(Created page with "Category:ECE Category:QE Category:CNSIP Category:problem solving Category:random variables Category:probability <center> <font size= 4> ECE_PhD_Qua...") |
|||
| Line 21: | Line 21: | ||
</center> | </center> | ||
---- | ---- | ||
| − | + | ===Solution=== | |
| + | <math> | ||
| + | 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) | ||
| + | </math> | ||
---- | ---- | ||
[[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]] | [[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]] | ||
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Revision as of 01:20, 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) $
