Counterexample in Convergence in Distribution

Question

I'm in my statistical inference course, and I've reached a problem related to convergence in distribution that I am slightly stuff on. Consider random variables $X_n,$ $Y_n$ who converge in distribution to $X$ and $Y$ respectively ($X_n \rightsquigarrow X \wedge Y_n \rightsquigarrow Y.$) I'm trying to find a construction fo $X_n$ and $Y_n$ such that $X_n+Y_n$ does not converge in distribution to $X+Y$. My mindset would be to find a distribution of $X_n$ that is symmetric on $0$, so that if $X_n = X \wedge Y_n = -X = X,$ $X_n \rightsquigarrow X \wedge Y_n \rightsquigarrow X$ but $X_n+Y_n \rightsquigarrow X-X = 0 \neq 2X.$ However, I am having issues finding this distribution for $X$. Any suggestions on how to go about finding this distribution?

Answer 1

Hint:

Consider $X:(-\frac{1}{2},\frac{1}{2})\to (-\frac{1}{2},\frac{1}{2})$ by $X(\omega)=\omega$, and $X':(-\frac{1}{2},-\frac{1}{2})\to (\frac{1}{2},\frac{1}{2})$ by $X'(\omega)=-X(\omega)=-\omega$, they have the same distribution.

Set $X_n=X,Y_n=X'$