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       <math> \therefore P(Y>i) = \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>

Revision as of 11:30, 3 December 2008


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}  $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva