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!


Both directions can be proved simply using definitions.

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 )$.

  • $\begingroup$ 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. $\endgroup$ – NotThatGrumpyAnymore Dec 19 '17 at 4:07
  • $\begingroup$ would that be $\leq Pr(|X_{n}-X|>\epsilon/2 \, \text{or} \, |Y_{n}-Y|>\epsilon/2)$? $\endgroup$ – NotThatGrumpyAnymore Dec 19 '17 at 4:09
  • $\begingroup$ Are you aware of the definition of convergence in probability for random variable and random vector ? $\endgroup$ – Statisfun Dec 19 '17 at 4:15
  • $\begingroup$ 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" $\endgroup$ – NotThatGrumpyAnymore Dec 19 '17 at 4:18
  • $\begingroup$ I am on my phone now but I think it is already quite obvious for the rest of the proof $\endgroup$ – Statisfun Dec 19 '17 at 4:18

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