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Proof: | Proof: | ||
| − | + | <math> | |
\begin{split} | \begin{split} | ||
E[Y_x] &= \sum^{+ \infty}_{k > 0} k \frac{e^{-\lambda_x}\lambda_x^k}{k!}\\ | E[Y_x] &= \sum^{+ \infty}_{k > 0} k \frac{e^{-\lambda_x}\lambda_x^k}{k!}\\ | ||
| Line 51: | Line 51: | ||
&= \lambda_x\\ | &= \lambda_x\\ | ||
\end{split} | \end{split} | ||
| − | + | </math> | |
---- | ---- | ||
Revision as of 16:11, 18 May 2017
Contents
ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)
August 2014, Problem 1
- Problem 1 , 2
Solution 1
a) $ E\left[Y_x\right] = \lambda_x $
Lack of proof. Should mention the property of Poisson distribution to show the equivalence. See the proof in solution 2.
b) Because the rate of absorption is proportional to the number of photons and the density of the material, so the attenuation of photons obeys the following equation
$ \frac{d\lambda_x}{dx} = -\mu(x)\lambda_x $
c) Solve the differential equation in b), we have
$ \lambda_x = \lambda_0e^{-\int^x_0\mu(t)dt} $
d) So the integral of the density, $ \int^T_0\mu(x)dx $ can be written as $ \int^T_0\mu(x)dx = -\log\left(\frac{\lambda_T}{\lambda_0}\right) $
e) $ \int^T_ \mu(x)dx \simeq -\log \left( \frac{Y_T}{Y_0} \right) $
Solution 2:
a). As we know $ P\left\{Y_x=k\right\} = \frac{e^{-\lambda_x}\lambda_x^k}{k!} $ is a Potion distribution, it is known that the expectation of a Poisson RV is $ \lambda_x $.
Proof:
$ \begin{split} E[Y_x] &= \sum^{+ \infty}_{k > 0} k \frac{e^{-\lambda_x}\lambda_x^k}{k!}\\ &= \sum^{+ \infty}_{k = 1} \frac{e^{-\lambda_x}\lambda_x^k}{(k-1)!}\\ &= \sum^{+ \infty}_{k = 1} \frac{e^{-\lambda_x}\lambda_x^{k-1}}{(k-1)!}\lambda_x\\ &= \lambda_xe^{-\lambda_x}\sum^{+ \infty}_{k = 0} \frac{\lambda_x^k}{k!}\\ &= \lambda_xe^{-\lambda_x}e^{\lambda_x}\\ &= \lambda_x\\ \end{split} $
