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 et al.(2023) proposed the following test statistic:
$$T_{ZZZ}=\frac{n_1 n_2}{np}(\bar{\boldsymbol{y}}_1-\bar{\boldsymbol{y}}_2)^{\top} \hat{\boldsymbol{D}}_n^{-1}(\bar{\boldsymbol{y}}_1-\bar{\boldsymbol{y}}_2),$$
where \(\bar{\boldsymbol{y}}_{i},i=1,2\) are the sample mean vectors, and \(\hat{\boldsymbol{D}}_n=\operatorname{diag}(\hat{\boldsymbol{\Sigma}}_1/n+\hat{\boldsymbol{\Sigma}}_2/n)\) with \(n=n_1+n_2\).
They showed that under the null hypothesis, \(T_{ZZZ}\) and a chi-squared-type mixture have the same limiting distribution.