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 $
