Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working on a question on "convergence in distribution" and I appreciate if you could guide me on how to approach this question:

Here is the question:

Let $X_n$ be integer-valued random variables. Show that $X_n \stackrel w{\longrightarrow} X_{\infty}$ converges in distribution if and only if $\mathrm{Pr}(X_n = m) \rightarrow \mathrm{Pr}(X_{\infty} = m)$ for each $m$.

I appreciate your help.

share|improve this question
How do you define convergence in distribution? Which ways of proving it do you know? –  Did Nov 28 '12 at 6:33
Hi @did, I'm very new to this concept. I don't know how many different ways exist. The one I know is to show that any for any continuous function f: R $\rightarrow R$, $f(X_n) \rightarrow f(X_{\infty})$ –  Eli Nov 28 '12 at 6:41
That is not the correct definition of weak convergence. For every continuous bounded $f$, you need $E_{X_n}[f]\rightarrow E_{X_\infty}[f]$ where $E_{X_n}$ is the expectation with respect to the empirical distribution of $X_n$. –  Learner Nov 28 '12 at 6:55
For an integer-valued random variable, you could define weak convergence as $Pr(X_n = m) \rightarrow Pr(X_{\infty} = m)$. Otherwise the Portmanteau theorem shows the definition of convergence of $E_{X_n}[f]\rightarrow E_{X_\infty}[f]$ is equivalent for an integer-valued variable to $Pr(X_n = m) \rightarrow Pr(X_{\infty} = m)$ (a consequence of theorem 29.1 in Billingsley). –  Learner Nov 28 '12 at 6:59
add comment

1 Answer

For the necessity, by Portmanteau theorem (theorem 29.1 in Billingsley or theorem 3.2.21 in Dembo's Notes), for any open interval $(a,b)$ which contains no integer, $$P(a<X_\infty<b) \le \liminf_n P(a<X_n<b) = 0$$ which implies that $$P(a<X_\infty<b)=0$$ and further that, for any non-integer $c$, $$P(X_\infty=c)=0$$ since $P(X_\infty=c) \le \lim_{\epsilon \to 0} P(c-\epsilon<X_n<c+\epsilon) = 0$.

Using the above two facts, for any integer $k$, we have \begin{align} P(X_\infty=k) & = P(k-0.5<X_\infty<k+0.5) \\ & \le \liminf_n P(k-0.5<X_n<k+0.5) \\ & = \liminf_n P(X_n=k) \\ & \le \limsup_n P(X_n=k) \\ & = \limsup_n P(k-0.5 \le X_n \le k+0.5) \\ & \le P(k-0.5 \le X_\infty \le k+0.5) \\ & = P(X_\infty=k)\\ \end{align} which implies that $P(X_\infty=k) = \lim_n P(X_n=k)$

For sufficiency, you should assume $X_\infty$ is also an integer valued random variable and then check the point-wise convergence of the distribution functions $P(X_n \le x)$at continuous points of $P(X_\infty \le x)$.

If $X_\infty$ is not integer valued, the converse direction doesn't hold. Here is a counter-example. $P(X_n=k)=1/n$ for $k=1,2,..n$ such that for any $k$, $\lim_{n\to \infty} P(X_n=k)=0$. Let $X_\infty \sim N(0,1)$. Then $P(X_\infty=k)=0$. However, $X_n$ do not converge weakly to $N(0,1)$.

share|improve this answer
add comment

Your Answer


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.