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===== <math>\color{blue}\text{Solution 1:}</math>  =====
 
===== <math>\color{blue}\text{Solution 1:}</math>  =====
<font color="#ff0000"><span style="font-size: 36px;">
 
 
<math> \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> \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>
 
</math>
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<math>\text{Define } Y(t_j+\tau) = \sum_{j=1}^{n}{\omega_jX(t_j+\tau)} \text{, so}
+
<math>
 +
\text{Define } Y(t_j+\tau) = \sum_{j=1}^{n}{\omega_jX(t_j+\tau)} \text{, so}
 
</math>  
 
</math>  
  
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<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}  
+
<font face="Times New Roman" font size="5"><math>
</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}  
</span></font>
+
</math></font>
 +
 
 +
 
 +
&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>
 +
var(Y(t_j+\tau)) = E \left [(\sum_{j=1}^{n}{w_j(X(t_j+\tau)-\mu)^2}  \right ]
 +
</math>
 +
 
 +
 
 +
&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>
 +
=\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>
 +
 
 +
 
 +
&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<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>
 +
 
 +
<font face="Times New Roman" font size="5"><math>
 +
\text{Which does not depend on } \tau.
 +
</math></font>
 +
 
  
 
----
 
----

Revision as of 14:37, 29 July 2012

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

Question 1, August 2011, Part 2

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:} $

$ \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:} $

$ \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: $


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

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