# Convergence in Probability, Product

I know this has been asked before on here, but none of the answers are satisfying to me, in the sense that the answers come up with set containments and bounds with no explanation. I was hoping to get some help in a way that any containments or inequalities stated have a little explanation.

Let $X_n, Y_n$ be sequences of r.v.'s defined on $(\Omega,\mathcal{F},P)$. Prove that if $X_n\rightarrow X$ in probability and $Y_n\rightarrow Y$ in probability, then:

$X_n Y_n\rightarrow XY$ in probability

Here is what I have so far. We use the usual plus minus trick to get, $|X_n Y_n -XY|=|X_n(Y_n-Y)+Y(X_n-X)|\leq |X_n||Y_n-Y|+|Y||X_n-X|$. This tells us that, $\{|X_nY_n-XY|\geq\epsilon\}\subseteq\{|X_n||Y_n-Y|+|Y||X_n-X|\geq\epsilon\}.$ Now in general, if $a+b\geq \epsilon$ we must have that either $a\geq \epsilon/2$ or $b\geq \epsilon/2$. For if both are smaller than $\epsilon/2$ we would get that $a+b<\epsilon$. This tells us that $\{|X_n||Y_n-Y|+|Y||X_n-X|\geq\epsilon\}\subseteq\{|X_n||Y_n-Y|\geq\epsilon/2\}\cup\{|Y||X_n-X|\geq\epsilon/2\}.$ So ultimately we have: $$P(\{|X_nY_n-XY|\geq\epsilon)\leq P(|X_n||Y_n-Y|\geq\epsilon/2\}\cup\{|Y||X_n-X|\geq\epsilon/2\})$$ $$P(|X_n||Y_n-Y|\geq\epsilon/2)+P(|Y||X_n-X|\geq\epsilon/2).$$

Here is where I get stuck. I know that you want to somehow use the finiteness of the probability measure, but I am unsure how.

Any help is greatly appreciated.

Since $X$ is a random variable, $|X|$ is also a random variable. Thus given $\epsilon>0$ there exists $M>0$ such that $P(|X|\geq M)<\epsilon$.

Since $X_n\to X$ in probability, first show that $|X_n|\to|X|$ in probability as well. Then, there exists $N\in\mathbb N$ such that $P(||X_n|-|X||\geq M)<\epsilon$. Now $P(|X_n|\geq 2M)=P(||X_n|-|X|+|X||\geq 2M)\leq P(||X_n|-|X||\geq M)+P(|X|\geq M)<2\epsilon$.

The second last inequality is true because $|A+B|>2C\implies |A|+|B|>2C\implies$ either $|A|>C$ or $|B|>C$.

Now similarly for $Y$ get $K>0$ such that $P(|Y|\geq K)<\epsilon$ and similarly obtain $P(|Y_n|\geq 2K)<2\epsilon$ for all $n\geq N_1\in\mathbb N$.

So we have stochastically bounded $X_n$ and $Y_n$.

Now take $n>\max\{N,N_1\}$. For any $c>0$, $P(|X_nY_n-XY|\geq c)=P(|X_nY_n-X_nY+X_nY-XY|\geq c)\leq P(|X_n||Y_n-Y|\geq c/2)+P(|Y||X_n-X|\geq c/2)$

I hope you understand this step.

Now, note that $A>0,B>0,AB\geq C$ gives that at least one of $A>C/u$ and $B>u$ for any $u>0$, otherwise if both $A<C/u$ and $B<u$ then $AB<C$ a contradiction.

So $P(|Y||X_n-X|\geq c/2)\leq P(|Y|\geq c/(2u))+P(|X_n-X|\geq u)$

Choose $u$ so that $c/(2u)=K$ and then $P(|Y|\geq c/(2u))<\epsilon$. Also, there exists some $N_2\in\mathbb N$ such that for all $n\geq N_2$ we have $P(|X_n-X|\geq u)<\epsilon$.

Now $P(|X_n||Y_n-Y|\geq c/2)\leq P(|X_n|\geq c/(2v))+P(|Y_n-Y|\geq v)$ for any $v>0$.

So choose $v$ such that $c/(2v)=2M$ then $P(|X_n|\geq c/(2v))<2\epsilon$ and there exists $N_3\in\mathbb N$ such that $P(|Y_n-Y|\geq v)<\epsilon$.

So as a whole, $P(|X_nY_n-XY|\geq c)<\epsilon+\epsilon+2\epsilon+\epsilon=5\epsilon$ for all $n\geq\max\{N_1,N_2,N_3,N\}$. Therefore, $X_nY_n\to XY$ in probability.