1
$\begingroup$

Let $X_n\xrightarrow{P} X$ and $Y_n\xrightarrow{P}Y$, then we have $(X_n,Y_n)\xrightarrow{P}(X,Y)$. In process of proof we have the following: $$ \mathbb{P}(\{||(X_n,Y_n)-(X,Y)||\geqslant\epsilon\})\leqslant\mathbb{P}(\{||X_n-X||+||Y_n-Y||\geqslant\epsilon\})\leqslant\mathbb{P}(\{||X_n-X||\geqslant\epsilon/2\})+\mathbb{P}(\{||Y_n-Y||\geqslant\epsilon/2\}) $$ I don't understand the above inequalities. Could someone explain?

$\endgroup$
2
$\begingroup$

$$\|(X_n, Y_n)-(X,Y)\|=\|(X_n-X, Y_n-Y)\|\le \|X_n-X\|+\|Y_n-Y\|$$ is true by triangle inequality.

Hence $$\|(X_n, Y_n) - (X,Y) \| \ge \epsilon \implies \|X_n-X\|+\|Y_n-Y\| \ge \epsilon$$

which explains the first inequality.

Also, $\|X_n - X\| + \|Y_n - Y\| \ge \epsilon$ implies that $\|X_n - X\| \ge \frac{\epsilon}2$ or $\|Y_n - Y\| \ge \frac{\epsilon}2$. Supppose on the contrary that this is not true, then we have $\|X_n - X\| < \frac{\epsilon}2$ and $\|Y_n - Y\| < \frac{\epsilon}2$ and summing them up would give us a contradiction.

Hence, we just have to use a union bound to get the second inequality.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.