Monotonicity of the optimal stopping region for the duration problem

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Let Un(x)U_n(x) be the expected duration of owning the relatively best object whose rank remains within the two when the time to go is nn, conditional on accepting the relatively best applicant with value xx. Define

Gn(x)=Un(x)−∑k=1n−1xk−1∫x1Un−k(y) dy.G_n(x)=U_n(x)-\sum_{k=1}^{n-1}x^{k-1}\int_x^1U_{n-k}(y)\,dy.

The stopping region is B={(n,x):Gn(x)≥0}B=\{(n,x):G_n(x)\geq 0\}. Monotonicity conjecture. If Gn(x)≥0G_n(x)\geq 0, then Gn(y)≥0G_n(y)\geq 0 for every y≥xy\geq x; equivalently, the optimal stopping rule is a threshold rule in the observed maximum. This would establish optimality of the 1-SLA stopping rule for the full-information duration problem when stopping is restricted to relatively best objects. The analogous implication Gn(x)≥0⇒Gn−k(x)≥0G_n(x)\geq 0\Rightarrow G_{n-k}(x)\geq 0 is stated to have been proved, while this monotonicity assertion remains open.

References

Primary source

Zdzisław Porosiński, Marek Skarupski and Krzysztof Szajowski, “Duration problem: basic concept and some extensions”, arXiv:1605.08364 (2016).

Progress summary

Refreshed
Claimed progress

A reader-submitted argument claims to prove the conjecture, but no independent verification is recorded, so the problem remains open.

The conjecture asks whether acceptance remains optimal whenever the observed maximum increases, which would justify the 11-SLA rule for the full-information duration problem. The related monotonicity in the remaining time is recorded as proved, but the value monotonicity was reported open in the published source.

Known results

  • The implication Gn(x)≥0⇒Gn−k(x)≥0G_n(x)\geq 0\Rightarrow G_{n-k}(x)\geq 0 is stated as proved; the implication in xx remains unresolved in the cited literature.

Community submission (unverified), August 26, 2026

A submitted proof argues for the conjecture using an explicit polynomial expansion and positivity of Bernstein-basis coefficients. It also claims that published formulas contain indexing errors and supplies corrected expressions for UnU_n and GnG_n. The argument has not been independently verified.

Current status (as of August 2026): The conjecture remains unverified; a community-submitted proof claim is the only reported new development, while the lower-time monotonicity implication is recorded as proved.

Sources

Solutions 1

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A Bernstein-basis proof of the stopping-region monotonicity

Source correction

There are two typographical indexing errors in the source formulas. They can be resolved from the probability model itself.

If there are n stages to go, including the present record of value x, its rank remains at most two at stage r+1 precisely when at most one of the first r future observations exceeds x. Consequently

Un(x)=∑r=0n−1(xr+r(1−x)xr−1)=2∑r=0n−2xr−(n−2)xn−1.(1)\begin{aligned} U_n(x) &=\sum_{r=0}^{n-1}\bigl(x^r+r(1-x)x^{r-1}\bigr)\\ &=2\sum_{r=0}^{n-2}x^r-(n-2)x^{n-1}. \tag{1} \end{aligned}

The summand with r=0 in the first line is interpreted as 1. In particular, U_1=1 and U_2=2. Kurushima--Ano's preceding probability sum is exactly the first line of (1), but both that paper and the later survey print -n x^{n-1} in the simplified formula. That cannot be correct: it would give U_1=-1, and direct simplification of the preceding line gives -(n-2)x^{n-1}.

We use the definition

Gn(x)=Un(x)−∑k=1n−1xk−1∫x1Un−k(y) dy.(2)G_n(x)=U_n(x)-\sum_{k=1}^{n-1}x^{k-1} \int_x^1U_{n-k}(y)\,dy. \tag{2}

This is the expected advantage of stopping over one-stage look-ahead.

Power coefficients

Let H_m=1+1/2+...+1/m, with H_0=0. Since

∫01Um(y) dy=2Hm−1,\int_0^1U_m(y)\,dy=2H_m-1,

expanding (2) gives

Gn(x)=∑j=0n−1(3−2Hn−j−1+2Hj)xj−2nxn−1.(3)\boxed{ G_n(x)=\sum_{j=0}^{n-1} \bigl(3-2H_{n-j-1}+2H_j\bigr)x^j-2n x^{n-1}.} \tag{3}

This is also the coefficient representation in Kurushima--Ano. Equivalently, the fully expanded formula is

Gn(x)=3∑k=1nxk−1−2nxn−1−2∑k=1n−1Hn−kxk−1+2∑k=1n−1Hkxk.(4)\begin{aligned} G_n(x)={}&3\sum_{k=1}^{n}x^{k-1}-2nx^{n-1} -2\sum_{k=1}^{n-1}H_{n-k}x^{k-1} +2\sum_{k=1}^{n-1}H_kx^k. \tag{4} \end{aligned}

The displayed threshold equation in both sources has H_{n-k-1} where (4) requires H_{n-k}. Formula (3), the probability calculation, and the defining integral all agree with (4).

Bernstein coefficients

Put d=n-1 and write G_n in the degree-d Bernstein basis:

Gn(x)=∑k=0dβk(dk)xk(1−x)d−k.(5)G_n(x)=\sum_{k=0}^{d}\beta_k {d\choose k}x^k(1-x)^{d-k}. \tag{5}

For k<d, put r=d-k. Then

βk=nr+1(3−2Hr),βd=n.(6)\boxed{\displaystyle \beta_k=\frac{n}{r+1}\bigl(3-2H_r\bigr),\qquad \beta_d=n.} \tag{6}

To verify (6), if c_j=3-2H_{d-j}+2H_j denotes the power coefficient before the final correction, the power-to-Bernstein conversion is

βk=∑j=0kcj(kj)(dj).\beta_k=\sum_{j=0}^{k}c_j \frac{{k\choose j}}{{d\choose j}}.

The two elementary identities

∑j=0k(kj)(dj)=d+1r+1,\sum_{j=0}^{k}\frac{{k\choose j}}{{d\choose j}} =\frac{d+1}{r+1},

and

∑j=0k(Hd−j−Hj)(kj)(dj)=Hrd+1r+1\sum_{j=0}^{k}(H_{d-j}-H_j) \frac{{k\choose j}}{{d\choose j}} =H_r\frac{d+1}{r+1}

give the first part of (6). They follow on putting j=k-l from the hockey-stick identity and its harmonic convolution. Finally, beta_d=G_n(1)=U_n(1)=n.

One sign change

For r>=3, H_r>3/2, while H_2=3/2 and H_1=1. Thus, in increasing order of k, the nonzero Bernstein coefficients consist of

  • negative coefficients through k=d-3;
  • beta_{d-2}=0;
  • beta_{d-1}=n/2>0 and beta_d=n>0.

There is exactly one sign variation when n>=4. Set t=x/(1-x). For 0<x<1, equation (5) becomes

Gn(x)=(1−x)d∑k=0dβk(dk)tk.(7)G_n(x)=(1-x)^d \sum_{k=0}^{d}\beta_k {d\choose k}t^k. \tag{7}

The polynomial in t has one coefficient sign change, so Descartes' rule of signs gives at most one root in (0,infinity), hence G_n has at most one root in (0,1).

For n>=4,

Gn(0)=3−2Hn−1<0,Gn(1)=n>0.G_n(0)=3-2H_{n-1}<0, \qquad G_n(1)=n>0.

Therefore G_n has exactly one root s_n in (0,1), is negative before s_n, and positive after it. The small cases are

G1(x)=1,G2(x)=1+x,G3(x)=3x.G_1(x)=1,\qquad G_2(x)=1+x,\qquad G_3(x)=3x.

It follows for every n and every 0<=x<=y<=1 that

Gn(x)≥0⟹Gn(y)≥0.G_n(x)\geq0\quad\Longrightarrow\quad G_n(y)\geq0.

This proves the monotonicity assertion in MathDB #333222 and establishes the threshold form of the one-stage-look-ahead stopping region for the intended expected-duration model.

Verification record

No indexed published resolution of this exact monotonicity assertion was located through 2026-08-18. This is an internally verified proof, not yet a claim of external peer review or publication.

Lean: https://github.com/antoshashakov/Principia-Math-In-Progress/blob/main/mathdb-open-problems/problems/333222/Problem333222.lean

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Models used: GPT 5.6 Sol, Fable