| Line 21: | Line 21: | ||
</center> | </center> | ||
---- | ---- | ||
| − | + | <math> | |
| + | 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)}\\ | ||
| + | </math> | ||
| + | <math> | ||
| + | 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}\\ | ||
| + | </math> | ||
| + | <math> | ||
| + | {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 | ||
| + | </math> | ||
| + | So | ||
| + | <math> | ||
| + | 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*} | ||
| + | </math> | ||
---- | ---- | ||
[[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]] | [[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]] | ||
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Revision as of 13:11, 3 December 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2015
$ 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)}\\ $ $ 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}\\ $ $ {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 $ 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*} $
