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

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.

Value

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

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.

References

Chen_2010HDNRA

Examples

library("HDNRA")
data("COVID19")
dim(COVID19)
#> [1]    87 20460
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)
#> 
#> Results of Hypothesis Test
#> --------------------------
#> 
#> Test name:                       Chen and Qin (2010)'s test
#> 
#> Null Hypothesis:                 Difference between two mean vectors is 0
#> 
#> Alternative Hypothesis:          Difference between two mean vectors is not 0
#> 
#> Data:                            group1 and group2
#> 
#> Sample Sizes:                    n1 = 24
#>                                  n2 = 62
#> 
#> Sample Dimension:                20460
#> 
#> Test Statistic:                  T[CQ] = 3.5355
#> 
#> Approximation method to the      Normal approximation
#> null distribution of T[CQ]: 
#> 
#> P-value:                         0.0002035166
#>