Revision as of 10:49, 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)\$ .

Solution 1 (retrived from here)

To find $P(min(X,Y)=k)\$ , let $Z = min(X,Y)\$ . Then finding the pmf of Z uses the fact that X and Y are iid

        $P(Z=k) = P(X \ge k,Y \ge k) = P(X \ge k)P(Y \ge k) = P(X \ge k)^2$

        $P(Z=k) = \left ( \sum_{i=k}^N \frac {1}{2^i} \right )^2 = \left ( \frac {1}{2^k} \right )^2 = \frac {1}{4^k}$

To find $P(X=Y)\$ , note that X and Y are iid and summing across all possible i,

        $P(X=Y) = \sum_{i=1}^\infty P(X=i, Y=i) = \sum_{i=1}^\infty P(X=i)P(Y=i) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3}$

To find $P(Y>X)\$ , again note that X and Y are iid and summing across all possible i,

        $P(Y>X) = \sum_{i=1}^\infty P(Y>i, X=i) = \sum_{i=1}^\infty P(Y>i)P(x=i)$

Next, find

P(Y<i)\

        $P(Y>i) = 1 - P(Y \le i)$

        $P(Y \le i) = \sum_{i=1}^\infty \frac {1}{2^i} = 1 + \frac {1}{2^i}$

      $\therefore P(Y>i) = \frac {1}{2^i}$

Plugging this result back into the original expression yields

        $P(Y<X) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3}$

To find $P(Y=kX)\$ , note that X and Y are iid and summing over all possible combinations one arrives at

        $P(Y=kX) = \sum_{i=1}^\infty i = 1^\infty P(Y=ki, X=i) = \sum_{i=1}^\infty P(Y=ki)P(X=i)$

Thus,

        $P(Y=kX) = \sum_{i=1}^\infty \frac {1}{2^{ki}} \frac {1}{2^i} = \sum_{i=1}^\infty \frac {1}{2^{(k+1)i}} = \frac {1}{2^{(k+1)}-1}$

Solution 2

Write it here.

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@@ Line 34: / Line 34: @@
 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>
-:'''Click [[ECE_PhD_QE_CNSIP_2007_Problem1.1|here]] to view student [[ECE_PhD_QE_CNSIP_2007_Problem1.1|answers and discussions]]'''
+        <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>
-----
+*To find <math> P(X=Y)\ </math>, note that  X and Y are iid and summing across all possible i,
-'''2. (25 Points)'''
+        <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>
-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}
+*To find <math> P(Y>X)\ </math>, again note that X and Y are iid and summing across all possible i,
-n\\
+        <math> P(Y>X) = \sum_{i=1}^\infty P(Y>i, X=i) = \sum_{i=1}^\infty P(Y>i)P(x=i)</math>
-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>
-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> .
+:Next, find <math> P(Y<i)\ </math>
+        <math> P(Y>i) = 1 - P(Y \le i) </math>
-'''Hint:'''
+        <math> P(Y \le i) = \sum_{i=1}^\infty \frac {1}{2^i} = 1 + \frac {1}{2^i} </math>
-You may find the following fact useful:
+      <math> \therefore P(Y>i) = \frac {1}{2^i} </math>
-<math class="inline">\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}.</math>
+:Plugging this result back into the original expression yields
+        <math> P(Y<X) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} </math>
-:'''Click [[ECE_PhD_QE_CNSIP_2007_Problem1.2|here]] to view student [[ECE_PhD_QE_CNSIP_2007_Problem1.2|answers and discussions]]''
+*To find <math> P(Y=kX)\ </math>, note that X and Y are iid and summing over all possible combinations one arrives at
+        <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==
+Write it here.
 ----
 [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]

Difference between revisions of "ECE PhD QE CNSIP 2007 Problem1" - Rhea

Revision as of 10:49, 10 March 2015

Question

Solution 1 (retrived from here)

Solution 2

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