# Linearity of convergence in distribution of random variables

if $X_n$ converges to $X$ and $Y_n$ converges to $Y$ in distribution, what about $X_n + Y_n$ would that converge to $X+Y$ in distribution ? any ideas how i could prove or disprove this

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It should be true if they are independent, see math.stackexchange.com/questions/591708/… – AlmostSureUser Nov 2 '15 at 16:50

Suppose that $X_n$ converges in distribution to $X$ where $X$ is a symmetric random variable, say $X \sim N(0,1)$. Then, trivially, $X_n$ also converges in distribution to $-X$ (since $X$ and $-X$ are identically distributed). However, $X_n + X_n$ does not converge in distribution to $X+(-X)=0$.
Convergence in distribution is a pretty weak concept.. Suppose you consider probability distributions on $[0,1]$. Let $X = Y = X_n$ for all $n$ have a density supported on $[0,1/2]$ alone, and let $Y_n$ be the same distribution except shifted to the right by $1/2$. Then all these random variables have the same distribution, so convergence in distribution is automatic. But also each $X_n + Y_n$ is the same, but different from $X + Y$, so you won't get convergence in distribution.
Maybe I have the definitions wrong... isn't convergence in distribution just that for all $a$ $P(Y_n > a)$ converges to $P(Y > a)$ as $n$ goes to infinity?. You can make them both Gaussian to ensure $X_n + Y_n$ isn't the same as $X + Y$. – Zarrax Aug 10 '11 at 14:51
You're not getting my definitions. The random variables can be unbounded, they're just defined on a measure space which I chose to be measurable subsets of $[0,1]$. At any rate, I'm done here. – Zarrax Aug 10 '11 at 15:25