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=e^{-(\lambda+\mu)}\dfrac{(\lambda+\mu)^z}{z!}</math><br> | =e^{-(\lambda+\mu)}\dfrac{(\lambda+\mu)^z}{z!}</math><br> | ||
b)<br> | b)<br> | ||
| − | when <math>x>n</math> <math>P_X(x)=0</math><br> | + | when <math>x>n</math><br> <math>P_X(x)=0</math><br> |
| − | when <math>0 | + | when <math>0\le x\le 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)!} </math><math> \dfrac{\lambda^x\mu^{n-x}}{(\lambda+\mu)^n}=\begin{pmatrix}n\\x\end{pmatrix}(\dfrac{\lambda}{\lambda+\mu})^x(\dfrac{\mu}{\lambda+\mu})^{(n-x)}</math> | + | <math>=\dfrac{n!}{x!(n-x)!} </math><math> \dfrac{\lambda^x\mu^{n-x}}{(\lambda+\mu)^n}=\begin{pmatrix}n\\x\end{pmatrix}(\dfrac{\lambda}{\lambda+\mu})^x(\dfrac{\mu}{\lambda+\mu})^{(n-x)}</math><br> |
| + | <math>=\begin{pmatrix}n\\x\end{pmatrix}(\dfrac{\lambda}{\lambda+\mu})^x(1-\dfrac{\lambda}{\lambda+\mu})^{(n-x)}</math><br> | ||
| + | Such that <math>x</math> on condition <math>z=n</math> is binomial distributed <math>n=n</math> <math>p=\dfrac{\lambda}{\lambda+\mu}</math><br> | ||
---- | ---- | ||
[[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]] | ||
Latest revision as of 15:33, 19 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\le x\le 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)!} $$ \dfrac{\lambda^x\mu^{n-x}}{(\lambda+\mu)^n}=\begin{pmatrix}n\\x\end{pmatrix}(\dfrac{\lambda}{\lambda+\mu})^x(\dfrac{\mu}{\lambda+\mu})^{(n-x)} $
$ =\begin{pmatrix}n\\x\end{pmatrix}(\dfrac{\lambda}{\lambda+\mu})^x(1-\dfrac{\lambda}{\lambda+\mu})^{(n-x)} $
Such that $ x $ on condition $ z=n $ is binomial distributed $ n=n $ $ p=\dfrac{\lambda}{\lambda+\mu} $
