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 $ .
