One of the motivations to study Robbins' problem is that with its solution all classical (four) secretary problems would be solved. But the major reason is to understand how to cope with full history dependence in a (deceptively easy-looking) problem. On the Ester's Book International Conference in Israel (2006) Robbins' problem was accordingly named one of the four most important problems in the field of optimal stopping and sequential analysis.
Herbert Robbins presented the above described problem at the International Conference on Search and Selection in Real Time8 in Amherst, 1990. He concluded his address with the words I should like to see this problem solved before I die. Scientists working in the field of optimal stopping have since called this problem Robbins' problem. Unfortunately, Herbert Robbins' wish did not become true. He died in 2001.
Another optimal stopping problem bearing Robbins' name (and not to be c onfused with Robbins' problem) is the Chow–Robbins game:910
Given an infinite sequence of IID random variables X 1 , X 2 , . . . {\displaystyle X_{1},X_{2},...} with distribution F {\displaystyle F} , how to decide when to stop, in order to maximize the sample average 1 n ( X 1 + ⋯ X n ) {\displaystyle {\frac {1}{n}}(X_{1}+\cdots X_{n})} where n {\displaystyle n} is the stopping time? The probability of eventually stopping must be 1 (that is, you are not allowed to keep sampling and never stop).
For any distribution F {\displaystyle F} with finite second moment, there exists an optimal strategy, defined by a sequence of numbers β 1 , β 2 , . . . {\displaystyle \beta _{1},\beta _{2},...} . The strategy is to keep sampling until 1 n ( X 1 + ⋯ X n ) ≥ β n {\displaystyle {\frac {1}{n}}(X_{1}+\cdots X_{n})\geq \beta _{n}} .1112
If F {\displaystyle F} has finite second moment, then after subtracting the mean and dividing by the standard deviation, we get a distribution with mean zero and variance one. Consequently it suffices to study the case of F {\displaystyle F} with mean zero and variance one.
With this, lim n β n / n ≈ α = 0.8399236757 {\displaystyle \lim _{n}\beta _{n}/{\sqrt {n}}\approx \alpha =0.8399236757} , where α {\displaystyle \alpha } is the solution to the equation13 α = ( 1 − α 2 ) ∫ 0 ∞ e λ α − λ 2 / 2 d λ {\displaystyle \alpha =\left(1-\alpha ^{2}\right)\int _{0}^{\infty }e^{\lambda \alpha -\lambda ^{2}/2}d\lambda } which can be proved by solving the same problem with continuous time, with a Wiener process. At the limit of n → ∞ {\displaystyle n\to \infty } , the discrete time problem becomes the same as the continuous time problem.
This was proved independently14 by.151617
When the game is a fair coin toss game, with heads being +1 and tails being -1, then there is a sharper result18 β n = α n − 1 / 2 + ( − 2 ζ ( − 1 / 2 ) ) α π n − 1 / 4 + O ( n − 7 / 24 ) {\displaystyle \beta _{n}=\alpha {\sqrt {n}}-1/2+{\frac {(-2\zeta (-1/2)){\sqrt {\alpha }}}{\sqrt {\pi }}}n^{-1/4}+O\left(n^{-7/24}\right)} where ζ {\displaystyle \zeta } is the Riemann zeta function.
When n is small, the asymptotic bound does not apply, and finding the value of β n {\displaystyle \beta _{n}} is much more difficult. Even the simplest case, where X 1 , X 2 , . . . {\displaystyle X_{1},X_{2},...} are fair coin tosses, is not fully solved.
For the fair coin toss, a strategy is a binary decision: after n {\displaystyle n} tosses, with k heads and (n-k) tails, should one continue or should one stop? Since 1D random walk is recurrent, starting at any k , ( n − k ) {\displaystyle k,(n-k)} , the probability of eventually having more heads than tails is 1. So, if k ≤ n − k {\displaystyle k\leq n-k} , one should always continue. However, if k > n − k {\displaystyle k>n-k} , it is tricky to decide whether to stop or continue.19
20 found an exact solution for all n ≤ 489241 {\displaystyle n\leq 489241} .
Elton21 found exact solutions for all n ≤ 9.06 × 10 7 {\displaystyle n\leq 9.06\times 10^{7}} , and it found an almost always optimal decision rule, of stopping as soon as k − ( n − k ) ≥ Δ k n {\displaystyle k-(n-k)\geq \Delta k_{n}} where Δ k n = ⌈ α n − 1 / 2 + ( − 2 ζ ( − 1 / 2 ) ) α π n − 1 / 4 ⌉ {\displaystyle \Delta k_{n}=\left\lceil {\alpha {\sqrt {n}}\,\,-1/2\,\,+\,\,{\frac {\left({-2\zeta (-1/2)}\right){\sqrt {\alpha }}}{\sqrt {\pi }}}{n^{-1/4}}}\right\rceil }
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The Joint Summer Research Conferences in the Mathematical Sciences were held at the University of Massachusetts from June 7 to July 4, 1990. These were sponsored by the AMS, SIAM, and the Institute for Mathematical Statistics (IMS). Topics in 1990 were: Probability models and statistical analysis for ranking data, Inverse scattering on the line, Deformation theory of algebras and quantization with applications to physics, Strategies for sequential search and selection in real time, Schottky problems, and Logic, fields, and subanalytic sets. ↩
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Elton, John H. (2023-06-06). "Exact Solution to the Chow-Robbins Game for almost all n, using the Catalan Triangle". arXiv:2205.13499 [math].{{cite arXiv}}: CS1 maint: date and year (link) /wiki/ArXiv_(identifier) ↩
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import numpy as np from scipy.integrate import quad from scipy.optimize import root def f(lambda_, alpha): return np.exp(lambda_ * alpha - lambda_**2 / 2) def equation(alpha): integral, error = quad(f, 0, np.inf, args=(alpha)) return integral * (1 - alpha**2) - alpha solution = root(equation, 0.83992, tol=1e-15) # Print the solution if solution.success: print(f"Solved α = {solution.x[0]} with a residual of {solution.fun[0]}") else: print("Solution did not converge") ↩
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Häggström, Olle; Wästlund, Johan (2013). "Rigorous Computer Analysis of the Chow–Robbins Game". The American Mathematical Monthly. 120 (10): 893. doi:10.4169/amer.math.monthly.120.10.893. /wiki/Olle_H%C3%A4ggstr%C3%B6m ↩
Christensen, Sören; Fischer, Simon (June 2022). "On the Sn/n problem". Journal of Applied Probability. 59 (2): 571–583. doi:10.1017/jpr.2021.73. ISSN 0021-9002. https://www.cambridge.org/core/journals/journal-of-applied-probability/article/abs/on-the-snn-problem/8CE3A5834AC35E7ADD137749F61E27A2 ↩