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

Question

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.

Answer 1

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 $n \to \infty$.

Answer 2

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..."

In stead of answering your question I will give a setup that might be less confusing.

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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 $X_{n}Y_{n}\stackrel{p}{\to}XY$ .