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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?

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  • $\begingroup$ Check out Slutsky's theorem $\endgroup$ Commented Mar 23, 2013 at 21:56
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    $\begingroup$ I know Slutsky's theorem, but does it apply here? It is more about convergence in distribution, isn't it? $\endgroup$
    – user68187
    Commented Mar 23, 2013 at 22:07

4 Answers 4

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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 $A>0$, $$ \{|X_n|\cdot|Y_n-Y|>\epsilon/2\}\subset \{|X_n-X|>1\}\cup\{|X|> A\}\cup\{|Y_n-Y|>\epsilon/2(A+1)\} $$ (Suppose all the three conditions on the right hold, then $|X_n| \cdot |Y_n-Y| > |X_n - X+X| \cdot |Y_n-Y| > (1+A) \cdot \epsilon/2(A+1) = \epsilon/2$) Hence, using the convegence in probability of $X_n$ to $X$ and of $Y_n$ to $Y$, we deduce that, for any $A>0$, $$ \limsup_{n\rightarrow\infty}\mathbb{P}(|X_n|\cdot|Y_n-Y|>\epsilon/2)\leq \mathbb{P}(|X|> A)\underset{A\rightarrow\infty}{\longrightarrow}0. $$ Similarily (in fact easier), $\mathbb{P}(|Y|\cdot|X_n-X|>\epsilon/2)$ goes to $0$ when $n\rightarrow\infty$. This concludes your proof!

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    $\begingroup$ Thank you. This one is the easiest for me. $\endgroup$
    – user68187
    Commented Mar 25, 2013 at 5:57
  • $\begingroup$ Can somebody prove the first statement in the proof above? $\endgroup$
    – CKM
    Commented Nov 3, 2016 at 5:32
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    $\begingroup$ The first statement comes from the following easy fact : if a+b, then a>=c/2 or b>=c/2. $\endgroup$
    – grodeni
    Commented Nov 6, 2016 at 18:08
  • $\begingroup$ do you mean $\frac ϵ {2(A+1)}$ cuz it seems like you wanted to say ϵ/2*(A+1) $\endgroup$ Commented Apr 4, 2018 at 2:10
  • $\begingroup$ i can edit that if that is right b/c o/w this does not hold $\endgroup$ Commented Apr 4, 2018 at 2:11
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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.

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  • $\begingroup$ Sorry, I don't really understand how this property applies here. Maybe you can make your point more directly? $\endgroup$
    – user68187
    Commented Mar 24, 2013 at 1:53
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    $\begingroup$ I explained the reasoning. Is it clear now? $\endgroup$
    – Siméon
    Commented Mar 24, 2013 at 16:56
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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 $\eta\gt0$, $$ \lim_{n\to\infty}\mathbb P(|X_nY_n-XY|\geqslant\eta)=0. $$

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  • $\begingroup$ Does $\inf_{u>0}P(|X|\ge u)$ imply that $X$ is integrable? Trying to see if there are any unstated conditions. Thanks for your answer! $\endgroup$
    – manofbear
    Commented Feb 7, 2017 at 14:52
  • $\begingroup$ Or are random variables necessarily finite a.e.? $\endgroup$
    – manofbear
    Commented Feb 7, 2017 at 14:57
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    $\begingroup$ @manofbear $$\inf\limits_{z>0}P(Z\geqslant z)=0\iff P(Z<\infty)=1$$ $\endgroup$
    – Did
    Commented Feb 7, 2017 at 15:04
  • $\begingroup$ Excellent, thanks for the quick response! And $X$ maps into $\mathbb{R}$ so $X$ is by definition finite a.e.. Got it! $\endgroup$
    – manofbear
    Commented Feb 7, 2017 at 17:00
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First, \begin{align} |X_nY_n-XY| &\le |(X_n-X)(Y_n-Y)|+|(X_n-X)Y|+|(Y_n-Y)X| \\ &\le \frac{1}{2}(X_n-X)^2+\frac{1}{2}(Y_n-Y)^2+|(X_n-X)Y|+|(Y_n-Y)X|. \end{align}

Then for any $\epsilon>0$ and $K>0$

\begin{align} P\{|X_nY_n-XY|>\epsilon\} &\le P\{|X_n-X|>\sqrt{\epsilon/2}\}+ P\{|Y_n-Y|>\sqrt{\epsilon/2}\} \\ &+P\{|X_n-X|>\epsilon/(4K)\}+P\{|Y_n-Y|>\epsilon/(4K)\} \\ &+P\{|X|>K\}+P\{|Y|>K\}; \\ \\ \because \{|(X_n-X)Y|>\epsilon/4\}&\subset\{|Y|>K\}\cup \{|X_n-X|>\epsilon/(4K)\},\\ \{|(Y_n-Y)X|>\epsilon/4\}&\subset\{|X|>K\}\cup \{|Y_n-Y|>\epsilon/(4K)\}. \end{align}

For any $\nu>0$ we can find $K$ s.t. $P\{|X|>K\}<\nu$ and $P\{|Y|>K\}<\nu$ so that

$$\limsup_{n\to\infty}P\{|X_nY_n-XY|>\epsilon\}\le 2\nu.$$

Now, send $\nu\downarrow 0$.

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