Zhang and Zhu (2022)'s test for testing equality of two-sample high-dimensional mean vectors without assuming that two covariance matrices are the same.
Arguments
- y1
The data matrix (p by n1) from the first population. Each column represents a \(p\)-dimensional observation.
- y2
The data matrix (p by n2) from the first population. Each column represents a \(p\)-dimensional observation.
Value
A (list) object of S3 class htest containing the following elements:
- p.value
the p-value of the test proposed by Zhang and Zhu (2022).
- statistic
the test statistic proposed by Zhang and Zhu (2022).
- beta0
parameter used in Zhang and Zhu (2022)'s test.
- beta1
parameter used in Zhang and Zhu (2022)'s test.
- df
estimated approximate degrees of freedom of Zhang and Zhu (2022)'s test.
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 $$H_{0}: \boldsymbol{\mu}_1 = \boldsymbol{\mu}_2\; \operatorname{versus}\; H_{1}: \boldsymbol{\mu}_1 \neq \boldsymbol{\mu}_2.$$ Zhang and Zhu (2022) proposed the following test statistic: $$T_{ZZ} = \|\bar{\boldsymbol{y}}_1 - \bar{\boldsymbol{y}}_2\|^2-\operatorname{tr}(\hat{\boldsymbol{\Omega}}_n),$$ where \(\bar{\boldsymbol{y}}_{i},i=1,2\) are the sample mean vectors and \(\hat{\boldsymbol{\Omega}}_n\) is the estimator of \(\operatorname{Cov}(\bar{\boldsymbol{y}}_1-\bar{\boldsymbol{y}}_2)\). They showed that under the null hypothesis, \(T_{ZZ}\) and a chi-squared-type mixture have the same normal or non-normal limiting distribution.
References
Zhang J, Zhu T (2022). “A further study on Chen-Qin’s test for two-sample Behrens--Fisher problems for high-dimensional data.” Journal of Statistical Theory and Practice, 16(1), 1. doi:10.1007/s42519-021-00232-w .
Examples
set.seed(1234)
n1 <- 20
n2 <- 30
p <- 50
mu1 <- t(t(rep(0, p)))
mu2 <- mu1
rho1 <- 0.1
rho2 <- 0.2
a1 <- 1
a2 <- 2
w1 <- (-2 * sqrt(a1 * (1 - rho1)) + sqrt(4 * a1 * (1 - rho1) + 4 * p * a1 * rho1)) / (2 * p)
x1 <- w1 + sqrt(a1 * (1 - rho1))
Gamma1 <- matrix(rep(w1, p * p), nrow = p)
diag(Gamma1) <- rep(x1, p)
w2 <- (-2 * sqrt(a2 * (1 - rho2)) + sqrt(4 * a2 * (1 - rho2) + 4 * p * a2 * rho2)) / (2 * p)
x2 <- w2 + sqrt(a2 * (1 - rho2))
Gamma2 <- matrix(rep(w2, p * p), nrow = p)
diag(Gamma2) <- rep(x2, p)
Z1 <- matrix(rnorm(n1 * p, mean = 0, sd = 1), p, n1)
Z2 <- matrix(rnorm(n2 * p, mean = 0, sd = 1), p, n2)
y1 <- Gamma1 %*% Z1 + mu1 %*% (rep(1, n1))
y2 <- Gamma2 %*% Z2 + mu2 %*% (rep(1, n2))
tsbf_zz2022(y1, y2)
#>
#>
#>
#> data:
#> statistic = 0.2389, df = 3.70891, beta0 = -2.31342, beta1 = 0.62375,
#> p-value = 0.3515
#>