Appendix 1
\(\mathbf{{A}}\) Under \({H_0}\) and conditions defined in \((a), (b_2)\) and \((c_2)\), the expectation and the variance of the statistic \({T_{n}}\) are
$$\begin{aligned} E(T_n) =\,&\text {tr}(D\Sigma _1/n_1) + \text {tr}(D\Sigma _2/n_2), \\ \text{ Var }({T_{n}}) =\,&\left[ \frac{2}{n_1^2} \text {tr}\{(D\Sigma _1)^2\} + \frac{2}{n_2^2} \text {tr}\{(D\Sigma _2)^2\} + \frac{4}{n_1n_2}\text {tr}(D\Sigma _2D\Sigma _1) \right] \{1 + o(1 )\}. \end{aligned}$$
In particular, assuming \(\Sigma = \Sigma _1 = \Sigma _2\), under \(H_0\) and conditions \((a), (b_1)\), \((c_1)\), the expectation and the variance of the statistic \(T_n\) simplify to
$$\begin{aligned} E(T_n) =\,&\tau \text {tr}(D\Sigma ), \\ \text{ Var }({T_{n}}) =\,&2{\tau ^2}\text {tr}{\{(D\Sigma )^2 \}}\{1 + o(1 )\}. \end{aligned}$$
Proof
The proposed statistic \({T_{n}}\) can be rewritten as
$$\begin{aligned} {T_{n}}&= ({{\bar{\textbf {X}}}_1}-{{\bar{\textbf {X}}}_2})^TD({{\bar{\textbf {X}}}_1}-{{\bar{\textbf {X}}}_2}) = {{\bar{\textbf {X}}}_1}^T D{{\bar{\textbf {X}}}_1} + {{\bar{\textbf {X}}}_2}^T D{{\bar{\textbf {X}}}_2 }-2{{\bar{\textbf {X}}}_1}^T D{{\bar{\textbf {X}}}_2}. \end{aligned}$$
Thus,
$$\begin{aligned} E(T_n) =\,&E({\bar{{\textbf {X}}}}_1^T D {\bar{{\textbf {X}}}}_1 + {\bar{{\textbf {X}}}}_2^T D {\bar{{\textbf {X}}}}_2 - 2{\bar{{\textbf {X}}}}_1^T D {\bar{{\textbf {X}}}}_2)\\ =\,&\text {tr}(D\Sigma _1)/n_1 + \varvec{\mu }_1^T D \varvec{\mu }_1 + \text {tr}(D\Sigma _2)/n_2 + \varvec{\mu }_2^T D \varvec{\mu }_2 - 2\varvec{\mu }_1^TD \varvec{\mu }_2\\ =\,&(\varvec{\mu }_1-\varvec{\mu }_2)^T D (\varvec{\mu }_1-\varvec{\mu }_2) + \text {tr}(D\Sigma _1)/n_1 + \text {tr}(D\Sigma _2)/n_2, \end{aligned}$$
where
$$\begin{aligned} E({\bar{{\textbf {X}}}}_1^T D {\bar{{\textbf {X}}}}_1) =\,&E({\bar{{\textbf {X}}}}_1 - \varvec{\mu }_1 +\varvec{\mu }_1)^T D({\bar{{\textbf {X}}}}_1 - \varvec{\mu }_1 +\varvec{\mu }_1)\\ =\,&E({\bar{{\textbf {X}}}}_1-\varvec{\mu }_1)^T D ({\bar{{\textbf {X}}}}_1-\varvec{\mu }_1) + E({\bar{{\textbf {X}}}}_1-\varvec{\mu }_1)^TD\varvec{\mu }_1 + \varvec{\mu }_1^TDE({\bar{{\textbf {X}}}}_1-\varvec{\mu }_1)+\varvec{\mu }_1^TD\varvec{\mu }_1\\ =\,&E\left[ \text {tr}\{D({\bar{{\textbf {X}}}}_1-\varvec{\mu }_1)({\bar{{\textbf {X}}}}_1-\varvec{\mu }_1)^T\}\right] +\varvec{\mu }_1^TD\varvec{\mu }_1\\ =\,&\text {tr}\left[ E\{D({\bar{{\textbf {X}}}}_1-\varvec{\mu }_1)({\bar{{\textbf {X}}}}_1-\varvec{\mu }_1)^T\} \right] + \varvec{\mu }_1^TD\varvec{\mu }_1\\ =\,&\text {tr}(D\Sigma _1)/n_1 + \varvec{\mu }_1^TD\varvec{\mu }_1. \end{aligned}$$
Similarly, \(E({\bar{{\textbf {X}}}}_2^T D {\bar{{\textbf {X}}}}_2) = \text {tr}(D\Sigma _2)/n_2 + \varvec{\mu }_2^TD\varvec{\mu }_2\). Since \({\bar{{\textbf {X}}}}_1\) and \({\bar{{\textbf {X}}}}_2\) are independent, we have that
$$\begin{aligned} E(2{\bar{{\textbf {X}}}}_1^TD{\bar{{\textbf {X}}}}_2) =\,&2 E \left\{ ({\bar{{\textbf {X}}}}_1 -\varvec{\mu }_1 +\varvec{\mu }_1)^T D ({\bar{{\textbf {X}}}}_2 -\varvec{\mu }_2 +\varvec{\mu }_2)\right\} \\ =\,&2E\left\{ ({\bar{{\textbf {X}}}}_1-\varvec{\mu }_1)^T D ({\bar{{\textbf {X}}}}_2-\varvec{\mu }_2)\right\} + 2E({\bar{{\textbf {X}}}}_1-\varvec{\mu }_1)^TD\varvec{\mu }_2\\&+ 2\varvec{\mu }_1^TDE({\bar{{\textbf {X}}}}_2-\varvec{\mu }_2) + 2\varvec{\mu }_1^TD\varvec{\mu }_2\\ =\,&2\varvec{\mu }_1^TD\varvec{\mu }_2. \end{aligned}$$
Hence, under \(H_0\), \(E(T_n) = \text {tr}(D\Sigma _1)/n_1 + \text {tr}(D\Sigma _2)/n_2\). In case of identical covariance matrices, \(E(T_n) = \tau \text {tr}(D\Sigma )\), where \(\tau = (n_1+n_2)/(n_1n_2)\).
Here, we derive the variance of \(T_n\) under some mild conditions. First, we write
$$\begin{aligned} \text{ Var }({T_{n}}) =\,&\text{ Var }({\bar{\textbf {X}}^T_1}D{{\bar{\textbf {X}}}_1}) + \text{ Var }({{\bar{\textbf {X}}}^T_2}D{{\bar{\textbf {X}}}_2}) + 4\text{ Var }({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_2}) \\&+ 2\text{ Cov }({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_1},{{\bar{\textbf {X}}}^T_2} D{{\bar{\textbf {X}}}_2})-4\text{ Cov }({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_1},{{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_2} ) \\&-4\text{ Cov }({{\bar{\textbf {X}}}^T_2}D{{\bar{\textbf {X}}}_2},{{\bar{\textbf {X}}}^T_1} D{{\bar{\textbf {X}}}_2}). \end{aligned}$$
We separately develop the six addends on the right hand of the above equation expressed as Step 1 to Step 6.
Step 1: We have
$$\begin{aligned} \text{ Var }({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_1}) =\,&E\{({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_1})^2\} - {E^2}({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_1}) \\ =\,&\frac{\Delta }{{n_1^3}}\sum \limits _{i = 1}^m {\gamma _{1ii}^2} + \frac{2}{{n_1^2}}\text {tr}\{(D\Sigma _1 )^2\} \nonumber \\&+ \frac{4}{{{n_1}}}{{{\varvec{\mu }^T_1}}}D\Sigma _1 D{{\varvec{\mu }_1}} + \frac{4}{{n_1^2}}(\Gamma _1^T D{{\varvec{\mu }_1}})\textrm{diag}(\Gamma _1^TD\Gamma _1 ){\beta _3^1}. \end{aligned}$$
where \(\beta _3^1 = {E( z_{1ij}^3)}\) and denoting by \({\bar{{\textbf {Z}}}}_i = \sum _{j=1}^{n_i} Z_{ij}/n_i\), it follows that
$$\begin{aligned} E\{({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_1})^2\} =\,&E\left[ \left\{ (\Gamma _1{\bar{{\textbf {Z}}}}_1 + \varvec{\mu }_1)^T D (\Gamma _1{\bar{{\textbf {Z}}}}_1 + \varvec{\mu }_1)\right\} ^2\right] \\ =\,&E\left\{ ({\bar{{\textbf {Z}}}}_1^T\Gamma _1^TD\Gamma _1{\bar{{\textbf {Z}}}}_1 + {\bar{{\textbf {Z}}}}_1^T\Gamma _1^TD\varvec{\mu }_1 + 2\varvec{\mu }_1^TD\Gamma _1{\bar{{\textbf {Z}}}}_1+ \varvec{\mu }_1^TD\varvec{\mu }_1)^2 \right\} \\ =\,&E\left\{ ({\bar{{\textbf {Z}}}}_1^T\Gamma _1^TD\Gamma _1{\bar{{\textbf {Z}}}}_1 + 2{\bar{{\textbf {Z}}}}_1^T\Gamma _1^TD\varvec{\mu }_1 + \varvec{\mu }_1^TD\varvec{\mu }_1)^2 \right\} \\ =\,&E\left\{ {({{\bar{\textbf {Z}}}^T_1}\Gamma _1^T D\Gamma _1 {{\bar{\textbf {Z}}}_1})^2} + 4{({{\bar{\textbf {Z}}}^T_1}\Gamma _1^TD{{\varvec{\mu }_1}})^2} + {({{{\varvec{ \mu }^T_1}}}D{{\varvec{\mu }_1}})^2}\right. \\&\quad \quad \left. +4{\bar{{\textbf {Z}}}}_1^T\Gamma _1^TD\Gamma _1{\bar{{\textbf {Z}}}}_1{\bar{{\textbf {Z}}}}_1^T\Gamma _1^TD\varvec{\mu }_1+ 2{{\bar{\textbf {Z}}}_1^T} \Gamma _1^TD\Gamma _1 {{\bar{\textbf {Z}}}_1}{{{\varvec{\mu }^T_1}}}D{{\varvec{\mu }_1}}\right. \\&\quad \quad \left. + 4{{\bar{\textbf {Z}}}^T_1}\Gamma _1^TD\varvec{\mu }_1\varvec{\mu }_1^TD{{\varvec{\mu }_1}} \right\} ; \end{aligned}$$
Since \(E\left( {\bar{{\textbf {Z}}}}_1\right) =0\) and denoting \(\Gamma _1^TD\Gamma _1 = {[{\gamma _{ij}^1}]_{m \times m}}\), then
$$\begin{aligned} E\{({\bar{\textbf {Z}}_1^T} \Gamma _1^TD\Gamma _1{\bar{\textbf {Z}}_1})^2\} = \sum \limits _{i = 1}^m {\sum \limits _{j = 1} ^m {\sum \limits _{k = 1}^m {\sum \limits _{l = 1}^m {E({\gamma ^1_{ij}}{{{\bar{z}} }_{1i}}{{{\bar{z}}}_{1j}} {\gamma ^1_{kl}} {{{\bar{z}}}_{1k}}{{{\bar{z}}}_{1l}})}}}}, \end{aligned}$$
where \({{\bar{z}}_{1k}} = \sum \nolimits _{j = 1}^{{n_1}} {{z_{1jk}}}/n_1,k = 1, \dots ,m\), and
$$\begin{aligned} E({{\bar{z}}_{1i}}{{\bar{z}}_{1j}}{{\bar{z}}_{1k}}{{\bar{z}}_{1l}}) = \left\{ \begin{array}{l} {{{\tilde{m}}}_4},\quad \quad \quad i = j = k = l;\\ \frac{1}{{n_1^2}},\quad \quad \quad i = j \ne k = l;i = k \ne j = l;i = l \ne j = k; \\ 0,\quad \quad \quad \;\; {\text {other}}. \\ \end{array} \right. \end{aligned}$$
According to the assumption (a), \({{\tilde{m}}_4} = E({\bar{z}}_{1k}^4) = \frac{\Delta }{{n_1^3}} + \frac{3}{{n_1^2}}\), where
$$\begin{aligned} E({\bar{z}}_{1k}^4) =\,&\frac{E \left\{ \left( \sum _{j=1}^{n_1}z_{1jk} \right) ^4\right\} }{n_1^4} = \frac{1}{{n_1^4}} \left\{ \sum \limits _{i = 1}^{{n_1}} {( 3 + \Delta ) + 3\mathop {\sum {\sum 1} }\limits _{1 \le i \ne j \le {n_1}}}\right\} \\ =\,&\frac{3n_1+\Delta n_1 + 3(n_1^2 -n_1) }{n_1^4}=\frac{\Delta }{{n_1^3}} + \frac{3}{{n_1^2}},\\ E({\bar{z}}_{1j}^2{\bar{z}}_{1k}^2 ) =\,&\text{ Cov }({\bar{z}}_{1j}^2, {\bar{z}}_{1k}^2) + E({\bar{z}}_{1j}^2) E({\bar{z}}_{1k}^2) =\frac{1}{n_1^2}\cdot \end{aligned}$$
Therefore,
$$\begin{aligned} E\{({{\bar{\textbf {Z}}}_1^T} \Gamma _1^TD\Gamma _1 {{\bar{\textbf {Z}}}_1})^2\} =\,&{{{\tilde{m}}}_4}\sum \limits _{i = 1 }^m {\gamma _{1ii}^2} + \frac{1}{{n_1^2}}\left( \sum \limits _{i \ne k} {{\gamma _{1ii}}{\gamma _ {1kk}} + \sum \limits _{i \ne j} {\gamma _{1ij}^2 + \sum \limits _{i \ne j} {{\gamma _{1ij}}{\gamma _{1ji }}}}}\right) \\ =\,&\tilde{m}_4\sum _{i=1}^{m}\gamma _{1ii}^2 - \frac{3}{n_1^2}\sum _{i=1}^{m}\gamma _{1ii}^2 + \frac{1}{n_1^2} \sum \limits _{i \ne k} {\gamma _{1ii}}{\gamma _ {1kk}} + \frac{1}{n_1^2} \sum _{i=1}^{m}\gamma _{1ii}^2 \\&+ \frac{1}{n_1^2} \sum \limits _{i \ne j} {\gamma _{1ij}^2} + \frac{1}{n_1^2} \sum _{i=1}^{m}\gamma _{1ii}^2 + \frac{1}{n_1^2} \sum \limits _{i \ne j} {\gamma _{1ij}}{\gamma _ {1ji}} + \frac{1}{n_1^2} \sum _{i=1}^{m}\gamma _{1ii}^2 \\ =\,&\frac{\Delta }{{n_1^3}}\sum \limits _{i = 1}^m {\gamma _{1ii}^ 2} + \frac{1}{{n_1^2}}\left[ \text {tr}^2(D\Sigma _1) + 2\text {tr}\{(D\Sigma _1 )^2\}\right] \cdot \end{aligned}$$
Moreover, we have \(E{({\bar{\textbf {Z}}^T_1}\Gamma _1^TD{{\varvec{\mu }_1}})^2} = \frac{1}{{{n_1}}}{{\varvec{\mu }^T_1}}D\Sigma _1 D{{\varvec{\mu }}_1}\) and \(E({\bar{\textbf {Z}}^T_1}\Gamma _1^TD\Gamma _1{\bar{\textbf {Z}}_1}{{\varvec{\mu }^T_1}}D{{\varvec{\mu }}_1}) = \frac{1}{{{n_1}}}\text {tr}(D\Sigma _1 ){{\varvec{\mu }^T_1}}D{ {\varvec{\mu }}_1}\), since \(Var({\bar{{\textbf {Z}}}}_1) = E({\bar{{\textbf {Z}}}}_1^2) = {\text {I}}_m/n_1\) \(\cdot\)
Now, we consider the calculation of \(E({\bar{\textbf {Z}}^T_1}\Gamma _1^TD\Gamma _1{\bar{\textbf {Z}}_1}{{\varvec{\mu }^T_1}}D\Gamma _1{\bar{\textbf {Z}}_1}) = \sum \nolimits _{i = 1}^m {\sum \nolimits _{j = 1}^m {\sum \nolimits _{k = 1}^m {E({\gamma _{1ij}}{b_{1k}}{{{\bar{z}}}_ {1i}}{{{\bar{z}}}_{1j}}{{{\bar{z}}}_{1k}})}}}\) with \({{\varvec{\mu }^T_1}}D\Gamma _1 = (b_{11}, \dots , b_{1m})\). Let
$$\begin{aligned} E({{\bar{z}}_{1i}}{{\bar{z}}_{1j}}{{\bar{z}}_{1k}}) = \left\{ \begin{array}{l} {{{\tilde{m}}}_3},\quad \quad \quad i = j = k; \\ 0,\quad \quad \quad \;\; {\text {other}}, \end{array} \right. \end{aligned}$$
where \({{\tilde{m}}_3} = E({\bar{z}}_{1i}^3) = \frac{1}{{n_1^3}}\sum \nolimits _{j = 1}^{{n_1}} {E( z_{1ij}^3)} = \frac{{{\beta _3^1}}}{{n_1^2}}\). Hence, we have
$$\begin{aligned} E({{\bar{\textbf {Z}}}^T_1}\Gamma _1^TD\Gamma _1{{\bar{\textbf {Z}}}_1}{{{\varvec{\mu }^T_1}}}D\Gamma _1{{\bar{\textbf {Z}}}_1 }) =\,&{{{\tilde{m}}}_3}\sum \limits _{i = 1}^m {{\gamma _{1ii}}{b_{1i}}} = \frac{{{\beta _3^1}}}{{n_1 ^2}}\sum \limits _{i = 1}^m {{\gamma _{1ii}}{b_{1i}}} \nonumber \\ =\,&\frac{{{\beta _3^1}}}{{n_1^2}}(\Gamma _1^T D{{\varvec{\mu }}_1})\textrm{diag}(\Gamma _1^TD\Gamma _1 ), \end{aligned}$$
where \(\textrm{diag}(\Gamma _1^TD\Gamma _1 )\) denotes a m-vector with entries, the diagonal elements of matrix \(\Gamma _1^TD\Gamma _1\). Thus,
$$\begin{aligned} E\{({{\bar{\textbf {X}}}_1^T} D{{\bar{\textbf {X}}}_1})^2\} =\,&\frac{\Delta }{{n_1^3}}\sum \limits _{i = 1}^m {\gamma _{1ii}^2} + \frac{1}{{n_1^2}}\left[ \text {tr}{^2}(D\Sigma _1 ) + 2\text {tr}\{(D\Sigma _1)^2\}\right] + \frac{4}{{{n_1}}}{{{\varvec{\mu }}}^T_1}D\Sigma _1 D{{\varvec{\mu }}_1} \\&+ {({{{\varvec{\mu }}}^T_1}D{{\varvec{\mu }}_1})^2}\; + \frac{2}{{{n_1}}}\text {tr}(D\Sigma _1 ){{{\varvec{\mu }}}^T_1}D{{\varvec{\mu }}_1} + \frac{4}{{n_1^2}}(\Gamma _1^T D{{\varvec{\mu }}_1})\textrm{diag}(\Gamma _1^TD\Gamma _1 ){\beta _3^1}; \\ E({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_1}) =\,&{{{\varvec{\mu }}}^T_1}D{{\varvec{\mu }}_1} + \frac{1}{{{n_1}}}\text {tr}(D\Sigma _1 );\\ {E^2}({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_1}) =\,&{({{{\varvec{\mu }}}^T_1}D{{\varvec{\mu }}_1})^2} + \frac{2}{{{n_1}}}{{{\varvec{\mu }}}^T_1}D{{\varvec{\mu }}_1}\text {tr}(D\Sigma _1) + \frac{1}{{n_1^2}}\text {tr}{^2}(D\Sigma _1 ). \end{aligned}$$
From the above result, we obtain that
$$\begin{aligned} Var({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_1}) =\,&\frac{\Delta }{{n_1^3}}\sum \limits _{i = 1}^m {\gamma _{1ii}^2} + \frac{2}{{n_1^2}}\text {tr}\{(D\Sigma _1 )^2\} + \frac{4}{{{n_1}}}{{{\varvec{\mu }}}^T_1}D\Sigma _1 D{{\varvec{\mu }}_1} \nonumber \\&+ \frac{4}{{n_1^2}}(\Gamma _1^T D{{\varvec{\mu }}_1})\textrm{diag}(\Gamma _1^T D\Gamma _1 ){\beta _3^1}. \end{aligned}$$
(4)
Step 2: with the same developments, we obtain
$$\begin{aligned} Var({\bar{\textbf {X}}^T_2}D{\bar{\textbf {X}}_2})\; =\,&\frac{\Delta }{{n_2^3}}\sum \limits _{i = 1}^m {\gamma _{2ii}^2} + \frac{2}{{n_2^2}}\text {tr}\{(D\Sigma _2 )^2\} + \frac{4}{{{n_2}}}{{\varvec{\mu }}^T_2}D\Sigma _2 D{{\varvec{\mu }}_2} \nonumber \\&+ \frac{4}{{n_2^2}}(\Gamma _2^TD{{\varvec{\mu }}_2})\textrm{diag}(\Gamma _2^TD\Gamma _2 ){\beta _3^2}, \end{aligned}$$
(5)
where \(\beta _3^2 = {E( z_{2ij}^3)}\) and \(\Gamma _2^TD\Gamma _2 = [\gamma _{2ij}]_{m \times m}\).
Step 3: We have that
$$\begin{aligned} Var({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2}) =\,&E\left\{ {({\bar{\textbf {X}}_1^T } D{\bar{\textbf {X}}_2})^2}\right\} -{E^2}({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2}) \\ =\,&\frac{1}{{{n_1}{n_2}}}\text {tr}{(D\Sigma _2 D \Sigma _1)} + \frac{1}{{{n_1}}}{{\varvec{\mu }}^T_2}D\Sigma _1 D{{\varvec{\mu }}_2} + \frac{1}{{{n_2}}}{{\varvec{\mu }}^T_1}D\Sigma _2 D{{\varvec{\mu }}_1}. \end{aligned}$$
where
$$\begin{aligned} E\left\{ ({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2})^2\right\} =\,&E\left\{ ({\bar{\textbf {Z}}^T_1}\Gamma _1^TD\Gamma _2 {\bar{\textbf {Z}}_2} + {\bar{\textbf {Z}}^T_1}\Gamma _1^TD{{\varvec{\mu }}_2} + {{\varvec{\mu }}^T_1}D\Gamma _2 {\bar{\textbf {Z}}_2} + {{\varvec{\mu }}^T_1 }D{{\varvec{\mu }}_2})^2\right\} \\ =\,&E\left\{ {({\bar{\textbf {Z}}^T_1}\Gamma _1^TD\Gamma _2 {\bar{\textbf {Z}}_2})^2 } + {({\bar{\textbf {Z}}^T_1}\Gamma _1^TD{{\varvec{\mu }}_2})^2} + {({{\varvec{\mu }}^T_1}D\Gamma _2 {\bar{\textbf {Z}}_2})^2} + {({{\varvec{\mu }}^T_1}D{{\varvec{\mu }}_2})^2}\right\} . \end{aligned}$$
Let \(\Gamma _1^TD\Gamma _2 = [\lambda _{ij}]_{m \times m}\), then
$$\begin{aligned} E{({\bar{\textbf {Z}}^T_1}\Gamma _1^TD\Gamma _2 {\bar{\textbf {Z}}_2})^2} = \sum \limits _{i = 1}^m {\sum \limits _{j = 1}^ m {\sum \limits _{k = 1}^m {\sum \limits _{l = 1}^m {E({\lambda _{ij}}{{{\bar{z}}} _{1i}}{{{\bar{z}}}_{2j}} {\lambda _{kl}} {{{\bar{z}}}_{2k}}{{{\bar{z}}}_{1l}})}}}}, \end{aligned}$$
where
$$\begin{aligned} E({{\bar{z}}_{1i}}{{\bar{z}}_{2j}}{{\bar{z}}_{2k}}{{\bar{z}}_{1l}}) = \left\{ \begin{array}{l} \frac{1}{{{n_1}{n_2}}},\quad \quad \quad i = j = k = l;i = l \ne j = k; \\ 0,\quad \quad \quad \quad \;\; \text {other} \end{array} \right. \end{aligned}$$
Since \(E({\bar{z}}_{1i}^2{\bar{z}}_{2i}^2) = E({\bar{z}}_{1i}^2)E({\bar{z}}_{2i}^2) = \frac{1}{{{n_1}{n_2}}}\) and \(E({\bar{z}}_{1i}^2{\bar{z}}_{2j}^2) = E({\bar{z}}_{1i}^2)E({\bar{z}}_{2j} ^2) = \frac{1}{{{n_1}{n_2}}}\), it follows that
$$\begin{aligned} E\left\{ ({{\bar{\textbf {Z}}}^T_1}\Gamma _1^TD\Gamma _2 {{\bar{\textbf {Z}}}_2})^2\right\} =\,&\frac{1}{{{n_1}{n_2}}}\left( \sum \limits _{i = 1}^m {\lambda _{ii}^2} + \mathop {\sum {\sum {\lambda _{ij}^2}} }\limits _{1 \le i \ne j \le m}\right) \\ =\,&\frac{1}{{{n_1}{n_2}}}\sum \limits _{i = 1}^m {\lambda _{ii}^2} + \frac{1}{{{n_1}{n_2}}}\sum \limits _{i = 1}^m {\sum \limits _{j = 1}^m {\lambda _{ ij}^2}}-\frac{1}{{{n_1}{n_2}}}\sum \limits _{i = 1}^m {\lambda _{ii}^2} \\ =\,&\frac{1}{n_1n_2} \text {tr}(\Gamma _1^T D \Gamma _2 \Gamma _2^T D \Gamma _1)\\ =\,&\frac{1}{{{n_1}{n_2}}}\text {tr}{(D\Sigma _2 D\Sigma _1)}, \end{aligned}$$
It is straightforward obtain \(E\left\{ ({\bar{\textbf {Z}}^T_1}\Gamma _1^TD{{\varvec{\mu }}_2})^2\right\} = \frac{1}{{{n_1}}}{{\varvec{\mu }}^T_2}D\Sigma _1D{{\varvec{\mu }}_2}\) and \(E\left\{ ({{\varvec{\mu }}^T_1}D\Gamma _2 {\bar{\textbf {{Z}}}_2})^2\right\} = \frac{1}{{{n_2}}}{{\varvec{\mu }}^T_1}D\Sigma _2D{{\varvec{\mu }}_1}\) and hence, we have
$$\begin{aligned} E\left\{ ({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2})^2\right\} = \frac{1}{{{n_1}{n_2}}}\text {tr}{(D\Sigma _2D\Sigma _1 )} + \frac{1}{{{n_1}}}{{\varvec{\mu }}^T_2}D\Sigma _1 D{{\varvec{\mu }}_2} + \frac{1}{{{n_2}} }{{\varvec{\mu }}^T_1}D\Sigma _2 D{{\varvec{\mu }}_1} + {({{\varvec{\mu }}^T_1}D{{\varvec{\mu }}_2})^2}, \end{aligned}$$
and
$$\begin{aligned} {E^2}({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2}) = {({{\varvec{\mu }}^T_1}D {{\varvec{\mu }}_2 })^2}. \end{aligned}$$
From the above results, we have
$$\begin{aligned} \text{ Var }({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2}) =\,&\frac{1}{{{n_1}{n_2}}}\text {tr}{(D\Sigma _2D\Sigma _1 )} + \frac{1}{{{n_1}}}{{\varvec{\mu }}^T_2}D\Sigma _1 D{{\varvec{\mu }}_2} + \frac{1}{{{n_2}}}{{\varvec{\mu }}^T_1}D\Sigma _2 D{{\varvec{\mu }}_1}. \end{aligned}$$
(6)
Step 4:
$$\begin{aligned} \text{ Cov }({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_1},{\bar{\textbf {X}}^T_2}D{\bar{\textbf {X}}_2}) = 0. \end{aligned}$$
(7)
Step 5: \(\text{ Cov }({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_1},{\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2}) = E({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_1}{\bar{\textbf {X}}^T_2}D{\bar{\textbf {X}}_1})-E({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_1})E({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2})\). Then, we rewrite
$$\begin{aligned} E({{\bar{\textbf {X}}}^T_1}D{{\bar{\textbf {X}}}_1}{{\bar{\textbf {X}}}^T_2}D{{\bar{\textbf {X}}}_1}) =\,&E({{\bar{\textbf {Z}}}^T _1}\Gamma _1^TD\Gamma _1{{\bar{\textbf {Z}}}_1}{{{\varvec{\mu }}}^T_2}D\Gamma _1{{\bar{\textbf {Z}}}_1} + {{\bar{\textbf {Z}}}^T_1 }\Gamma _1^TD\Gamma _1 {{\bar{\textbf {Z}}}_1}{{{\varvec{\mu }}}^T_2}D{{\varvec{\mu }}_1} \\&+ 2{{\bar{\textbf {Z}}}^T_1}\Gamma _1^TD{{\varvec{\mu }}_1}{{{\varvec{\mu }}}^T_2}D\Gamma _1 {{\bar{\textbf {Z}}}_1} + {{{\varvec{\mu }}}^T_1}D{{\varvec{\mu }}_1}{{{ \varvec{\mu }}}^T_2}D{{\varvec{\mu }}_1}), \end{aligned}$$
where, from Step1
$$\begin{aligned} E({\bar{\textbf {Z}}_1^T} \Gamma _1^T D\Gamma _1 {\bar{\textbf {Z}}_1}{{\varvec{\mu }}^T_2}D\Gamma _1 {\bar{\textbf {Z}}_1}) =\,&\frac{\beta _3^1}{{n_1^2}}(\Gamma _1^TD{{\varvec{\mu }}_2})\textrm{diag}(\Gamma _1^TD\Gamma _1 );\\ E({\bar{\textbf {Z}}^T_1}\Gamma _1^TD\Gamma _1 {\bar{{\textbf {Z}}}}_1 \varvec{\mu }_2^T D \varvec{\mu }_1) =\,&\frac{1}{{{n_1}}} \text {tr}(D\Sigma _1) {{\varvec{\mu }}^T_2}D{{\varvec{\mu }}_1};\\ E({\bar{\textbf {Z}}^T_1}\Gamma _1^TD{{\varvec{\mu }}_1}{{\varvec{\mu }}^T_2}D\Gamma _1 {\bar{\textbf {Z}}_1}) =\,&\frac{1}{{{n_1}}}{{\varvec{\mu }}^T_1}D\Sigma _1 D{{\varvec{\mu }}_2}. \end{aligned}$$
Then,
$$\begin{aligned} E({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_1}{\bar{\textbf {X}}^T_2}D{\bar{\textbf {X}}_1}) =\,&\frac{{{\beta _3^1}}}{{n_1^2}}( \Gamma _1^TD{{\varvec{\mu }}_2})\textrm{diag}(\Gamma _1^TD\Gamma _1) + \frac{1}{{{n_1}}}\text {tr}(D\Sigma _1 ){{\varvec{\mu }}^T_2}D{{\varvec{\mu }}_1} \\&+ \frac{2}{{{n_1}}}{{\varvec{\mu }}^T_1 }D\Sigma _1 D{{\varvec{\mu }}_2} + {{\varvec{\mu }}^T_1}D{{\varvec{\mu }}_1}{{\varvec{\mu } }^T_2}D{{\varvec{\mu }}_1} \end{aligned}$$
and
$$\begin{aligned} E({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_1})E({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2}) = {{\varvec{\mu }}^T_1}D{{ \varvec{\mu }}_1}{{\varvec{\mu }}^T_2}D{{\varvec{\mu }}_1} + \frac{1}{{{n_1}}}\text {tr}(D\Sigma _1 ){{\varvec{\mu }}^T_2}D{{\varvec{\mu }}_1}. \end{aligned}$$
Therefore, we have
$$\begin{aligned} \text{ Cov }({\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_1},{\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2}) = \frac{{\beta _3^1}}{{n_1^2}}(\Gamma _1^TD {{\varvec{\mu }}_2})\textrm{diag}(\Gamma _1^TD\Gamma _1 ) + \frac{2}{{{n_1}}}{{\varvec{\mu }}^T_1 }D\Sigma _1 D{{\varvec{\mu }}_2}. \end{aligned}$$
(8)
Step 6: with similar developments we have
$$\begin{aligned} \text{ Cov }({\bar{\textbf {X}}^T_2}D{\bar{\textbf {X}}_2},{\bar{\textbf {X}}^T_1}D{\bar{\textbf {X}}_2}) = \frac{{\beta _3^2}}{{n_2^2}}(\Gamma _2^TD {{\varvec{\mu }}_1})\textrm{diag}(\Gamma _2^TD\Gamma _2 ) + \frac{2}{{{n_2}}}{{\varvec{\mu }}^T_1 }D\Sigma _2 D{{\varvec{\mu }}_2}. \end{aligned}$$
(9)
Using the results of (A1.1)-(A1.6), we obtain
$$\begin{aligned} \text{ Var }({T_{n}}) =\,&\frac{\Delta }{{n_1^3}}\sum \limits _{i = 1}^m {\gamma _{1ii}^2} + \frac{2 }{{n_1^2}}\text {tr}\{(D\Sigma _1 )^2\} + \frac{4}{{{n_1}}}{{{\varvec{\mu }}}^T_1}D\Sigma _1 D{ {\varvec{\mu }}_1} + \frac{4}{{n_1^2}}(\Gamma _1^TD{{\varvec{\mu }}_1})\textrm{diag}(\Gamma _1^TD\Gamma _1 ){ \beta _3^1} \\&+ \frac{\Delta }{{n_2^3}}\sum \limits _{i = 1}^m {\gamma _{2ii}^ 2} + \frac{2}{{n_2^2}}\text {tr}\{(D\Sigma _2 )^2\} + \frac{4}{{{n_2}}}{{{\varvec{\mu }}}^T _2}D\Sigma _2 D{{\varvec{\mu }}_2} + \frac{4}{{n_2^2}}(\Gamma _2^TD{{\varvec{\mu }}_2})\textrm{diag}(\Gamma _2^TD\Gamma _2 ){\beta _3^2} \\&+ 4\left\{ \frac{1}{{{n_1}{n_2}}}\text {tr}{(D\Sigma _2D\Sigma _1 )} + \frac{1 }{{{n_1}}}{{{\varvec{\mu }}}^T_2}D\Sigma _1 D{{\varvec{\mu }}_2} + \frac{1}{{{n_2}}} {{{\varvec{\mu }}}^T_1}D\Sigma _2 D{{\varvec{\mu }}_1}\right\} \\&-4\left\{ \frac{1}{{n_1^2}}(\Gamma _1^TD{{\varvec{\mu }}_2})\textrm{diag}(\Gamma _1^T D\Gamma _1 ){\beta _3^1} + \frac{2}{{{n_1}}}{{{\varvec{\mu ^T}}}_1}D\Sigma _1 D{{\varvec{\mu }}_2}\right\} \\&-4\left\{ \frac{1}{{n_2^2}}(\Gamma _2^TD{{\varvec{\mu }}_1})\textrm{diag}(\Gamma _2^T D\Gamma _2 ){\beta _3^2} + \frac{2}{{{n_2}}}{{{\varvec{\mu }}}^T_1}D\Sigma _2 D{{\varvec{\mu }}_2}\right\} \\ =\,&\frac{\Delta }{{n_1^3}}\sum \limits _{i = 1}^m {\gamma _{1ii}^2} + \frac{\Delta }{{n_2^3}}\sum \limits _{i = 1}^m {\gamma _{2ii}^2}+ \frac{2 }{{n_1^2}}\text {tr}\{(D\Sigma _1 )^2\} + \frac{2 }{{n_2^2}}\text {tr}\{(D\Sigma _2 )^2\}+ \frac{4}{{{n_1}{n_2}}}\text {tr}{(D\Sigma _2D\Sigma _1 )} \\&+ \frac{4}{n_1} (\varvec{\mu }_1 - \varvec{\mu }_2)^T D \Sigma _1 D (\varvec{\mu }_1 - \varvec{\mu }_2) + \frac{4}{n_2} (\varvec{\mu }_1 - \varvec{\mu }_2)^T D \Sigma _2 D (\varvec{\mu }_1 - \varvec{\mu }_2)\\&+ \frac{4}{{n_1^2}}\Gamma _1^TD({{\varvec{\mu }}_1} - \varvec{\mu }_2)\textrm{diag}(\Gamma _1^TD\Gamma _1 ){ \beta _3^1} + \frac{4}{{n_2^2}}\Gamma _2^TD({{\varvec{\mu }}_2} -\varvec{\mu }_1)\textrm{diag}(\Gamma _2^TD\Gamma _2 ){\beta _3^2}. \end{aligned}$$
Under \(H_0: \varvec{\mu }_1 = \varvec{\mu }_2\) and the condition \((b_2)\), we have
$$\begin{aligned} \text{ Var }(T_n) = \frac{2 }{{n_1^2}}\text {tr}\{(D\Sigma _1 )^2\} + \frac{2 }{{n_2^2}}\text {tr}\{(D\Sigma _2 )^2\}+ \frac{4}{{{n_1}{n_2}}}\text {tr}{(D\Sigma _2D\Sigma _1 )} \{1+o(1)\}. \end{aligned}$$
In particular, when two covariance matrices are identical, it is easy to derive
$$\begin{aligned} \text{ Var }({T_{n}}) =\,&\frac{\Delta }{{n_1^3}}\sum \limits _{i = 1}^m {\gamma _{ii}^2} + \frac{2 }{{n_1^2}}\text {tr}\{(D\Sigma )^2\} + \frac{4}{{{n_1}}}{{{\varvec{\mu }}}^T_1}D\Sigma D{ {\varvec{\mu }}_1} + \frac{4}{{n_1^2}}(\Gamma ^TD{{\varvec{\mu }}_1})\textrm{diag}(\Gamma ^TD\Gamma ){ \beta _3^1} \\&+ \frac{\Delta }{{n_2^3}}\sum \limits _{i = 1}^m {\gamma _{ii}^ 2} + \frac{2}{{n_2^2}}\text {tr}\{(D\Sigma )^2\} + \frac{4}{{{n_2}}}{{{\varvec{\mu }}}^T _2}D\Sigma D{{\varvec{\mu }}_2} + \frac{4}{{n_2^2}}(\Gamma ^TD{{\varvec{\mu }}_2})\textrm{diag}(\Gamma ^TD\Gamma ){\beta _3^2} \\&+ 4\left\{ \frac{1}{{{n_1}{n_2}}}\text {tr}\{(D\Sigma )^2\} + \frac{1 }{{{n_1}}}{{{\varvec{\mu }}}^T_2}D\Sigma D{{\varvec{\mu }}_2} + \frac{1}{{{n_2}}} {{{\varvec{\mu }}}^T_1}D\Sigma D{{\varvec{\mu }}_1}\right\} \\&-4\left\{ \frac{1}{{n_1^2}}(\Gamma ^TD{{\varvec{\mu }}_2})\textrm{diag}(\Gamma ^T D\Gamma ){\beta _3^1} + \frac{2}{{{n_1}}}{{{\varvec{\mu ^T}}}_1}D\Sigma D{{\varvec{\mu }}_2}\right\} \\&-4\left\{ \frac{1}{{n_2^2}}(\Gamma ^TD{{\varvec{\mu }}_1})\textrm{diag}(\Gamma ^T D\Gamma ){\beta _3^2} + \frac{2}{{{n_2}}}{{{\varvec{\mu }}}^T_1}D\Sigma D{{\varvec{\mu }}_2}\right\} . \end{aligned}$$
Under the null hypothesis and the assumption \((b_1)\), the above equation is simplified as
$$\begin{aligned} \text{ Var }({T_{n}}) = 2{\tau ^2}\text {tr}\{(D\Sigma )^2\}\{1 + o(1)\}. \end{aligned}$$
\(\square\)
