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]]''
 
----
 
----
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]

Revision as of 10:52, 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

Back to ECE Qualifying Exams (QE) page

Alumni Liaison

Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal