# Problem 1

X and Y are iid

   $P(X=i) = P(Y=i) = \frac {1}{2^i}\ ,i = 1,2,3,...$


## Part A

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

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


## Part B

• Find $P(X=Y)\$
   Noting that X and Y are iid and summing across all possible i,
$P(X=Y) = \sum_{i=1}^\infty P(X=i, Y=i) = \sum_{i=1}^\infty P(X=i)P(Y=i) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3}$


## Part C

• Find $P(Y>X)\$
   Again noting 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
$P(Y<X) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3}$


## Part D

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


# Problem 4

## Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett