ECE Ph.D. Qualifying Exam: Automatic Control (AC)- Question 1, August 2007

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,

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

Write it here.