Line 34: Line 34:
  
 
d) Find <math> P(Y=kX)\ </math>.
 
d) Find <math> P(Y=kX)\ </math>.
----
 
=Solution 1 (retrived from [[Automatic_Controls:Linear_Systems_(HKNQE_August_2007)_Problem_1|here]])=
 
  
*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
 
        <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>
 
  
        <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>
+
:'''Click [[ECE_PhD_QE_CNSIP_2007_Problem1.1|here]] to view student [[ECE_PhD_QE_CNSIP_2007_Problem1.1|answers and discussions]]'''
  
*To find <math> P(X=Y)\ </math>, note that  X and Y are iid and summing across all possible i,
+
----
        <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>
+
'''2. (25 Points)'''
  
*To find <math> P(Y>X)\ </math>, again note that X and Y are iid and summing across all possible i,
+
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(Y>X) = \sum_{i=1}^\infty P(Y>i, X=i) = \sum_{i=1}^\infty P(Y>i)P(x=i)</math>
+
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).</math>  
  
:Next, find <math> P(Y<i)\ </math>
+
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> .
        <math> P(Y>i) = 1 - P(Y \le i) </math>
+
  
        <math> P(Y \le i) = \sum_{i=1}^\infty \frac {1}{2^i} = 1 + \frac {1}{2^i} </math>
+
'''Hint:'''
  
      <math> \therefore P(Y>i) = \frac {1}{2^i} </math>
+
You may find the following fact useful:
  
:Plugging this result back into the original expression yields
+
<math class="inline">\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}.</math>  
        <math> P(Y<X) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} </math>
+
  
  
*To find <math> P(Y=kX)\ </math>, note that X and Y are iid and summing over all possible combinations one arrives at
+
:'''Click [[ECE_PhD_QE_CNSIP_2007_Problem1.2|here]] to view student [[ECE_PhD_QE_CNSIP_2007_Problem1.2|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>
+
:Thus,
+
        <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>
+
 
----
 
----
==Solution 2==
+
'''3. (25 Points)'''
Write it here.
+
 
 +
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> .
 +
 
 +
'''(a)'''
 +
 
 +
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> .
 +
 
 +
ref.
 +
 
 +
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> .
 +
 
 +
'''Solution'''
 +
 
 +
<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> .
 +
 
 +
'''(b)'''
 +
 
 +
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> .
 +
 
 +
 
 +
:'''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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Revision as of 09:52, 10 March 2015


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2007



Question

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

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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood