Revision as of 08:26, 27 June 2012

ECE Ph.D. Qualifying Exam: COMMUNICATIONS, NETWORKING, SIGNAL AND IMAGE PROESSING (CS)- Question 1, 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.

Back to ECE Qualifying Exams (QE) page