Revision as of 07:40, 23 November 2010 by Nelder (Talk | contribs)

7.10 QE 2005 August

1. (30 Points)

Assume that $ \mathbf{X} $ is a binomial distributed random variable with probability mass function (pmf) given by $ p_{n}\left(k\right)=\left(\begin{array}{c} n\\ k \end{array}\right)p^{k}\left(1-p\right)^{n-k}\;,\qquad k=0,1,2,\cdots,n, $$ where <math>0<p<1 $ .

(a)

Find the characteristic function of $ \mathbf{X} $ . (You must show how you derive the characteristic function.)

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

(b)

Compute the standard deviation of $ \mathbf{X} $ .

$ E\left[\mathbf{X}\right]=\frac{d}{di\omega}\Phi_{\mathbf{X}}\left(\omega\right)\Bigl|_{i\omega=0}=n\left(pe^{i\omega}+1-p\right)^{n-1}\cdot pe^{i\omega}\Bigl|_{i\omega=0}=np. $

$ E\left[\mathbf{X}^{2}\right]=\frac{d^{2}}{d\left(i\omega\right)^{2}}\Phi_{\mathbf{X}}\left(\omega\right)\Bigl|_{i\omega=0}=\frac{d}{di\omega}npe^{i\omega}\left(pe^{i\omega}+1-p\right)^{n-1}\Bigl|_{i\omega=0} $$ =np\left[\frac{d}{di\omega}e^{i\omega}\left(pe^{i\omega}+1-p\right)^{n-1}\Bigl|_{i\omega=0}\right] $$ =np\left[e^{i\omega}\left(pe^{i\omega}+1-p\right)^{n-1}+e^{i\omega}(n-1)\left(pe^{i\omega}+1-p\right)^{n-2}\cdot pe^{i\omega}\Bigl|_{i\omega=0}\right] $$ =np\left[1+\left(n-1\right)p\right]=np+n\left(n-1\right)p^{2}. $

$ Var\left[\mathbf{X}\right]=E\left[\mathbf{X}^{2}\right]-\left(E\left[\mathbf{X}\right]\right)^{2}=np+n\left(n-1\right)p^{2}-\left(np\right)^{2}=np+n^{2}p^{2}-np^{2}-n^{2}p^{2}=np\left(1-p\right). $

$ \therefore\sigma_{\mathbf{X}}=\sqrt{np\left(1-p\right)}. $

(c)

Find the value or values of $ k $ for which $ p_{n}\left(k\right) $ is maximum, and express the answer in terms of $ p $ and $ n $ . Give the most complete answer to this question that you can.

$ \frac{p_{n}\left(k\right)}{p_{n}\left(k-1\right)}=\frac{\left(\begin{array}{c} n\\ k \end{array}\right)p^{k}\left(1-p\right)^{n-k}}{\left(\begin{array}{c} n\\ k-1 \end{array}\right)p^{k-1}\left(1-p\right)^{n-k+1}}=\frac{\frac{n!}{\left(n-k\right)!k!}p^{k}\left(1-p\right)^{n-k}}{\frac{n!}{\left(n-k+1\right)!\left(k-1\right)!}p^{k-1}\left(1-p\right)^{n-k+1}} $$ =\frac{n!}{\left(n-k\right)!k!}\cdot\frac{\left(n-k+1\right)!\left(k-1\right)!}{n!}\cdot\frac{p}{1-p}=\frac{n-k+1}{k}\cdot\frac{p}{1-p}. $

$ \frac{p_{n}\left(k\right)}{p_{n}\left(k-1\right)} $ is monotonically descreasing function of $ k $ . Hence, $ p_{n}\left(k\right) $ is maximized by the largest $ k $ such that $ \frac{p_{n}\left(k\right)}{p_{n}\left(k-1\right)}\geq1 $ .$ \left(n-k+1\right)p\geq k\left(1-p\right)\Longrightarrow np-kp+p\geq k-kp\Longrightarrow\left(n+1\right)p\geq k $. $ k $ should be the integer. Thus, $ k=\left\lfloor \left(n+1\right)p\right\rfloor $ maximize $ p_{n}\left(k\right) $ . If $ p_{n}\left(k\right)=p_{n}\left(k-1\right) $ , both $ k=\left(n+1\right)p $ and $ k=\left(n+1\right)p-1 $ maximize $ p_{n}\left(k\right) $ .

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 $ .

3. (40 Points)

Consider a homogeneous Poisson point process with rate $ \lambda $ and points (event occurrence times) $ \mathbf{T}_{1},\mathbf{T}_{2},\cdots,\mathbf{T}_{n},\cdots $ .

(a)

Derive the pdf $ f_{k}\left(t\right) $ of the $ k $ -th point $ \mathbf{T}_{k} $ for arbitrary $ k $ .

• The cdf is $ F_{\mathbf{T}_{k}}\left(t\right)=P\left(\mathbf{T}_{k}\leq t\right)=P\left(\text{at least }k\text{ points within }t\right)=\sum_{j=k}^{\infty}\frac{e^{-\lambda t}\cdot\left(\lambda t\right)^{j}}{j!} $$ =1-P\left(\mathbf{T}_{k}>t\right)=1-P\left(\text{less than }k\text{ points within }t\right)=1-\sum_{j=0}^{k-1}\frac{e^{-\lambda t}\cdot\left(\lambda t\right)^{j}}{j!}. $

• The pdf by differentiating the cdf $ isf_{\mathbf{T}_{k}}\left(t\right)=\frac{dF_{\mathbf{T}_{k}}}{dt}=-\sum_{j=0}^{k-1}\frac{\left(-\lambda\right)e^{-\lambda t}\cdot\left(\lambda t\right)^{j}}{j!}-\sum_{j=0}^{k-1}\frac{e^{-\lambda t}\cdot j\left(\lambda t\right)^{j-1}\cdot\lambda}{j!} $$ =\lambda e^{-\lambda t}\sum_{j=0}^{k-1}\frac{\left(\lambda t\right)^{j}}{j!}-\lambda e^{-\lambda t}\sum_{j=1}^{k-1}\frac{j\left(\lambda t\right)^{j-1}}{j!}=\lambda e^{-\lambda t}\sum_{j=0}^{k-1}\frac{\left(\lambda t\right)^{j}}{j!}-\lambda e^{-\lambda t}\sum_{j=1}^{k-1}\frac{\left(\lambda t\right)^{j-1}}{\left(j-1\right)!} $$ =\lambda e^{-\lambda t}\sum_{j=0}^{k-1}\frac{\left(\lambda t\right)^{j}}{j!}-\lambda e^{-\lambda t}\sum_{j=0}^{k-2}\frac{\left(\lambda t\right)^{j}}{j!}=\lambda e^{-\lambda t}\frac{\left(\lambda t\right)^{k-1}}{\left(k-1\right)!}. $

– This is Erlang distribution.

(b)

What kind of distribution does $ \mathbf{T}_{1} $ have?

• If $ k=1 $ , then $ f_{\mathbf{T}_{1}}\left(t\right)=\lambda e^{-\lambda t} $ . Thus $ \mathbf{T}_{1} $ is a exponential random variable with parameter $ \lambda $ .

(c)

What is the conditional pdf of $ \mathbf{T}_{k} $ given $ \mathbf{T}_{k-1}=t_{0} $ , where $ t_{0}>0 $ ? (You can give the answer without derivation if you know it.)

• The conditional cdf is

$ F_{\mathbf{T}_{k}}\left(t_{k}|\mathbf{T}_{k-1}=t_{0}\right)=P\left(\mathbf{T}_{k}\leq t_{k}|\mathbf{T}_{k-1}=t_{0}\right)=P\left(N\left(t_{k},t_{0}\right)\geq1\right) $$ =1-P\left(N\left(t_{k},t_{0}\right)=0\right)=1-e^{-\lambda\left(t_{k}-t_{0}\right)}. $

• The conditional pdf by differentiating the conditional cdf is

$ \therefore f_{\mathbf{T}_{k}}\left(t_{k}|\mathbf{T}_{k-1}=t_{0}\right)=\lambda e^{-\lambda\left(t_{k}-t_{0}\right)}. $

(d)

Suppose you have a random number generator that produces independent, identically distributed (i.i.d. ) random variables $ \mathbf{X}_{1},\mathbf{X}_{2},\cdots,\mathbf{X}_{n},\cdots $ that are uniformaly distributed on the interval $ \left(0,1\right) $ . Explain how you could use these to simulate the Poisson points $ \mathbf{T}_{1},\mathbf{T}_{2},\cdots,\mathbf{T}_{n},\cdots $ describe above. Provide as complete an explanation as possible.

• This problem is similar to QE 2003 August Problem 2.


Back to ECE600

Back to ECE 600 QE

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett