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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 http://math.stackexchange.com/questions/591708/sum-of-two-independent-random-variables-converges-in-distribution2015-11-02

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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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    In this example, $X$ and $-X$ are not independent random variables2018-09-26
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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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    Right. Let me precise and simplify this. First, one endows the set Omega=[0,1] with the Borel sigma-algebra and the Lebesgue measure P. Second, one can simply choose f(t)=1 for every t, hence X(t)=X_n(t)=Y(t) is 1 if t<1/2 and 0 if t>1/2 in [0,1]. Then X is what is called$a$Bernoulli random variable (this means that P(X=0 or 1)=1) and P(X=0)=P(X=1)=1/2. Let Y_n=1-X. Then$Y_n$is also Bernoulli hence (X_n) converges to X, (Y_n) converges to Y but P(X_n+Y_n=1)=1 and P(X+Y=0 or 2)=1 hence (X_n+Y_n) cannot converge to X+Y (all convergences in distribution). We happy.2011-08-10