Difference between revisions of "ECE PhD QE CNSIP 2015 Problem1.1" - Rhea

Revision as of 13:14, 3 December 2015

First of all, the conditional distribution can be written as:

$ 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)} $

And

$ 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} $

Also

$ {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 $

So, we get $ 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*} $

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