Convergence in distribution of sum implies marginal convergence? Let $X_n, X, Y$ be random variables such that $X_n + cY \stackrel{d}{\rightarrow} X + cY $ for every positive constant $c$. Prove that $X_n \stackrel{d}{\rightarrow} X$. 
I know if only we have joint convergence we can show this by the continuous mapping theorem, but even that is not assumed. Just what am I missing?
 A: I may prove it under the assumption that $\mathrm{E}\left|Y\right|<\infty$. In order
to prove $X_{n}\rightarrow_{d}X$, we wish to show that $\mathrm{E}f\left(X_{n}\right)\rightarrow\mathrm{E}f\left(X\right)$
for all bounded, Lipschitz functions $f$ (this is Portmanteau lemma).
Hereafter let $f$ be an arbitrary bounded Lipschitz function satisfying $\left|f\left(x\right)-f\left(y\right)\right|\leq L\left|x-y\right|$
for some finite constant $L$ ($L$ could depend on $f$).
We have
\begin{align*}
 & \left|\mathrm{E}f\left(X_{n}\right)-\mathrm{E}f\left(X\right)\right|\\
= & \left|\mathrm{E}f\left(X_{n}\right)-\mathrm{E}f\left(X_{n}+cY\right)+\mathrm{E}f\left(X_{n}+cY\right)-\mathrm{E}f\left(X+cY\right)+\mathrm{E}f\left(X+cY\right)-\mathrm{E}f\left(X\right)\right|\\
\leq & \left|\mathrm{E}f\left(X_{n}\right)-\mathrm{E}f\left(X_{n}+cY\right)\right|+\left|\mathrm{E}f\left(X_{n}+cY\right)-\mathrm{E}f\left(X+cY\right)\right|+\left|\mathrm{E}f\left(X+cY\right)-\mathrm{E}f\left(X\right)\right|.
\end{align*}The term $\mathrm{E}f\left(X_{n}+cY\right)-\mathrm{E}f\left(X+cY\right)\rightarrow0$
for $X_{n}+cY\rightarrow_{d}X+cY$. Moreover,
\begin{eqnarray*}
\left|\mathrm{E}f\left(X_{n}\right)-\mathrm{E}f\left(X_{n}+cY\right)\right|\leq\mathrm{E}\left|f\left(X_{n}\right)-f\left(X_{n}+cY\right)\right| & \leq & Lc\mathrm{E}\left|Y\right|
\end{eqnarray*}
for every positive constant $c$. Let $c\downarrow0$, $Lc\mathrm{E}\left|Y\right|\rightarrow0$
when $\mathrm{E}\left|Y\right|<\infty$. Thus, we conclude $\left|\mathrm{E}f\left(X_{n}\right)-\mathrm{E}f\left(X_{n}+cY\right)\right|\rightarrow0$.
Similarly, $\left|\mathrm{E}f\left(X+cY\right)-\mathrm{E}f\left(X\right)\right|\rightarrow0$.
Hence we have shown $\left|\mathrm{E}f\left(X_{n}\right)-\mathrm{E}f\left(X\right)\right|\rightarrow0$.
Thus, $\mathrm{E}f\left(X_{n}\right)\rightarrow\mathrm{E}f\left(X\right)$
and $X_{n}\rightarrow_{d}X$.
A: It is a particular case of the accompanying law theorem. 
Let $f$ be a bounded uniformly continuous function on $\mathbf R$. Since 
$$|E[f(X_n)]-E[f(X)]|\leqslant |E[f(X_n)]-E[f(X_n+cY)]|+|E[f(X_n+cY)]-E[f(X+cY)]|+|E[f(X+cY)-E[f(X)]]|$$
and $X_n+cY\to X+cY$ in distribution, we obtain for each positive $c$, 
$$\limsup_{n\to +\infty}|E[f(X_n)]-E[f(X)]|\leqslant \limsup_{n\to +\infty} |E[f(X_n)]-E[f(X_n+cY)]|+|E[f(X+cY)-E[f(X)]]|.$$
Fix a positive $\varepsilon$ and pick $\delta$ such that $|f(x+y)-f(y)|\leqslant \varepsilon$ if $|x|\lt\delta$. Then 
$$|f(X_n)-f(X_n+cY)|\chi_{\{|cY|\lt \delta\}}\leqslant\varepsilon,\mbox{ and }$$
$$E\left[|f(X_n)-f(X_n+cY)|\chi_{\{|cY|\geqslant \delta\}}\right]\leqslant 2\sup_t|f(t)|\cdot \mathbb P\{|Y|\geqslant \delta/c\},$$
and we deduce that for each positive $\varepsilon$ and each positive $c$, 
$$\limsup_{n\to +\infty}|E[f(X_n)]-E[f(X)]|\leqslant \varepsilon+2\sup_t|f(t)|\cdot P\{|Y|\geqslant \delta/c\}+|E[f(X+cY)-E[f(X)]]|.$$
Letting $c\to 0$ then $\varepsilon\to 0$ we get $X_n\to X$ in distribution.
