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− | = [[ECE- | + | = [[ECE-QE_CS5-2011|Question 5, August 2011]], Part 1 = |
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− | "Communication, Networks, Signal, and Image Processing" (CS)- Question | + | "Communication, Networks, Signal, and Image Processing" (CS)- Question 5, August 2011 |
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− | *Part 1: [[ECE- | + | *Part 1: [[ECE-QE_CS5-2011_solusion-1|solutions and discussions]] |
− | *Part 2: [[ECE-QE | + | *Part 2: [[ECE-QE CS5-2011 solusion-2|solutions and discussions]] |
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− | [[Category:ECE]] [[Category:QE]] [[Category: | + | [[Category:ECE]] [[Category:QE]] [[Category:CS]] [[Category:Problem_solving]] |
Revision as of 17:50, 30 July 2012
ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS)
Question 5, August 2011, Part 1
- Part 1,2]
$ \color{blue}\text{1. } \left( \text{25 pts} \right) \text{ Let X, Y, and Z be three jointly distributed random variables with joint pdf} f_{XYZ}\left ( x,y,z \right )= \frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy} exp \left [ -\frac{1}{2}\left ( \frac{x-y}{z}\right )^{2} \right ] \cdot 1_{\left[0,\infty \right )}\left(y \right )\cdot1_{\left[1,2 \right]} \left ( z \right) $
$ \color{blue}\left( \text{a} \right) \text{ Find the joint probability density function } f_{YZ}(y,z). $
$ \color{blue}\text{Solution 1:} $
$ f_{YZ}\left (y,z \right )=\int_{-\infty}^{+\infty}f_{XYZ}\left(x,y,z \right )dx $
$ =\frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy}\int_{-\infty}^{+\infty}exp\left[-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2} \right ]dx\cdot 1_{[0,\infty)} \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) $
$ \text{But}\int_{-\infty}^{+\infty}exp\left[-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2} \right ]dx \text{looks like the Gaussian pdf, so} $
$ =\frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy} \underset{\sqrt[]{2\pi}z}{\underbrace{\frac{7\sqrt[]{2\pi}z}{7\sqrt[]{2\pi}z} \int_{-\infty}^{+\infty}exp\left[-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2} \right ]dx}}\cdot 1_{[0,\infty)} \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) $
$ =\frac{3z^{2}}{7}e^{-zy}\cdot 1_{[0,\infty)} \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) $
$ \color{blue}\text{Solution 2:} $
here put sol.2
$ \color{blue}\left( \text{b} \right) \text{Find} f_{x}\left( x|y,z\right ) $
$ \color{blue}\text{Solution 1:} $
$ = \frac{f_{XYZ}\left( x,y,z\right )}{f_{YZ}\left(y,z \right )} $
$ = \frac{e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}}{\sqrt[]{2\pi}z} $
$ \color{blue}\text{Solution 2:} $
sol2 here
$ \color{blue}\left( \text{c} \right) \text{Find} f_{Z}\left( z\right ) $
$ \color{blue}\text{Solution 1:} $
$ =\int_{0}^{+\infty}{f_{YZ}\left(y,z \right )dy} $
$ =\frac{3z^{2}}{7}\cdot1_{\left[1,2 \right ]}(z) $
$ \color{blue}\text{Solution 2:} $
sol2 here
$ \color{blue}\left( \text{d} \right) \text{Find} f_{Y}\left(y|z \right ) $
$ \color{blue}\text{Solution 1:} $
$ =\frac{f_{YZ}\left(y,z \right )}{f_{Z}(z)} $
$ =e^{-zy}z\cdot1_{\left[0,\infty \right )}(y) $
$ \color{blue}\text{Solution 2:} $
sol2 here
$ \color{blue}\left( \text{e} \right) \text{Find} f_{XY}\left(x,y|z \right ) $
$ \color{blue}\text{Solution 1:} $
$ =\frac{f_{XYZ}\left(x,y,z \right )}{f_{Z}(z)} $
$ =\frac{e^{-zy}}{\sqrt[]{2\pi}}e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}\cdot1_{\left[0,\infty \right )}(y) $
$ \color{blue}\text{Solution 2:} $
sol2 here
"Communication, Networks, Signal, and Image Processing" (CS)- Question 5, August 2011
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- Part 1: solutions and discussions
- Part 2: solutions and discussions