Classification of repeated measurements using bias corrected Euclidean distance discriminant function

Ngailo, Edward Kanuti; Nadarajah, Saralees

doi:10.1007/s42952-023-00246-z

Classification of repeated measurements using bias corrected Euclidean distance discriminant function

Research Article
Published: 12 December 2023

Volume 53, pages 281–305, (2024)
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Edward Kanuti Ngailo¹ &
Saralees Nadarajah²

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Abstract

This paper introduces a novel approach for approximating misclassification probabilities in Euclidean distance classifier when the group means exhibit a bilinear structure such as in the growth curve model first proposed by Potthoff and Roy (Biometrika 51:313–326, 1964). Initially, by leveraging certain statistical relationships, we establish two general results for the improved Euclidean discriminant function in both weighted and unweighted growth curve mean structures. We derive these approximations for the expected misclassification probabilities with respect to the distribution of the improved Euclidean discriminant function. Additionally, we compare the misclassification probabilities of the improved Euclidean discriminant function, the standard Euclidean discriminant function, and the linear discriminant function. It is important to note that in cases where the mean structure is weighted, a higher number of repeated measurements yields better classification results with the improved Euclidean discriminant function and the standard Euclidean discriminant function, allowing for more information to be acquired, as opposed to the linear discriminant function, which performs well with a smaller number of repeated measurements. Furthermore, we evaluate the accuracy of the suggested approximations by Monte Carlo simulations.

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Department of Physics, Mathematics and Informatics, University of Dar es Salaam, Dar es Salaam, Tanzania
Edward Kanuti Ngailo
Department of Mathematics, University of Manchester, Manchester, M13 9PL, UK
Saralees Nadarajah

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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).

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Ngailo, E.K., Nadarajah, S. Classification of repeated measurements using bias corrected Euclidean distance discriminant function. J. Korean Stat. Soc. 53, 281–305 (2024). https://doi.org/10.1007/s42952-023-00246-z

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Received: 15 June 2023
Accepted: 26 November 2023
Published: 12 December 2023
Issue Date: March 2024
DOI: https://doi.org/10.1007/s42952-023-00246-z

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Classification of repeated measurements using bias corrected Euclidean distance discriminant function

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Classification of repeated measurements using bias corrected Euclidean distance discriminant function

Abstract

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Using the growth curve model in classification of repeated measurements

Generalized Gini linear and quadratic discriminant analyses

Unequal Priors in Linear Discriminant Analysis

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