Weak convergence of the sum of dependent variables, question

Hi guys,

Problem: Let {Xn},{Yn} - real-valued random variables.

{Xn}-->{X} - weakly; {Yn}-->{Y} weakly.

Assume that Xn and Yn - independent for all n and that X and Y - are independent.

Fact that {Xn+Yn}-->{X+Y} weakly, can be shown using characteristic functions and Levy's theorem.

Question:

If independence does not hold, can you construct a counterexample?

I appreciate any help in advance.

Re: Weak convergence of the sum of dependent variables, question

Hello,

You can just consider a symmetric rv : Xn such that Xn has the same distribution as -Xn.

So we can consider Xn and Yn=-Xn, which is not independent of Xn. It is true to say that Xn converges to a rv X and Yn converges to a rv Y that follows the same distribution as X.

But Xn+Yn=0 doesn't converge to X+Y.

For example $\displaystyle P(X_n=1)=P(X_n=-1)=\frac 12$

It converges weakly to X, such that $\displaystyle P(X=1)=P(X=-1)=\frac 12$.

Same for $\displaystyle Y_n=-X_n$ and Y.

Re: Weak convergence of the sum of dependent variables, question