# An Asymptotic Analysis on Generalized Secretary Problem

As a famous result, the “37% Law” for Secretary Problem has widely influenced peoples' perception on online decision strategies about choice. However, using this strategy, too many attractive candidates may be rejected in the first 37%, and in practice people also tend to stop earlier<cit.>. In this paper, we argued that in most cases, the best-only optimization does not obtain an optimal outcome, while the optimal cutoff should be O(√(n)). And we also showed that in some strict objective that only cares several best candidates, Θ(n) skips are still needed.

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## 1 Introduction

As a fundamental online decision problem, the Secretary Problem requires the Decision Maker(DM) go through a sequence of

applicants, and decide to accept or reject immediately after observing one applicant, and the target is to maximize the probability to choose the best applicant

[Ferguson1989].

In the original problem, the optimal solution is to skip the first applicants and accept one immediately afterwards if it is the best so far. the maximized probability is .

However, in practice, it is often more desirable to choose a “good” applicant, not exactly the “best”. It was discovered [Bearden2006] that the optimal cutoff point decreases to if the target is to minimize the expected ranking. As the result differs from the original problem, it suggests that different targets make a great difference of the optimal cutoff point. The “37% rule” has wide impression on people’s perception of choice, but it is actually misleading because most people are not stubborn for the very best. Therefore, our motivation is to go further and make it general for arbitrary utility functions, and explore the high-level rule for the generalized Secretary Problem.

In this paper, we inherited the method to compute with the “actual values” rather than ranks, but as the payoff function can be non-linear, the mapping is asymptotic when , more subtle with concentration properties, and only works for Lipschitz continuous functions of relative ranks. Then we proved the optimal cutoff is generally in most cases, as a strong challenge against “37% rule”.

In the last part, we go to the case where the target is to choose one of the best applicants. In this case, though, skips are still required.

## 2 Asymptotic evaluation

### 2.1 Randomized relative ranking technique

The generalized Secretary Problem is:

P1: applicants with different ranking are interviewed by DM one by one. After applicant is interviewed, DM only knows whether he is the best so far, and must make a decision to accept or reject. If the accepted applicant is in rank , the utility is (as a decreasing funtion) and the game ends; if rejects, DM goes on to applicant . Applicant must be accepted if reached.

Inspired by Bearden(2006), we can consider their “randomized relative ranking”, assume them to be uniformly distributed in interval

, and define to be the utility as a function of relative ranking. This randomized relative ranking differs from the actual relative ranking, but it can be proved that when is fixed and , with high probability the difference converges to 0.

Therefore, we firstly consider the problem P2:

P2: applicants with i.i.d types uniformly distributed in are interviewed by DM one by one, and there is a decreasing utility function . After applicant is interviewed, DM only knows whether , and must make a decision to accept or reject. If he accepts, the utility is and the game ends; if rejects, DM goes on to applicant . Applicant must be accepted if reached.

Fix and . When the cutoff point is , for the probability of accepting the -th applicant is:

 Pc(t) = (t−1∏s=cs−1s)⋅1t = c−1t(t−1),

and the expected utility of the best among is:

 Ec(t)=t∫10w(x)(1−x)t−1dx.

The probability of accepting the th applicant is:

 Pc(n) = (n−1∏s=cs−1s) = c−1n−1,

and the last is reached iff the best is in the first places or the best is in place and the second best is in the first places, with probability and respectively. Therefore

 Ec(n)=n−1nE−+1nEM,

with denoting the expectation of any non-maximum , and denoting the expectation of the maximum of Because in each instance of , there are non-maximum and 1 maximum, is exactly the expectation of i.e.

 Ec(n)=∫10w(x)dx.

Therefore, the overall expectation is

 Ec = n−1∑t=cPc(t)Ec(t)+Pc(n)Ec(n) (1) = (n−1∑t=cc−1t−1∫10w(x)(1−x)t−1dx) +c−1n−1∫10w(x)dx.

### 2.2 Equivalence for asymptoticity

We denote be the th smallest in Now we compute a high-probability bound to the range of . If , then at least samples among in are less than , and if , then at most samples among in are less than . By Chernoff Bound,

 Pr[sn,i
 Pr[sn,i>i+1n+ϵ]

so for large , with high probability,

 ∀i∈[n],∣∣∣sn,i−in∣∣∣≤lnn√n.

Therefore, for fixed uniformly continuous and large

, (1) is an asymptotic estimation of the expected utility in Problem P1.

## 3 Finding the optimum

### 3.1 Concavity of expected utility

By intuition, should be a single-peak function, because initially as more applicants are observed, the DM has more experience to identify good candidates, but if too many have been rejected, probably the best is also skipped. Now we prove this.

Define , , then

 ΔEc = (n−1∑t=c1t−1∫10w(x)(1−x)t−1dx) (2) −∫10w(x)(1−x)c−2dx+∫10w(x)dxn−1,
 Δ2Ec=1c−2∫10w(x)⋅(1−x)c−3⋅((c−1)x−1)dx. (3)

For because

and

 w(x)⋅(1−x)c−3⋅((c−1)x−1){<0,x∈(0,1c−1)>0,x∈(1c−1,1),

wlog we can assume , then as is a decreasing function, . Therefore,

 Δ2Ec≤0. (4)

So is a concave function of . To find the maximum of , we only need to find (via binary search) the largest s.t.

### 3.2 General estimation of optimal cutoff

As Bearden has proved that for linear , it shows that in some cases the optimal cutoff is much less than , which differs much from “37% rule”. Here, we prove that it is the general case for all “smooth” utility functions.

#### Theorem 1

In problem P2, the optimal cutoff if is bounded and satisfies Lipschitz condition in for some .

#### Proof of Theorem 1

Wlog we assume . Let for fixed .

From (2), we have

 ΔEc= ∫10w(x)n−1dx + ∫ϵ0(n−1∑t=c(1−x)t−1t−1−(1−x)c−2)w(x)dx + ∫1ϵ(n−1∑t=c(1−x)t−1t−1−(1−x)c−2)w(x)dx.

For fixed , we have

 ∫10w(x)n−1dx<0,∫10w(x)n−1dx=Θ(1n). (5)

Because satisfies Lipschitz condition in , there exists a constant s.t , .

Therefore,

 ∫ϵ0(n−1∑t=c(1−x)t−1t−1−(1−x)c−2)w(x)dx (6) < −∫ϵ0(1−x)c−2w(x)dx ≤ L∫ϵ0(1−x)c−2xdx = O(∫ϵ0(1−x)λ√nxdx) = O(1−(1−ϵ)λ√n+1(1+ϵ+ϵλ√n)2+3λ√n+λ2n) = O(1λ2n).

Assume is bounded to . Then

 ∫1ϵ(n−1∑t=c(1−x)t−1t−1−(1−x)c−2)w(x)dx (7) ≤ −∫1ϵ(1−x)c−2w(x)dx ≤ M∫1ϵ(1−x)c−2dx ≤ M(1−ϵ)(1−ϵ)c−2 = O(e−ϵλ√n) = o(1n).

From (5)(6)(7), for any fixed satisfying the conditions, we can find (big enough) constant to make for large enough. Therefore, based on the concavity of , implying

 cOPT=O(√n),

and an asymptotic upper bound is

 cOPT≲√L^wn,

in which

 ^w=∫10(w(0)−w(x))dx.

## 4 When are Θ(n) skips still needed?

Theorem 1 suggests in most cases the optimal cutoff point is , but it needs the DM to only care about “relative” ranking and do not care much about the top ones. However, there does exist cases where only the best one (or several of best ones) matters, e.g. competing for a medal in Olympic Games.

In the original Secretary Problem, the asymptotic optimal cutoff point is when the top one is desired. What if the target is to accept one of top applicants no matter how large is? Intuitively, skips are still needed. And we prove this.

#### Theorem 2

To maximize the probability to accept one of the best participants among , , and the success probability

#### Proof of Theorem 2

If the th participant is among the best ones (denoted as “good ones”) and picked and , then for , the probability that any good one is skipped converges to 0. So with probability , he must be the first of good ones, and the best one in first must be in the first .

Therefore, the success probability is

 P(c) = n∑i=cCk−1n−iCkn⋅c−1i−1+o(1) ∼ n∑i=c(1−in)k−1nk⋅c−1i−1,

so

 P(c)−P(c−1) ∼ 1nkn∑i=c(1−in)k−1i−1−c−2nk(1−c−1n)k−1c−2 = 1nk(n∑i=c(1−in)k−1i−1−(1−c−1n)k−1).

Because , for large enough,

 n∑i=c(1−in)k−1i−1 > 3c∑i=c(1−in)k−1i−1 > (1−cn)k−13c∑i=c1i−1 = (1−o(1))(ln3+o(1)) = ln3+o(1),

and

 (1−c−1n)k−1<1.

Therefore, for and ,

 P(c)−P(c−1)>0

always holds, indicating . Therefore,

 cOPT=Ω(n).

And because , we have

 cOPT=Θ(n).

Because ,

 P(cOPT)=Θ(1).

## 5 Conclusion

Theorem 1 indicates that, in either of following cases, the optimal cutoff point is :

1. Applicants’ abilities are i.i.d. drawn from a bounded distribution with density function greater than near the top 111It cannot be altered to “a distribution without long tails”, e.g. has no long tail, but violates the condition of Theorem 1. In fact, it does extremely prioritize the top ones. , and the DM wants to maximize the expected ability of the accepted one.

2. The DM cares about the relative ranking of the accepted in any way that “does not extremely prioritize the top ones”.

However, the normal distribution does not satisfy condition 1), but in practice, deviations beyond a few

s are extremely unlikely, so if we do not expect them to come, or their excellence does not make a big difference actually, then the rule still makes sense.

This result suggests that in most cases, the decision maker does not need to skip so many applicants for a best candidate. Intuitively, if the 37% rule is adopted, with 37% probability the desired best one is rejected at the beginning, resulting in a bad outcome of choosing the random one at the last position. But if only are skipped, with high () probability the end is not reached and the accepted one is probably fairly good. That also explains why people tend to stop early [Bearden, Rapoport, and Murphy2006]. It is not merely an irrational psychological phenomenon or due to “cost of observing”, but does improve the expected outcome when they are not too particular.

The asymptotic upper bound indicates that the more the DM cares about top ones (larger compared to ), the more skips are needed. As the condition of Theorem 1 requires the DM not to “extremely prioritize” top ones, it suggests that if the DM does have some strict objective, e.g. only cares several top ones instead of relative rankings, skips are still required (as in the original Secretary Problem), and Theorem 2 shows it.

## References

• [Bearden, Rapoport, and Murphy2006] Bearden, J. N.; Rapoport, A.; and Murphy, R. O. 2006. Sequential observation and selection with rank-dependent payoffs: An experimental study. Management Science 52(9):1437–1449.
• [Bearden2006] Bearden, J. N. 2006. A new secretary problem with rank-based selection and cardinal payoffs. Journal of Mathematical Psychology 50(1):58 – 59.
• [Ferguson1989] Ferguson, T. S. 1989. Who solved the secretary problem? Statist. Sci. 4(3):282–289.