从样本看先知不平等:越多越好吗?

Tomer Ezra
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引用次数: 0

摘要

我们研究了单选预言家不等式问题的一个变体,在这个变体中,决策者不知道基本分布,只能从分布中获取一组样本。鲁宾斯坦等人[2020]的研究表明,通过观察一组 $n$ 样本(每种分布都有一个样本),竟然可以得到 $\frac{1}{2}$ 的最优竞争比。在本文中,我们将证明当决策者获得一组更多的样本时,$\frac{1}{2}$的竞争比率将变得无法实现。我们研究了顺序静态阈值算法的自然类,该算法选择排名最高的第 i 个样本,将该样本设为静态阈值,然后选择超过该阈值的第一个值。我们证明,该类算法中的最佳算法可实现 0.433 美元的竞争比率。在此过程中,我们利用论文中开发的工具,为鲁宾斯坦等人的主要结果提供了另一种证明。[2020].
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Prophet Inequality from Samples: Is the More the Merrier?
We study a variant of the single-choice prophet inequality problem where the decision-maker does not know the underlying distribution and has only access to a set of samples from the distributions. Rubinstein et al. [2020] showed that the optimal competitive-ratio of $\frac{1}{2}$ can surprisingly be obtained by observing a set of $n$ samples, one from each of the distributions. In this paper, we prove that this competitive-ratio of $\frac{1}{2}$ becomes unattainable when the decision-maker is provided with a set of more samples. We then examine the natural class of ordinal static threshold algorithms, where the algorithm selects the $i$-th highest ranked sample, sets this sample as a static threshold, and then chooses the first value that exceeds this threshold. We show that the best possible algorithm within this class achieves a competitive-ratio of $0.433$. Along the way, we utilize the tools developed in the paper and provide an alternative proof of the main result of Rubinstein et al. [2020].
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