(New page: = ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS) = = Question 1, August 2011, Part 1 = :[[ECE...)
 
 
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= [[ECE PhD Qualifying Exams|ECE Ph.D. Qualifying Exam]] in "Communication, Networks, Signal, and Image Processing" (CS)  =
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[[Category:ECE]]
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[[Category:QE]]
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[[Category:CNSIP]]
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[[Category:problem solving]]
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[[Category:random variables]]
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[[Category:probability]]
  
= [[ECE-QE_CS1-2011|Question 1, August 2011]], Part 1 =
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<center>
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<font size= 4>
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[[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]
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</font size>
  
:[[ECE-QE_CS1-2011_solusion-1|Part 1]],[[ECE-QE CS1-2011 solusion-2|2]]]
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<font size= 4>
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Communication, Networking, Signal and Image Processing (CS)
  
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Question 1: Probability and Random Processes
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</font size>
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August 2011
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</center>
 
----
 
----
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----
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=Part 2 =
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Jump to [[ECE-QE_CS1-2011_solusion-1|Part 1]],[[ECE-QE CS1-2011 solusion-2|2]]
 +
----
 +
&nbsp;<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{Show that if a continuous-time Gaussian random process } \mathbf{X}(t) \text{ is wide-sense stationary, it is also strict-sense stationary.}
 +
</math></span></font>
  
&nbsp;<font color="#ff0000"><span style="font-size: 19px;"><math>\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) </math></span></font>  
+
===== <math>\color{blue}\text{Solution 1:}</math>  =====
 +
<math>
 +
{\color{green} \text{Recall Should be added:}}
 +
</math>
  
'''<math>\color{blue}\left( \text{a} \right) \text{ Find the joint probability density function } f_{YZ}(y,z).</math>'''<br>  
+
<math>
 +
{\color{green} \text{A random process is wide sense stationary (WSS) if}}
 +
</math>
  
===== <math>\color{blue}\text{Solution 1:}</math> =====
+
<math>
 +
{\color{green} i) \text{ its mean is constant.}}
 +
</math>
  
<math> f_{YZ}\left (y,z \right )=\int_{-\infty}^{+\infty}f_{XYZ}\left(x,y,z \right )dx </math>&nbsp;
+
<math>
 +
{\color{green} ii) \text{ its correlation only depends on time deference.}}
 +
</math>
  
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<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)}
+
<math>
\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )</math><br>
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{\color{green} \text{A random process is Strict Sense Stationary (SSS) if its cdf only depends on time deference.}}
 +
</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{This Also true for the Moment Generating Function of the process, so we can use this function for our proof:}}
 +
</math>
  
<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)}
+
<font face="serif"><span style="font-size: 19px;"><math>
\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )
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\mathbf{X}(t) \text{ is SSS if } F_{(t_1+\tau)...(t_n+\tau)}(x_1,...,x_n) \text{ does not depend on } \tau. \text{ To show that, we can show that } \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) \text{ does not depend on } \tau:
 +
</math></span></font>
 +
 
 +
<math>
 +
\Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) = E \left [e^{i\sum_{j=1}^{n}{\omega_jX(t_j+\tau)}} \right ]
 
</math>
 
</math>
 +
 +
 +
<math>
 +
\text{Define } Y(t_j+\tau) = \sum_{j=1}^{n}{\omega_jX(t_j+\tau)} \text{, so}
 +
</math>
 +
  
 
<math>
 
<math>
=\frac{3z^{2}}{7}e^{-zy}\cdot 1_{[0,\infty)}
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\Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) = E \left [e^{Y(t_j+\tau)} \right ] = \Phi_{(t_1+\tau)...(t_n+\tau)}(1)
\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )
+
 
</math>
 
</math>
  
----
 
  
<math>\color{blue}\text{Solution 2:}</math>  
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<font face="serif"><span style="font-size: 19px;"><math>
 +
\text{Since } Y(t) \text{ is Gaussian, it is characterized just by its mean and variance. So, we just need to show that mean and variance of } Y(t) \text{ do not depend on } \tau. \text{ Since } Y(t) \text{ is  WSS, its mean is constant and does not depend on . For variance}
 +
</math></span></font>
  
here put sol.2
 
----
 
  
<math>\color{blue}\left( \text{b} \right) \text{Find}
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<math>  
f_{x}\left( x|y,z\right )
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var(Y(t_j+\tau)) = E \left [(\sum_{j=1}^{n}{w_j(X(t_j+\tau)-\mu)^2} \right ]
</math><br>  
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</math>
  
<math>\color{blue}\text{Solution 1:}</math>
 
  
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
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<math>  
= \frac{f_{XYZ}\left( x,y,z\right )}{f_{YZ}\left(y,z \right )}
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=\sum_{j=1}^{n}{\omega_j^2E \left [ (X(t_j+\tau)-\mu)^2 \right ]} + \sum_{i,j=1}^{n}{\omega_i \omega_j E \left[ (X(t_i+\tau)-\mu)(X(t_j+\tau)-\mu) \right]}  
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
+
</math>
</span></font>  
+
  
'''<font face="serif"><math>
 
= \frac{e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}}{\sqrt[]{2\pi}z}
 
</math>&nbsp;&nbsp;</font>'''
 
  
----
+
<math>
 +
=\sum_{i,j=1}^{n}{\omega_j^2 cov(t_j,t_j)}  + \sum_{i,j=1}^{n}{\omega_i \omega_j cov(t_j,t_j)}
 +
</math>
  
<math>\color{blue}\text{Solution 2:}</math><br>
 
  
sol2 here
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<font face="serif"><span style="font-size: 19px;"><math>
 +
\text{Which does not depend on } \tau.
 +
</math></span></font>
 +
 
 
----
 
----
  
<math>\color{blue}\left( \text{c} \right) \text{Find}
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===== <math>\color{blue}\text{Solution 2:}</math> =====
f_{Z}\left( z\right )
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</math><br>
+
  
<math>\color{blue}\text{Solution 1:}</math>  
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<math> \text{Suppose } \mathbf{X}(t) \text{ is a Gaussian Random Process}
 +
</math>
  
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
 
=\int_{0}^{+\infty}{f_{YZ}\left(y,z \right )dy}
 
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
 
</span></font>
 
  
'''<font face="serif"><math>
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<font face="serif"><span style="font-size: 19px;"><math>  
=\frac{3z^{2}}{7}\cdot1_{\left[1,2 \right ]}(z)
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\Rightarrow f(x(t_1),x(t_2),...,x(t_k)) = \frac{1}{2\pi^{(\frac{k}{2})} |\Sigma |^{\frac{1}{2}}} exp(-\frac{1}{2}(\overrightarrow{x} - \overrightarrow{m})^T \Sigma ^{-1}(\overrightarrow{x} - \overrightarrow{m}))
</math>&nbsp;&nbsp;</font>'''
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</math></span></font>
  
----
 
  
<math>\color{blue}\text{Solution 2:}</math><br>
 
  
sol2 here
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<font face="serif"><span style="font-size: 19px;"><math>
----
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\text{for any number of time instances.}
 +
</math></span></font>
  
<math>\color{blue}\left( \text{d} \right) \text{Find}
 
f_{Y}\left(y|z \right )
 
</math><br>
 
  
<math>\color{blue}\text{Solution 1:}</math>  
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<math>  
 +
\text{If } \mathbf{X}(t) \text{is WSS}
 +
</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;">
 
</span></font>
 
  
'''<font face="serif"><math>
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<font face="serif"><span style="font-size: 19px;"><math>  
=e^{-zy}z\cdot1_{\left[0,\infty \right )}(y)
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\Rightarrow \text{ (1) } m_X(t_1) = m_X(t_2) = ... = m_X(t_K) = m
</math>&nbsp;&nbsp;</font>'''
+
</math></span></font>
  
----
 
  
<math>\color{blue}\text{Solution 2:}</math><br>  
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<font face="serif"><span style="font-size: 19px;"><math>  
 +
\text{ (2) } R_X(t_i,t_i) = R_X(t_i + \tau, t_j + \tau)
 +
</math></span></font>
  
sol2 here
 
----
 
<math>\color{blue}\left( \text{e} \right) \text{Find}
 
f_{XY}\left(x,y|z \right )
 
</math><br>
 
  
<math>\color{blue}\text{Solution 1:}</math>  
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<math>
 +
\Sigma = \begin{bmatrix}
 +
&R_X(t_1,t_1)  &... &R_X(t_1,t_k)\\
 +
&\vdots        &                \\
 +
&R_X(t_k,t_1)  &... &R_X(t_k,t_k)\\
 +
\end{bmatrix}
 +
</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)}
 
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
 
</span></font>
 
  
'''<font face="serif"><math>
+
<font face="serif"><span style="font-size: 19px;"><math>
=\frac{e^{-zy}}{\sqrt[]{2\pi}}e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}\cdot1_{\left[0,\infty \right )}(y)
+
\text{From (1): } \overrightarrow{m}' = (m_X(t_1+\tau) , m_X(t_2+\tau) , ... , m_X(t_K+\tau)) = \overrightarrow{m}
</math>&nbsp;&nbsp;</font>'''
+
</math></span></font>
  
----
 
  
<math>\color{blue}\text{Solution 2:}</math><br>  
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<math>
 +
\text{From (2): }
 +
\Sigma' = \begin{bmatrix}
 +
&R_X(t_1,t_1)  &... &R_X(t_1,t_k)\\
 +
&\vdots        &                \\
 +
&R_X(t_k,t_1)  &... &R_X(t_k,t_k)\\
 +
\end{bmatrix} = \Sigma
 +
</math>
 +
 
 +
 
 +
 
 +
<font face="serif"><span style="font-size: 19px;"><math>
 +
{\color{green} \text{Should be clarified that:}}
 +
</math></span></font>
 +
 
 +
 
 +
<math>{ \color{green}
 +
\text{From (2): }
 +
\Sigma' = \begin{bmatrix}
 +
&R_X(t_1+\tau,t_1+\tau)  &... &R_X(t_1+\tau,t_k+\tau)\\
 +
&\vdots        &                \\
 +
&R_X(t_k+\tau,t_1+\tau)  &... &R_X(t_k+\tau,t_k+\tau)\\
 +
\end{bmatrix} = \begin{bmatrix}
 +
&R_X(t_1,t_1)  &... &R_X(t_1,t_k)\\
 +
&\vdots        &                \\
 +
&R_X(t_k,t_1)  &... &R_X(t_k,t_k)\\
 +
\end{bmatrix} = \Sigma
 +
}</math>
 +
 
 +
 
 +
<font face="serif"><span style="font-size: 19px;"><math>
 +
\text{So }  f(x(t_1+\tau),x(t_2+\tau),...,x(t_k+\tau)) \text{ is not related to } \tau.
 +
</math></span></font>
 +
 
 +
 
 +
<font face="serif"><span style="font-size: 19px;"><math>
 +
f(x(t_1+\tau),x(t_2+\tau),...,x(t_k+\tau))
 +
</math></span></font>
 +
 
 +
 
 +
<font face="serif"><span style="font-size: 19px;"><math>
 +
= f(x(t_1),x(t_2),...,x(t_k))
 +
</math></span></font>
 +
 
 +
 
 +
<math>
 +
\Rightarrow \mathbf{X}(t) \text{ is Strict Sense Stationary. }
 +
</math>
 +
 
  
sol2 here
 
 
----
 
----
  

Latest revision as of 10:30, 13 September 2013


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2011



Part 2

Jump to Part 1,2


 $ \color{blue}\text{Show that if a continuous-time Gaussian random process } \mathbf{X}(t) \text{ is wide-sense stationary, it is also strict-sense stationary.} $

$ \color{blue}\text{Solution 1:} $

$ {\color{green} \text{Recall Should be added:}} $

$ {\color{green} \text{A random process is wide sense stationary (WSS) if}} $

$ {\color{green} i) \text{ its mean is constant.}} $

$ {\color{green} ii) \text{ its correlation only depends on time deference.}} $

$ {\color{green} \text{A random process is Strict Sense Stationary (SSS) if its cdf only depends on time deference.}} $

$ {\color{green} \text{This Also true for the Moment Generating Function of the process, so we can use this function for our proof:}} $


$ \mathbf{X}(t) \text{ is SSS if } F_{(t_1+\tau)...(t_n+\tau)}(x_1,...,x_n) \text{ does not depend on } \tau. \text{ To show that, we can show that } \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) \text{ does not depend on } \tau: $

$ \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) = E \left [e^{i\sum_{j=1}^{n}{\omega_jX(t_j+\tau)}} \right ] $


$ \text{Define } Y(t_j+\tau) = \sum_{j=1}^{n}{\omega_jX(t_j+\tau)} \text{, so} $


$ \Phi_{(t_1+\tau)...(t_n+\tau)}(\omega_1,...,\omega_n) = E \left [e^{Y(t_j+\tau)} \right ] = \Phi_{(t_1+\tau)...(t_n+\tau)}(1) $


$ \text{Since } Y(t) \text{ is Gaussian, it is characterized just by its mean and variance. So, we just need to show that mean and variance of } Y(t) \text{ do not depend on } \tau. \text{ Since } Y(t) \text{ is WSS, its mean is constant and does not depend on . For variance} $


$ var(Y(t_j+\tau)) = E \left [(\sum_{j=1}^{n}{w_j(X(t_j+\tau)-\mu)^2} \right ] $


$ =\sum_{j=1}^{n}{\omega_j^2E \left [ (X(t_j+\tau)-\mu)^2 \right ]} + \sum_{i,j=1}^{n}{\omega_i \omega_j E \left[ (X(t_i+\tau)-\mu)(X(t_j+\tau)-\mu) \right]} $


$ =\sum_{i,j=1}^{n}{\omega_j^2 cov(t_j,t_j)} + \sum_{i,j=1}^{n}{\omega_i \omega_j cov(t_j,t_j)} $


$ \text{Which does not depend on } \tau. $


$ \color{blue}\text{Solution 2:} $

$ \text{Suppose } \mathbf{X}(t) \text{ is a Gaussian Random Process} $


$ \Rightarrow f(x(t_1),x(t_2),...,x(t_k)) = \frac{1}{2\pi^{(\frac{k}{2})} |\Sigma |^{\frac{1}{2}}} exp(-\frac{1}{2}(\overrightarrow{x} - \overrightarrow{m})^T \Sigma ^{-1}(\overrightarrow{x} - \overrightarrow{m})) $


$ \text{for any number of time instances.} $


$ \text{If } \mathbf{X}(t) \text{is WSS} $


$ \Rightarrow \text{ (1) } m_X(t_1) = m_X(t_2) = ... = m_X(t_K) = m $


$ \text{ (2) } R_X(t_i,t_i) = R_X(t_i + \tau, t_j + \tau) $


$ \Sigma = \begin{bmatrix} &R_X(t_1,t_1) &... &R_X(t_1,t_k)\\ &\vdots & \\ &R_X(t_k,t_1) &... &R_X(t_k,t_k)\\ \end{bmatrix} $


$ \text{From (1): } \overrightarrow{m}' = (m_X(t_1+\tau) , m_X(t_2+\tau) , ... , m_X(t_K+\tau)) = \overrightarrow{m} $


$ \text{From (2): } \Sigma' = \begin{bmatrix} &R_X(t_1,t_1) &... &R_X(t_1,t_k)\\ &\vdots & \\ &R_X(t_k,t_1) &... &R_X(t_k,t_k)\\ \end{bmatrix} = \Sigma $


$ {\color{green} \text{Should be clarified that:}} $


$ { \color{green} \text{From (2): } \Sigma' = \begin{bmatrix} &R_X(t_1+\tau,t_1+\tau) &... &R_X(t_1+\tau,t_k+\tau)\\ &\vdots & \\ &R_X(t_k+\tau,t_1+\tau) &... &R_X(t_k+\tau,t_k+\tau)\\ \end{bmatrix} = \begin{bmatrix} &R_X(t_1,t_1) &... &R_X(t_1,t_k)\\ &\vdots & \\ &R_X(t_k,t_1) &... &R_X(t_k,t_k)\\ \end{bmatrix} = \Sigma } $


$ \text{So } f(x(t_1+\tau),x(t_2+\tau),...,x(t_k+\tau)) \text{ is not related to } \tau. $


$ f(x(t_1+\tau),x(t_2+\tau),...,x(t_k+\tau)) $


$ = f(x(t_1),x(t_2),...,x(t_k)) $


$ \Rightarrow \mathbf{X}(t) \text{ is Strict Sense Stationary. } $



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

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