sum of Random variable doesn't converge in distribution

Question

Could you tell me a case That $X(n)$ converge to $X$ in distrubution and $Y(n)$ to $Y$ in distribution but $X(n)+Y(n)$ doesn't converge to $X+Y$ in distribution? I can't find such case..

Answer 1

What is the definition of convergence in distribution? The answer lies in there. The convergence does not hold simply because the random variables uniquely determine a distribution but a distribution does not uniquely determine the random variable.

One definition which is most often used is that $X_{n}\to X$ in distribution if $F_{n}(x)\to F(x)$ for all points $x$ such that $F(x)$ is continuous. Here $F_{n}$ and $F$ denote the cdf's of $X_{n}$ and $X$ respectively.

So now the problem is that two different random variables can have the same cdf. For example if $B\sim Bern(\frac{1}{2})$ . Then $1-B\sim Bern(\frac{1}{2})$ . So $B$ and $1-B$ have the same distribution and hence the same cdf's.

So if $X_{n}=X$ for all $n$.

Then $X_{n}\to X$ in distribution. But also $X_{n}\to 1-X$ in distribution.

Thus you can let $X_{n}=Y_{n}=B$ . But let $X=1-B$ and $Y=B$. Then $X_{n}+Y_{n}= 2B$. But $X+Y=1$ . But obviously the random variable $2B$ is a constant like $1$ is .

Like this you can construct many counter-examples.

Answer 2

Take $X_n=Y_n\sim N(0,1)$ and $X=-Y\sim N(0,1)$.