Difference between revisions of "ECE PhD QE CNSIP 2015 Problem1.3" - Rhea

Latest revision as of 21:00, 7 March 2016

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 $

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