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[[Category:random variables]] | [[Category:random variables]] | ||
+ | [[Category:probability]] | ||
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+ | <center> | ||
+ | <font size= 4> | ||
+ | [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] | ||
+ | </font size> | ||
+ | |||
+ | <font size= 4> | ||
+ | Communication, Networking, Signal and Image Processing (CS) | ||
+ | |||
+ | Question 1: Probability and Random Processes | ||
+ | </font size> | ||
+ | |||
+ | August 2000 | ||
+ | </center> | ||
+ | ---- | ||
+ | ---- | ||
+ | =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> | ||
Latest revision as of 09:34, 13 September 2013
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2000
Contents
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.
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.