CQ2010.TSBF.NABT: Normal-approximation-based test for two-sample BF problem...

View source: R/CQ2010.TSBF.NABT.R

CQ2010.TSBF.NABTR Documentation

Normal-approximation-based test for two-sample BF problem proposed by Chen and Qin (2010)

Description

Chen and Qin (2010)'s test for testing equality of two-sample high-dimensional mean vectors without assuming that two covariance matrices are the same.

Usage

CQ2010.TSBF.NABT(y1, y2)

Arguments

y1

The data matrix (n_1 \times p) from the first population. Each row represents a p-dimensional observation.

y2

The data matrix (n_2 \times p) from the second population. Each row represents a p-dimensional observation.

Details

Suppose we have two independent high-dimensional samples:

\boldsymbol{y}_{i1},\ldots,\boldsymbol{y}_{in_i}, \;\operatorname{are \; i.i.d. \; with}\; \operatorname{E}(\boldsymbol{y}_{i1})=\boldsymbol{\mu}_i,\; \operatorname{Cov}(\boldsymbol{y}_{i1})=\boldsymbol{\Sigma}_i,i=1,2.

The primary object is to test

H_{0}: \boldsymbol{\mu}_1 = \boldsymbol{\mu}_2\; \operatorname{versus}\; H_{1}: \boldsymbol{\mu}_1 \neq \boldsymbol{\mu}_2.

Chen and Qin (2010) proposed the following test statistic:

T_{CQ} = \frac{\sum_{i \neq j}^{n_1} \boldsymbol{y}_{1i}^\top \boldsymbol{y}_{1j}}{n_1 (n_1 - 1)} + \frac{\sum_{i \neq j}^{n_2} \boldsymbol{y}_{2i}^\top \boldsymbol{y}_{2j}}{n_2 (n_2 - 1)} - 2 \frac{\sum_{i = 1}^{n_1} \sum_{j = 1}^{n_2} \boldsymbol{y}_{1i}^\top \boldsymbol{y}_{2j}}{n_1 n_2}.

They showed that under the null hypothesis, T_{CQ} is asymptotically normally distributed.

Value

A list of class "NRtest" containing the results of the hypothesis test. See the help file for NRtest.object for details.

References

\insertRef

Chen_2010HDNRA

Examples

library("HDNRA")
data("COVID19")
dim(COVID19)
group1 <- as.matrix(COVID19[c(2:19, 82:87), ]) ## healthy group
group2 <- as.matrix(COVID19[-c(1:19, 82:87), ]) ## COVID-19 patients
CQ2010.TSBF.NABT(group1,group2)



HDNRA documentation built on Oct. 30, 2024, 9:28 a.m.