Line 34: Line 34:
  
 
d) Find <math> P(Y=kX)\ </math>.
 
d) Find <math> P(Y=kX)\ </math>.
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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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*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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        <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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        <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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:'''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>
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'''2. (25 Points)'''
  
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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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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}
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        <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>  
  
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:Next, find <math> P(Y<i)\ </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> .
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        <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>
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'''Hint:'''
  
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      <math> \therefore P(Y>i) = \frac {1}{2^i} </math>
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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]]''
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        <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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−
----
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==Solution 2==
+
−
Write it here.
+
 
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[[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 09:45, 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

Back to ECE Qualifying Exams (QE) page

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