Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2015
$ \begin{align*} P(X=x|X+Y=n) &=\frac{P(X=x, X+Y=n)}{P(X+Y=n)}\\ &=\frac{P(X=x, Y=n-x)}{P(X+Y=n)} \end{align*} \begin{align*} P(X=x, Y=n-x) &=P(X=x)P(Y=n-x)\\ &=\frac{e^{-\lambda_1}\lambda^x}{x!}\times \frac{e^{-\lambda_2}\lambda^(n-x)}{(n-x)!}\\ &=\frac{e^{-(\lambda_1+\lambda_2)}}{x!} \left( \begin{array}{c} n\\x \end{array} \right) \lambda_1^x\lambda_2^{n-x} \end{align*} \begin{align*} {P(X+Y=n)} &={\sum_{k=0}^{k=n}P(X=k,Y=n-k)}\\ &={\sum_{k=0}^{k=n}P(X=k)P(Y=n-k)}\\ &=\frac{e^{-(\lambda_1+\lambda_2)}}{n!}\sum_{k=0}^{k=n} \left( \begin{array}{c} n\\k \end{array} \right) \lambda_1^k\lambda_2^{n-k} &=\frac{e^{-(\lambda_1+\lambda_2)}}{n!}(\lambda_1+\lambda_2)^n \end{align*} So \begin{align*} P(X=x|X+Y=n) &= \left( \begin{array}{c} n\\k \end{array} \right) (\frac{\lambda_1}{\lambda_1+\lambda_2})^x(\frac{\lambda_2}{\lambda_1+\lambda_2})^{n-x} \end{align*} $
