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==Question from [[ECE_PhD_QE_CNSIP_2000_Problem1|ECE QE CS Q1 August 2000]]==  
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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]]
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[[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]
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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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August 2000
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=Part 4=
 
A RV is given by <math class="inline">\mathbf{Z}=\sum_{n=0}^{8}\mathbf{X}_{n}</math> where <math class="inline">\mathbf{X}_{n}</math> 's are i.i.d.  RVs with characteristic function given by <math class="inline">\Phi_{\mathbf{X}}\left(\omega\right)=\frac{1}{1-j\omega/2}.</math>  
 
A RV is given by <math class="inline">\mathbf{Z}=\sum_{n=0}^{8}\mathbf{X}_{n}</math> where <math class="inline">\mathbf{X}_{n}</math> 's are i.i.d.  RVs with characteristic function given by <math class="inline">\Phi_{\mathbf{X}}\left(\omega\right)=\frac{1}{1-j\omega/2}.</math>  
  
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'''(b)''' Determine the pdf of <math class="inline">\mathbf{Z}</math> . You can leave your answer in integral form.
 
'''(b)''' Determine the pdf of <math class="inline">\mathbf{Z}</math> . You can leave your answer in integral form.
:'''Click [[ECE_PhD_QE_CNSIP_2000_Problem1.4|here]] to view student [[ECE_PhD_QE_CNSIP_2000_Problem1.4|answers and discussions]]'''
 
 
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==Share and discuss your solutions below.==
 
==Share and discuss your solutions below.==

Latest revision as of 10:34, 13 September 2013


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2000



Part 4

A RV is given by $ \mathbf{Z}=\sum_{n=0}^{8}\mathbf{X}_{n} $ where $ \mathbf{X}_{n} $ 's are i.i.d. RVs with characteristic function given by $ \Phi_{\mathbf{X}}\left(\omega\right)=\frac{1}{1-j\omega/2}. $

(a) Determine the characteristic function of $ \mathbf{Z} $ .

(b) Determine the pdf of $ \mathbf{Z} $ . You can leave your answer in integral form.


Share and discuss your solutions below.


Solution 1 (retrived from here)

(a)

$ \Phi_{\mathbf{Z}}\left(\omega\right)=E\left[e^{i\omega\mathbf{Z}}\right]=E\left[e^{i\omega\sum_{n=0}^{8}\mathbf{X}_{n}}\right]=E\left[\prod_{n=0}^{8}e^{i\omega\mathbf{X}_{n}}\right]=\prod_{n=0}^{8}E\left[e^{i\omega\mathbf{X}_{n}}\right]=\left(\frac{1}{1-j\omega/2}\right)^{9}. $

(b)

$ f_{\mathbf{Z}}\left(z\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\Phi_{\mathbf{Z}}\left(\omega\right)e^{-i\omega z}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\frac{1}{1-j\omega/2}\right)^{9}e^{-i\omega z}d\omega. $


Solution 2

Write it here.


Back to QE CS question 1, August 2000

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Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva