# weak convergence of independent sequence

I want to show that if $X_n\to^w X$ and $Y_n\to^w Y$ which is 'weak convergence'

and the $X_n,Y_n$ are independent RV's on the same probability space,

Then we also have weak convergence of the random vector $(X_n,Y_n)\to^w (X,Y)$ Apparantly he independence condition is crucial here ..

I only know that the joint probability distribution is the product of both distributions. I'm not sure how this implies weak convergence of the random vector..

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Are the random variables real valued? –  Davide Giraudo Nov 10 '12 at 22:06
yes sorry, forgot to mention that. –  DinkyDoe Nov 11 '12 at 16:09
Use the Cramer Wold then compute the characteristic function using independence. –  Pk.yd Nov 12 '12 at 8:23
Are $X$ and $Y$ assumed to be independent? –  Davide Giraudo Nov 12 '12 at 14:00
No, so aparantly this follows...isn't it true that $X_n,Y_n$ independant implies $X,Y$ independant.? –  DinkyDoe Nov 12 '12 at 18:12

If you do not assume that $X$ and $Y$ are independent, this is false.
Take $\Omega = [0,1]^{\mathbb{N}} \times [0,1]^{\mathbb{N}}$. And let $$X_n(x,y) = x_n \quad\text{and}\quad Y_n(x,y) = y_n.$$ Then, $X_n \xrightarrow{w} X_1$ and $Y_n \xrightarrow{w} X_1$ (not a typo!).
But $(X_n, Y_n)$ is identically distributed, so $(X_n, Y_n) \xrightarrow{w} (X_1, Y_1)$ which has a different distribution from $(X_1, X_1)$.
For the result to be true, you will have to assume that $X$ and $Y$ are independent. That is, $(X_n, Y_n) \not \xrightarrow{w} (X_1, X_1)$.