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− | + | [[Category:ECE]] | |
+ | [[Category:QE]] | ||
+ | [[Category:CNSIP]] | ||
+ | [[Category:problem solving]] | ||
+ | [[Category:random variables]] | ||
+ | [[Category:probability]] | ||
− | = [[ECE | + | <center> |
+ | <font size= 4> | ||
+ | [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] | ||
+ | </font size> | ||
− | + | <font size= 4> | |
+ | Communication, Networking, Signal and Image Processing (CS) | ||
+ | Question 1: Probability and Random Processes | ||
+ | </font size> | ||
+ | |||
+ | August 2011 | ||
+ | </center> | ||
+ | ---- | ||
+ | ---- | ||
+ | =Part 1 = | ||
+ | Jump to [[ECE-QE_CS1-2011_solusion-1|Part 1]],[[ECE-QE CS1-2011 solusion-2|2]] | ||
---- | ---- | ||
Line 13: | Line 31: | ||
===== <math>\color{blue}\text{Solution 1:}</math> ===== | ===== <math>\color{blue}\text{Solution 1:}</math> ===== | ||
− | <math> f_{YZ}\left (y,z \right )=\int_{-\infty}^{+\infty}f_{XYZ}\left(x,y,z \right )dx </math> | + | <math> f_{YZ}\left (y,z \right )=\int_{-\infty}^{+\infty}f_{XYZ}\left(x,y,z \right )dx </math> |
− | + | ||
+ | <math> =\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 )</math><br> | \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )</math><br> | ||
− | <math>\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} </math> | + | |
+ | <math>\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} </math> | ||
+ | |||
+ | <math> | ||
+ | \color{green}\text{It should be added: Based on the Axioms of Probability, this integral over R will be 1.} | ||
+ | </math> | ||
+ | |||
+ | <math> | ||
+ | \color{green}\text{So we can then replace this integral with one.} | ||
+ | </math> | ||
+ | |||
<math> =\frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy} | <math> =\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} | + | \underset{{\color{Orange}\sqrt[]{2\pi}z}} {\underbrace{ {\color{Orange} \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 ) | \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) | ||
</math> | </math> | ||
+ | |||
<math> | <math> | ||
Line 29: | Line 59: | ||
\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) | \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) | ||
</math> | </math> | ||
+ | |||
---- | ---- | ||
Line 35: | Line 66: | ||
<math> | <math> | ||
− | f_{YZ}(y,z) = \int_{-\infty}{\infty}{f_{XYZ}(x,y,z)dx} | + | f_{YZ}(y,z) = \int_{-\infty}^{\infty}{f_{XYZ}(x,y,z)dx} |
</math> | </math> | ||
+ | |||
<math> | <math> | ||
Line 44: | Line 76: | ||
<math> | <math> | ||
− | = \int_{-\infty}^{\infty}{\frac{1}{7\sqrt[]{2\pi}z} e^{-\frac{(x-y)^2}{2z^2}} \cdot \frac{3}{7}z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) dx} | + | = \int_{-\infty}^{\infty}{\frac{1}{7\sqrt[]{2\pi}z} e^{-\frac{(x-y)^2}{2z^2}} \cdot \frac{3}{7}z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) dx} \color{green}\text{ Here the student wants to form a Gaussian pdf.} |
</math> | </math> | ||
+ | |||
+ | <math> | ||
+ | \color{green}\text{Based on the Axioms of Probability, this integral over R will be 1.} | ||
+ | </math> | ||
+ | |||
+ | <math> | ||
+ | \color{green}\text{So he wants to replace this integral with 1:} | ||
+ | </math> | ||
<math> | <math> | ||
= \frac{3}{7}z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) | = \frac{3}{7}z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) | ||
</math> | </math> | ||
+ | |||
+ | <math> | ||
+ | {\color{red}\text{Here, the student forgot to discard one } z. | ||
+ | }</math> | ||
+ | |||
+ | <math> | ||
+ | {\color{red}\text{The correct answer is:} | ||
+ | }</math> | ||
+ | |||
+ | <math> | ||
+ | {\color{red}=\frac{3z^{2}}{7}e^{-zy}\cdot 1_{[0,\infty)} | ||
+ | \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) | ||
+ | }</math> | ||
+ | |||
---- | ---- | ||
− | <math>\color{blue}\left( \text{b} \right) \text{Find} | + | <math>\color{blue}\left( \text{b} \right) \text{Find } |
f_{x}\left( x|y,z\right ) | f_{x}\left( x|y,z\right ) | ||
</math><br> | </math><br> | ||
<math>\color{blue}\text{Solution 1:}</math> | <math>\color{blue}\text{Solution 1:}</math> | ||
+ | |||
+ | <math> | ||
+ | \color{green}\text{It should be added: According to the Bayes rule:} | ||
+ | </math> | ||
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math> | <font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math> | ||
− | = \frac{f_{XYZ}\left( x,y,z\right )}{f_{YZ}\left(y,z \right )} | + | f_X(x|y,z) = \frac{f_{XYZ}\left( x,y,z\right )}{f_{YZ}\left(y,z \right )} |
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;"> | </math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;"> | ||
</span></font> | </span></font> | ||
Line 73: | Line 131: | ||
<math>\color{blue}\text{Solution 2:}</math><br> | <math>\color{blue}\text{Solution 2:}</math><br> | ||
− | + | <math> | |
+ | f_X(x|y,z) = \frac{f_{XYZ}(x,y,z)}{f_{YZ}(y,z)} = \frac{\frac{3z^2}{7\sqrt[]{2\pi}} e^{-zy} \cdot e^{- \frac{(x-y)^2}{2z^2}} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)} {\frac{3}{7} z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)} | ||
+ | </math> | ||
+ | |||
+ | |||
+ | <math> | ||
+ | \frac{1}{\sqrt[]{2\pi}z} e^{- \frac{(x-y)^2}{2z^2}} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) | ||
+ | </math> | ||
+ | |||
+ | <math> | ||
+ | {\color{red}\text{Here, the student forgot to discard } cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) | ||
+ | }</math> | ||
+ | |||
+ | <math> | ||
+ | {\color{red}\text{Correct answer: } \frac{1}{\sqrt[]{2\pi}z} e^{- \frac{(x-y)^2}{2z^2}} | ||
+ | }</math> | ||
+ | |||
---- | ---- | ||
− | <math>\color{blue}\left( \text{c} \right) \text{Find} | + | <math>\color{blue}\left( \text{c} \right) \text{Find } |
f_{Z}\left( z\right ) | f_{Z}\left( z\right ) | ||
</math><br> | </math><br> | ||
Line 83: | Line 157: | ||
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math> | <font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math> | ||
− | =\int_{0}^{+\infty}{f_{YZ}\left(y,z \right )dy} | + | f_Z(z) = \int_{0}^{+\infty}{f_{YZ}\left(y,z \right )dy} |
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;"> | </math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;"> | ||
</span></font> | </span></font> | ||
Line 95: | Line 169: | ||
<math>\color{blue}\text{Solution 2:}</math><br> | <math>\color{blue}\text{Solution 2:}</math><br> | ||
− | + | <math> | |
+ | f_Z(z) = \int_{-\infty}^{\infty}{f_{YZ}(y,z)dy} | ||
+ | </math> | ||
+ | |||
+ | |||
+ | <math> | ||
+ | = \int_{0}^{\infty}{\frac{3z^3}{7} e^{-zy} \cdot 1_{[1,2]}(z) dy} | ||
+ | </math> | ||
+ | |||
+ | |||
+ | <math> | ||
+ | = \frac{3z^2}{7} \cdot \int_{0}^{\infty} z e^{-zy} dy \cdot 1_{[1,2]}(z) | ||
+ | </math> | ||
+ | |||
+ | |||
+ | <math> | ||
+ | = -\frac{3z^2}{7} \cdot e^{-zy} |_{0}^{\infty} \cdot 1_{[1,2]}(z) | ||
+ | </math> | ||
+ | |||
+ | |||
+ | <math> | ||
+ | = \frac{3}{7} z^2 \cdot 1_{[1,2]}(z) | ||
+ | </math> | ||
+ | |||
+ | |||
---- | ---- | ||
− | <math>\color{blue}\left( \text{d} \right) \text{Find} | + | <math>\color{blue}\left( \text{d} \right) \text{Find } |
f_{Y}\left(y|z \right ) | f_{Y}\left(y|z \right ) | ||
</math><br> | </math><br> | ||
Line 105: | Line 203: | ||
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math> | <font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math> | ||
− | =\frac{f_{YZ}\left(y,z \right )}{f_{Z}(z)}</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;"> | + | f_Y(y|z) = \frac{f_{YZ}\left(y,z \right )}{f_{Z}(z)}</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;"> |
</span></font> | </span></font> | ||
Line 116: | Line 214: | ||
<math>\color{blue}\text{Solution 2:}</math><br> | <math>\color{blue}\text{Solution 2:}</math><br> | ||
− | + | <math> | |
+ | f_Y(y|z) = \frac{f_{YZ}(y,z)}{f_Z(z)} = \frac{\frac{3}{7} z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)}{\frac{3}{7} z^2 \cdot 1_{[1,2]}(z)} | ||
+ | </math> | ||
+ | |||
+ | |||
+ | <math> | ||
+ | = ze^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) | ||
+ | </math> | ||
+ | |||
+ | |||
+ | <math> | ||
+ | {\color{red}\text{Here, the student forgot to discard } \cdot 1_{[1,2]}(z) | ||
+ | }</math> | ||
+ | |||
+ | <math> | ||
+ | {\color{red}\text{Correct answer: } = ze^{-zy} \cdot 1_{[0,\infty)}(y) | ||
+ | }</math> | ||
+ | |||
---- | ---- | ||
− | <math>\color{blue}\left( \text{e} \right) \text{Find} | + | <math>\color{blue}\left( \text{e} \right) \text{Find } |
f_{XY}\left(x,y|z \right ) | f_{XY}\left(x,y|z \right ) | ||
</math><br> | </math><br> | ||
Line 125: | Line 240: | ||
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math> | <font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math> | ||
− | =\frac{f_{XYZ}\left(x,y,z \right )}{f_{Z}(z)} | + | f_{XY}(x,y|z) = \frac{f_{XYZ}\left(x,y,z \right )}{f_{Z}(z)} |
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;"> | </math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;"> | ||
</span></font> | </span></font> | ||
Line 137: | Line 252: | ||
<math>\color{blue}\text{Solution 2:}</math><br> | <math>\color{blue}\text{Solution 2:}</math><br> | ||
− | + | <math> | |
+ | f_{XY}(x,y|z) = \frac{f_{XYZ}(x,y,z)}{f_Z(z)} | ||
+ | </math> | ||
+ | |||
+ | |||
+ | <math> | ||
+ | = \frac{\frac{3z^2}{7\sqrt[]{2\pi}} e^{-zy} \cdot e^{-\frac{1}{2} (\frac{x-y}{z})^2} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)}{\frac{3}{7} z^2 \cdot 1_{[1,2]}(z)} | ||
+ | </math> | ||
+ | |||
+ | |||
+ | <math> | ||
+ | = \frac{1}{\sqrt[]{2\pi}} e^{-zy} \cdot e^{-\frac{1}{2} (\frac{x-y}{z})^2} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) | ||
+ | </math> | ||
+ | |||
+ | |||
+ | <math> | ||
+ | {\color{red}\text{Here, the student forgot to discard } \cdot 1_{[1,2]}(z) | ||
+ | }</math> | ||
+ | |||
+ | <math> | ||
+ | {\color{red}\text{Correct answer: } = \frac{1}{\sqrt[]{2\pi}} e^{-zy} \cdot e^{-\frac{1}{2} (\frac{x-y}{z})^2} \cdot 1_{[0,\infty)}(y) | ||
+ | }</math> | ||
---- | ---- | ||
Line 149: | Line 285: | ||
---- | ---- | ||
− | [[ECE PhD Qualifying Exams|Back to ECE Qualifying Exams (QE) page | + | [[ECE PhD Qualifying Exams|Back to ECE Qualifying Exams (QE) page]] |
− | + | ||
− | + |
Latest revision as of 10:29, 13 September 2013
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2011
Part 1
Jump to 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} $
$ \color{green}\text{It should be added: Based on the Axioms of Probability, this integral over R will be 1.} $
$ \color{green}\text{So we can then replace this integral with one.} $
$ =\frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy} \underset{{\color{Orange}\sqrt[]{2\pi}z}} {\underbrace{ {\color{Orange} \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:} $
$ f_{YZ}(y,z) = \int_{-\infty}^{\infty}{f_{XYZ}(x,y,z)dx} $
$ = \int_{-\infty}^{\infty}{\frac{3z^2}{7\sqrt[]{2\pi}} e^{-zy} \cdot e^{-\frac{1}{2} \frac{(x-y)^2}{z^2}} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) dx} $
$ = \int_{-\infty}^{\infty}{\frac{1}{7\sqrt[]{2\pi}z} e^{-\frac{(x-y)^2}{2z^2}} \cdot \frac{3}{7}z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) dx} \color{green}\text{ Here the student wants to form a Gaussian pdf.} $
$ \color{green}\text{Based on the Axioms of Probability, this integral over R will be 1.} $
$ \color{green}\text{So he wants to replace this integral with 1:} $
$ = \frac{3}{7}z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) $
$ {\color{red}\text{Here, the student forgot to discard one } z. } $
$ {\color{red}\text{The correct answer is:} } $
$ {\color{red}=\frac{3z^{2}}{7}e^{-zy}\cdot 1_{[0,\infty)} \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) } $
$ \color{blue}\left( \text{b} \right) \text{Find } f_{x}\left( x|y,z\right ) $
$ \color{blue}\text{Solution 1:} $
$ \color{green}\text{It should be added: According to the Bayes rule:} $
$ f_X(x|y,z) = \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:} $
$ f_X(x|y,z) = \frac{f_{XYZ}(x,y,z)}{f_{YZ}(y,z)} = \frac{\frac{3z^2}{7\sqrt[]{2\pi}} e^{-zy} \cdot e^{- \frac{(x-y)^2}{2z^2}} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)} {\frac{3}{7} z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)} $
$ \frac{1}{\sqrt[]{2\pi}z} e^{- \frac{(x-y)^2}{2z^2}} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) $
$ {\color{red}\text{Here, the student forgot to discard } cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) } $
$ {\color{red}\text{Correct answer: } \frac{1}{\sqrt[]{2\pi}z} e^{- \frac{(x-y)^2}{2z^2}} } $
$ \color{blue}\left( \text{c} \right) \text{Find } f_{Z}\left( z\right ) $
$ \color{blue}\text{Solution 1:} $
$ f_Z(z) = \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:} $
$ f_Z(z) = \int_{-\infty}^{\infty}{f_{YZ}(y,z)dy} $
$ = \int_{0}^{\infty}{\frac{3z^3}{7} e^{-zy} \cdot 1_{[1,2]}(z) dy} $
$ = \frac{3z^2}{7} \cdot \int_{0}^{\infty} z e^{-zy} dy \cdot 1_{[1,2]}(z) $
$ = -\frac{3z^2}{7} \cdot e^{-zy} |_{0}^{\infty} \cdot 1_{[1,2]}(z) $
$ = \frac{3}{7} z^2 \cdot 1_{[1,2]}(z) $
$ \color{blue}\left( \text{d} \right) \text{Find } f_{Y}\left(y|z \right ) $
$ \color{blue}\text{Solution 1:} $
$ f_Y(y|z) = \frac{f_{YZ}\left(y,z \right )}{f_{Z}(z)} $
$ =e^{-zy}z\cdot1_{\left[0,\infty \right )}(y) $
$ \color{blue}\text{Solution 2:} $
$ f_Y(y|z) = \frac{f_{YZ}(y,z)}{f_Z(z)} = \frac{\frac{3}{7} z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)}{\frac{3}{7} z^2 \cdot 1_{[1,2]}(z)} $
$ = ze^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) $
$ {\color{red}\text{Here, the student forgot to discard } \cdot 1_{[1,2]}(z) } $
$ {\color{red}\text{Correct answer: } = ze^{-zy} \cdot 1_{[0,\infty)}(y) } $
$ \color{blue}\left( \text{e} \right) \text{Find } f_{XY}\left(x,y|z \right ) $
$ \color{blue}\text{Solution 1:} $
$ f_{XY}(x,y|z) = \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:} $
$ f_{XY}(x,y|z) = \frac{f_{XYZ}(x,y,z)}{f_Z(z)} $
$ = \frac{\frac{3z^2}{7\sqrt[]{2\pi}} e^{-zy} \cdot e^{-\frac{1}{2} (\frac{x-y}{z})^2} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)}{\frac{3}{7} z^2 \cdot 1_{[1,2]}(z)} $
$ = \frac{1}{\sqrt[]{2\pi}} e^{-zy} \cdot e^{-\frac{1}{2} (\frac{x-y}{z})^2} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) $
$ {\color{red}\text{Here, the student forgot to discard } \cdot 1_{[1,2]}(z) } $
$ {\color{red}\text{Correct answer: } = \frac{1}{\sqrt[]{2\pi}} e^{-zy} \cdot e^{-\frac{1}{2} (\frac{x-y}{z})^2} \cdot 1_{[0,\infty)}(y) } $
"Communication, Networks, Signal, and Image Processing" (CS)- Question 1, August 2011
Go to
- Part 1: solutions and discussions
- Part 2: solutions and discussions