Weak convergence of the sum of dependent variables, question

[vovchik]

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.

 

[bpet]

How about simply Yn = -Xn?

 

[mathman]

How about simply Yn = -Xn?

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

 

[bpet]

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).

 

[blue_raver22]

Yes, it is possible to construct a counterexample where the weak convergence of the sum of dependent variables does not hold. Consider the following scenario:

Let {Xn} and {Yn} be two sequences of random variables such that Xn = Yn for all n. In other words, Xn and Yn are simply two different names for the same random variable. Now, let us define a new sequence of random variables {Zn} = {Xn + Yn}.

Since Xn and Yn are the same random variable, Zn = 2Xn for all n. Therefore, Zn is a constant random variable for all n.

Now, let X and Y be two independent random variables with X ~ N(0,1) and Y ~ N(0,1).

Using the Central Limit Theorem, it can be shown that {Xn} --> X and {Yn} --> Y weakly. However, {Zn} does not converge weakly to X + Y, since it is a constant random variable and X + Y is a non-constant random variable.

This counterexample shows that even if Xn and Yn converge weakly to X and Y respectively, the sum of dependent variables may not converge weakly to the sum of the limiting variables. Therefore, the assumption of independence is necessary for the weak convergence of the sum of dependent variables.