Secretary Problem Secretary Problem Mohammad Mahdian R. Preston McAfee David Pennock
Secretary Admin problem • Observe a sequence of candidates • Rejected candidates can’t be recalled • Examples: jobs, air fare, spouse • Examples: jobs, air fare, spouse • T potential candidates
Classic Solution • Dynkin, 1963 • Goal: Maximize probability of finding the best • Observe k = T/e candidates and reject them • Set an aspiration level max { v , …, v } • Set an aspiration level max { v 1 , …, v k } • Search until meeting or exceeding the aspiration level • Nothing beats this on every distribution in the limit as T gets large.
Dynkin Applied to Spouse Search • 70 years to search • Search for 25¾ years • Best observed set as aspiration level • Best observed set as aspiration level • Continue search until finding one or exhaust candidates – 51½ years of search – 37% (1/ e ) chance of failure
Problems • Reject excellent candidates at T -1 –Consequence of maximizing probability of identifying top candidate • Time of acquisition doesn’t matter • Once and for all decision • Extreme distribution
Hire Secretary • Period of need T . • Hire at t, employ secretary from t to T • Discount factor δ ≤ 1 • Discount factor δ ≤ 1 • If you hire in period j a secretary of value v , you obtain − δ j − δ T T 1 ∑ δ = t v v − δ 1 = t j
Assumptions • Distribution F of values, iid draws • F (0)=0 • Aspiration strategy is used • Aspiration strategy is used –Observe distribution for k periods then set max as a min target
Expected Payoff • The expected payoff of the aspiration strategy is ∞ ∞ − δ j − δ T − T 1 1 ∑ ∑ ∫ ∫ ∫ ∫ − − − π π = = k j k kF kF y y f f y y F F y y xf xf x x dx dx 1 1 ( ( ) ) ( ( ) ) ( ( ) ) ( ( ) ) − δ 1 = + j k 1 y 0 ∞ ∫ + δ − − − T T k F y x f x dx dy 1 1 ( ) ( ) 0
Reduction − − δ T δ j − δ T − k T 1 1 1 1 1 1 ∑ ∫ ∫ − − − π = + j F z dz k F z z dz 1 1 1 ( ) ( ) k − − − δ T j 1 1 1 = + j k 1 0 0 δ − − δ − − δ T − j j T z 1 T 1 1 1 1 ∑ ∑ ∫ ∫ = = − + + F F z z k k dz dz 1 ( ( ) ) − − − δ T j 1 1 1 = + j k 1 0
Difference − δ − − δ j j T z 1 m 1 1 ∑ ∫ π − π = − F z k 1 ( ) k m − − δ j 1 1 = + j k 1 0 δ − − δ − − δ T j j T z m 1 1 1 ∑ ∑ − − − − + + m m k k dz dz ( ( ) ) − − − − − − δ δ T T j j 1 1 1 1 1 1 = + j k 1 • For m>k , β =[ • ] satisfies –Negative at 0 –If decreasing, stays decreasing
π m , m>k Maximize π π π k – π π π π • Strategy: Minimize π k – π m and conclude it is negative • Note F -1 is non-decreasing 1 1 • Lemma: Suppose for all a , ∫ β ≤ z dz ( ) 0 a Then π k – π m ≥ 0.
Proof of Lemma 1 ∫ π − π = − β F z z dz 1 ( ) ( ) k m 0 y 1 ∫ ∫ ∫ ∫ ≤ ≤ − β β + + − β β F F z z z z dz dz F F z z z z dz dz 1 1 ( ( ) ) ( ( ) ) ( ( ) ) ( ( ) ) x y y 1 1 ∫ ∫ ∫ ≤ − β + − β = − β F y z dz F y z dz F y z dz 1 1 1 ( ) ( ) ( ) ( ) ( ) ( ) x y x
Standard Trick 1 ∫ β ≤ z dz ( ) 0 a • is equivalent to β (1) ≤ 0, or δ j − δ T δ j − δ T − − T T 1 1 ∑ ∑ ≤ k m j j = = j k j m
Theorem δ j − δ T − T 1 ∑ k • Suppose k* maximizes j = j k then π k* – π m ≤ 0 for all m>k*. − − T T j 1 ∑ Corollary: for δ =1, k* = arg max k j = j k and k*/ T ≈ 0.203
Conclusions: Spouse Search • T = 70 years • 10% annual discount • Search for 4.1 years • Search for 4.1 years
Hazard Rate • Standard approach permits very long tails − • F 1 ( ) • Impose hazard rate: nondecreas ing ( • • f f ( ) ) • No discounting, one time acquistion
Same Starting Attack • The expected payoff of the aspiration strategy is ∞ ∞ − T 1 ∑ ∑ ∫ ∫ ∫ ∫ − − − π π = = k j k kF kF y y f f y y F F y y xf xf x x dx dx 1 1 ( ( ) ) ( ( ) ) ( ( ) ) ( ( ) ) k k = + j k 1 y ∞ 0 ∫ + − − T k F y x f x dx dy 1 ( ) ( ) 0 − j z 1 T 2 1 ∑ ∫ = − + F z k dz 1 ( ) − T j 1 = j k 0
Integrate by Parts − j z 1 T 2 1 ∑ ∫ π − π = − + − k F z z dz 1 ( ) + k k − T j 1 1 = + j k 1 0 ′ + + k − j z z z 1 T 1 1 2 ∑ ∑ ∫ ∫ = = − − − − − − − F F z z dz dz 1 ( ( ) ) + − + k T j j 1 1 ( 1 ) = + j k 1 0
Integrate by Parts − j z 1 T 2 1 ∑ ∫ π − π = − + − k F z z dz 1 ( ) + k k − T j 1 1 = + j k 1 0 ′ + + k − j z z z 1 T 1 1 2 ∑ ∑ ∫ ∫ = = − − − − − − − F F z z dz dz 1 ( ( ) ) + − + k T j j 1 1 ( 1 ) = + j k 1 0 ′ + + − i − i z z 1 T T 1 1 3 3 ∑ ∑ ∫ = − − − F z z dz 1 ( )( 1 ) + − i T 1 1 = = i k i 0 0 − T 2 1 1 1 ∑ = + • Use the fact + − + k T j j 1 1 ( 1 ) = + j k 1
Hazard Rate ′ − F x 1 ( ) − − = F z z 1 ( )( 1 ) f x ( ) + + i i − − T z T z 1 1 3 3 ∑ ∑ ∑ ∑ β β = = − − z z • Let • Let ( ( ) ) + − i T 1 1 = = i k i 0 ′ 1 ∫ π − π = − − β F z z z dz 1 • Then ( )( 1 ) ( ) + k k 1 0
Hazard Rate ′ − F x 1 ( ) − − = F z z 1 ( )( 1 ) f x ( ) + + i i − − T z T z 1 1 3 3 ∑ ∑ ∑ ∑ β β = = − − z z • Let • Let ( ( ) ) + − i T 1 1 = = i k i 0 ′ 1 ∫ π − π = − − β F z z z dz 1 • Then ( )( 1 ) ( ) + k k 1 0 nondecreasing
Properties of β • If k<( T -1)/ e , β (1)>0 • If β (1)>0, β (z)>0 iff z>z* • Lemma: Suppose β (z)>0 iff z>z*. 1 1 ∫ ∫ β ≤ β ≤ If z dz then x z z dz ( ) 0 , max ( ) ( ) 0 ′ ≥ x 0 0 0
Applying the Lemma 1 ∫ β ≤ π − π ≤ If z dz then ( ) 0 , 0 + k k 1 0 − T 1 1 1 1 1 ∑ ∑ ∫ ∫ β β = = − − z z dz dz • But • But ( ( ) ) + + − − k k T T i i 1 1 1 1 = i 1 0 − T 1 ≥ − π − π ≤ If k 1 , 0 + k k − T 1 1 ∑ 1 i = i 1
Observations − 1 − T • Maximal discovery is 1 Log T ( ) • At T =10 6 , 6.9% • Formula exact for exponential • Formula exact for exponential
Comparison � � � �� ���� � � ������� ������ � ����� ������ ����� ������ �� �� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� ��� �� ��� ��� ��� ���� � � �� ��� ��� ��� �!��� � � � � ��� ��� ��� ��!��� � � ��� ��� ��� ���
Optimal Search • Given distribution, choose so that ∞ ∫ = + + v F c v xf x dx max ( ) ( ) t t 1 c c • So that c=v t+1, and ∞ ∫ = + − v v F x dx 1 ( ) + t t 1 v t + 1
Conclusions • For problem of which ad to run • Ads are durable • Discounting fairly irrelevant • Discounting fairly irrelevant • Examine distribution, compute bound • Uniform distribution ( ) = + − ≈ − k T T * ½ 4 1 3 2
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