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} $
And $ \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=\frac{1}{(1+iu\mu)^2} $
Solution 2
$ \phi_X(w)=E[e^{iwX}]=\int_0^{+\infty}e^{iwX}\frac{1}{\mu}e^{-\frac{x}{\mu}}dx=e^{X(iw-\frac{1}{\mu})}\frac{1}{\mu}\frac{1}{iw-\frac{1}{\mu}}|_0^{+\infty}\\ = 0 - \frac{1}{\mu}\cdot\frac{1}{iw-\frac{1}{\mu}}=\frac{1}{1-iw\mu}\\ \phi_{X+Y}(w)=E[e^{iw(X+Y)}]\\ =\int\int e^{iw(X+Y)}f_X(x)f_Y(y)dxdy = \int e^{iwx}f_X(x)dx \cdot \int e^{iwy}f_Y(y)dy=\phi_X(w)\phi_Y(w)\\ =\frac{1}{(1+iw\mu)^2} $
Solution 3
For this problem, it is very useful to note that for any independent random variables $ X $ and $ Y $ and their characteristic functions $ \phi_X(\omega), \phi_Y(\omega) $ we have the following property:
$ \phi_{X+Y}(\omega) = \phi_X(\omega)\phi_Y(\omega) $.
We then note that the characteristic function of an exponential random variable $ Z $ is written as
$ \phi_{Z}(\omega) = \frac{\lambda}{\lambda - i\omega} $
where $ \lambda $ parameterizes the exponential distribution. As such, we can write the characteristic function of $ X + Y $ as
$ \phi_{X+Y}(\omega) = \phi_X(\omega)\phi_Y(\omega) = \left(\frac{\lambda}{\lambda - i\omega}\right)^2 $.
Next, we recall that the mean of an exponential random variable is equal to the inverse of its parameter, i.e. $ \frac{1}{\lambda} $. Then the above expression becomes
$ \phi_{X+Y}(\omega) = \left(\frac{\frac{1}{\mu}}{\frac{1}{\mu} - i\omega}\right)^2. $.
Multiplying by $ \frac{\mu}{\mu} $ gives
$ \phi_{X+Y}(\omega) = \left(\frac{1}{1-i\omega\mu}\right)^2 $
Similar Problem
Let $ X(t) $ be a Poisson random process with mean function $ \mu(t) $ and covariance function $ C_{xx}(t_1,t_2) $. Find the $ n^{th} $-order characteristic function of $ X(t) $.
