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</center> | </center> | ||
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| − | 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>. |
| + | |||
| + | <math> | ||
| + | \phi_{X+Y}=E[e^{it(X+Y)}]=\int_{X}\int_{Y}e^{it(X+Y)}p(x,y)dxdy | ||
| + | </math> | ||
| + | |||
| + | As X and Y are independent | ||
| + | |||
| + | <math> | ||
| + | \phi_{X+Y}=\int_{X}\int_{Y}e^{it(X+Y)}p(x)p(y)dxdy | ||
| + | </math> | ||
---- | ---- | ||
Revision as of 15:55, 3 December 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2015
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 $
