Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2015

## Contents

### Solution 1

$P((Z(t)=0) = P(Z(0)=0, N(t)=Even) + P(Z(0)=1, N(t)=Odd)\\ = pP( N(t)=Even) + (1-p)P( N(t)=Odd)\\ =p\sum_{m=0,1, 2, ...}P(N(t) = 2m)+ (1-p)\sum_{n=0,1,2,...}P(N(t)=2n-1)\\ =p\sum_{m=0,1,2,...}\frac{1}{1+\lambda t}(\frac{\lambda t}{1+\lambda t})^2m + (1-p)\sum_{n=0,1,2,...}\frac{1}{1+\lambda t}(\frac{\lambda t}{1+\lambda t})^{2n-1}\\ =p\cdot\frac{1}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2} + (1-p)\cdot\frac{\lambda t}{1+\lambda t}\cdot\frac{1}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}\\ =\frac{p+\lambda t}{1+2\lambda t}$

$P((Z(t)=1) = 1 - P((Z(t)=0) = \frac{1+\lambda t - p}{1+2\lambda t}$

### Solution 2

$P(Z(t)=0)=P(Z(t)=0|N(t)=even)P(N(t)=even)+P(Z(t)=0|N(t)=odd)P(N(t)=odd)$

Note that $\{Z(t)=0|N(t)=odd\}=\{Z(0)=1\}$ and $\{Z(t)=0|N(t)=even\}=\{Z(0)=0\}$, therefore,

$P(Z(t)=0)=P(Z(0)=0)P(N(t)=even)+P(Z(0)=1)P(N(t)=odd)\\ =p\sum_{k=0}^{\infty}\frac{1}{1+\lambda t}\cdot (\frac{\lambda t}{1+\lambda t})^{2k}+(1-p)\sum_{k=0}^{\infty}\frac{1}{1+\lambda t}\cdot (\frac{\lambda t}{1+\lambda t})^{2k+1}\\ =\frac{p}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}+ \frac{(1-p)\lambda t}{(1+\lambda t)^2}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}\\ =\frac{p+\lambda t}{1+2\lambda t}\\ P(Z(t)=1) = 1- P(Z(t)=0) = 1-\frac{p+\lambda t}{1+2\lambda t} = \frac{1-p+\lambda t}{1+2\lambda t}\\ P(Z(t)=k)=\left\{ \begin{array}{cc} \frac{p+\lambda t}{1+2\lambda t}, k =0 \\ \frac{1-p+\lambda t}{1+2\lambda t}, k =1\\ 0, else \end{array} \right.$

The solution is correct. However the else case is not necessary. K can only be 0 or 1.

### Solution 3

We know that $Z(t)$ can only take on the values 0 and 1, so we set out to find the probability that $Z(t)$ = 0; if we subtract this probability from 1, we will have found the probability that $Z(t) = 1$, and thus we will have described the entire pmf. We also know that if $Z(0)$ = 0, $Z(t)$ must be equal to 0 if $N(t)$ is even (i.e., if $k$ is even). Similarly, if $Z(0)\neq 0, Z(t)$ must be equal to 0 if $k$ is odd. As such, we can write the expression

$P(Z(t) = 0) = P(Z(0) = 0, \,N(t)\,\,is\,\,even) + P(Z(0) = 1, \,N(t)\,\,is\,\,odd) = P(Z(0) = 0)P(N(t)\,\,is\,\,even) + P(Z(0) = 1)P(N(t)\,\,is\,\,odd) \\= p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j+1}.$

We now recall that the sum of an infinite geometric series can be expressed as

$\sum_{k = 0}^\infty ar^k = \frac{a}{1-r}.$

We can use this to simplify the preceding equation:

$P(Z(t) = 0) = p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j+1} \\ = p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\frac{\lambda t}{1 + \lambda t}\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j} \\ = p\cdot\frac{1}{1+\lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} + (1-p)\cdot\frac{1}{1+\lambda t}\cdot\frac{\lambda t}{1 + \lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2}.$

We first combine terms:

$P(Z(t) = 0) =p\cdot\frac{1}{1+\lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} + (1-p)\cdot\frac{1}{1+\lambda t}\cdot\frac{\lambda t}{1 + \lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} \\ = \frac{p}{(1+\lambda t)\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} + \frac{\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} - \frac{p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ = \frac{p + p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} + \frac{\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} - \frac{p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ = \frac{p + \lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)}.$

Then we simplify the denominator:

$P(Z(t) = 0) = \frac{p + \lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ = \frac{p + \lambda t}{(1+\lambda t)^2 - (\lambda t)^2} \\ = \frac{p + \lambda t}{1 + 2\lambda t}.$

Now that we have found $P(Z(t) = 0)$, we can easily find $P(Z(t) = 1)$ by subtracting our result from 1:

$P(Z(t) = 1) = 1 - P(Z(t) = 0) \\ = 1 - \frac{p + \lambda t}{1 + 2\lambda t}\\ = \frac{1+2\lambda t}{1+ 2\lambda t} - \frac{p + \lambda t}{1 + 2\lambda t} \\ = \frac{1 + \lambda t-p}{1 + 2\lambda t}.$

### Similar Problem

Find the mean function $\mu(t)$ and covariance function $C_{zz}(t_1,t_2)$ of the process $Z(t)$.

ECE462 Survivor

Seraj Dosenbach