Difference between revisions of "ECE PhD QE CNSIP 2004 Problem1.1" - Rhea

Latest revision as of 00:35, 10 March 2015

Question

1. (20 pts.)

A probability space $ \left(\mathcal{S},\mathcal{F},\mathcal{P}\right) $ has a sample space consisting of all pairs of positive integers: $ \mathcal{S}=\left\{ \left(k,m\right):\; k=1,2,\cdots;\; m=1,2,\cdots\right\} $ . The event space $ \mathcal{F} $ is the power set of $ \mathcal{S} $ , and the probability measure $ \mathcal{P} $ is specified by the pmf $ p\left(k,m\right)=p^{2}\left(1-p\right)^{k+m-2},\qquad p\in\left(0,1\right) $.

(a)

Find $ P\left(\left\{ \left(k,m\right):\; k\geq m\right\} \right) $ .

$ P\left(\left\{ \left(k,m\right):\; k\geq m\right\} \right)=\sum_{k=1}^{\infty}\sum_{m=1}^{k}p\left(k,m\right)=\sum_{k=1}^{\infty}\sum_{m=1}^{k}p^{2}\left(1-p\right)^{k+m-2}=\frac{p^{2}}{\left(1-p\right)^{2}}\cdot\sum_{k=1}^{\infty}\left(1-p\right)^{k}\sum_{m=1}^{k}\left(1-p\right)^{m} $$ =\frac{p^{2}}{\left(1-p\right)^{2}}\cdot\sum_{k=1}^{\infty}\left(1-p\right)^{k}\cdot\frac{\left(1-p\right)\left(1-\left(1-p\right)^{k}\right)}{1-\left(1-p\right)}=\frac{p}{1-p}\cdot\sum_{k=1}^{\infty}\left(1-p\right)^{k}\cdot\left(1-\left(1-p\right)^{k}\right) $$ =\frac{p}{1-p}\cdot\left[\sum_{k=1}^{\infty}\left(1-p\right)^{k}-\sum_{k=1}^{\infty}\left(1-p\right)^{2k}\right]=\frac{p}{1-p}\cdot\left[\frac{1-p}{1-\left(1-p\right)}-\frac{\left(1-p\right)^{2}}{1-\left(1-p\right)^{2}}\right] $$ =\frac{p}{1-p}\cdot\left[\frac{1-p}{p}-\frac{\left(1-p\right)^{2}}{p\left(2-p\right)}\right]=1-\frac{1-p}{2-p}=\frac{2-p-1+p}{2-p}=\frac{1}{2-p}. $

(b)

Find $ P\left(\left\{ \left(k,m\right):\; k+m=r\right\} \right) $ , for $ r=2,3,\cdots $ .

$ P\left(\left\{ \left(k,m\right):\; k+m=r\right\} \right)=\sum_{r=2}^{\infty}\sum_{k=1}^{r-1}p\left(k,r-k\right)=\sum_{r=2}^{\infty}\sum_{k=1}^{r-1}p^{2}\left(1-p\right)^{r-2} $$ =\frac{p^{2}}{\left(1-p\right)^{2}}\cdot\sum_{r=2}^{\infty}\left(r-1\right)\left(1-p\right)^{r}=\frac{p^{2}}{\left(1-p\right)^{2}}\cdot\sum_{r=1}^{\infty}r\left(1-p\right)^{r+1} $$ =\frac{p^{2}}{1-p}\cdot\sum_{r=1}^{\infty}r\left(1-p\right)^{r}=\frac{p^{2}}{1-p}\cdot\frac{1-p}{\left(1-\left(1-p\right)\right)^{2}}=1. $

Note

We use Taylor Series: $ \sum_{r=1}^{\infty}r\left(1-p\right)^{r}=\frac{1-p}{\left(1-\left(1-p\right)\right)^{2}} $ .

(c)

Find $ P\left(\left\{ \left(k,m\right):\; k\text{ is an odd number}\right\} \right) $ .

$ P\left(\left\{ \left(k,m\right):\; k\text{ is an odd number}\right\} \right)=1-P\left(\left\{ \left(k,m\right):\; k\text{ is an even number}\right\} \right) $$ =1-\sum_{i=1}^{\infty}\sum_{m=1}^{\infty}p\left(2i,m\right)=1-\sum_{i=1}^{\infty}\sum_{m=1}^{\infty}p^{2}\left(1-p\right)^{2i+m-2} $$ =1-\frac{p^{2}}{\left(1-p\right)^{2}}\cdot\sum_{i=1}^{\infty}\left(1-p\right)^{2i}\sum_{m=1}^{\infty}\left(1-p\right)^{m}=1-\frac{p^{2}}{\left(1-p\right)^{2}}\cdot\sum_{i=1}^{\infty}\left(1-p\right)^{2i}\cdot\frac{1-p}{1-\left(1-p\right)} $$ =1-\frac{p}{1-p}\cdot\sum_{i=1}^{\infty}\left(1-p\right)^{2i}=1-\frac{p}{1-p}\cdot\frac{\left(1-p\right)^{2}}{1-\left(1-p\right)^{2}}=1-\frac{p}{1-p}\cdot\frac{\left(1-p\right)^{2}}{p\left(2-p\right)} $$ =1-\frac{1-p}{2-p}=\frac{2-p-1+p}{2-p}=\frac{1}{2-p}. $