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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… – 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 $X+(-X)=0$.

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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.

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Sorry but I do not get it: the measure obtained by shifting a distribution mu by a nonzero t cannot be mu itself. – Did Aug 10 '11 at 9:06
Sure but, in the example of your post, the random variables Y_n (almost surely in [1/2,1]) do not converge in distribution to the random variable Y (almost surely in [0,1/2]), although you sem to say they do. So, one wonders why X_n+Y_n should converge in distribution to X+Y anyway... – Did Aug 10 '11 at 14:39
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
Precisely, P(Y_n>1/2)=1 for every n and P(Y>1/2)=0 hence the former does not converge to the latter. (Convergence in distribution is not exactly equivalent to what you write… but let us ignore that.) About the second sentence of your comment: Gaussian random variables are clearly out of the scope of your post since no (nondegenerate) Gaussian is almost surely in [0,1]. – Did Aug 10 '11 at 15:18
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

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