Convergence in probability of the product of two random variables

Suppose $\{X_n\}$ and $\{Y_n\}$ converge in probability to $X$ and $Y$, respectively. Will $X_n Y_n$ converge in probability to $X Y$?

I know the answer is yes. If we treat $(X_n,Y_n)$ as a random vector, and it converges in probability to $(X,Y)$ by the assumption. Then $g(x,y) = xy$ is a continuous function and according to the continuous mapping theorem, $g(X_n,Y_n)$ converges in probability to $g(X,Y)$.

My question is how to go from the definition without using the continuous mapping theorem. My attempt is as follows.

$$P(|X_nY_n-XY|>\epsilon)=P(|X_nY_n-X_nY+X_nY-XY|>\epsilon)$$ $$\leq P(|X_n(Y_n-Y)|+|Y(X_n-X)|>\epsilon)$$

It seems tempting to conclude that the last term goes to zero as $n$ goes to infinity. But I am not sure about it. Am I right or did I miss something?

Solution 1:

It seems tempting to conclude that the last term goes to zero as n goes to infinity. But I am not sure about it. Am I right or did I miss something?

You're right, this can be done directly and we only need a little bit more work to control the right term. The following inclusion of events is easy to check: $$ \{|X_n(Y_n-Y)|+|Y(X_n-X)|>\epsilon\}\subset \{|X_n|\cdot|Y_n-Y|>\epsilon/2\}\cup\{|Y|\cdot|X_n-X|>\epsilon/2\}. $$ Now, for any $A>0$, $$ \{|X_n|\cdot|Y_n-Y|>\epsilon/2\}\subset \{|X_n-X|>1\}\cup\{|X+1|> A\}\cup\{|Y_n-Y|>\epsilon/2(A+1)\} $$ Hence, using the convegence in probability of $X_n$ to $X$ and of $Y_n$ to $Y$, we deduce that, for any $A>0$, $$ \limsup_{n\rightarrow\infty}\mathbb{P}(|X_n|\cdot|Y_n-Y|>\epsilon/2)\leq \mathbb{P}(|X+1|> A)\xrightarrow[A\rightarrow\infty]{}0. $$ Similarily (in fact easier), $\mathbb{P}(|Y|\cdot|X_n-X|>\epsilon/2)$ goes to $0$ when $n\rightarrow\infty$. This concludes your proof!

Solution 2:

This is pretty straightforward if you use that

$X_n$ tends to $X$ in probability if, and only if, every subsequence of $X_n$ has a sub(sub)sequence that tends to $X$ a.s.

This lemma follows from:

Fact 1. If $X_n$ tends to $X$ a.s., then $X_n$ tends to $X$ in probability.

Fact 2. If $X_n$ tends to $X$ in probability, it has a subsequence that tends to $X$ a.s.

Fact 3. Let $(a_n)$ be a sequence of real numbers. Then $(a_n)$ converges to $a \in \Bbb R$ if, and only if, every subsequence of $(a_n)$ has a sub(sub)sequence that tends to $a$.

Application

Let $(X_{\phi(n)}Y_{\phi(n)})$ be a subsequence of $(X_nY_n)$. We need to show that it admits a subsequence converging to $XY$ a.s. Since $X_n$ tends to $X$ in probability, there exists $\psi$ such that $X_{\phi(\psi(n))}$ tends to $X$ a.s. Since $Y_n$ tends to $Y$ in probability, there exists $\chi$ such that $Y_{\phi(\psi(\chi(n)))}$ tends to $Y$ a.s. Now, remark that $X_{\phi(\psi(\chi(n)))}Y_{\phi(\psi(\chi(n)))}$ tends to $XY$ a.s.

Solution 3:

For every $\varepsilon\gt0$ and $u\geqslant0$, let $\alpha_{u,\varepsilon}=\varepsilon(u+2\varepsilon)$. Then $$ [|X_nY_n-XY|\geqslant\alpha_{u,\varepsilon}]\subseteq[|X_n-X|\geqslant\varepsilon]\cup[|Y_n-Y|\geqslant\varepsilon]\cup[|X|\geqslant u]\cup[|Y|\geqslant u]. $$ (Proof: If $|x_n-x|\lt\varepsilon$, $|y_n-y|\lt\varepsilon$, $|x|\lt u$ and $|y|\lt u$, then $|x_ny_n-xy|\lt\varepsilon(u+2\varepsilon)$.)

Hence, $$ \mathbb P(|X_nY_n-XY|\geqslant\alpha_{u,\varepsilon})\leqslant\mathbb P(|X_n-X|\geqslant\varepsilon)+\mathbb P(|Y_n-Y|\geqslant\varepsilon)+\mathbb P(|X|\geqslant u)+\mathbb P(|Y|\geqslant u). $$ Consider the limit $n\to\infty$. One gets $$ \limsup_{n\to\infty}\mathbb P(|X_nY_n-XY|\geqslant\alpha_{u,\varepsilon})\leqslant\mathbb P(|X|\geqslant u)+\mathbb P(|Y|\geqslant u). $$ For every $\eta\gt0$ and $u\gt0$, there exists $\varepsilon$ such that $\eta\geqslant\alpha_{u,\varepsilon}$, thus $$ \limsup_{n\to\infty}\mathbb P(|X_nY_n-XY|\geqslant\eta)\leqslant\inf\limits_{u\gt0}\left(\mathbb P(|X|\geqslant u)+\mathbb P(|Y|\geqslant u)\right). $$ The infimum on the RHS is zero hence, for every $\eta\gt0$, $$ \lim_{n\to\infty}\mathbb P(|X_nY_n-XY|\geqslant\eta)=0. $$