对于未知但有界的误差,区间估计通常比平均要好

G. Walster, V. Kreinovich
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引用次数: 16

摘要

对于许多测量设备,我们所知道的唯一信息就是它们的最大可能误差。比;0. 换句话说,我们知道误差Δx = x - x(即测量值x与实际值x之间的差值)是随机的,这个误差有时会变得像ε或者-,但是我们没有任何关于不同误差值的概率的信息。统计方法使我们能够通过多次测量x1,…来对x做出更好的估计。例如,如果平均误差为0 (E(Δx) = 0),那么在n次测量后,我们可以取平均值x = (x1 +…+ xn)/n,得到一个标准差(和相应的置信区间)小于n倍的估计值。另一个估计来自区间分析:对于每个测量xi,我们知道实际值x属于区间[xi-ε, xi+ε]。所以x属于所有这些区间的交点。从某种意义上说,这种估计比基于传统工程统计(即平均)的估计要好:区间估计得到了保证。在本文中,我们证明了在许多情况下,这个交集在某种意义上也更好,因为它对x给出了比平均更准确的估计:即,在某些合理的条件下,这个区间估计的误差比平均的误差减少得更快(1/n)(只减少为1/ √n)。当我们测量几个辅助量,并使用测量结果来估计所需量y的值时,在多维情况下证明了类似的结果。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
For unknown-but-bounded errors, interval estimates are often better than averaging
For many measuring devices, the only information that we have about them is their biggest possible error ε > 0. In other words, we know that the error Δx = x - x (i.e., the difference between the measured value x and the actual values x) is random, that this error can sometimes become as big as ε or - ε, but we do not have any information about the probabilities of different values of error.Methods of statistics enable us to generate a better estimate for x by making several measurements x1, ..., xn. For example, if the average error is 0 (Ex) = 0), then after n measurements, we can take an average x = (x1 + ... + xn)/n, and get an estimate whose standard deviation (and the corresponding confidence intervals) are √n times smaller.Another estimate comes from interval analysis: for every measurement xi, we know that the actual value x belongs to an interval [xi-ε, xi+ε]. So, x belongs to the intersection of all these intervals. In one sense, this estimate is better than the one based on traditional engineering statistics (i.e., averaging): interval estimation is guaranteed. In this paper, we show that for many cases, this intersection is also better in the sense that it gives a more accurate estimate for x than averaging: namely, under certain reasonable conditions, the error of this interval estimate decreases faster (as 1/n) than the error of the average (that only decreases as 1/ √n).A similar result is proved for a multi-dimensional case, when we measure several auxiliary quantities, and use the measurement results to estimate the value of the desired quantity y.
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