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[[Category:probability]]
 
[[Category:probability]]
  
= [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]: COMMUNICATIONS, NETWORKING, SIGNAL AND IMAGE PROESSING (CS)- Question 1, August 2007=
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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 2007
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==Question==
 
==Question==
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'''1. (25 Points)'''
  
 
X and Y are iid random variable with
 
X and Y are iid random variable with
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d) Find <math> P(Y=kX)\ </math>.
 
d) Find <math> P(Y=kX)\ </math>.
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:'''Click [[ECE_PhD_QE_CNSIP_2007_Problem1.1|here]] to view student [[ECE_PhD_QE_CNSIP_2007_Problem1.1|answers and discussions]]'''
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----
 
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=Solution 1 (retrived from [[Automatic_Controls:Linear_Systems_(HKNQE_August_2007)_Problem_1|here]])=
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'''2. (25 Points)'''
  
*To find <math> P(min(X,Y)=k)\ </math>, let <math> Z = min(X,Y)\ </math>. Then finding the pmf of Z uses the fact that X and Y are iid
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Let <math class="inline">\left\{ \mathbf{X}_{n}\right\} _{n\geq1}</math>  be a sequence of binomially distributed random variables, with the <math class="inline">n</math> -th random variable <math class="inline">\mathbf{X}_{n}</math> having pmf <math class="inline">p_{\mathbf{X}_{n}}\left(k\right)=P\left(\left\{ \mathbf{X}_{n}=k\right\} \right)=\left(\begin{array}{c}
        <math> P(Z=k) = P(X \ge k,Y \ge k) = P(X \ge k)P(Y \ge k) = P(X \ge k)^2 </math>
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n\\
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k
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\end{array}\right)p_{n}^{k}\left(1-p_{n}\right)^{n-k}\;,\qquad k=0,\cdots,n,\quad p_{n}\in\left(0,1\right).</math>  
  
        <math> P(Z=k) = \left ( \sum_{i=k}^N \frac {1}{2^i} \right )^2 = \left ( \frac {1}{2^k} \right )^2 = \frac {1}{4^k} </math>
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Show that, if the <math class="inline">p_{n}</math>  have the property that <math class="inline">np_{n}\rightarrow\lambda</math>  as <math class="inline">n\rightarrow\infty</math> , where <math class="inline">\lambda</math>  is a positive constant, then the sequence <math class="inline">\left\{ \mathbf{X}_{n}\right\} _{n\geq1}</math>  converges in distribution to a Poisson random variable <math class="inline">\mathbf{X}</math> with mean <math class="inline">\lambda</math> .
  
*To find <math> P(X=Y)\ </math>, note that  X and Y are iid and summing across all possible i,
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'''Hint:'''
        <math> P(X=Y) = \sum_{i=1}^\infty P(X=i, Y=i) = \sum_{i=1}^\infty P(X=i)P(Y=i) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} </math>
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*To find <math> P(Y>X)\ </math>, again note that X and Y are iid and summing across all possible i,
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You may find the following fact useful:
        <math> P(Y>X) = \sum_{i=1}^\infty P(Y>i, X=i) = \sum_{i=1}^\infty P(Y>i)P(x=i)</math>
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:Next, find <math> P(Y<i)\ </math>
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<math class="inline">\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}.</math>  
        <math> P(Y>i) = 1 - P(Y \le i) </math>
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        <math> P(Y \le i) = \sum_{i=1}^\infty \frac {1}{2^i} = 1 + \frac {1}{2^i} </math>
 
  
      <math> \therefore P(Y>i) = \frac {1}{2^i} </math>
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:'''Click [[ECE_PhD_QE_CNSIP_2007_Problem1.2|here]] to view student [[ECE_PhD_QE_CNSIP_2007_Problem1.2|answers and discussions]]''
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----
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'''3. (25 Points)'''
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Let <math class="inline">\mathbf{X}\left(t\right)</math> be a real Gaussian random process with mean function <math class="inline">\mu\left(t\right)</math>  and autocovariance function <math class="inline">C_{\mathbf{XX}}\left(t_{1},t_{2}\right)</math> .
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'''(a)'''
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Write the expression for the <math class="inline">n</math> -th order characteristic function of <math class="inline">\mathbf{X}\left(t\right)</math>  in terms of <math class="inline">\mu\left(t\right)</math>  and <math class="inline">C_{\mathbf{XX}}\left(t_{1},t_{2}\right)</math> .
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ref.
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There are the note about the [[ECE 600 General Concepts of Stochastic Processes Definitions|n-th order characteristic function of Gaussians random process]] . The only difference between the note and this problem is that this problem use the <math class="inline">\mu\left(t\right)</math>  rather than <math class="inline">\eta_{\mathbf{X}}\left(t\right)=E\left[\mathbf{X}\left(t\right)\right]</math> .
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'''Solution'''
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<math class="inline">\Phi_{\mathbf{X}\left(t_{1}\right)\cdots\mathbf{X}\left(t_{n}\right)}\left(\omega_{1},\cdots,\omega_{n}\right)=\exp\left\{ i\sum_{k=1}^{n}\mu_{\mathbf{X}}\left(t_{k}\right)\omega_{k}-\frac{1}{2}\sum_{j=1}^{n}\sum_{k=1}^{n}C_{\mathbf{XX}}\left(t_{j},t_{k}\right)\omega_{j}\omega_{k}\right\}</math> .
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'''(b)'''
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Show that the probabilistic description of <math class="inline">\mathbf{X}\left(t\right)</math>  is completely characterized by <math class="inline">\mu\left(t\right)</math>  and autocovariance function <math class="inline">C_{\mathbf{XX}}\left(t_{1},t_{2}\right)</math> .
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:'''Click [[ECE_PhD_QE_CNSIP_2007_Problem1.3|here]] to view student [[ECE_PhD_QE_CNSIP_2007_Problem1.3|answers and discussions]]''
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----
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'''4. (25 Points)'''
  
:Plugging this result back into the original expression yields
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Let <math class="inline">\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots</math>  be a sequence of independent, identically distributed random variables, each having Cauchy pdf <math class="inline">f\left(x\right)=\frac{1}{\pi\left(1+x^{2}\right)}\;,\qquad-\infty<x<\infty. Let \mathbf{Y}_{n}=\frac{1}{n}\sum_{i=1}^{n}\mathbf{X}_{i}.</math> Find the pdf of <math class="inline">\mathbf{Y}_{n}</math> . Describe how the pdf of <math class="inline">\mathbf{Y}_{n}</math>  depends on <math class="inline">n</math> . Does the sequence <math class="inline">\mathbf{Y}_{1},\mathbf{Y}_{2},\mathbf{Y}_{3},\cdots</math> converge in distribution? If yes, what is the distribution of the random variable it converges to?
        <math> P(Y<X) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} </math>
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*To find <math> P(Y=kX)\ </math>, note that X and Y are iid and summing over all possible combinations one arrives at
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:'''Click [[ECE_PhD_QE_CNSIP_2007_Problem1.4|here]] to view student [[ECE_PhD_QE_CNSIP_2007_Problem1.4|answers and discussions]]''
        <math> P(Y=kX) = \sum_{i=1}^\infty i = 1^\infty P(Y=ki, X=i) = \sum_{i=1}^\infty P(Y=ki)P(X=i) </math>
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:Thus,
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        <math> P(Y=kX) = \sum_{i=1}^\infty \frac {1}{2^{ki}} \frac {1}{2^i} = \sum_{i=1}^\infty \frac {1}{2^{(k+1)i}} = \frac {1}{2^{(k+1)}-1} </math>
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==Solution 2==
 
Write it here.
 
 
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[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]
 
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Latest revision as of 09:57, 10 March 2015


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2007



Question

1. (25 Points)

X and Y are iid random variable with

$ P(X=i) = P(Y=i) = \frac {1}{2^i}\ ,i = 1,2,3,... $

a) Find $ P(min(X,Y)=k)\ $.

b) Find $ P(X=Y)\ $.

c) Find $ P(Y>X)\ $.

d) Find $ P(Y=kX)\ $.


Click here to view student answers and discussions

2. (25 Points)

Let $ \left\{ \mathbf{X}_{n}\right\} _{n\geq1} $ be a sequence of binomially distributed random variables, with the $ n $ -th random variable $ \mathbf{X}_{n} $ having pmf $ p_{\mathbf{X}_{n}}\left(k\right)=P\left(\left\{ \mathbf{X}_{n}=k\right\} \right)=\left(\begin{array}{c} n\\ k \end{array}\right)p_{n}^{k}\left(1-p_{n}\right)^{n-k}\;,\qquad k=0,\cdots,n,\quad p_{n}\in\left(0,1\right). $

Show that, if the $ p_{n} $ have the property that $ np_{n}\rightarrow\lambda $ as $ n\rightarrow\infty $ , where $ \lambda $ is a positive constant, then the sequence $ \left\{ \mathbf{X}_{n}\right\} _{n\geq1} $ converges in distribution to a Poisson random variable $ \mathbf{X} $ with mean $ \lambda $ .

Hint:

You may find the following fact useful:

$ \lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}. $


'Click here to view student answers and discussions

3. (25 Points)

Let $ \mathbf{X}\left(t\right) $ be a real Gaussian random process with mean function $ \mu\left(t\right) $ and autocovariance function $ C_{\mathbf{XX}}\left(t_{1},t_{2}\right) $ .

(a)

Write the expression for the $ n $ -th order characteristic function of $ \mathbf{X}\left(t\right) $ in terms of $ \mu\left(t\right) $ and $ C_{\mathbf{XX}}\left(t_{1},t_{2}\right) $ .

ref.

There are the note about the n-th order characteristic function of Gaussians random process . The only difference between the note and this problem is that this problem use the $ \mu\left(t\right) $ rather than $ \eta_{\mathbf{X}}\left(t\right)=E\left[\mathbf{X}\left(t\right)\right] $ .

Solution

$ \Phi_{\mathbf{X}\left(t_{1}\right)\cdots\mathbf{X}\left(t_{n}\right)}\left(\omega_{1},\cdots,\omega_{n}\right)=\exp\left\{ i\sum_{k=1}^{n}\mu_{\mathbf{X}}\left(t_{k}\right)\omega_{k}-\frac{1}{2}\sum_{j=1}^{n}\sum_{k=1}^{n}C_{\mathbf{XX}}\left(t_{j},t_{k}\right)\omega_{j}\omega_{k}\right\} $ .

(b)

Show that the probabilistic description of $ \mathbf{X}\left(t\right) $ is completely characterized by $ \mu\left(t\right) $ and autocovariance function $ C_{\mathbf{XX}}\left(t_{1},t_{2}\right) $ .


'Click here to view student answers and discussions

4. (25 Points)

Let $ \mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots $ be a sequence of independent, identically distributed random variables, each having Cauchy pdf $ f\left(x\right)=\frac{1}{\pi\left(1+x^{2}\right)}\;,\qquad-\infty<x<\infty. Let \mathbf{Y}_{n}=\frac{1}{n}\sum_{i=1}^{n}\mathbf{X}_{i}. $ Find the pdf of $ \mathbf{Y}_{n} $ . Describe how the pdf of $ \mathbf{Y}_{n} $ depends on $ n $ . Does the sequence $ \mathbf{Y}_{1},\mathbf{Y}_{2},\mathbf{Y}_{3},\cdots $ converge in distribution? If yes, what is the distribution of the random variable it converges to?


'Click here to view student answers and discussions


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