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===Solution 2=== | ===Solution 2=== | ||
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| − | \phi_X(w)=E[e^{iwX}]=\int_0^{\infty}e^{iwX}\frac{1}{\mu}e^{-\frac{x}{\mu}dx= | + | \phi_X(w)=E[e^{iwX}]=\int_0^{\infty}e^{iwX}\frac{1}{\mu}e^{-\frac{x}{\mu}}dx= |
</math> | </math> | ||
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Revision as of 11:35, 7 December 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2015
Solution 1
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} $
So $ \phi_{X+Y}=E[e^{it(X+Y)}]=\phi_{X}\phi_{Y} =( \frac{\lambda}{\lambda-iu})^2 $
Solution 2
$ \phi_X(w)=E[e^{iwX}]=\int_0^{\infty}e^{iwX}\frac{1}{\mu}e^{-\frac{x}{\mu}}dx= $
