Take the 2-minute tour ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
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 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 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 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 at 12:33
    
Thanks, it all makes sense now, great answer. –  Stacy Jul 18 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 $X_{n}\rightarrow_{d}X$.

share|improve this answer
    
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 at 12:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.