三智独立随机漫步可以稍微无界

Pub Date : 2022-01-03 DOI:10.1002/rsa.21075
Shyam Narayanan
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引用次数: 0

摘要

最近,许多流算法都利用了这样一个事实,即任何4 - wise独立随机漫步在n步线上的期望最大距离为O(n)。$$ O\left(\sqrt{n}\right) $$ . 在本文中,我们通过构造一个期望最大距离Ω(nlgn)的3 - wise独立随机漫步来证明所有这些算法都需要4 - wise独立性。$$ \Omega \left(\sqrt{n}\lg n\right) $$ 从原点开始。我们证明了这个界对于一阶矩和二阶矩是紧的,并从这些结果中提取了一个令人惊讶的矩阵不等式。$$ {X}_i $$ 是具有有界PTH矩的k独立随机变量。我们强调k=4 p=2的情况$$ k=4,p=2 $$ 在这里,我们证明了最远距离的第二弯矩为O∑Xi2$$ O\left(\sum {X}_i^2\right) $$ . 这暗示了一个比只需要4个独立随机变量的Kolmogorov极大不等式更强的渐近命题,并推广了Błasiok的最新结果。
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Three‐wise independent random walks can be slightly unbounded
Recently, many streaming algorithms have utilized generalizations of the fact that the expected maximum distance of any 4‐wise independent random walk on a line over n steps is O(n)$$ O\left(\sqrt{n}\right) $$ . In this paper, we show that 4‐wise independence is required for all of these algorithms, by constructing a 3‐wise independent random walk with expected maximum distance Ω(nlgn)$$ \Omega \left(\sqrt{n}\lg n\right) $$ from the origin. We prove that this bound is tight for the first and second moment, and also extract a surprising matrix inequality from these results. Next, we consider a generalization where the steps Xi$$ {X}_i $$ are k‐wise independent random variables with bounded pth moments. We highlight the case k=4,p=2$$ k=4,p=2 $$ : here, we prove that the second moment of the furthest distance traveled is O∑Xi2$$ O\left(\sum {X}_i^2\right) $$ . This implies an asymptotically stronger statement than Kolmogorov's maximal inequality that requires only 4‐wise independent random variables, and generalizes a recent result of Błasiok.
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