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<math>\color{blue}\text{Solution 2:}</math>  
 
<math>\color{blue}\text{Solution 2:}</math>  
  
−
<math>d\in\Re^{2}, d\neq0 \text{ is a feasible direction at } x^{*}</math>&nbsp;<br>
+
here put sol.2
−
 
+
−
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\text{ if } \exists \alpha_{0} \text{ that } \left[ \begin{array}{c} \frac{1}{2} \\ 0 \end{array} \right] + \alpha\left[ \begin{array}{c} d_{1} \\ d_{2} \end{array} \right] \in\Omega \text{ for all } 0\leq\alpha\leq\alpha_{0}</math>
+
−
 
+
−
'''<math>\because \begin{Bmatrix}x\in\Omega: x_{1}\geq0, x_{2}\geq0\end{Bmatrix}</math>'''
+
−
 
+
−
<br> <math>\therefore d=
+
−
\left[ \begin{array}{c} d_{1} \\ d_{2} \end{array} \right], d_{1}\in\Re, d_{2}\geq0</math>
+
−
 
+
 
----
 
----
  
−
<math>\color{blue}\left( \text{ii} \right) \text{Write down the second-order necessary condition for } x^{*} \text{. Does the point } x^{*} \text{ satisfy this condition?}</math><br>  
+
<math>\color{blue}\left( \text{b} \right) \text{Find}  
 +
f_{x}\left( x|y,z\right )
 +
</math><br>  
  
 
<math>\color{blue}\text{Solution 1:}</math>  
 
<math>\color{blue}\text{Solution 1:}</math>  
  
−
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>\text{Let } f\left(x\right)=x_{1}^{2}-x_{1}+x_{2}+x_{1}x_{2} \text{ , } g_{1}\left(x\right)=-x_{1} \text{ , } g_{2}\left(x\right)=-x_{2}</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
+
<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 )}
 +
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
 
</span></font>  
 
</span></font>  
  
−
'''<font face="serif"><math>\text{It is equivalent to minimize } f\left(x\right) \text{,  }</math>&nbsp;&nbsp;</font>'''
+
'''<font face="serif"><math>
−
 
+
= \frac{e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}}{\sqrt[]{2\pi}z}
−
'''<font face="serif"></font>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\text{ subject to } g_{1}(x)\leq0, g_{2}(x)\leq0</math>'''
+
</math>&nbsp;&nbsp;</font>'''  
−
 
+
−
<font color="#ff0000" style="font-size: 17px;">'''<math>\left\{\begin{matrix}
+
−
l\left(x,\mu \right) = \nabla f(x)+\mu_{1}\nabla g_{1}(x)+ \mu_{2}\nabla g_{2}(x) \\ =\left( \begin{array}{c} 2x_{1}-1+x_{2} \\ 1+x_{1} \end{array} \right) + \left( \begin{array}{c} -\mu_{1} \\ 0 \end{array} \right) +\left( \begin{array}{c} 0 \\ -\mu_{2} \end{array} \right) =0\\
+
−
-\mu_{1}x_{1}-\mu_{2}x_{2} = 0 \\
+
−
x_{1} = \frac{1}{2},x_{2} = 0
+
−
\end{matrix}\right.</math>'''</font><br><math>\Rightarrow \mu_{1}=0 , \mu_{2}=3/2</math>&nbsp; &nbsp;  
+
−
 
+
−
<math>\therefore x^{*} \text{ satisfies FONC}</math>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;
+
−
 
+
−
<math>\color{green} \text{There exist } \mu \text{ which make point } x^{*} \text{ satisfies FONC.}</math>
+
−
 
+
−
<math>\text{SONC: } L(x^{*},\mu^{*}) = \nabla l(x^{*},\mu^{*})=\left( \begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array} \right)</math>
+
−
 
+
−
<font color="#ff0000">'''<math>T(x^{*},\mu^{*}): \begin{cases} y^{T}\nabla g_{1}(x)=0 \\ y^{T}\nabla g_{2}(x)=0 \end{cases} : \begin{cases} y^{T}\left( \begin{array}{c} -1 \\ 0 \end{array} \right)=0 \\ y^{T}\left( \begin{array}{c} 0 \\-1 \end{array} \right)=0 \end{cases} \Rightarrow y=\left( \begin{array}{c} 0 \\0 \end{array} \right)</math><br>'''</font>
+
−
 
+
−
<math>\color{green} \text {Here not using formal set expression. }</math>&nbsp;&nbsp;<math>\color{red} T\left( x^{* },\mu^{* } \right) \text{ should be } T\left( x^{* } \right)</math>
+
−
 
+
−
<math>\text{The SONC condition is for all } y\in T \left(x^{*},\mu^{*} \right) , y^{T}L\left(x^{*},\mu^{*} \right)y \geq 0</math>
+
−
 
+
−
<math>y^{T}L\left(x^{*},\mu^{*} \right)y =0 \geq 0 \text{. So } x^{*} \text{satisfies SONC.}</math><br>
+
−
 
+
−
<math>\color{red} \text{For SONC, } T\left( x^{* } \right)= \left \{ y\in\Re^{n}: Dh\left( x^{*} \right)y=0, Dg_{j}\left( x^{*} \right)y=0, j\in J\left( x^{*} \right)  \right \}</math>
+
−
 
+
−
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\color{red}  J\left(x^{*}\right)= \left \{  j:g_{j}\left(x^{*}\right)=0 \right \}</math>
+
−
 
+
−
<math>\color{red} \text{For SOSC, }  \tilde{T}\left( x^{* },\mu^{*} \right)= \left \{ y: Dh\left( x^{*} \right)y=0, Dg_{i}\left( x^{*} \right)y=0, i\in \tilde{J}\left( x^{*},\mu^{*} \right)  \right \}</math>
+
−
 
+
−
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\color{red} \tilde{J}\left ( x^{\ast },\mu ^{\ast } \right )= \left \{ i:g_{i}\left ( x^{\ast } \right ) = 0,\mu_{i}^{\ast }> 0\right \}</math><br>
+
−
 
+
−
<math>\color{red} \tilde{J}\left ( x^{\ast },\mu ^{\ast } \right ) \subset 
+
−
J\left(x^{*}\right)</math>, &nbsp; &nbsp;&nbsp;<math>\color{red} T\left( x^{* } \right) \subset \tilde{T}\left( x^{* },\mu^{*} \right)</math>
+
  
 
----
 
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Line 91: Line 56:
 
<math>\color{blue}\text{Solution 2:}</math><br>  
 
<math>\color{blue}\text{Solution 2:}</math><br>  
  
−
<math>\text{The problem is equivalent to  min} f\left(x_{1},x_{2}\right) = x_{1}^{2}-x_{1}+x_{2}+x_{1}x_{2}</math>&nbsp;&nbsp;
+
sol2 here
−
 
+
−
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\text{subject to  }  x_{1}\leq0, x_{2}\leq0</math>
+
−
 
+
−
<math>Df\left ( x \right )=\left ( \nabla f\left ( x \right ) \right )^{T} = \left [ \frac{\partial f}{\partial x_{1}}\left ( x \right ),\frac{\partial f}{\partial x_{2}}\left ( x \right ) \right ]=\left [ 2x_{1}-1+x_{2},1+x_{1} \right ]</math><br>
+
−
 
+
−
<math>F\left ( x \right ) =D^{2}f\left ( x \right )=\begin{bmatrix}
+
−
\frac{\partial^{2} f}{\partial x_{1}^{2}}\left ( x \right ) & \frac{\partial^{2} f}{\partial x_{2}\partial x_{1}}\left ( x \right )\\
+
−
\frac{\partial^{2} f}{\partial x_{1}\partial x_{2}}\left ( x \right ) & \frac{\partial^{2} f}{\partial x_{2}^{2}}\left ( x \right )
+
−
\end{bmatrix}=\left [ \begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array} \right ]</math>
+
−
 
+
−
<math>\text{SONC for local minimizer } x^{*}=\begin{bmatrix} \frac{1}{2}\\0 \end{bmatrix}</math>
+
−
 
+
−
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>d^{T} \nabla f\left ( x^{*} \right )=0  \cdots \left ( 1 \right )</math>&nbsp; &nbsp; &nbsp;
+
−
 
+
−
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>d^{T} F\left ( x^{*} \right )d\geq 0  \cdots \left ( 2\right )</math><br>
+
−
 
+
−
<math>\text{For (1), } \begin{bmatrix} d_{1} & d_{2} \end{bmatrix}\begin{bmatrix} 0\\ \frac{3}{2}\end{bmatrix} =0 \Rightarrow  d_{1}\in\Re, d_{2}=0</math><br>
+
−
 
+
−
<math>\text{For (2), } F\left ( x \right ) = \begin{bmatrix} 2 &1 \\ 1 &0\end{bmatrix}>0</math>&nbsp; &nbsp; &nbsp; &nbsp;<math>\color{green}  A=\begin{bmatrix} a &b \\ c &d\end{bmatrix} \text{ is positive definite when } a>0 \text{ and } ac-b^{2}>0</math><br>
+
−
 
+
−
<math>\therefore \text{ for all } d\in\Re^{n}, d^{T}F\left ( x^{*} \right )d\geq 0</math>
+
−
 
+
−
<font face="serif"><math>\text{The point } x^{*}=\begin{bmatrix} \frac{1}{2}\\0 \end{bmatrix} \text{ satisfies SONC for local minimizer.}</math><br></font>
+
−
 
+
−
----
+
−
----
+
−
<font face="serif"></font><math>\color{blue}\text{Related Problem: For function }</math>
+
−
 
+
−
&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>f\left( x_{1},x_{2}  \right) =\frac{1}{3} x_{1}^{3} + \frac{1}{3} x_{2}^{3} -x_{1}x_{2}</math>
+
−
 
+
−
<math>\color{blue} \text{Find point(s) that satisfy FONC and check if they are strict local minimizers.}</math>
+
−
 
+
−
<math>\color{blue}\text{Solution:}</math>
+
−
 
+
−
<math>\text{Applying FONC gives } \nabla f\left ( x \right )=\begin{bmatrix}
+
−
x_{1}^{2}-x_{2}\\
+
−
x_{2}^{2}-x_{1}
+
−
\end{bmatrix}=0</math>
+
−
 
+
−
&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\Rightarrow x^{\left ( 1 \right )}=\begin{bmatrix}
+
−
0\\
+
−
0
+
−
\end{bmatrix} \text{ and }x^{\left ( 2 \right )}=\begin{bmatrix}
+
−
1\\
+
−
1
+
−
\end{bmatrix}</math>
+
−
 
+
−
<math>\text{The Hessian matrix: } F\left ( x \right )=\begin{bmatrix}
+
−
2x_{1} & -1\\
+
−
-1 & 2x_{2}
+
−
\end{bmatrix}</math>
+
−
 
+
−
&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\text{The matrix } F\left ( x^{\left ( 1 \right )} \right )=\begin{bmatrix}
+
−
0 & -1\\
+
−
-1 & 0
+
−
\end{bmatrix} \text{ is indefinite. The point is not a minimizer.}</math>
+
−
 
+
−
&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\text{The matrix } F\left ( x^{\left ( 2\right )} \right )=\begin{bmatrix}
+
−
0 & -1\\
+
−
-1 & 0
+
−
\end{bmatrix} \text{ is positive definite. }</math>
+
−
 
+
−
<math>\therefore x^{\left ( 2 \right )}=\begin{bmatrix}
+
−
1\\
+
−
1
+
−
\end{bmatrix} \text{ satisfies SOSC to be a strict local minimizer.}</math>
+
−
 
+
 
----
 
----
  
−
Automatic Control (AC)- Question 3, August 2011  
+
"Communication, Networks, Signal, and Image Processing" (CS)- Question 1, August 2011  
  
 
Go to  
 
Go to  
  
−
*Part 1: [[ECE-QE_AC3-2011_solusion-1|solutions and discussions]]  
+
*Part 1: [[ECE-QE_CS1-2011_solusion-1|solutions and discussions]]  
−
*Part 2: [[ECE-QE AC3-2011 solusion-2|solutions and discussions]]
+
*Part 2: [[ECE-QE CS1-2011 solusion-2|solutions and discussions]]  
−
*Part 3: [[ECE-QE AC3-2011 solusion-3|solutions and discussions]]
+
−
*Part 4: [[ECE-QE AC3-2011 solusion-4|solutions and discussions]]
+
−
*Part 5: [[ECE-QE AC3-2011 solusion-5|solutions and discussions]]
+
  
 
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Revision as of 14:20, 24 July 2012

ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS)

Question 1, 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

"Communication, Networks, Signal, and Image Processing" (CS)- Question 1, August 2011

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