Secretary Problem Secretary Problem Mohammad Mahdian R. Preston - - PowerPoint PPT Presentation

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 {v1, …, vk}
• 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

• f 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 δ δ δ − − =

− =

∑

1

1 T j T j t t

v v

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

∫ ∫ ∑

∞ − − ∞ − − −

   − =

k j T T j k

dx x xf y F y f y kF ) ( ) ( ) ( ) (

1 1 1 1

δ δ π

∫ ∫ ∑

+ =

   − =

y k j

dx x xf y F y f y kF ) ( ) ( 1 ) ( ) (

1

δ π

dy dx x f x y F

k T T

    +

∫

∞ − − −

) ( ) (

1 1

δ

Reduction

∫ ∑ ∫

− − − − + = − −

− − − + − =

1 1 1 1 1 1 1 1 1

) ( 1 1 1 ) ( 1 dz z z F j k dz z F T k

j T j T k j T k

δ δ δ δ π

∫ ∑

      − + =

− − − − − 1 1 1 1 1 1

) ( dz z k z F

T j T j T

δ δ δ

∫ ∑

      − − + − =

+ = 1

1 1 1 ) ( dz j T k z F

k j

δ

Difference

∫ ∑

   − − − = −

− + = − − 1 1 1 1 1

1 1 ) ( δ δ δ π π

T j m k j j m k

j z k z F dz j z T k m

T j m j T

        − − − + − − −

− − −

∑

δ δ δ δ 1 1 1 ) (

1 1 1

• For m>k, β=[•] satisfies

–Negative at 0 –If decreasing, stays decreasing

dz j T k m

k j

         − − + − − −

+ =

∑

δ 1 1 1 ) (

1

Maximize π π π πk–π π π πm, m>k

• Strategy: Minimize πk–πm and conclude

it is negative

• Note F-1 is non-decreasing

1

• Lemma: Suppose for all a,

Then πk–πm ≥ 0.

∫

≤

1

) (

a

dz z β

Proof of Lemma

dz z z F

m k

) ( ) (

1 1

β π π

∫

−

= − dz z z F dz z z F

y

) ( ) ( ) ( ) (

1 1 1

β β

∫ ∫

− −

+ ≤ dz z z F dz z z F

y x

) ( ) ( ) ( ) ( β β

∫ ∫

+ ≤

∫ ∫ ∫

− − −

= + ≤

1 1 1 1 1

) ( ) ( ) ( ) ( ) ( ) (

x y y x

dz z y F dz z y F dz z y F β β β

Standard Trick

• is equivalent to β(1)≤0, or

∫

≤

1

) (

a

dz z β

∑ ∑

− = − =

− ≤ −

1 1 T m j T j T k j T j

j m j k δ δ δ δ

Theorem

• Suppose k* maximizes

then πk*–πm ≤ 0 for all m>k*.

∑

− =

−

1 T k j T j

j k δ δ

Corollary: for δ=1, k* = arg max and k*/T ≈0.203

∑

− =

−

1 T k j

j j T k

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

• Impose hazard rate:

ing nondecreas ) ( ) ( 1

• −

f F

• No discounting, one time acquistion

) (• f

Same Starting Attack

• The expected payoff of the aspiration

strategy is

∫ ∫ ∑

∞ ∞ − − − −

   =

T k j k k

dx x xf y F y f y kF ) ( ) ( ) ( ) (

1 1 1

π

∫ ∫ ∑

+ =

   =

y k j k

dx x xf y F y f y kF ) ( ) ( ) ( ) (

1

π

dy dx x f x y F

k T

    +

∫

∞ − −

) ( ) (

1

∫ ∑

        + − =

− = − 1 2 1

1 1 ) ( dz j z T k z F

T k j j

Integrate by Parts

∫ ∑

        − + − = −

− + = − + 1 2 1 1 1

1 1 ) ( dz z j z T z F

T k j k j k k

π π

∫ ∑

      − − ′ − =

− + + − 1 2 1 1 1

) ( dz z z z z F

T j k

∫ ∑

      + − − − + − =

+ = 1

) 1 ( 1 1 ) ( dz j j T k z F

k j

Integrate by Parts

∫ ∑

        − + − = −

− + = − + 1 2 1 1 1

1 1 ) ( dz z j z T z F

T k j k j k k

π π

∫ ∑

      − − ′ − =

− + + − 1 2 1 1 1

) ( dz z z z z F

T j k

• Use the fact

∫ ∑

      + − − − + − =

+ = 1

) 1 ( 1 1 ) ( dz j j T k z F

k j

∑

− + =

+ + − = +

2 1

) 1 ( 1 1 1 1 1

T k j

j j T k

∫ ∑ ∑

        − − + − ′ =

− = + − = + − 1 3 1 3 1 1

1 1 ) 1 )( ( dz T z i z z z F

T i i T k i i

Hazard Rate

• Let

) ( ) ( 1 ) 1 )( (

1

x f x F z z F − = − ′

−

∑ ∑

− + − +

− =

3 1 3 1

) (

T i T i

z z z β

• Let
• Then

∑ ∑

= =

− − + = 1 1 ) (

i k i

T i z β

∫

− ′ = −

− + 1 1 1

) ( ) 1 )( ( dz z z z F

k k

β π π

Hazard Rate

• Let

) ( ) ( 1 ) 1 )( (

1

x f x F z z F − = − ′

−

∑ ∑

− + − +

− =

3 1 3 1

) (

T i T i

z z z β

• Let
• Then

∑ ∑

= =

− − + = 1 1 ) (

i k i

T i z β

∫

− ′ = −

− + 1 1 1

) ( ) 1 )( ( dz z z z F

k k

β π π

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

) ( ) ( max , ) ( dz z z x then dz z If

x

β β

Applying the Lemma

• But

, ) (

1 1

≤ − ≤

+

∫

k k

then dz z If π π β

∑ ∫

−

− − + =

1 1

1 1 1 1 1 ) (

T

i T k dz z β

• But

∑ ∫

=

− − + =

1

1 1 ) (

i

i T k dz z β , 1 1 1

1 1 1

≤ − − − ≥

+ − =

∑

k k T i

i T k If π π

Observations

• Maximal discovery is
• At T=106, 6.9%
• Formula exact for exponential

1 ) ( 1 − − T Log T

• Formula exact for exponential

Comparison

• !
• !

Optimal Search

• Given distribution, choose so that

∫

∞ + +

=

c t c t

dx x xf v c F v ) ( ) ( max

1

• So that c=vt+1, and

∫

∞ +

+

− + =

1

) ( 1

1

t

v t t

dx x F v v

Conclusions

• For problem of which ad to run
• Ads are durable
• Discounting fairly irrelevant
• Discounting fairly irrelevant
• Examine distribution, compute bound
• Uniform distribution

( )

2 3 1 4 ½ * − ≈ − + = T T k