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when <math>0<=x<=n</math><br> | when <math>0<=x<=n</math><br> | ||
<math>P_{X|Z}(x|n) = P_{X,Y}(X=x,Y=n-x|Z=n)=\dfrac{e^{-\lambda}\dfrac{\lambda^x}{x!}e^{-\mu}\dfrac{\mu^{n-x}}{(n-x)!}}{e^{-(\lambda+\mu)}\dfrac{(\lambda+\mu)^n}{n!}}</math><br> | <math>P_{X|Z}(x|n) = P_{X,Y}(X=x,Y=n-x|Z=n)=\dfrac{e^{-\lambda}\dfrac{\lambda^x}{x!}e^{-\mu}\dfrac{\mu^{n-x}}{(n-x)!}}{e^{-(\lambda+\mu)}\dfrac{(\lambda+\mu)^n}{n!}}</math><br> | ||
| − | <math>=\dfrac{n!}{x!(n-x)!}\dot \dfrac{\lambda^x\dot \mu^{n-x}}{(\lambda+\mu)^n}=\begin{pmatrix}n\\x\end{pmatrix}(\dfrac{\lambda}{\lambda+\mu})^x\dot (\dfrac{\mu}{\lambda+\mu})^{(n-x)}</math> | + | <math>=\dfrac{n!}{x!(n-x)!}\dot</math><math> \dfrac{\lambda^x\dot \mu^{n-x}}{(\lambda+\mu)^n}=\begin{pmatrix}n\\x\end{pmatrix}(\dfrac{\lambda}{\lambda+\mu})^x\dot (\dfrac{\mu}{\lambda+\mu})^{(n-x)}</math> |
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[[ECE-QE_CS1-2016|Back to QE CS question 1, August 2016]] | [[ECE-QE_CS1-2016|Back to QE CS question 1, August 2016]] | ||
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Revision as of 23:46, 18 February 2019
Communication Signal (CS)
Question 1: Random Variable
August 2016 Problem 3
Solution
a)
Because $ X, Y $ are independent jointly distribute Poisson random variable.
$ P_{X+Y}(x,y)=P_X(x)\dot P_Y(y) $
Such that $ P_Z(z)=\sum_{x=0}^{z} e^{-\lambda}\dfrac{\lambda^x}{x!}e^{-\mu}\dfrac{\mu^{(z-x)}}{(z-x)!} =\dfrac{e^{-(\lambda+\mu)}}{z!}\sum_{x=0}^{z} \begin{pmatrix} z \\ x \end{pmatrix} \lambda^x\mu^{(z-x)} =e^{-(\lambda+\mu)}\dfrac{(\lambda+\mu)^z}{z!} $
b)
when $ x>n $ $ P_X(x)=0 $
when $ 0<=x<=n $
$ P_{X|Z}(x|n) = P_{X,Y}(X=x,Y=n-x|Z=n)=\dfrac{e^{-\lambda}\dfrac{\lambda^x}{x!}e^{-\mu}\dfrac{\mu^{n-x}}{(n-x)!}}{e^{-(\lambda+\mu)}\dfrac{(\lambda+\mu)^n}{n!}} $
$ =\dfrac{n!}{x!(n-x)!}\dot $$ \dfrac{\lambda^x\dot \mu^{n-x}}{(\lambda+\mu)^n}=\begin{pmatrix}n\\x\end{pmatrix}(\dfrac{\lambda}{\lambda+\mu})^x\dot (\dfrac{\mu}{\lambda+\mu})^{(n-x)} $
