| Line 25: | Line 25: | ||
P((Z(t)=0) = P(Z(0)=0, N(t)=Even) + P(Z(0)=1, N(t)=Odd)\\ | 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)\\ | = 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, ...}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}\\ |
| + | = | ||
</math> | </math> | ||
---- | ---- | ||
Revision as of 01:30, 4 December 2015
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}\\ = $
