Weak convergence of the sum of dependent variables, question

vovchik
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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.
 
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How about simply Yn = -Xn?
 
bpet said:
How about simply Yn = -Xn?

This is not a counterexample Xn -> X, Yn -> Y (=-X) Xn + Yn -> 0 (= X + Y).
 
mathman said:
This is not a counterexample Xn -> X, Yn -> Y (=-X) Xn + Yn -> 0 (= X + Y).

With weak convergence you could set Y iid to -X instead of Y=-X (so that X+Y <> 0 if X is non-trivial).
 
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