# Prove that the vector $(X_{n},Y_{n})$ converges in probability if and only if each component converges in probability

For a bivariate, real, continuous function $f:\mathbb{R}^{2}$, in order to show that $f(X_{n},Y_{n}) \to^{P} f(X,Y)$ as $n\to \infty$ whenever $X_{n} \to^{P} X$ and $Y_{n} \to^{P} Y$ as $n \to \infty$, I need first to establish the following lemma:

Lemma: A sequence of vectors $\{(X_{n},Y_{n})\} \in \mathbb{R}^{2}$ converges in probability to the vector $(X,Y) \in \mathbb{R}^{2}$ if and only if $X_{n} \to^{P} X$ and $Y_{n} \to^{P} Y$.

I am having difficulty proving this, however.

For example, in the $(\Rightarrow)$ direction, it seems obvious to me that if $(X_{n},Y_{n}) \to^{P} (X,Y)$, then we should be able to just "pick out" each of the components and say that they converge as well, but I doubt it's that easy. So, how would I prove this direction?

In the $(\Leftarrow)$ direction, I'm completely lost, and would appreciate whatever help you could give me!

Thank you ahead of time for your time and patience!

For the $\Rightarrow$ direction, use $\Pr\left(|X_n-X| > \epsilon\right) \le \Pr\left(\sqrt{|X_n-X|^2+|Y_n-Y|^2}> \epsilon\right)$.
For the $\Leftarrow$ direction, note $\Pr\left(\sqrt{|X_n-X|^2+|Y_n-Y|^2}> 2\epsilon \right)\le \Pr(|X_n-X| > \epsilon$ or$|Y_n-Y|> \epsilon )$.
• I've never actually done one of these types of proofs by myself before. Could you show me all the details for one of the directions, and then hopefully, I can figure out how everything works and apply it to the other direction myself? For instance, in the $\Rightarrow$ direction, I don't how I can break up the $Pr(\sqrt{|X_{n}-X|^{2}+|Y_{n}-Y|^{2}}>\epsilon)$ in order to get a further bound. Dec 19, 2017 at 4:07
• would that be $\leq Pr(|X_{n}-X|>\epsilon/2 \, \text{or} \, |Y_{n}-Y|>\epsilon/2)$? Dec 19, 2017 at 4:09
• in my notes, it says "for a r.v. $X$ and a sequence of r.v.'s $X_{n}$ on the probability space $(\Omega, \mathcal{F}, P)$ if for any $\epsilon > 0$, $Pr(|X_{n}-X|<\epsilon) \to 1$ as $n \to \infty$ then $X_{n}$ is said to converge to $X$ in probability" Dec 19, 2017 at 4:18