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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).

Andrea Fuzzi
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2 Answers2

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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].

Kavi Rama Murthy
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  • Sorry to revive this one year later; but in the end you've concluded negatively, whereas the poster beneath you has concluded positively. Both conclusions, unfortunately, seem reasonable to me. Any update? – TrostAft Apr 01 '20 at 17:31
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    @TrostAft: What I show is that if $X$ and $Y$ are independent (ie if you replace $X$ and $Y$ with independent copies) — then $X_n+Y_n \rightarrow X+Y$ in distribution. What this answer shows is that the statement is false in general. – Aphelli Apr 01 '20 at 18:14
  • @Mindlack, oh I see, I misunderstood. Thanks! – TrostAft Apr 01 '20 at 18:20
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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 $Y$.

Aphelli
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  • 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 00:00