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 $n=n$ $p=\dfrac{\lambda}{\lambda+\mu}$