# 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?

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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.

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Thanks, but I don't get how $|E[f(X_n+cY)]-E[f(X)]|$ disappeared in the second step. I also don't understand what you mean when you multiply by $\chi_{\{|cY|\lt \delta\}}$. From what I see it seems like an indicator function? Sorry I'm still learning :) –  Stacy Jul 18 '14 at 11:23
You are right, I missed the term $|E[f(X+cY)-E[f(X)]]|$. Fixed now. And yes, $\chi$ denotes an indicator function. –  Davide Giraudo Jul 18 '14 at 12:16
Great, and shouldn't $2\sup_t|f(t)|$ still be in the final bounding equation or can we somehow get rid of it? –  Stacy Jul 18 '14 at 12:29
I'm not sure because everything has to be homogeneous (if $f$ is replaced by $\lambda f$), so I have to add it. –  Davide Giraudo Jul 18 '14 at 12:33
Thanks, it all makes sense now, great answer. –  Stacy Jul 18 '14 at 12:39

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 $X_{n}\rightarrow_{d}X$.

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Good answer under the assumption. Really sorry but I had to pick the answer which solved it without the assumption. Still good insight though! –  Stacy Jul 18 '14 at 12:40