# Show that $X_n \overset{p}{\to} X$ and $Y_n \overset{p}{\to} Y$ implies $(X_n, Y_n) \overset{p}{\to} (X,Y)$.

How can I show that if $X_n \overset{p}{\to} X$ and $Y_n \overset{p}{\to} Y$, then $(X_n, Y_n) \overset{p}{\to} (X,Y)$? that is, if $X_n$ converges in probability to $X$, and $Y_n$ converges in probability to $Y$, then we have convergence in the joint sense. I've tried to use an epsilon approach, but cannot seem to get a bound. Does anyone have any hints? Thanks!

For any $\epsilon>0$
$$P\left\{d\left((X_n,Y_n),(X,Y)\right)>\epsilon\right\}\le P\{|X_n-X|>\epsilon/\sqrt{2}\}+P\{|Y_n-Y|>\epsilon/\sqrt{2}\}\to 0.$$
Here, $d(z, w)=\|z-w\|_2$.
• Could I ask where you got the $\sqrt{2}$ in the epsilons? – user136503 Nov 23 '15 at 4:52
• @user136503 $$P\left\{d\left((X_n,Y_n),(X,Y)\right)>\epsilon\right\}=P\{(X_n-X)^2+(Y_n-Y)^2>\epsilon^2\}\le \cdots$$ – d.k.o. Nov 23 '15 at 5:01
• Nope. Dots are for the second term in the inequality: $$P\{A^2+B^2>\epsilon^2\}\le P\{A^2>\epsilon^2/2\}+P\{B^2>\epsilon^2/2\}$$ – d.k.o. Nov 23 '15 at 5:12