}}$$ of $$n_{k}$$ such that $$X_{n_{k_{m}}}\xrightarrow{a.s.}X$$ and $$Y_{n_{k_{m}}}\xrightarrow{a.s.} Y$$

Hence $$X_{n_{k_{m}}}Y_{n_{k_{m}}}\xrightarrow{a.s.}XY$$ and hence $$X_{n_{k_{m}}}Y_{n_{k_{m}}}\xrightarrow{P}XY$$

Thus each subsequence has a further subsequence that converges in probability to $$XY$$ and thus, the whole sequence converges in Probability to $$XY$$.

Proof of the claim I am using.

One way is clear that convergence in probability implies convergence for all subsequences.

So assume that $$X_{n}$$ does not converge in probability to $$X$$. Then there exists $$X_{n_{k}}$$ such that $$P(|X_{n_{k}}-X|>\epsilon)\geq \delta_{0}$$ for some $$\epsilon>0$$, for some $$\delta_{0}>0$$ and all $$k\geq 1$$.

Then, for this subsequence, we can find $$X_{n_{k_{l}}}\xrightarrow{P}X$$ and hence $$P(|X_{n_{k_{l}}}-X|>\epsilon)\xrightarrow{l\to\infty} 0$$. But this is not possible as $$P(|X_{n_{k}}-X|>\epsilon)\geq \delta_{0}$$ for all $$k\geq 1$$.

Hence by contradiction, $$X_{n}\xrightarrow{P}X$$

• Thank you for the answer and for additionally proving your claim. What remains non-trivial is the proposition $X_n \overset{P}{\rightarrow} X$ implies that there exists a subsequence with $X_{n_k} \overset{a.s.}{\rightarrow} X$. Commented Jun 6, 2023 at 19:11
• Inductively find $n_{k}$ such that $P(|X_{n_{k}}-X|\geq \frac{1}{k})\leq \frac{1}{2^{k}}$ . Now, let $A_{k}=\{|X_{n_{k}}-X|>\frac{1}{k}\}$ Then, by Borel-Cantelli, you have that only finitely many $A_{k}$'s occur almost surely. So, almost surely, there exists $N_k$ such that $k\geq N_{k}$ means $|X_{n_{k}}-X|\leq \frac{1}{k}$ for all $k\geq N_{k}$. So let $\epsilon>0$ . Now find $k_{0}$ such that $\frac{1}{k_{0}}<\epsilon$ . So for $k\geq N_{k_{0}}$ , you have $|X_{n_{k}}-X|<\epsilon$ and this holds for all $\epsilon$. Thus $X_{n_k}\xrightarrow{a.s}X$ . There, done in the comments itself. Commented Jun 6, 2023 at 19:59
• @Lime91 It's not exactly "non-trivial" and it's a Borel-Cantelli argument/technique that you should always keep in mind and in your arsenal. Commented Jun 6, 2023 at 20:02
• I think you are using the following: By Borel-Cantelli, we have $P(\limsup A_k) = 0$. Thus, $P((\limsup A_k)^c) = P(\liminf A_k^c) = 1$. After this, I'm trying to digest how you conclude "almost surely, there exists $N_k$ such that $k \geq N_k$ means $| X_{n_k} − X | \leq \frac{1}{k}$". There are just too many indices involved :-) I'll continue tomorrow. Commented Jun 6, 2023 at 21:59
• What does $x\in\lim\sup$ of a sequence of sets $A_{k}$ mean. It precisely means that $x\in A_{k}$ for infinitely many $k$. And $x\in\lim\inf A_{k}$ means $x\in A_{k}$ for all but finitely many $k$ (In fact this is the definition in set-builder form). So if $P(\lim\sup A_{k})=0$. Then for all $\omega\in(\lim\sup A_{k})^{c}$ (this set has probab. 1 btw), you have that $|X_{n_{k}}(\omega)-X(\omega)|>\frac{1}{k}$ for finitely many $k$. So, there must exist $N_{k}(\omega)$ such that $k\geq N_{k}(\omega)\implies |X_{n_{k}}(\omega)-X(\omega)|\leq\frac{1}{k}$. @Lime91 Commented Jun 7, 2023 at 6:14