Let $X_n$ have the binomial distribution with parameter $(n,p_n)$, and suppose that $n\,p_n\rightarrow\lambda\geq0$. Prove that $X_n$ converges in dist. to the Poisson d.f. with parameter $\lambda$. (In the old days this was called the law of small numbers.)
$\bullet$ Proof.
We know that if the complex numbers have a limit, $c_n\rightarrow c$, then $$\underset{n\rightarrow\infty}{\lim}\left(1+\frac{c_n}{n}\right)^n=e^c.$$The ch.f. for $X_n$ is $$f_{X_n}(t)=\left(1-p_n+p_n\,e^{it}\right)^n=\left(1+\frac{n\,p_n}{n}(e^{it}-1)\right)^n.$$Since $n\,p_n\rightarrow\lambda$, we have $$\underset{n\rightarrow\infty}{\lim}f_{X_n}(t)=\underset{n\rightarrow\infty}{\lim}\left(1+\frac{n\,p_n}{n}(e^{it}-1)\right)^n=e^{\lambda(e^{it}-1)},$$which is the ch.f. of Poisson distribution with parameter $\lambda$.
$\Box$
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