Computer Engineering(CE)

Question 1: Algorithms

August 2015

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 $\frac{\mu^2}{\mu^2}$ gives

$\phi_{X+Y}(\omega) = \left(\frac{1}{1-i\omega\mu}\right)^2$