# Proof of convergence in distribution of a discrete random variable

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 $m$.

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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 $\rightarrow R$, $f(X_n) \rightarrow f(X_{\infty})$ – user48405 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

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 $N(0,1)$.