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 
 A: 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 $X+(-X)=0$.
A: 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. 
A: The result works in the case that $Y_n \rightarrow c$ in distribution. This is because in this case  $Y_n \rightarrow c$ in probability. Thus $P(|Y_n-c|>\epsilon)=P(|(X_n+Y_n)-(X_n+c)|>\epsilon) \rightarrow 0$. We deduce that $X_n+Y_n \rightarrow X_n+c$ in probability and hence also in distribution.
But $X_n+c \rightarrow X+c$ in distribution (the proof is straightforward). 
So we have $X_n+Y_n \rightarrow X+c$ in distribution.
