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Unformatted text preview: 1 Great (Conditional) Expectations X and Y are jointly discrete random variables Recall, conditional expectation of X given Y = y: Analogously, jointly continuous random variables: x Y X x y x p x y Y x X P x y Y X E )  ( )  ( ]  [  dx y x f x y Y X E Y X )  ( ]  [  Computing Probabilities by Conditioning X = indicator variable for event A: E[X] = P(A) Similarly, E[X  Y = y] = P(A  Y = y) for any Y So: E[X] = E Y [ E X [X  Y] ] = E[ E[X  Y] ] = E[ P(A  Y) ] In discrete case: o Also holds analogously in continuous case Generalize, defining indicator variable F i = (Y = y i ): o Called Law of total probability otherwise occurs if 1 A X ) ( ) ( )  ( ] [ A P y Y P y Y A P X E y n i i i F P F A P A P 1 ) ( )  ( ) ( Hiring Software Engineers Interviewing n software engineer candidates All n ! orderings equally likely, but only hiring 1 candidate o Claim: There is ato1 factor difference in productivity between the best and average software engineer o Depending on who you talk to, usually: 10 < a < 100 Right after each interview must decide hire/no hire Feedback from interview of candidate i is just relative ranking with respect to previous i 1 candidates o Strategy: first interview k (of n ) candidates, then hire next candidate better than all of first k candidates o P k (best) = probability that best of all n candidates is hired o X = position of best candidate (1, 2, ..., n ) n i k n i k k i X P i X P i X P P n 1 1 )  Best ( ) ( )  Best ( ) Best ( 1 Hiring Software Engineers (cont.) Note: We will select best candidate (in position i ) if best of first i 1 candidates is among the first k interviewed o To maximize, differentiate P k (Best) with respect to k : o Set g ( k ) = 0 and solve for k : o Interview n/e candidates, then pick best: P...
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 Spring '09

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