# Convergence in distribution of sum of independent random variable, with no independent limit

Take $$X_n$$ and $$Y_n$$ to be two sequence of r.v., and $$X,Y$$ r.v. such that

1) $$X_n \Rightarrow X$$ and $$Y_n \Rightarrow Y$$ in distribution.

2) $$X_n$$ is independent of $$Y_n$$ for each $$n \in \mathbb{N}$$

is that true that $$X_n+Y_n\Rightarrow X+Y$$. Note that i'm not requiring $$X$$ to be independent on $$Y$$, as assumed in the case treated in this post (Sum of two independent random variables converges in distribution).

## 2 Answers

Let $$(X_n)$$ and $$(Y_n)$$ be i.i.d. standard normal, independent of each other. Then $$X_n \to X_1$$ in distribution and $$Y_n \to -X_1$$ in distribution but $$X_n+Y_n$$ does not tend to $$X_1-X_1=0$$ in distribution. [ $$X_n+Y_n$$ has $$N(0,2)$$ distribution].

Well, you can see that the characteristic functions of $$X_n+Y_n$$ converge pointwise to $$\varphi_X\varphi_Y=\varphi_{X’+Y’}$$, where $$X’$$ and $$X$$ have the same law, and $$Y’$$ and $$Y$$ have tre same law, and $$X’$$ and $$Y’$$ are independent. From Levi’s theorem, $$X_n+Y_n \rightarrow X’+Y’$$ in distribution, so $$X_n+Y_n \rightarrow X+Y$$ in distribution iff $$X’+Y’$$ and $$X+Y$$ have the same law.

In a nutshell, $$X_n+Y_n$$ converges to the “independent sum” of $$X$$ and $$Y$$.

• Why is $\varphi_X \varphi_Y = \varphi_Z$ for some $Z$? is always true that the product of two characteristic is a characteristic of some r.v. ? also i don't understand how this agree with Kavi's answer – Andrea Fuzzi Dec 15 '18 at 0:00