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

Revision as of 13:08, 3 December 2015

Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2015

waht

@@ Line 21: / Line 21: @@
 </center>
 ----
-<math>
+waht
-\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*}
-</math>
 ----
 [[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]]
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