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

Revision as of 11:54, 7 December 2015

Solution 1

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} $

Solution 2

Let $ Z=X+Y $,

$ P_Z(k)=P_Z(z=k)=P_Z(x+yk)\\ =\sum_{i=0}^{k}P(x=i)(y=k-i)=\sum_{i=0}^{k}\frac{\lambda_1^i e^{\lambda_1}}{i!}\cdot\frac{\lambda_2^{k-i}e^{-\lambda_2}}{(k-i)!} =e^{-\lambda_1-\lambda_2}\cdot \frac{(\lambda_1+\lambda_2)^k}{k!} $

Using $ (a+b)^k=\sum_{i=0}^{k}a^ib^{(k-i)}\cdot \frac{k!}{i!(k-i)!} ---- [[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]] [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] $