I've been recently studying some probability theory and came across this problem in the lecture note, which states:
if $X_{n}$ converges to $X$ in distribution, $Y_{n}$ converges to $Y$ in distribution, for all n, $X_{n}$ and $Y_{n}$ are independent, $X$ and $Y$ are independent, then does $X_{n}+Y_{n}$ converge to $X+Y$ in distribution?
I was asked to prove this using characteristic functions. So I tried to use exponential distribution to show that when $X_{n}$ and $Y_{n}$ follow Exp(n), $\phi_{X_{n}}\phi_{Y_{n}}$
would converge to 1 but $\phi_{X}\phi_{Y}$ would be 0 since $d\mu_{X_{n}}$ would converge to 0 a.e. But I'm not sure if this proof itself is correct. I'd appreciate any sort of help about this problem. Much thanks.
Best Answer
The answer is yes, as shown in the following computation. For each $t \in \mathbb{R}$ we have, by using independence twice: $$\phi_{X_n + Y_n}(t) = \phi_{X_n}(t) \phi_{Y_n}(t) \to \phi_X (t) \phi_Y(t) = \phi_{X+Y}(t)$$ showing that the characteristic function of $X_n + Y_n$ converges pointwise to the characteristic function of $X+Y$. By Lévy's continuity theorem, this implies that $X_n + Y_n$ converges in distribution to $X+Y$.