# Proof (confusion) of the product of two random variables of two sequences converge to XY

If $X_n \to X$ $Y_n \to Y$ in probability, then $X_nY_n \to XY$.

The textbook proof is as follows:

By the triangle inequality, for $\varepsilon > 0$, we have $$\{|X_nY_n − XY | > \varepsilon\} \subset \left\{|(X_n − X)Y_n| > \frac {\varepsilon} {2}\right\} \cup \left\{|(Y_n − Y )X| > \frac {\varepsilon} {2}\right\}$$

So

$$P(|X_nY_n − XY | > \varepsilon) \leq P \left(|(X_n − X)Y_n| > \frac {\varepsilon} {2}\right) + P \left(|(Y_n − Y )X| > \frac {\varepsilon} {2}\right)$$

Fix $\delta > 0$. We want to show that eventually, the right hand side is less than $4\delta$. Choose $K$ so that $P (|X| > K) < \delta/2$ and $P (|Y | > K) < \delta/2$. Since, $\{|Y_n| > 2K\} \subset \{|Y_n −Y | > K\} \cup \{|Y | > K\}$, we also get that eventually, $P (|Y_n| > 2K) < \delta$. Considering whether $|Y_n| > 2K$ or $|Y_n| \leq 2K$, we thus get

$$P (|(X_n − X)Y_n| > \varepsilon/2) \leq P (|Y_n| > 2K) + P (|X_n − X| > \varepsilon/(2K)) < 2\delta$$

for large enough n. Dealing with the second term is similar (and simpler), which finishes the proof.

My confusion:

First part of the proof is clear but the second part of fixing $\delta > 0$ and finding a $K$ is very confusing.

1) What is $K$ here?

2) Can anyone break down

$$P (|Y_n| > 2K) < δ$$

and

$$P (|(X_n − X)Y_n| > \varepsilon/2) \leq P (|Y_n| > 2K) + P (|X_n − X| > \varepsilon/(2K)) < 2\delta$$

Because i beleive

$$P (|Y_n| > 2K) < 0 + \delta/2 = \delta/2$$

and

$$P (|(X_n − X)Y_n| > \varepsilon/2) \leq P (|Y_n| > 2K) + P (|X_n − X| > \varepsilon/(2K)) < \delta/2 + 0 = \delta/2$$

3) if you would be so kind please write out the second term as well which the solution skipped (this is trivial, my main confusion is the above two.)

Thank you.

It is not a matter of searching such $K$ but of knowing that such $K$ exists. That is expressed in words like: "we can find/choose a $K$ such that..."

For every $\epsilon>0$ and $k>0$ and positive integer $n$ it is true for $x_{n},y_{n}x,y\in\mathbb{R}$ that:

\begin{aligned}\left|x_{n}-x\right|\leq\frac{\epsilon}{4k}\wedge\left|y_{n}\right|\leq2k\wedge\left|y_{n}-y\right|\leq\frac{\epsilon}{2k}\wedge\left|x\right|\leq k & \implies\left|x_{n}-x\right|\left|y_{n}\right|+\left|y_{n}-y\right|\left|x\right|\leq\epsilon\\ & \implies\left|\left(x_{n}-x\right)y_{n}\right|+\left|\left(y_{n}-y\right)x\right|\leq\epsilon\\ & \implies\left|\left(x_{n}-x\right)y_{n}+\left(y_{n}-y\right)x\right|\leq\epsilon\\ & \implies\left|x_{n}y_{n}-xy\right|\leq\epsilon \end{aligned}

and also that:

\begin{aligned}\left|y_{n}-y\right|\leq k\wedge\left|y\right|\leq k & \implies\left|y_{n}-y\right|+\left|y\right|\leq2k\\ & \implies\left|\left(y_{n}-y\right)+y\right|\leq2k\\ & \implies\left|y_{n}\right|\leq2k \end{aligned}

Turning the inequalities around we find equivalently:

$\left|x_{n}y_{n}-xy\right|>\epsilon\implies\left|x_{n}-x\right|>\frac{\epsilon}{4k}\vee\left|y_{n}\right|>2k\vee\left|y_{n}-y\right|>\frac{\epsilon}{2k}\vee\left|x\right|>k$

and

$\left|y_{n}\right|>2k\implies\left|y_{n}-y\right|>k\vee\left|y\right|>k$

justifying the inequalities:

$$\mathsf{P}\left(\left|X_{n}Y_{n}-XY\right|>\epsilon\right)\leq$$$$\mathsf{P}\left(\left|X_{n}-X\right|>\frac{\epsilon}{4k}\right)+\mathsf{P}\left(\left|Y_{n}\right|>2k\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>\frac{\epsilon}{2k}\right)+\mathsf{P}\left(\left|X\right|>k\right)$$

and

$$\mathsf{P}\left(\left|Y_{n}\right|>2k\right)\leq\mathsf{P}\left(\left|Y_{n}-Y\right|>k\right)+\mathsf{P}\left(\left|Y\right|>k\right)$$

Combining both inequalities we find:

$$\mathsf{P}\left(\left|X_{n}Y_{n}-XY\right|>\epsilon\right)\leq$$$$\mathsf{P}\left(\left|X_{n}-X\right|>\frac{\epsilon}{4k}\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>k\right)+\mathsf{P}\left(\left|Y\right|>k\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>\frac{\epsilon}{2k}\right)+\mathsf{P}\left(\left|X\right|>k\right)$$

For a fixed $\delta>0$ a $k_{\delta}>0$ exists with $\mathsf{P}\left(\left|Y\right|>k_{\delta}\right)<\frac{1}{3}\delta$ and $\mathsf{P}\left(\left|X\right|>k_{\delta}\right)<\frac{1}{3}\delta$

Substituting $k=k_{\delta}$ (which is allowed!) we find:

$$\mathsf{P}\left(\left|X_{n}Y_{n}-XY\right|>\epsilon\right)\leq$$$$\mathsf{P}\left(\left|X_{n}-X\right|>\frac{\epsilon}{4k_{\delta}}\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>k_{\delta}\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>\frac{\epsilon}{2k_{\delta}}\right)+\frac{2}{3}\delta$$

Since $X_{n}\stackrel{p}{\to}X$ and $Y_{n}\stackrel{p}{\to}Y$ some $n_{\delta}$ exists with $$\mathsf{P}\left(\left|X_{n}-X\right|>\frac{\epsilon}{2k_{\delta}}\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>k_{\delta}\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>\frac{\epsilon}{k_{\delta}}\right)<\frac{1}{3}\delta$$ for every $n>n_{\delta}$.

So for $n>n_{\delta}$ we found actually that: $$\mathsf{P}\left(\left|X_{n}Y_{n}-XY\right|>\epsilon\right)\leq\delta$$

Proved is now that $X_{n}Y_{n}\stackrel{p}{\to}XY$ .

For any random variable $X$, $P\{|X|>K\} \to 0$ as $K \to \infty$ because the sets $\{|X|>K\}$ decrease to the empty set as $K \to \infty$. About the other confusions you have you are writing $P\{|X_n-X| > \alpha\}=0$ etc, but what you know is that $P\{|X_n-X| > \alpha\} \to 0$ as $n \to \infty$.