# Weak convergence of the sum of dependent variables, question

• Dec 13th 2011, 03:27 PM
vovchik
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.
• Dec 13th 2011, 05:16 PM
Moo
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 $P(X_n=1)=P(X_n=-1)=\frac 12$
It converges weakly to X, such that $P(X=1)=P(X=-1)=\frac 12$.
Same for $Y_n=-X_n$ and Y.
• Dec 13th 2011, 05:46 PM
vovchik
Re: Weak convergence of the sum of dependent variables, question
Thank you!