2
$\begingroup$

How to proof $X_n\rightarrow^{a.s} X, Y_n\rightarrow^{a.s} Y$, thus $X_n Y_n\rightarrow^{a.s} XY$ (2) The proof One:

I try to show $P(\bigcup_{n=m}^{+\infty}|X_nY_n-XY|>\epsilon)=0$ as $m\rightarrow\infty$, then we can conclude that $X_n Y_n\rightarrow^{a.s} XY$

$P(\bigcup_{n=m}^{+\infty}|X_nY_n-XY|>\epsilon)=P(\bigcup_{n=m}^{+\infty}|X_nY_n-X_nY+X_nY-XY|>\epsilon)=P(\bigcup_{n=m}^{+\infty}|X_n(Y_n-Y)+(X_n-X)Y|>\epsilon)=P(\bigcup_{n=m}^{+\infty}|(X_n-X)(Y_n-Y)+X(Y_n-Y)+(X_n-X)Y|>\epsilon)\leq P(\bigcup_{n=m}^{+\infty}|(X_n-X)(Y_n-Y)|>\epsilon/3)+P(|X(Y_n-Y)|>\epsilon/3)+P(|(X_n-X)Y|>\epsilon/3)$

We need to show $P(\bigcup_{n=m}^{+\infty}|X(Y_n-Y)|>\epsilon/3)\rightarrow 0$ as $m \rightarrow \infty$, that is $Y_n\rightarrow^{a.s} Y$ and $X$ is a random variable, we want to show $Y_n X\rightarrow^{a.s.} YX$

I use truncate to see this part $P(\bigcup_{n=m}^{+\infty}|X(Y_n-Y)|>\epsilon/3)\leq P(|X|> M, \bigcup_{n=m}^{+\infty}|X(Y_n-Y)|>\epsilon/3)+P(|X|\geq M, \bigcup_{n=m}^{+\infty}|X(Y_n-Y)|>\epsilon/3) \leq P(|X|> M, \bigcup_{n=m}^{+\infty}|X(Y_n-Y)|>\epsilon/3)+P(\bigcup_{n=m}^{+\infty} |M(Y_n-Y)|>\epsilon/3) \leq P(\bigcup_{n=m}^{+\infty}|(Y_n-Y)|>\epsilon/(3M))+P(\bigcup_{n=m}^{+\infty} |M(Y_n-Y)|>\epsilon/3)\rightarrow 0$ as $m\rightarrow \infty$

The similar to show, $P(\bigcup_{n=m}^{+\infty}|(X_n-X)Y|>\epsilon/3)\rightarrow 0$ as $m\rightarrow \infty$

As , we have $P(\bigcup_{n=m}^{+\infty}|(X_n-X)(Y_n-Y)|>\epsilon/3)\leq P(\bigcup_{n=m}^{+\infty}|(X_n-X)|>\epsilon/3)+P(\bigcup_{n=m}^{+\infty}|Y_n-Y)|>\epsilon/3)$ if n is large enough, we can bound $|Y_n-Y|$ and |X_n-X| by $1$ . Thus we also have $P(\bigcup_{n=m}^{+\infty}|(X_n-X)(Y_n-Y)|>\epsilon/3) \rightarrow 0$ as $m\rightarrow \infty$

Thus we prove the inequality. I only know the definition of convergence in probability and almost surely. Can anyone give some suggestions, either for the proof based on the basic definition or other much simpler proof?

I am a self-learner. These questions are from Chung Kai-lai's book. Thanks a lot!

(2) The proof two: The suggestion proof is very simple and clear. It shows as below. $$\mathbb{P}(X_n Y_n \not\to XY) \leq \mathbb{P}(\{X_n \not\to X\}\cup \{Y_n \not\to Y\}) \leq \mathbb{P}(X_n \not\to X) + \mathbb{P}(Y_n \not\to Y) = 0 + 0 = 0$$

Here the first inequality follows because if $X_n(\omega) \to X(\omega)$ and $Y_n(\omega) \to Y(\omega)$ for some $\omega $, then $X_n(\omega)Y_n(\omega) \to X(\omega) Y(\omega)$. By contraposition, we get

$$\{X_nY_n \not\to XY\} \subseteq \{X_n \not\to X\}\cup \{Y_n \not\to Y\}$$

$\endgroup$
1

1 Answer 1

5
$\begingroup$

You overcomplicate this.

$$\mathbb{P}(X_n Y_n \not\to XY) \leq \mathbb{P}(\{X_n \not\to X\}\cup \{Y_n \not\to Y\}) \leq \mathbb{P}(X_n \not\to X) + \mathbb{P}(Y_n \not\to Y) = 0 + 0 = 0$$

Here the first inequality follows because if $X_n(\omega) \to X(\omega)$ and $Y_n(\omega) \to Y(\omega)$ for some $\omega $, then $X_n(\omega)Y_n(\omega) \to X(\omega) Y(\omega)$. By contraposition, we get

$$\{X_nY_n \not\to XY\} \subseteq \{X_n \not\to X\}\cup \{Y_n \not\to Y\}$$

$\endgroup$
15
  • $\begingroup$ Thanks so much! Your suggestion is very precious! $\endgroup$
    – Olivia
    Dec 11, 2019 at 18:38
  • $\begingroup$ Very welcome! Good luck with measure theory/probability. These are very fun topics to study! $\endgroup$
    – J. De Ro
    Dec 11, 2019 at 18:40
  • $\begingroup$ This answer also has the advantage that it works for general measures, while your approach only generalises to finite measure spaces. $\endgroup$
    – J. De Ro
    Dec 11, 2019 at 18:41
  • 1
    $\begingroup$ Thanks a lot! I will try to apply if I meet the similar problem. $\endgroup$
    – Olivia
    Dec 11, 2019 at 18:44
  • $\begingroup$ If this answers your question, you might consider accepting it so other users know your question has been answered :) $\endgroup$
    – J. De Ro
    Dec 11, 2019 at 18:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .