ECE PhD QE CNSIP 2015 Problem1.2 - Rhea

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.

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