(13 intermediate revisions by one other user not shown)
Line 1: Line 1:
−
= [[ECE PhD Qualifying Exams|ECE Ph.D. Qualifying Exam]] in "Communication, Networks, Signal, and Image Processing" (CS)  =
+
[[Category:ECE]]
 +
[[Category:QE]]
 +
[[Category:CNSIP]]
 +
[[Category:problem solving]]
 +
[[Category:random variables]]
 +
[[Category:probability]]
  
−
= [[ECE-QE_CS1-2011|Question 1, August 2011]], Part 1 =
+
<center>
 +
<font size= 4>
 +
[[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]
 +
</font size>
  
−
:[[ECE-QE_CS1-2011_solusion-1|Part 1]],[[ECE-QE CS1-2011 solusion-2|2]]]
+
<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 21: Line 39:
  
 
<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} \int_{-\infty}^{+\infty}exp\left[-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2} \right ]dx}}\cdot 1_{[0,\infty)}
+
\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>
Line 33: 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 49: 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>
 
<math>
−
= \frac{3}{7}z^2 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)
+
\color{green}\text{Based on the Axioms of Probability, this integral over R will be 1.}  
 
</math>
 
</math>
 +
 +
<math>
 +
\color{green}\text{So he wants to replace this integral with 1:}
 +
</math>
 +
 +
<math>
 +
= \frac{3}{7}z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)
 +
</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>
 +
  
 
----
 
----
Line 64: Line 113:
  
 
<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>
Line 84: Line 137:
  
 
<math>
 
<math>
−
\frac{1}{\sqrt[]{2\pi}z} e^{- \frac{(x-y)^2}{2z^2}}
+
\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>
 +
 +
<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>
  
 
----
 
----
Line 159: Line 220:
  
 
<math>
 
<math>
−
= ze^{-zy} \cdot 1_{[0,\infty)}(y)
+
= ze^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)
 
</math>
 
</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>
  
 
----
 
----
Line 190: Line 258:
  
 
<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}
+
= \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>
  
Line 198: Line 266:
 
</math>
 
</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 209: Line 285:
 
----
 
----
  
−
[[ECE PhD Qualifying Exams|Back to ECE Qualifying Exams (QE) page]]
+
[[ECE PhD Qualifying Exams|Back to ECE Qualifying Exams (QE) page]]
−
 
+
−
[[Category:ECE]] [[Category:QE]] [[Category:Automatic_Control]] [[Category:Problem_solving]]
+

Latest revision as of 10:29, 13 September 2013


ECE Ph.D. Qualifying Exam

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

Back to ECE Qualifying Exams (QE) page

Retrieved from "https://www.projectrhea.org/rhea/index.php?title=ECE-QE_CS1-2011_solusion-1&oldid=58235"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010