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| − | + | Computer Engineering(CE) | |
| − | Question 1: | + | Question 1: Algorithms |
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| − | + | ===Solution 3=== | |
| + | For this problem, it is very useful to note that for any independent random variables <math>X</math> and <math>Y</math> and their characteristic functions <math>\phi_X(\omega),\,\phi_Y(\omega)</math> we have the following property: | ||
| − | ---- | + | <math> |
| − | [[ECE- | + | \phi_{X+Y}(\omega) = \phi_X(\omega)\phi_Y(\omega) |
| + | </math> | ||
| + | |||
| + | We then note that the characteristic function of an exponential random variable <math>Z</math> is written as | ||
| + | |||
| + | <math> | ||
| + | \phi_Z (\omega) = \frac{\lambda}{\lambda - i\omega} | ||
| + | </math> | ||
| + | |||
| + | where <math>\lambda</math> parameterizes the exponential distribution. As such, we can write the characteristic function of <math>X+Y</math> as | ||
| + | |||
| + | <math> | ||
| + | \phi_{X+Y}(\omega) = \phi_X(\omega)\phi_Y(\omega) \\ | ||
| + | = \left(\frac{\lambda}{\lambda-i\omega}\right)^2 | ||
| + | </math> | ||
| + | |||
| + | Next, we recall that the mean of an exponential random variable is equal to the inverse of its parameter, i.e. <math>\frac{1}{\lambda}</math>. Then the above expression becomes | ||
| + | |||
| + | <math> | ||
| + | \phi_{X+Y}(\omega) = \left(\frac{\frac{1}{\mu}}{\frac{1}{\mu}-i\omega}\right)^2 | ||
| + | </math> | ||
| + | |||
| + | Multiplying by <math>\frac{\mu^2}{\mu^2}</math> gives | ||
| + | |||
| + | <math> | ||
| + | \phi_{X+Y}(\omega) = \left(\frac{1}{1-i\omega\mu}\right)^2 | ||
| + | </math> | ||
| + | |||
| + | |||
| + | [[ECE-QE_CE1-2015|Back to QE CE question 1, August 2015]] | ||
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Latest revision as of 21:00, 7 March 2016
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 $
