# Example. Sequence of binomially distributed random variables

Let $\left\{ \mathbf{X}_{n}\right\} _{n\geq1}$ be a sequence of binomially distributed random variables, with the $n_{th}$ random variable $\mathbf{X}_{n}$ having pmf

$P_{\mathbf{X}_{n}}(k)=P\left(\left\{ \mathbf{X}_{n}=k\right\} \right)=\left(\begin{array}{c} n\\ k \end{array}\right)p_{n}^{k}\left(1-p_{n}\right)^{n-k}\;,\; k=0,1,\cdots,n,\; p_{n}\in\left(0,1\right).$ Show that, if the $p_{n}$ have the property that $np_{n}\rightarrow\lambda$ as $n\rightarrow\infty$ , where $\lambda$ is a positive constant, then the sequence $\left\{ \mathbf{X}_{n}\right\} _{n\leq1}$ converges in distribution to a Poisson random variable $\mathbf{X}$ with mean $\lambda$ .

Hint:

You may find the following fact useful:

$\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}.$

Solution

If $\mathbf{X}_{n}$ converges to $\mathbf{X}$ in distribution, then $F_{\mathbf{X}_{n}}(x)\rightarrow F_{\mathbf{X}}(x) $$\forall x\in\mathbf{R} , where F_{\mathbf{X}}(x) is continuous. This occurs iff \Phi_{\mathbf{X}_{n}}(\omega)\rightarrow\Phi_{\mathbf{X}}(\omega)$$ \forall x\in\mathbf{R}$ . We will show that $\Phi_{\mathbf{X}_{n}}(\omega)$ converges to $e^{-\lambda\left(1-e^{i\omega}\right)}$ as $n\rightarrow\infty$ , which is the characteristic function of a Poisson random variable with mean $\lambda$ .

$\Phi_{\mathbf{X}_{n}}(\omega)=E\left[e^{i\omega\mathbf{X}_{n}}\right]=\sum_{k=0}^{n}e^{i\omega k}\left(\begin{array}{c} n\\ k \end{array}\right)p_{n}^{k}\left(1-p_{n}\right)^{n-k}=\sum_{k=0}^{n}\left(\begin{array}{c} n\\ k \end{array}\right)\left(p_{n}e^{i\omega}\right)^{k}\left(1-p_{n}\right)^{n-k}$$=\left(p_{n}e^{i\omega}+1-p_{n}\right)^{n}=\left(1+p_{n}\left(e^{i\omega}-1\right)\right)^{n}.$

Now as $n\rightarrow\infty$ , $np_{n}\rightarrow\lambda\Rightarrow p_{n}\rightarrow\frac{\lambda}{n}$ .

$\lim_{n\rightarrow\infty}\Phi_{\mathbf{X}_{n}}(\omega)=\lim_{n\rightarrow\infty}\left(1+p_{n}\left(e^{i\omega}-1\right)\right)^{n}=\lim_{n\rightarrow\infty}\left(1+\frac{\lambda}{n}\left(e^{i\omega}-1\right)\right)^{n}=e^{\lambda\left(e^{i\omega}-1\right)}=e^{-\lambda\left(1-e^{i\omega}\right)},$

which is the characteristic function of Poisson random variable with mean $\lambda$ .

c.f.

The problem 2 of the August 2007 QE is identical to this example.

## Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin