A two sample test in high dimensional data
A two sample test in high dimensional data
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DOI：
10.1016/j.jmva.2012.08.014
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发表时间：
2013-02-01
影响因子：
1.6
通讯作者：
Kano, Yutaka
中科院分区：
文献类型：
--
作者：
Srivastava, Muni S.;Katayama, Shota;Kano, Yutaka
本文基于N-1和N-2个独立分布的p维观测向量，提出了一种检验两组具有不等协方差矩阵的均值向量相等性的方法。将假设来自第一组的N-1个观测向量是正态分布的，具有平均向量mu（1）和协方差矩阵Sigma（1）。类似地，来自第二组的N-2个观测向量是正态分布的，具有均值向量mu（2）和协方差矩阵Sigma（2）。我们提出了一个检验mu（1）= mu（2）的假设的检验。这个检验在p × p非奇异对角矩阵群下是不变的。渐近分布为（N-1，N-2，p）->无穷大，N-1/（N-1 + N-2）-> k是（0，1）的一个元素，但N-1/p和N-2/p可能趋于零或无穷大。它是比较最近提出的非不变的测试。结果表明，所提出的测试执行最好的。(C)2012 Elsevier Inc. All rights reserved.
In this paper we propose a test for testing the equality of the mean vectors of two groups with unequal covariance matrices based on N-1 and N-2 independently distributed p-dimensional observation vectors. It will be assumed that N-1 observation vectors from the first group are normally distributed with mean vector mu(1) and covariance matrix Sigma(1). Similarly, the N-2 observation vectors from the second group are normally distributed with mean vector mu(2) and covariance matrix Sigma(2). We propose a test for testing the hypothesis that mu(1) = mu(2). This test is invariant under the group of p x p nonsingular diagonal matrices. The asymptotic distribution is obtained as (N-1, N-2, p) -> infinity and N-1/(N-1 + N-2) -> k is an element of (0, 1) but N-1/p and N-2/p may go to zero or infinity. It is compared with a recently proposed non-invariant test. It is shown that the proposed test performs the best. (C) 2012 Elsevier Inc. All rights reserved.