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!} $
