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\phi_{X}=E[e^{itX}]=\int_{-\infty}^{\infty}e^{itx}\lambda e^{-\lambda x} dx \\ | \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\\ | = \lambda \int_{-\infty}^{\infty}e^{-(\lambda -iu)x} dx = -\frac{\lambda}{\lambda-iu}e^{-(\lambda-iu)x}\|_0^\infty\\ | ||
| − | =frac{\lambda}{\lambda-iu} | + | =\frac{\lambda}{\lambda-iu} |
</math> | </math> | ||
---- | ---- | ||
Revision as of 23:13, 3 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} $
