Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2005

2. (30 Points)

Let $\mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n},\cdots$ be a sequence of binomially distributed random variables, with $\mathbf{X}_{n}$ having probability mass function $p_{n}\left(k\right)=\left(\begin{array}{c} n\\ k \end{array}\right)p_{n}^{k}\left(1-p_{n}\right)^{n-k}\;,\qquad k=0,1,2,\cdots,n,$ where $0<p_{n}<1$ for all $n=1,2,3,\cdots$ . Show that if $np_{n}\rightarrow\lambda\text{ as }n\rightarrow\infty,$ then the random sequence $\mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n},\cdots$ converges in distribution to a Poisson random variable having mean $\lambda$ .

$\Phi_{\mathbf{X}_{n}}\left(\omega\right)=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(e^{i\omega}p_{n}\right)^{k}\left(1-p_{n}\right)^{n-k}$$=\left(e^{i\omega}p_{n}+1-p_{n}\right)^{n}=\left(1-p_{n}\left(e^{i\omega}-1\right)\right)^{n}.$

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

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

$e^{\lambda\left(e^{i\omega}-1\right)}$ is the characteristic function of a Poisson random variable with mean $\lambda$ . Thus, as $n\rightarrow\infty$ , $\mathbf{X}_{n}$ converges in distribution to a Poisson random variable with mean $\lambda$ .

## Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett