Appendix
In this section, we present the proofs for theorems.
A1: Proof of the theorem when the mean parameter of EDF is unweighted
Theorem 5.1
Consider the Euclidean discriminant function in (5). If \({\widehat{{\textbf{b}}}}_i,\) \(i\in \{1, 2\}\) is unweighted, then
$$\begin{aligned} {\mathbb{E}}\left[ \widehat{D}_E|{\varvec{x}}\in \pi _i\right] = (-1)^{i+1}\frac{1}{2}{\boldsymbol{\Delta }}_E^{2}+\frac{1}{2}\Bigg (\frac{1}{n_2}- \frac{1}{n_1}\Bigg )\text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) . \end{aligned}$$
Proof
We present the proof for the moments of \(U_i.\) Recall that
$$\begin{aligned} U_i=\left( {{\textbf{b}}}^{i}-\frac{1}{2}\left( \widehat{{{\textbf{b}}}}_{1}+ \widehat{{\textbf{b}}}_{2}\right) \right) ' {{\textbf{A}}}'{{\textbf{A}}}\left( \widehat{{\textbf{b}}}_{1}- \widehat{{{\textbf{b}}}}_{2}\right) . \end{aligned}$$
The expectation of \(U_i\) can be calculated as
$$\begin{aligned} {\mathbb{E}}\left[ U_i\right]&=\text{tr}\left\{ {\mathbb{E}}\left[ {{\textbf{A}}} \left( \widehat{{\textbf{b}}}_{1}-\widehat{{{\textbf{b}}}}_{2}\right) \left( \left( {{\textbf{A}}}{\textbf{b}}^{i}\right) '- \frac{1}{2} \left( \widehat{{\textbf{b}}}_{1}+\widehat{{{\textbf{b}}}}_{2} \right) '{{\textbf{A}}}'\right) \right] \right\} \\ &= \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{b}}_{1} -{{\textbf{b}}}_{2}\right) {{\textbf{b}}}^{i}{{\textbf{A}}}'- \frac{1}{2}{{\textbf{A}}}\widehat{{{\textbf{b}}}}_{1} \widehat{{\textbf{b}}}'_{1}{{\textbf{A}}}'+ \frac{1}{2}{{\textbf{A}}}\widehat{{\textbf{b}}}_{2} \widehat{{{\textbf{b}}}}'_{2}{{\textbf{A}}}' \right\} \\ &= \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{b}}_{1} -{{\textbf{b}}}_{2}\right) {{\textbf{b}}}^{i}{{\textbf{A}}}'- \frac{1}{2}{{\textbf{A}}}{\mathbb{E}}\left[ \widehat{{\textbf{b}}}_{1} \widehat{{{\textbf{b}}}}'_{1}\right] {{\textbf{A}}}'+ \frac{1}{2}{\textbf{A}}{\mathbb{E}}\left[ \widehat{{{\textbf{b}}}}_{2} \widehat{{\textbf{b}}}'_{2}\right] {{\textbf{A}}}' \right\} \\ &= \text{tr}\Big \{{{\textbf{A}}}\left( {{\textbf{b}}}_{1} -{\textbf{b}}_{2}\right) {{\textbf{b}}}^{i}{{\textbf{A}}}'- \frac{1}{2}{\textbf{A}}\left( \frac{1}{n_1}\left( {{\textbf{A}}}' {{\textbf{A}}}\right) ^{-1} {\textbf{A}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}} \left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}+{\textbf{b}}_{1} {{\textbf{b}}}'_{1}\right) {{\textbf{A}}}'\\ &\quad +\frac{1}{2}{{\textbf{A}}}\left( \frac{1}{n_2} \left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}{{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}}\left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1} +{{\textbf{b}}}_{2}{\textbf{b}}'_{2}\right) {{\textbf{A}}}'\Big \}\\ &= \text{tr}\Big \{{{\textbf{A}}}\left( {{\textbf{b}}}_{1} -{\textbf{b}}_{2}\right) {{\textbf{b}}}^{i}{{\textbf{A}}}'- \frac{1}{2}\frac{1}{n_1}{{\textbf{A}}}\left( {{\textbf{A}}}' {{\textbf{A}}}\right) ^{-1} {\textbf{A}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}} \left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}- \frac{1}{2}{{\textbf{A}}}{{\textbf{b}}}_{1}{{\textbf{b}}}'_{1}{{\textbf{A}}}' \end{aligned}$$
$$\begin{aligned} &\quad +\frac{1}{2}\frac{1}{n_2}{\textbf{A}}\left( {{\textbf{A}}}' {{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{\textbf{A}} \left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}+ \frac{1}{2}{{\textbf{A}}}{\textbf{b}}_{2}{{\textbf{b}}}'_{2}{{\textbf{A}}}' \Big \}\\ &= \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{b}}_{1}-{{\textbf{b}}}_{2}\right) {{\textbf{b}}}^{i}{{\textbf{A}}}'{{\boldsymbol{\Sigma }}}^{-1}- \frac{1}{2}{{\textbf{A}}}{{\textbf{b}}}'_{1}{{\textbf{b}}}'_{1} {{\textbf{A}}}'+ \frac{1}{2}{{\textbf{A}}}{{\textbf{b}}}_{2} {{\textbf{b}}}'_{2}{{\textbf{A}}}'\right\} \\ &\quad -\frac{1}{2}\left( \dfrac{1}{n_1}- \dfrac{1}{n_2}\right) \text{tr}\left\{ {{\textbf{A}}}\left( {{\textbf{A}}}' {\textbf{A}}\right) ^{-1}{{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}} \left( {{\textbf{A}}}'{\textbf{A}}\right) ^{-1}{{\textbf{A}}}'\right\} \\ &= \text{tr}\left\{ {{\textbf{A}}}\left( \left( {\textbf{b}}_{1} -{{\textbf{b}}}_{2}\right) {{\textbf{b}}}^{i}- \frac{1}{2} \left( {\textbf{b}}_{1}-{{\textbf{b}}}_{2}\right) \left( {{\textbf{b}}}_{1}+{\textbf{b}}_{2}\right) '\right) {{\textbf{A}}}'\right\} \\ &\quad -\frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{A}}^{\prime } {{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{\textbf{A}} \left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}{{\textbf{A}}}'\right\} \\ &= \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{b}}_{1}- {{\textbf{b}}}_{2}\right) \left( {{\textbf{b}}}^{i}-\frac{1}{2} \left( {{\textbf{b}}}_{1}+{{\textbf{b}}}_{2}\right) '\right) {\textbf{A}}'\right\} - \frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \end{aligned}$$
$$\begin{aligned} &\quad \cdot \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{A}}^{\prime } {{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{\textbf{A}} \left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}{{\textbf{A}}}'\right\} \\ &= \frac{1}{2}\left( {{\textbf{b}}}_{1}-{\textbf{b}}_{2}\right) ' {{\textbf{A}}}'{{\textbf{A}}}\left( {{\textbf{b}}}_{1}- {\textbf{b}}_{2}\right) (-1)^{i-1} - \frac{1}{2}\left( \dfrac{1}{n_1}- \dfrac{1}{n_2}\right) \\ &\quad \cdot \text{tr}\left\{ {{\textbf{A}}} \left( {\textbf{A}}^{\prime }{{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{\textbf{A}}\left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}{{\textbf{A}}}'\right\} \\ &= \frac{1}{2}\Delta ^2(-1)^{i-1}-\frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \text{tr}\left( {{\textbf{P}}}_{{{\textbf{A}}}}{{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{\textbf{A}}}\right) \\ &= \frac{1}{2}\Delta ^2(-1)^{i-1}-\frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}{{\textbf{P}}}_{{\textbf{A}}}\right) \\ &= \frac{1}{2}\Delta ^2(-1)^{i-1}-\frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) , \quad {\textbf{P}}_{{{\textbf{A}}}}{{\textbf{P}}}_{{{\textbf{A}}}} ={{\textbf{P}}}_{{{\textbf{A}}}}, \quad i=1, 2. \end{aligned}$$
\(\square\)
A2: Proof of the theorem when the mean parameter of EDF is weighted
Theorem 5.2
Consider the Euclidean discriminant function in (5). If \({\widehat{{\textbf{b}}}}_i,\) \(i\in \{1, 2\}\) is weighted, then
$$\begin{aligned} {\mathbb{E}}\left[ \widehat{D}_E|{\varvec{x}}\in \pi _i\right] = \frac{1}{2}(-1)^{(i-1)}{\boldsymbol{\Delta }}^2_E+\frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \frac{n-r({\textbf{C}})-1}{n-r({\textbf{C}})-p+r({\textbf{A}})-1}\text{tr}\left( {\textbf{A}}\left( {\textbf{A}}'{{\boldsymbol{\Sigma }}}^{-1}{\textbf{A}}\right) ^{-1}{\textbf{A}}'\right) . \end{aligned}$$
Proof
For moments of \(U_i,\) \({\mathbb{E}}\left[ U_i\right]\) can be calculated as
$$\begin{aligned} {\mathbb{E}}\left[ U_i\right]&= \text{tr}\left\{ {\mathbb{E}}\left[ \left( \widehat{{\textbf{b}}}_{1}- \widehat{{{\textbf{b}}}}_{2}\right) ' {{\textbf{A}}}'{\varvec{x}}_i- \frac{1}{2}\left( \widehat{{{\textbf{b}}}}_{1}-\widehat{{\textbf{b}}}_{2} \right) ' {{\textbf{A}}}'{{\textbf{A}}}\left( \widehat{{\textbf{b}}}_{1}+ \widehat{{{\textbf{b}}}}_{2}\right) \right] \right\} {\mathbb{E}}_{{\textbf{S}}}\left[ {\mathbb{E}}\left[ U_i|{\widehat{{\textbf{b}}}}_1, \widehat{{\textbf{b}}}_2, {\textbf{S}}\right] \right] \\ &= {\mathbb{E}}_{{\textbf{S}}} \text{tr}\left\{ {\mathbb{E}}\left[ \left( {{\textbf{A}}}\widehat{{{\textbf{b}}}}_{1}-{{\textbf{A}}} \widehat{{{\textbf{b}}}}_{2}\right) \left( \left( {{\textbf{A}}} {{\textbf{b}}}^{i}-\frac{1}{2}\left( {{\textbf{A}}} \widehat{{{\textbf{b}}}}_{1}+ {\textbf{A}}\widehat{{{\textbf{b}}}}_{2}\right) ' \right) | \widehat{{{\textbf{b}}}}_{1}, \widehat{{{\textbf{b}}}}_{2}, {{\textbf{S}}}\right) \right] \right\} \\ &= \left( {{\textbf{b}}}_{1}-{{\textbf{b}}}_{2}\right) ' {\textbf{A}}'{{\textbf{A}}}{{\textbf{b}}}^{i} - \frac{1}{2}\frac{1}{n_1} \text{tr} \left\{ {\mathbb{E}}\left[ {{\textbf{A}}}\left( {{\textbf{A}}}' {\textbf{S}}^{-1}{{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}' {{\textbf{S}}}^{-1} {{\boldsymbol{\Sigma }}}{\textbf{S}}^{-1}{{\textbf{A}}} \left( {{\textbf{A}}}^{\prime }{{\textbf{S}}}^{-1}{\textbf{A}}\right) ^{-1}{{\textbf{A}}}' \right] \right\} \end{aligned}$$
$$\begin{aligned} &\quad +\frac{1}{2}\frac{1}{n_2}\text{tr}\left\{ {\mathbb{E}}\left[ {{\textbf{A}}}\left( {\textbf{A}}'{{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1}{{\textbf{A}}}' {{\textbf{S}}}^{-1} {{\boldsymbol{\Sigma }}}{{\textbf{S}}}^{-1} {{\textbf{A}}}\left( {{\textbf{A}}}^{\prime }{\textbf{S}}^{-1}{{\textbf{A}}} \right) ^{-1}{{\textbf{A}}}'\right] \right\} \\ &\quad -\frac{1}{2}\left( \widehat{{\textbf{b}}}_{1}-\widehat{{{\textbf{b}}}}_{2}\right) ' {{\textbf{A}}}'{\textbf{A}}\left( \widehat{{{\textbf{b}}}}_{1}+\widehat{{{\textbf{b}}}}_{2}\right) \\ &= \left( {{\textbf{b}}}_{1}-{{\textbf{b}}}_{2}\right) '{\textbf{A}}'{{\textbf{A}}} \left( {\textbf{b}}_i-\frac{1}{2}\left( {\textbf{b}}_1+{\textbf{b}}_2\right) \right) - \frac{1}{2}\left( \frac{1}{n_1}-\frac{1}{n_2}\right) \\ &\quad \cdot \dfrac{(p-q-1)(p-1)}{(n-q-1)(p-n+q-1)} \text{tr}\left( \left( {\textbf{A}}'{{\boldsymbol{\Sigma }}}^{-1}{\textbf{A}}\right) ^{-1}\right) \\ &= \frac{1}{2}(-1)^{i+1}\Delta ^2 - \frac{1}{2}\left( \frac{1}{n_1}-\frac{1}{n_2}\right) \text{tr}{\mathbb{E}}\left[ {{\textbf{A}}}\left( {{\textbf{A}}}'{\textbf{S}}^{-1}{{\textbf{A}}} \right) ^{-1} {{\textbf{A}}}' {{\textbf{S}}}^{-1} {{\boldsymbol{\Sigma }}}{{\textbf{S}}}^{-1}{{\textbf{A}}} \left( {{\textbf{A}}}^{\prime }{\textbf{S}}^{-1}{{\textbf{A}}} \right) ^{-1}{{\textbf{A}}}' \right] \\ &= (-1)^{i+1}\frac{1}{2}\Delta ^2-\frac{1}{2}\left( \frac{1}{n_1}-\frac{1}{n_2}\right) \frac{n-r({\textbf{C}})-1}{n-r({\textbf{C}})-p+r({\textbf{A}})-1}\text{tr}\left( {\textbf{A}}\left( {\textbf{A}}'{{\boldsymbol{\Sigma }}}^{-1}{\textbf{A}}'\right) ^{-1}{\textbf{A}}'\right) . \end{aligned}$$
\(\square\)
A3: Proof of Theorem 3.3, bias corrected Euclidean discriminant function (unweighted case)
We first present the proof for the moments of \(U_i.\) Recall that
$$\begin{aligned} U_i=\left( {{\textbf{b}}}^{i}-\frac{1}{2}\left( \widehat{{{\textbf{b}}}}_{1}+\widehat{{\textbf{b}}}_{2}\right) \right) ' {{\textbf{A}}}'{{\textbf{A}}} \left( \widehat{{\textbf{b}}}_{1}-\widehat{{{\textbf{b}}}}_{2}\right) + \left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) . \end{aligned}$$
Then \({\mathbb{E}}\left[ U_i\right]\) can be calculated as
$$\begin{aligned} {\mathbb{E}}\left[ U_i\right]&= \text{tr}\left\{ {\mathbb{E}}\left[ {{\textbf{A}}}\left( \widehat{{\textbf{b}}}_{1}-\widehat{{{\textbf{b}}}}_{2}\right) \left( \left( {{\textbf{A}}}{\textbf{b}}^{i}\right) ' - \frac{1}{2}\left( \widehat{{\textbf{b}}}_{1}+\widehat{{{\textbf{b}}}}_{2}\right) '{{\textbf{A}}}'\right) \right] \right\} + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) \\ &= \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{b}}_{1}-{{\textbf{b}}}_{2}\right) {{\textbf{b}}}^{i}{{\textbf{A}}}' - \frac{1}{2}{{\textbf{A}}}\widehat{{{\textbf{b}}}}_{1}\widehat{{\textbf{b}}}'_{1}{{\textbf{A}}}' + \frac{1}{2}{{\textbf{A}}}\widehat{{\textbf{b}}}_{2}\widehat{{{\textbf{b}}}}'_{2}{{\textbf{A}}}' \right\} + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) \\ &= \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{b}}_{1}-{{\textbf{b}}}_{2}\right) {{\textbf{b}}}^{i}{{\textbf{A}}}' - \frac{1}{2}{{\textbf{A}}}{\mathbb{E}} \left( \widehat{{\textbf{b}}}_{1}\widehat{{{\textbf{b}}}}'_{1}\right) {{\textbf{A}}}'+ \frac{1}{2}{\textbf{A}}{\mathbb{E}} \left( \widehat{{{\textbf{b}}}}_{2}\widehat{{{\textbf{b}}}}'_{2} \right) {{\textbf{A}}}' \right\} + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) \\ &= \text{tr}\Big \{{{\textbf{A}}}\left( {{\textbf{b}}}_{1}-{\textbf{b}}_{2}\right) {{\textbf{b}}}^{i}{{\textbf{A}}}'- \frac{1}{2}{\textbf{A}}\left( \frac{1}{n_1}\left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1} {\textbf{A}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}} \left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}+{\textbf{b}}_{1}{{\textbf{b}}}'_{1}\right) {{\textbf{A}}}' \\ &\quad +\frac{1}{2}{{\textbf{A}}}\left( \frac{1}{n_2}\left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1} {\textbf{A}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}}\left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}+ {\textbf{b}}_{2}{{\textbf{b}}}'_{2}\right) {{\textbf{A}}}'\Big \} + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{\textbf{P}}_{{{\textbf{A}}}}\right) \\ &= \text{tr}\Big \{{{\textbf{A}}}\left( {{\textbf{b}}}_{1}-{\textbf{b}}_{2}\right) {{\textbf{b}}}^{i}{{\textbf{A}}}'- \frac{1}{2}\frac{1}{n_1}{{\textbf{A}}}\left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1} {\textbf{A}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}}\left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1} - \frac{1}{2}{{\textbf{A}}}{{\textbf{b}}}_{1}{{\textbf{b}}}'_{1}{{\textbf{A}}}' \\ &\quad +\frac{1}{2}\frac{1}{n_2}{{\textbf{A}}}\left( {{\textbf{A}}}'{\textbf{A}}\right) ^{-1} {{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}}\left( {{\textbf{A}}}'{\textbf{A}}\right) ^{-1} + \frac{1}{2}{{\textbf{A}}}{{\textbf{b}}}_{2}{\textbf{b}}'_{2}{{\textbf{A}}}' \Big \} + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) \\ &= \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{b}}_{1}-{{\textbf{b}}}_{2}\right) {{\textbf{b}}}^{i}{{\textbf{A}}}'{{\boldsymbol{\Sigma }}}^{-1} - \frac{1}{2}{{\textbf{A}}}{{\textbf{b}}}'_{1}{{\textbf{b}}}'_{1}{{\textbf{A}}}' + \frac{1}{2}{{\textbf{A}}}{{\textbf{b}}}_{2}{{\textbf{b}}}'_{2}{{\textbf{A}}}'\right\} \\ &\quad -\frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \text{tr}\left\{ {{\textbf{A}}}\left( {{\textbf{A}}}'{\textbf{A}}\right) ^{-1} {{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}}\left( {{\textbf{A}}}'{\textbf{A}}\right) ^{-1}{{\textbf{A}}}'\right\} + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{\textbf{P}}_{{{\textbf{A}}}}\right) \\ &= \text{tr}\left\{ {{\textbf{A}}}\left( \left( {\textbf{b}}_{1}-{{\textbf{b}}}_{2}\right) {{\textbf{b}}}^{i} - \frac{1}{2}\left( {\textbf{b}}_{1}-{{\textbf{b}}}_{2}\right) \left( {{\textbf{b}}}_{1}+{\textbf{b}}_{2}\right) '\right) {{\textbf{A}}}'\right\} \\ &\quad -\frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{A}}^{\prime }{{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}} \left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}{{\textbf{A}}}'\right\} + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) \\ &= \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{b}}_{1}-{{\textbf{b}}}_{2}\right) \left( {{\textbf{b}}}^{i}-\frac{1}{2}\left( {{\textbf{b}}}_{1}+{{\textbf{b}}}_{2}\right) '\right) {\textbf{A}}'\right\} - \frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \\ &\quad \cdot \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{A}}^{\prime }{{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{\textbf{A}}\left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1}{{\textbf{A}}}'\right\} + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) \\ &= \frac{1}{2}\left( {{\textbf{b}}}_{1}-{\textbf{b}}_{2}\right) ' {{\textbf{A}}}'{{\textbf{A}}} \left( {{\textbf{b}}}_{1}-{\textbf{b}}_{2}\right) (-1)^{i-1}- \frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \\ &\quad \cdot \text{tr}\left\{ {{\textbf{A}}}\left( {\textbf{A}}^{\prime }{{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{\textbf{A}}\left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}'\right\} + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) \\ &= \frac{1}{2}\Delta ^2(-1)^{i-1}-\frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \text{tr}\left( {{\textbf{P}}}_{{{\textbf{A}}}}{{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{\textbf{A}}}\right) + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) \\ &= \frac{1}{2}\Delta ^2(-1)^{i-1}-\frac{1}{2}\left( \dfrac{1}{n_1}-\dfrac{1}{n_2}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}{{\textbf{P}}}_{{\textbf{A}}}\right) + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) \\ &= \frac{1}{2}(-1)^{i+1}\Delta ^2-\frac{1}{2}\left( \frac{1}{n_1}-\frac{1}{n_2}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) - \frac{1}{2}\left( \frac{1}{n_1}-\frac{1}{n_2}\right) \text{tr}\left( {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right) , \end{aligned}$$
(18)
where, in (18), \({{\textbf{P}}}_{{{\textbf{A}}}}{{\textbf{P}}}_{{\textbf{A}}}={{\textbf{P}}}_{{{\textbf{A}}}},\) \(i=1, 2.\)
Next we provide a derivation for \({\mathbb{E}}\left[ V_0\right]\):
$$\begin{aligned} {\mathbb{E}}\left[ V_0\right]&= {\mathbb{E}}\left[ \left( \widehat{{{\textbf{b}}}}_1-\widehat{{\textbf{b}}}_2\right) ' {{\textbf{A}}}'{{\textbf{A}}} \left( \widehat{{\textbf{b}}}_1-\widehat{{{\textbf{b}}}}_2\right) \right] \\ &= {\mathbb{E}}\left[ \text{tr}\left\{ {{\textbf{A}}}'{\textbf{A}} \left( \widehat{{{\textbf{b}}}}_1-\widehat{{{\textbf{b}}}}_2\right) \left( \widehat{{{\textbf{b}}}}_1-\widehat{{{\textbf{b}}}}_2\right) ' \right\} \right] \\ &= \text{tr}\left\{ {{\textbf{A}}}'{{\textbf{A}}}{\mathbb{E}} \left[ \left( \widehat{{{\textbf{b}}}}_1-\widehat{{{\textbf{b}}}}_2\right) \left( \widehat{{{\textbf{b}}}}_1-\widehat{{{\textbf{b}}}}_2\right) ' \right] \right\} \\ &= \text{tr}\left\{ {{\textbf{A}}}'{{\textbf{A}}}\left( \left( \frac{1}{n_1}+\frac{1}{n_2}\right) \left( {{\textbf{A}}}'{\textbf{A}}\right) ^{-1}{{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}} \left( {{\textbf{A}}}'{\textbf{A}}\right) ^{-1}+ \left( {{\textbf{b}}}_1-{{\textbf{b}}}_2\right) \left( {\textbf{b}}_1-{{\textbf{b}}}_2\right) '\right) \right\} \\ &= \text{tr}\left\{ {{\textbf{A}}}'{{\textbf{A}}}\left( \frac{1}{n_1}+\frac{1}{n_2}\right) \left( {{\textbf{A}}}'{\textbf{A}}\right) ^{-1}{{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}} \left( {{\textbf{A}}}'{\textbf{A}}\right) ^{-1}\right\} + \left( {{\textbf{b}}}_1-{{\textbf{b}}}_2\right) '{\textbf{A}}'{{\textbf{A}}} \left( {{\textbf{b}}}_1-{{\textbf{b}}}_2\right) \\ &= \left( \frac{1}{n_1}+\frac{1}{n_2}\right) \text{tr}\left\{ {{\textbf{A}}}\left( {{\textbf{A}}}'{{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}'{{\boldsymbol{\Sigma }}}{{\textbf{A}}} \left( {{\textbf{A}}}'{\textbf{A}}\right) ^{-1}{{\textbf{A}}}'\right\} + \left( {{\textbf{b}}}_1-{\textbf{b}}_2\right) '{{\textbf{A}}}'{{\textbf{A}}} \left( {{\textbf{b}}}_1-{{\textbf{b}}}_2\right) \\ &= \left( \frac{1}{n_1}+\frac{1}{n_2}\right) \text{tr}\left\{ {{\textbf{P}}}_{{{\textbf{A}}}}{{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{\textbf{A}}}\right\} + \left( {{\textbf{b}}}_1-{{\textbf{b}}}_2\right) '{{\textbf{A}}}'{{\textbf{A}}} \left( {{\textbf{b}}}_1-{{\textbf{b}}}_2\right) \\ &= \left( \frac{1}{n_1}+\frac{1}{n_2}\right) \text{tr}\left\{ {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}{{\textbf{P}}}_{{\textbf{A}}}\right\} + \left( {{\textbf{b}}}_1-{{\textbf{b}}}_2\right) '{{\textbf{A}}}'{{\textbf{A}}} \left( {{\textbf{b}}}_1-{{\textbf{b}}}_2\right) \\ &= \left( \frac{1}{n_1}+\frac{1}{n_2}\right) \text{tr}\left\{ {{\boldsymbol{\Sigma }}}{{\textbf{P}}}_{{{\textbf{A}}}}\right\} + \left( {\textbf{b}}_1-{{\textbf{b}}}_2\right) '{{\textbf{A}}}'{{\textbf{A}}} \left( {{\textbf{b}}}_1-{\textbf{b}}_2\right) , \quad {{\textbf{P}}}_{{{\textbf{A}}}}{\textbf{P}}_{{{\textbf{A}}}}={{\textbf{P}}}_{{{\textbf{A}}}}. \end{aligned}$$
A4: Proof of Theorem 3.3
We have
$$\begin{aligned} \left| e(i|j)-\Phi \left( \gamma ^{(i)}_0\right) \right|&= \left| {\mathbb{E}}\left[ \Phi \left( U_i/\sqrt{V_0}\right) \right] - \Phi \left( \gamma ^{(i)}_0\right) \right| , \quad i\in \{1, 2\}, \ i\ne j\\ &= \left| {\mathbb{E}}\left[ \Phi \left( U_i/\sqrt{V_0}\right) -\Phi \left( \gamma ^{(i)}_0\right) \right] \right| \\ &\le {\mathbb{E}}\left[ \left| \Phi \left( U_i/\sqrt{V_0}\right) -\Phi \left( \gamma ^{(i)}_0\right) \right| \right] . \end{aligned}$$
Using Lebesgue’s dominated convergence theorem and Lemma 3.2, we see that the sequence \(\left| \Phi \left( U_i/\sqrt{V_0}\right) - \Phi \left( \gamma ^{(i)}_0\right) \right|\) is uniformly integrable. Consequently, under Assumptions A1–A2 and Lemma 3.3, we have
$$\begin{aligned} {\mathbb{E}}\left[ \left| \Phi \left( U_i/\sqrt{V_0}\right) -\Phi \left( \gamma ^{(i)}_0 \right) \right| \right] \rightarrow 0. \end{aligned}$$
Therefore, we can assert that \(\Phi \left( \gamma ^{(i)}_0\right)\) provides an asymptotic approximation to e(i|j).
A5: Proof of Theorem 3.5, bias corrected Euclidean discriminant function when the mean parameter \({\widehat{\textbf{B}}}\) is weighted
For moments of \({\widehat{D}}_{EBC_2} \left( {\varvec{x}}\right) ,\) \({\mathbb{E}} \left[ \widehat{D}_{EBC_2}\left( {\varvec{x}}\right) \right]\) can be calculated as
$$\begin{aligned}&{\mathbb{E}}_{{\textbf{S}}}\left[ {\mathbb{E}}\left[ \widehat{D}_{EBC_2}|{\widehat{{\textbf{b}}}}_1, {\widehat{{\textbf{b}}}}_2, {\textbf{S}}\right] \right] \\ &\quad = {\mathbb{E}}_{{\textbf{S}}} \text{tr}\left\{ {\mathbb{E}} \left[ \left( {{\textbf{A}}}\widehat{{{\textbf{b}}}}_{1}-{\textbf{A}}\widehat{{{\textbf{b}}}}_{2}\right) \left( \left( {{\textbf{A}}}{\textbf{b}}^{i}-\frac{1}{2} \left( {{\textbf{A}}}\widehat{{{\textbf{b}}}}_{1}+{\textbf{A}}\widehat{{{\textbf{b}}}}_{2}\right) ' \right) | \widehat{{{\textbf{b}}}}_{1}, \widehat{{{\textbf{b}}}}_{2}, {{\textbf{S}}}\right) \right] \right\} \\ &\qquad +\frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \frac{n-r({\textbf{C}})-1}{n-r({\textbf{C}})-p+r({\textbf{A}})-1} \text{tr}\left( {\textbf{A}}\left( {\textbf{A}}'{{\boldsymbol{\Sigma }}}^{-1}{\textbf{A}}\right) ^{-1}{\textbf{A}}'\right) \\ &\quad = \left( {{\textbf{b}}}_{1}-{{\textbf{b}}}_{2}\right) '{\textbf{A}}'{{\textbf{A}}}{{\textbf{b}}}^{i}- \frac{1}{2}\frac{1}{n_1}\text{tr} {\mathbb{E}}\left[ {{\textbf{A}}}\left( {{\textbf{A}}}'{{\textbf{S}}}^{-1}{\textbf{A}}\right) ^{-1} {{\textbf{A}}}' {{\textbf{S}}}^{-1} {{\boldsymbol{\Sigma }}}{{\textbf{S}}}^{-1}{\textbf{A}} \left( {{\textbf{A}}}^{\prime }{{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1}{{\textbf{A}}}' \right] \\ &\qquad +\frac{1}{2}\frac{1}{n_2}\text{tr} \left\{ {\mathbb{E}}\left[ {{\textbf{A}}}\left( {{\textbf{A}}}'{{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1} {{\textbf{A}}}' {{\textbf{S}}}^{-1} {{\boldsymbol{\Sigma }}}{{\textbf{S}}}^{-1}{\textbf{A}} \left( {{\textbf{A}}}^{\prime }{{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1}{{\textbf{A}}}'\right] \right\} \\ &\qquad -\frac{1}{2}\left( {{\textbf{b}}}_{1}-{\textbf{b}}_{2}\right) ' {{\textbf{A}}}'{{\textbf{A}}}\left( {{\textbf{b}}}_{1}+{\textbf{b}}_{2}\right) + \frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \frac{n-r({\textbf{C}})-1}{n-r({\textbf{C}})-p+r({\textbf{A}})-1} \\ &\qquad \cdot \text{tr}\left( {\textbf{A}}\left( {\textbf{A}}'{{\boldsymbol{\Sigma }}}^{-1}{\textbf{A}}\right) ^{-1}{\textbf{A}}'\right) \\ &\quad = \left( {{\textbf{b}}}_{1}-{{\textbf{b}}}_{2}\right) '{\textbf{A}}'{{\textbf{A}}} \left( {\textbf{b}}_i-\frac{1}{2}\left( {\textbf{b}}_1+{\textbf{b}}_2\right) \right) - \frac{1}{2}\left( \frac{1}{n_1}-\frac{1}{n_2}\right) \\ &\qquad \cdot \frac{n-r({\textbf{C}})-1}{n-r({\textbf{C}})-p+r({\textbf{A}})-1} \text{tr}{\mathbb{E}}\left[ {{\textbf{A}}}\left( {{\textbf{A}}}'{\textbf{S}}^{-1}{{\textbf{A}}}\right) ^{-1} {{\textbf{A}}}' {{\textbf{S}}}^{-1} {{\boldsymbol{\Sigma }}}{\textbf{S}}^{-1}{{\textbf{A}}} \left( {{\textbf{A}}}^{\prime }{{\textbf{S}}}^{-1}{\textbf{A}}\right) ^{-1}{{\textbf{A}}}'\right] \\ &\qquad +\frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \frac{n-r({\textbf{C}})-1}{n-r({\textbf{C}})-p+r({\textbf{A}})-1} \text{tr}\left( {\textbf{A}}\left( {\textbf{A}}'{{\boldsymbol{\Sigma }}}^{-1}{\textbf{A}}\right) ^{-1}{\textbf{A}}'\right) \\ &\quad = \frac{1}{2}(-1)^{i+1}\Delta ^2-\frac{1}{2}\left( \frac{1}{n_1}-\frac{1}{n_2}\right) \text{tr}{\mathbb{E}}\left[ {{\textbf{A}}}\left( {{\textbf{A}}}'{\textbf{S}}^{-1}{{\textbf{A}}} \right) ^{-1} {{\textbf{A}}}' {{\textbf{S}}}^{-1} {{\boldsymbol{\Sigma }}}{{\textbf{S}}}^{-1}{{\textbf{A}}} \left( {{\textbf{A}}}^{\prime }{\textbf{S}}^{-1}{{\textbf{A}}} \right) ^{-1}{{\textbf{A}}}' \right] \\ &\qquad +\frac{1}{2}\left( \frac{1}{n_2}-\frac{1}{n_1}\right) \frac{n-r({\textbf{C}})-1}{n-r({\textbf{C}})-p+r({\textbf{A}})-1} \text{tr}\left( {\textbf{A}}\left( {\textbf{A}}'{{\boldsymbol{\Sigma }}}^{-1}{\textbf{A}}\right) ^{-1}{\textbf{A}}'\right) \\ &\quad = (-1)^{i+1}\frac{1}{2}\Delta ^2-\frac{1}{2}\left( \frac{1}{n_1}-\frac{1}{n_2}\right) \frac{n-r({\textbf{C}})-1}{n-r({\textbf{C}})-p+r({\textbf{A}})-1} \text{tr}\left( {\textbf{A}}\left( {\textbf{A}}'{{\boldsymbol{\Sigma }}}^{-1}{\textbf{A}}'\right) ^{-1}{\textbf{A}}'\right) . \end{aligned}$$
(19)
The result from von Rosen (2018, page 447) has been used to simplify (19).
Next, we calculate the variance of either group \(\pi _1\) or \(\pi _2\) conditioned on \(\widehat{{{\textbf{b}}}}_{1}\) and \(\widehat{{\textbf{b}}}_{2}.\) The conditional expectation of \({\mathbb{E}}[V_1]\) given \({{\textbf{S}}}\) is
$$\begin{aligned} {\mathbb{E}}\left[ V_1\right] ={\mathbb{E}}_{{\textbf{S}}}\left[ {\mathbb{E}}\left[ V_1|{\textbf{S}}\right] \right]&= {\mathbb{E}}_{{\textbf{S}}}\left[ {\mathbb{E}}\left[ \left( \widehat{{\textbf{b}}}_1-\widehat{{{\textbf{b}}}}_2\right) ^{\prime } {{\textbf{A}}}^{\prime } {{\boldsymbol{\Sigma }}}{{\textbf{A}}}\left( \widehat{{{\textbf{b}}}}_1-\widehat{{\textbf{b}}}_2\right) |{{\textbf{S}}}\right] \right] \\ &= {\mathbb{E}}_{{\textbf{S}}}\left[ \text{tr}\left( {{\textbf{A}}}^{\prime } {{\boldsymbol{\Sigma }}}{{\textbf{A}}} {\mathbb{E}}\left[ \left( \widehat{{\textbf{b}}}_1-\widehat{{{\textbf{b}}}}_2\right) \left( \widehat{{\textbf{b}}}_1-\widehat{{{\textbf{b}}}}_2\right) ^{\prime }|{{\textbf{S}}} \right] \right) \right] . \end{aligned}$$
(20)
From (20),
$$\begin{aligned}&{\mathbb{E}}_{{\textbf{S}}}\left[ {\mathbb{E}}\left[ V_1|{\textbf{S}}\right] \right] \\ &\quad = {\mathbb{E}}_{{\textbf{S}}} \Big [\text{tr}\Big \{{\textbf{A}}^{\prime } {{\boldsymbol{\Sigma }}}{{\textbf{A}}} \Big ( \frac{1}{n_1}\left( {\textbf{A}}^{\prime } {{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1} {{\textbf{A}}}^{\prime } {{\textbf{S}}}^{-1} {{\boldsymbol{\Sigma }}}{{\textbf{S}}}^{-1}{{\textbf{A}}} \left( {\textbf{A}}^{\prime }{{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1}\\ &\qquad +\frac{1}{n_2}\left( {{\textbf{A}}}^{\prime } {{\textbf{S}}}^{-1} {{\textbf{A}}} \right) ^{-1} {{\textbf{A}}}^{\prime } {\textbf{S}}^{-1} {{\boldsymbol{\Sigma }}}{{\textbf{S}}}^{-1}{{\textbf{A}}} \left( {\textbf{A}}^{\prime }{{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1}\Big )+ \left( {\textbf{b}}_1-{{\textbf{b}}}_2\right) \left( {{\textbf{b}}}_1-{\textbf{b}}_2\right) ^{\prime }\Big \}\Big ]\\ &\quad = \frac{n_1+n_2}{n_1 n_2} {\mathbb{E}}_{{\textbf{S}}}\left[ \text{tr}\left\{ {{\textbf{A}}}^{\prime } {{\boldsymbol{\Sigma }}}{{\textbf{A}}}\left( {{\textbf{A}}}^{\prime }{{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1} {{\textbf{A}}}^{\prime } {{\textbf{S}}}^{-1} {{\boldsymbol{\Sigma }}}{{\textbf{S}}}^{-1} {{\textbf{A}}}\left( {{\textbf{A}}}^{\prime }{{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1}\right\} \right] \\ &\qquad +\left( {{\textbf{b}}}_1-{\textbf{b}}_2\right) ^{\prime } {{\textbf{A}}}^{\prime }{{\textbf{A}}} \left( {\textbf{b}}_1-{{\textbf{b}}}_2\right) \\ &\quad = \frac{n_1+n_2}{n_1 n_2} \text{tr}\left\{ {{\boldsymbol{\Sigma }}}{\mathbb{E}}_{{\textbf{S}}} \left[ {{\textbf{A}}} \left( {\textbf{A}}^{\prime }{{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1} {{\textbf{A}}}^{\prime } {{\textbf{S}}}^{-1} {{\boldsymbol{\Sigma }}}{{\textbf{S}}}^{-1}{{\textbf{A}}} \left( {\textbf{A}}^{\prime }{{\textbf{S}}}^{-1}{{\textbf{A}}} \right) ^{-1}{\textbf{A}}^{\prime }\right] \right\} \\ &\qquad +\left( {{\textbf{b}}}_1-{\textbf{b}}_2\right) ^{\prime } {{\textbf{A}}}^{\prime }{{\textbf{A}}} \left( {\textbf{b}}_1-{{\textbf{b}}}_2\right) \\ &\quad = \frac{n_1+n_2}{n_1 n_2} \frac{n-r({\textbf{C}})-1}{n-r({\textbf{C}})-p+r({\textbf{A}})-1} \text{tr}\left\{ {{\boldsymbol{\Sigma }}}{\textbf{A}}\left( {\textbf{A}}'{{\boldsymbol{\Sigma }}}^{-1}{\textbf{A}}'\right) ^{-1}\right\} \\ &\qquad +\left( {{\textbf{b}}}_1-{\textbf{b}}_2\right) ^{\prime } {{\textbf{A}}}^{\prime }{{\textbf{A}}} \left( {\textbf{b}}_1-{{\textbf{b}}}_2\right) . \end{aligned}$$
A6: Proof of Theorem 3.5
We have
$$\begin{aligned} \left| e(i|j)-\Phi \left( \gamma ^{(i)}\right) \right|&= \left| {\mathbb{E}}\left[ \Phi \left( U_i/\sqrt{V_1}\right) \right] - \Phi \left( \gamma ^{(i)}\right) \right| = \left| {\mathbb{E}}\left[ \Phi \left( U_i/\sqrt{V_1}\right) -\Phi \left( \gamma ^{(i)}\right) \right] \right| \\ &\le {\mathbb{E}}\left[ \left| \Phi \left( U_i/\sqrt{V_1}\right) -\Phi \left( \gamma ^{(i)}\right) \right| \right] . \end{aligned}$$
By Lebesgue’s dominated convergence theorem, Theorem 3.5 yields that the sequence \(\left\{ \left| \Phi \left( U_i/\sqrt{V_1}\right) -\Phi \left( \gamma ^{(i)} \right) \right| \right\}\) is uniformly integrable. Hence, under the Assumptions A1–A2 and Lemma 3.4
$$\begin{aligned} {\mathbb{E}}\left[ \left| \Phi \left( U_i/\sqrt{V_1}\right) -\Phi \left( \gamma ^{(i)} \right) \right| \right] \longrightarrow 0. \end{aligned}$$
Thus, we can see that \(\Phi \left( \gamma _0\right)\) provides an asymptotic approximation to e(i|j).