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..

Solution 1:

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, $(X_n, Y_n) \not \xrightarrow{w} (X_1, X_1)$.