Does $X\stackrel{d}\to X_1$ and $Y\stackrel{d}\to Y_1$ imply $X+Y\stackrel{d}\to X_1+Y_1$? Let $X,X_1, Y, Y_1$ be random variables. If $X\to X_1$ and $Y\to Y_1$ converge in distribution, does $X+Y\to X_1+Y_1$ in distribution?
 A: What $X+Y$ and $X_1+Y_1$ converge to depends upon the joint distributions of $(X,Y)$ and $(X_1,Y_1)$; what $X, X_1, Y, Y_1$ converge to individually depends upon the marginal distributions.  You can have the same marginal distributions with different joint distributions.  Because the joint distributions may be different, functions of both variables may have different distributions as well.  As convergence of the marginal distributions to the same distribution does not imply convergence of the joint distributions to the same distribution, the answer to your question is "no".
For a simple example using equality of distributions rather than convergence, let $X, X_1$ be i.i.d. $\text{Normal}(0,1)$, with $Y = X$ and $Y_1 = -X_1$.  Then $X = X_1$ in distribution and $Y = Y_1$ in distribution, all four being $\text{Normal}(0,1)$.  However, $X+Y \neq X_1+Y_1$ in distribution,  the former being distributed $\text{Normal}(0,2)$ and the latter being identically equal to $0$.
Edit:
As @GordonSmyth observes in comments, if the limiting dependence between $X$ and $Y$ is the same as that between $X_1$ and $Y_1$, then the conjecture in the question holds, but not necessarily otherwise. This holds for the case where $X$ and $Y$ are independent and so are $X_1$ and $Y_1$, among others.
