Line 27: Line 27:
 
=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, ...}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\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}
+
=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}
 +
</math>
 +
 
 +
<math>
 +
P((Z(t)=1) = 1 - P((Z(t)=0) = \frac{1+\lambda t - p}{1+2\lambda t}
 
</math>
 
</math>
 
----
 
----

Revision as of 00:46, 4 December 2015


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2015


Solution

$ 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} $


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