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<math>P_Z(k)=P_Z(z=k)=P_Z(x+yk)\\ | <math>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)!} | =\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!} | ||
</math> | </math> | ||
---- | ---- | ||
Revision as of 11:53, 7 December 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 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!} $
