Suppose that $\{(X_n,Y_n)\}^\infty_{n=1}$ is a sequence of pairs of real-valued random variables that converge in distribution to $(X,Y)$. Show that $X_n + Y_n$ converges in distribution to X+Y.

Attempt at a solution: If $h(x,y) = x+y$, then $h(x,y)$ is continuous, and so is any $f(h(x,y))$ where $f$ is a continuous function. I know that the class of bounded, continuous functions is measure-determining, which I think would be useful, but I just can't wrap my head around how to apply it.


You are very close.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ an arbitrary continuous and bounded function.

Note that the function $h:\mathbb{R}^{2}\rightarrow\mathbb{R}$ prescribed by $\langle x,y\rangle\mapsto x+y$ is continuous so that the composition $f\circ h:\mathbb{R}^{2}\rightarrow\mathbb{R}$ is continuous.

Also $f\circ h$ is bounded, since $f$ is bounded.

Then $\lim_{n\rightarrow\infty}\mathbb{E}f\left(h\left(X_{n},Y_{n}\right)\right)=\mathbb{E}f\left(h\left(X,Y\right)\right)$ as a consequence of $\left(X_{n},Y_{n}\right)\stackrel{d}{\rightarrow}\left(X,Y\right)$.

This comes to the same as $\lim_{n\rightarrow\infty}\mathbb{E}f\left(X_{n}+Y_{n}\right)=\mathbb{E}f\left(X+Y\right)$.

This is true for any continuous and bounded function $f:\mathbb{R}\rightarrow\mathbb{R}$, so we are allowed to conclude that $X_{n}+Y_{n}\stackrel{d}{\rightarrow}X+Y$.

  • $\begingroup$ Why was this answer downvoted? At least leave an explanation of that! $\endgroup$ – drhab Mar 20 '18 at 8:10


According to this wikipedia article, this statement is false. I don't know what the counterexample would be. It might take me a little bit to think of a counterexample.

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    $\begingroup$ I know that it is false in the case that $X_n \to X$ and $Y_n \to Y$ in distribution, but I think that when the pair converges, it is true... but, I could be wrong. $\endgroup$ – poppy3345 Mar 13 '15 at 19:34
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    $\begingroup$ @poppy3345 what's the difference between a sequence of pairs and two "seperate" sequences of each coordinate? $\endgroup$ – GuySa Jun 25 '19 at 6:44
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    $\begingroup$ Let $X_n\sim \text{Ber}(0.5)$ and $Y_n = 1-X_n$ for $n=0,1,2,\dots$. Then both $X_n\rightarrow X_0$ and $Y_n \rightarrow X_0$, but $X_n + Y_n \rightarrow 0$. Jointly, it is not true that $(X_n, Y_n) \rightarrow (X_0, X_0)$, but it is true that $(X_n, Y_n)\rightarrow (X_0, Y_0)$ and $X_n + Y_n \rightarrow X_0 + Y_0 = 0$ in distribution. $\endgroup$ – Olivier Feb 6 '20 at 23:08

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