| Line 21: | Line 21: | ||
</center> | </center> | ||
---- | ---- | ||
| + | ===Solution=== | ||
Let <math>\lambda = \frac{1}{\mu}</math>, then <math>E(X)=E(Y)=\frac{1}{\lambda}</math>. | Let <math>\lambda = \frac{1}{\mu}</math>, then <math>E(X)=E(Y)=\frac{1}{\lambda}</math>. | ||
| Line 33: | Line 34: | ||
</math> | </math> | ||
| + | <math> | ||
| + | \phi_{X}=E[e^{itX}]=\int_{-\infty}^{\infty}e^{itx}\lambda e^{-\lambda x) dx \\ | ||
| + | = \lambda \int_{-\infty}^{\infty}e^{-(\lambda -iu)x) dx = -\frac{\lambda}{\lambda-iu}e^{-(\lambda-iu)x}\|_0^\infty\\ | ||
| + | =frac{\lambda}{\lambda-iu} | ||
| + | </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 00:09, 4 December 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2015
Solution
Let $ \lambda = \frac{1}{\mu} $, then $ E(X)=E(Y)=\frac{1}{\lambda} $.
$ \phi_{X+Y}=E[e^{it(X+Y)}]=\int_{X}\int_{Y}e^{it(X+Y)}p(x,y)dxdy $
As X and Y are independent
$ \phi_{X+Y}=\int_{X}\int_{Y}e^{it(x+y)}p(x)p(y)dxdy = \int_{X}e^{itx}p(x)dx\int_{Y}e^{ity}p(y)dy=\phi_{X}\phi_{Y} $
$ \phi_{X}=E[e^{itX}]=\int_{-\infty}^{\infty}e^{itx}\lambda e^{-\lambda x) dx \\ = \lambda \int_{-\infty}^{\infty}e^{-(\lambda -iu)x) dx = -\frac{\lambda}{\lambda-iu}e^{-(\lambda-iu)x}\|_0^\infty\\ =frac{\lambda}{\lambda-iu} $
