View source for Automatic Controls:Linear Systems (HKNQE August 2007) - Rhea

[[Category: ECE]]
[[Category: HKN]]
[[Category: QE]]
[[Category: Automatic Controls]]
[[Category: Linear Systems]]
[[Category: August]]
[[Category: 2007]]

=Problem 1=
X and Y are iid

    <math> P(X=i) = P(Y=i) = \frac {1}{2^i}\ ,i = 1,2,3,... </math>

==Part A==
*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>

==Part B==
*Find <math> P(X=Y)\ </math>

    Noting 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>

==Part C==
*Find <math> P(Y>X)\ </math>

    Again noting that X and Y are iid and summing across all possible i,
        <math> P(Y>X) = \sum_{i=1}^\infty P(Y>i, X=i) = \sum_{i=1}^\infty P(Y>i)P(x=i)</math>

    Next, find <math> P(Y<i)\ </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>

      <math> \therefore P(Y>i) = \frac {1}{2^i} </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>

==Part D==
*Find <math> P(Y=kX)\ </math>

    Noting 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>