Appendix 1
The sample estimate of the standard deviation, \(s_{X}\), can be written using the generalized exp-log notation (Eq. 5) as
$$\begin{aligned} s_{X}={\mathbf{A}}_{5}\cdot \mathrm {exp}\left( {\mathbf{A}}_{4}\cdot \mathrm {log}\left( {\mathbf{A}}_{3}\cdot \mathrm {exp}\left( {\mathbf{A}}_{2}\cdot \mathrm {log}\left( {\mathbf{A}}_{1}\cdot {\hat{\mathbf{m}}} \right) \right) \right) \right) . \end{aligned}$$
(43)
Let \({\mathbf{y}}^{{(2)}}\) be the vector containing the squares of the elements in \({\mathbf{y}}\). Then, the \(4 \times k\) matrix \({\mathbf{A}}_{1}\) equals
$$\begin{aligned} {\mathbf{A}}_{1}=\left[ {\begin{array}{llll} {\mathbf{r}} &{} {\mathbf{r}}^{{(}2{)}} &{} {\mathbf {1}}_{k} &{} {\mathbf {1}}_{k}\\ \end{array} } \right] ', \end{aligned}$$
(44)
the \(4 \times 4\) matrix \({\mathbf{A}}_{2}\) equals
$$\begin{aligned} {\mathbf{A}}_{2}=\left[ {\begin{array}{cccc} 2 &{} 0 &{} -1 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 1 &{} -1\\ \end{array} } \right] , \end{aligned}$$
(45)
the \(2 \times 4\) matrix \({\mathbf{A}}_{3}\) equals
$$\begin{aligned} {\mathbf{A}}_{3}=\left[ {\begin{array}{cccc} -1 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 1 &{} -1\\ \end{array} } \right] , \end{aligned}$$
(46)
the \(1 \times 2\) vector \({\mathbf{A}}_{4}\) equals
$$\begin{aligned} {\mathbf{A}}_{4}=\left[ {\begin{array}{cc} \frac{1}{2} &{} -\frac{1}{2}\\ \end{array} } \right] , \end{aligned}$$
(47)
and \({\mathbf{A}}_{5}\) equals the scalar 1. It follows that \({\mathbf{g}}_{0}={\hat{\mathbf{m}}}\), the \(4 \times 1\) vector \({\mathbf{g}}_{1}\) equals
$$\begin{aligned} {\mathbf{g}}_{1}=\mathrm {log}\left( {\mathbf{A}}_{1}\cdot {\mathbf{g}}_{0} \right) =\log \left( \left[ {\begin{array}{llll} {\mathbf{r}} &{} {\mathbf{r}}^{\left( 2 \right) } &{} {\mathbf {1}}_{k} &{} {\mathbf {1}}_{k}\\ \end{array} } \right] ^{\prime } \cdot {\hat{\mathbf{m}}} \right) =\left[ {\begin{array}{c} \mathrm {log}\left( \sum X_{i} \right) \\ \mathrm {log}\left( \sum X_{i}^{2} \right) \\ \mathrm {log}\left( N \right) \\ \mathrm {log}\left( N \right) \\ \end{array} } \right] , \end{aligned}$$
(48)
the \(4 \times 1\) vector \({\mathbf{g}}_{2}\) equals
$$\begin{aligned} {\mathbf{g}}_{2}=\mathrm {exp}\left( {\mathbf{A}}_{2}\cdot {\mathbf{g}}_{1} \right) = \mathrm {exp}\left( \left[ {\begin{array}{cccc} 2 &{} 0 &{} -1 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 1 &{} -1\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} \mathrm {log}\left( \sum X_{i} \right) \\ \mathrm {log}\left( \sum X_{i}^{2} \right) \\ \mathrm {log}\left( N \right) \\ \mathrm {log}\left( N \right) \\ \end{array} } \right] \right) =\left[ {\begin{array}{c} \frac{\left( \sum X_{i} \right) ^{2}}{N}\\ \sum X_{i}^{2} \\ N\\ \mathrm {1}\\ \end{array} } \right] , \end{aligned}$$
(49)
the \(2 \times 1\) vector \({\mathbf{g}}_{3}\) equals
$$\begin{aligned} {\mathbf{g}}_{3}= & {} \log \left( {\mathbf{A}}_{3}\cdot {\mathbf{g}}_{2} \right) =\log \left( \left[ {\begin{array}{cccc} -1 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 1 &{} -1\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} \frac{\left( \sum X_{i} \right) ^{2}}{N}\\ \sum X_{i}^{2} \\ N\\ \mathrm {1}\\ \end{array} } \right] \right) = \left[ {\begin{array}{c} \mathrm {log}\left( SS \right) \\ \mathrm {log} (N-1)\\ \end{array} } \right] , \end{aligned}$$
(50)
where \(SS = \sum X_{i}^{2} -\frac{\left( \sum X_{i} \right) ^{2}}{N}\), and
$$\begin{aligned} {\mathbf{g}}_{4}=\exp \left( {\mathbf{A}}_{4}\cdot {\mathbf{g}}_{3} \right) =\exp \left( \left[ {\begin{array}{ll} \frac{1}{2} &{} -\frac{1}{2}\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} \mathrm {log}\left( SS \right) \\ \log \left( N-1 \right) \\ \end{array} } \right] \right) =\sqrt{\frac{SS}{N-1}}, \end{aligned}$$
(51)
because \({\mathbf{A}}_{5}=1\), \({\mathbf {g}}({\hat{\mathbf {m}}})={\mathbf{g}}_{5}={\mathbf{g}}_{4}=s_{X}\).
After some tedious algebra, it may be verified that \({\mathbf{G}}_{0}={\mathbf{I}}_{k}\),
$$\begin{aligned} {\mathbf{G}}_{1}= & {} {\mathbf{D}}^{-1}\left[ {\mathbf{A}}_{1}\cdot {\mathbf{g}}_{\mathrm {0}} \right] \cdot {\mathbf{A}}_{\mathrm {1}} \cdot {\mathbf{G}}_{0} \nonumber \\= & {} {\mathbf{D}}^{-1}\left[ {\begin{array}{c} {\sum }_{i=1}^N X_{i} \\ {\sum }_{i=1}^N X_{i}^{2} \\ N\\ N\\ \end{array} } \right] \cdot \left[ {\begin{array}{llll} {\mathbf{r}} &{} {\mathbf{r}}^{\left( 2 \right) } &{} {\mathbf {1}}_{k} &{} {\mathbf {1}}_{k}\\ \end{array} } \right] ^{\prime } \cdot {\mathbf{I=}}\left[ {\begin{array}{c} \frac{{\mathbf{r}}^{\prime }}{\sum X_{i} }\\ \frac{{\mathbf{r}}^{\left( 2 \right) \prime }}{\sum X_{i}^{2} }\\ \frac{{\mathbf {1}}_{k}^{{\prime }}}{N}\\ \frac{{\mathbf {1}}_{k}^{{\prime }}}{N}\\ \end{array} } \right] , \end{aligned}$$
(52)
$$\begin{aligned} {\mathbf{G}}_{2}= & {} {\mathbf{D}}\left[ \exp \left( {\mathbf{A}}_{2}\cdot {\mathbf{g}}_{\mathrm {1}} \right) \right] \cdot {\mathbf{A}}_{\mathrm {2}} \cdot {\mathbf{G}}_{1} \nonumber \\= & {} {\mathbf{D}}\left[ {\begin{array}{c} \frac{\left( \sum X_{i} \right) ^{2}}{N}\\ \sum X_{i}^{2} \\ N\\ \mathrm {1}\\ \end{array} } \right] \cdot \left[ {\begin{array}{cccc} 2 &{} 0 &{} -1 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 1 &{} -1\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} \frac{{\mathbf{r}}^{\prime }}{\sum X_{i} }\\ \frac{{\mathbf{r}}^{\left( 2 \right) \prime }}{\sum X_{i}^{2} }\\ \frac{{\mathbf {1}}_{k}^{{\prime }}}{N}\\ \frac{{\mathbf {1}}_{k}^{{\prime }}}{N}\\ \end{array} } \right] =\left[ {\begin{array}{c} 2{\mathbf{r}}^{{\prime }}{\bar{X}}-{\bar{X}}^{2}\\ {\mathbf{r}}^{\left( 2 \right) \prime }\\ {\mathbf {1}}_{k}^{{\prime }}\\ {\mathbf {0}}_{k}^{{\prime }}\\ \end{array} } \right] . \end{aligned}$$
(53)
Then, \({\mathbf{G}}_{3}\) is a \(2 \times k\) matrix,
$$\begin{aligned} {\mathbf{G}}_{3}= & {} {\mathbf{D}}^{-1}\left[ {\mathbf{A}}_{3}\cdot {\mathbf{g}}_{\mathrm {2}} \right] \cdot {\mathbf{A}}_{\mathrm {3}} \cdot {\mathbf{G}}_{2} \nonumber \\= & {} {\mathbf{D}}^{-1}\left[ {\begin{array}{c} SS\\ N-1\\ \end{array} } \right] \cdot \left[ {\begin{array}{cccc} -1 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 1 &{} -1\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} 2{\mathbf{r}}^{{\prime }}{\bar{X}}-{\bar{X}}^{2}\\ {\mathbf{r}}^{\left( 2 \right) {\prime }}\\ {\mathbf {1}}_{k}^{{\prime }}\\ {\mathbf {0}}_{k}^{{\prime }}\\ \end{array} } \right] =\left[ {\begin{array}{c} \frac{{\mathbf{d}}^{\left( 2 \right) {\prime }}}{SS}\\ {\mathbf{e}}^{\prime }\\ \end{array} } \right] , \end{aligned}$$
(54)
where the \(k \times 1\) vector \({\mathbf{d}}={\mathbf{r}}-{\bar{X}}\) and \({\mathbf{e}}={\mathbf {1}}_{k}/(N-1)\),
and \({\mathbf{G}}_{4}\) is the \(1 \times k\) vector
$$\begin{aligned} {\mathbf{G}}_{4}= & {} {\mathbf{D}}\left[ \exp \left( {\mathbf{A}}_{4}\cdot {\mathbf{g}}_{\mathrm {3}} \right) \right] \cdot {\mathbf{A}}_{\mathrm {4}} \cdot {\mathbf{G}}_{3} \nonumber \\= & {} {\mathbf{D}}\left[ s_{X} \right] \cdot \left[ {\begin{array}{cc} \frac{1}{2} &{} -\frac{1}{2}\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} \frac{{\mathbf{d}}^{\left( 2 \right) }}{SS}\\ {\mathbf{e}}\\ \end{array} } \right] =0.5s_{X}\left( \frac{{\mathbf{d}}^{2}}{SS}-{\mathbf{e}} \right) . \end{aligned}$$
(55)
Finally, one can derive that \({\mathbf{G}=}{\mathbf{G}}_{\mathrm {5}} = {\mathbf{G}}_{\mathrm {4}}\).
Because the sample estimate of the standard deviation is obtained by dividing the sum of squared deviation scores by \(N-1, {\mathbf {g}}({\hat{\mathbf {m}}})\) (Eq. 51) is not homogeneous of order 0. As a result, the asymptotic variance is approximated by \(V_{S_{X}}\approx {\mathbf{GD}}({\hat{\mathbf{m}}}){\mathbf{G}}^{\prime } {\mathbf{-G}}{\hat{\mathbf{m}}}N^{-1}{\hat{\mathbf{m}}}^{\prime } {\mathbf{G}}^{\prime }\) (cf. Eq. 2). Inserting the Jacobian \({\mathbf{G}}\) (Eq. 55) in Eq. 2 results in the sample estimate of the asymptotic variance of \(s_{X}\),
$$\begin{aligned} V_{s_{X}}\approx .25s_{X}^{2}\cdot {\sum }_{\varvec{i}} {\sum }_{\varvec{j}} {\left( \frac{d_{i}^{2}}{SS}-e \right) \left( \frac{d_{j}^{2}}{SS}-e \right) \left( {\delta }_{ij}{\hat{m}}_{i}-\frac{{\hat{m}}_{i}{\hat{m}}_{j}}{N} \right) }. \end{aligned}$$
(56)
For large N,
$$\begin{aligned} V_{s_{X}}\approx .25s_{X}^{2}\cdot {\sum }_{\varvec{i}} {\hat{m}}_{i} \left( \frac{d_{i}^{2}}{SS}-\frac{1}{N} \right) ^{2}. \end{aligned}$$
(57)
Appendix 2
The sample estimates of the percentile ranks, collected in vector \({\mathbf{PR}} ={\begin{array}{lll} {(PR}_{r_{1}} &{} \cdots &{} {PR}_{r_{k}})\\ \end{array}}^{\prime }\) (Eq. 26), can be written as
$$\begin{aligned} {\mathbf{PR}}={\mathbf{A}}_{3}\cdot \mathrm {exp}\left( {\mathbf{A}}_{2}\cdot \mathrm {log} \left( {{\mathbf{A}}}_{1}\cdot {\hat{\mathbf{m}}}\right) \right) . \end{aligned}$$
(58)
Let \({\mathbf{L}}_{p}\) be a lower triangular matrix of ones. The \((k+1) \times k\) matrix \({\mathbf{A}}_{1}\) equals
$$\begin{aligned} {\mathbf{A}}_{1}=\left[ {\begin{array}{c} {\mathbf{L}}_{k}\\ {\mathbf {1}}_{k}^{\prime } \\ \end{array} } \right] , \end{aligned}$$
(59)
and the \(k \times (k+1)\) matrix \({\mathbf{A}}_{2}\) equals
$$\begin{aligned} {\mathbf{A}}_{2}=\left[ {\begin{array}{ll} {\mathbf{I}}_{k} &{} {-{\mathbf{1}}}_{k}\\ \end{array} } \right] . \end{aligned}$$
(60)
Let \({\mathbf{A}}_{3}\) be a lower bidiagonal \(k \times k\) matrix in which each nonzero element equals 50; that is,
$$\begin{aligned} {\mathbf{A}}_{3}=50\left( {\mathbf{I}}_{k} +\left[ {\begin{array}{cc} {\mathbf {0}}_{k-1}^{\prime } &{} 0\\ {\mathbf{I}}_{k-1} &{} {\mathbf {0}}_{k-1}\\ \end{array} } \right] \right) . \end{aligned}$$
(61)
It follows that \({\mathbf{g}}_{0}={\hat{\mathbf{m}}}\). Let \({\hat{\mathbf{m}}}^{*}= (\sum _{i=1}^1 {\hat{\mathrm{m}}}_{i} , \sum _{i=1}^2 {\hat{\mathrm{m}}}_{i} , \ldots , \sum _{i=1}^k {\hat{\mathrm{m}}}_{i} )\) be a vector of cumulative frequencies; that is, \({\hat{\mathbf{m}}}^{*}={\mathbf{L}}_{k}\cdot {\hat{\mathbf{m}}}\), and let \({\mathbf{P}}^{*}={\hat{\mathbf{m}}}^{*}/N=\left( P\left( X\le r_{1} \right) ,\ldots , P\left( X\le r_{k} \right) \right) ^{\prime }\). Function \({\mathbf{g}}_{1}\) is a \((k+1) \times 1\) vector,
$$\begin{aligned} {\mathbf{g}}_{1}=\log \left( {\mathbf{A}}_{1}\cdot {\hat{\mathbf{m}}} \right) =\mathrm {log}\left( \left[ {\begin{array}{c} {\mathbf{L}}_{k}\\ {\mathbf {1}}_{k}^{\prime } \\ \end{array} } \right] \cdot {\hat{\mathbf{m}}} \right) = \mathrm {log}\left[ {\begin{array}{c} {\hat{\mathbf{m}}}^{*}\\ N\\ \end{array} } \right] , \end{aligned}$$
(62)
\({\mathbf{g}}_{2}\) is a \(k \times 1\) vector,
$$\begin{aligned} {\mathbf{g}}_{2}=\mathrm {exp}\left( {\mathbf{A}}_{2}\cdot {\mathbf{g}}_{1} \right) =\mathrm {exp}\left( \left[ {\begin{array}{ll} {\mathbf{I}}_{k} &{} {-{\mathbf{1}}}_{k}\\ \end{array} } \right] \cdot \mathrm {log}\left[ {\begin{array}{c} {\hat{\mathbf{m}}}^{*}\\ N\\ \end{array} } \right] \right) ={\mathbf{P}}^{*}, \end{aligned}$$
(63)
and \({\mathbf{g}}_{3}\) is a \(k \times 1\) vector,
$$\begin{aligned} {\mathbf {g}}({\hat{\mathbf {m}}})={\mathbf{g}}_{3}={\mathbf{A}}_{3}\cdot {\mathbf{g}}_{2}=50\left( {\mathbf{I}}_{{\varvec{k}}} + \left[ {\begin{array}{cc} {\mathbf {0}}_{\varvec{k-1}}^{{\prime }} &{} {\mathbf {0}}\\ {\mathbf{I}}_{\varvec{k-1}} &{} {\mathbf {0}}_{\varvec{k-1}}\\ \end{array} } \right] \right) \cdot {\mathbf{P}}^{*} = {\mathbf{PR}}. \end{aligned}$$
(64)
It follows that \({\mathbf{G}}_{0}={\mathbf{I}}\), the \((k+1) \times k\) matrix \({\mathbf{G}}_{1}\) equals
$$\begin{aligned} {\mathbf{G}}_{1}={\mathbf{D}}^{-1}\left[ {\mathbf{A}}_{1}{{\mathbf{\cdot g}}}_{0} \right] \cdot {\mathbf{A}}_{1}\cdot {\mathbf{G}}_{0}{{\mathbf{=D}}}^{-1}\left( \left[ {\begin{array}{c} {\hat{\mathbf{m}}}^{*}\\ N\\ \end{array} } \right] \right) \cdot \left[ {\begin{array}{c} {\mathbf{L}}_{k}\\ {\mathbf {1}}_{k}^{\prime }\\ \end{array} } \right] , \end{aligned}$$
(65)
the \(k \times k\) matrix \({\mathbf{G}}_{2}\) equals
$$\begin{aligned} {\mathbf{G}}_{2}={\mathbf{D}}\left[ \exp ({\mathbf{A}}_{2}\cdot {\mathbf{g}}_{\mathrm {1}}) \right] \cdot {\mathbf{A}}_{\mathrm {2}} \cdot {\mathbf{G}}_{1}={\mathbf{D}}\left( {\mathbf{P}}^{*} \right) \cdot \left[ {\begin{array}{ll} {\mathbf{I}}_{k} &{} {-{\mathbf{1}}}_{k}\\ \end{array} } \right] \cdot {\mathbf{D}}^{-1}\left( \left[ {\begin{array}{c} {\hat{\mathbf{m}}}^{*}\\ N\\ \end{array} } \right] \right) \cdot \left[ {\begin{array}{c} {\mathbf{L}}_{k}\\ {\mathbf {1}}_{k}^{\prime }\\ \end{array} } \right] , \end{aligned}$$
(66)
and the \(k \times k\) matrix \({\mathbf{G}}_{3}\) equals
$$\begin{aligned} {\mathbf{G}}= & {} {\mathbf{G}}_{3}={\mathbf{A}}_{3}\cdot {\mathbf{G}}_{2} \nonumber \\= & {} 50\left( {\mathbf{I}}_{k} +\left[ {\begin{array}{cc} {\mathbf {0}}_{k-1}^{\prime } &{} 0\\ {\mathbf{I}}_{k-1} &{} {\mathbf {0}}_{k-1}\\ \end{array} } \right] \right) \cdot {\mathbf{D}}\left( \left[ {\mathbf{P}}^{*} \right] \right) \cdot \left[ {\begin{array}{ll} {\mathbf{I}}_{k} &{} {-{\mathbf{1}}}_{k}\\ \end{array} } \right] \cdot {\mathbf{D}}^{-1}\left( \left[ {\begin{array}{c} {\hat{\mathbf{m}}}^{*}\\ N\\ \end{array} } \right] \right) \cdot \left[ {\begin{array}{c} {\mathbf{L}}_{k}\\ {\mathbf {1}}_{k}^{\prime }\\ \end{array} } \right] . \end{aligned}$$
(67)
Some tedious algebra shows that the elements of \({\mathbf{G}}\) equal
$$\begin{aligned} G_{gi}=\frac{50}{N}\cdot \left\{ {\begin{array}{ll} {2-P}_{g-1}^{*}{-P}_{g}^{*} &{} \hbox {if } g>i\\ 1{-P}_{g-1}^{*}{-P}_{g}^{*} &{} \hbox {if } g=i\\ 0{-P}_{g-1}^{*}{-P}_{g}^{*} &{} \hbox {if } g<i\\ \end{array} } \right. , \end{aligned}$$
(68)
where \(P_{x}^{*}=P\left( X\le x \right) \). Because the percentile ranks are homogeneous of order 0, the asymptotic variance is approximated by \(V_{{\mathbf{PR}}}\approx {\mathbf{GD}}({\hat{\mathbf{m}}}){\mathbf{G}}^{\prime }\) (cf. Eq. 3). The sample estimates of the elements of the asymptotic covariance matrix of \({\mathbf{PR}}\) are
$$\begin{aligned} V_{{{\mathbf{PR}}}_{gh}}\approx \frac{2500}{N^{2}} {\sum }_i {{\hat{\mathrm{m}}}_{i}\big ({{\gamma }_{gi}-P}_{g-1}^{*}{-P}_{g}^{*} \big )\left( {\gamma _{hi}-P}_{h-1}^{*}{-P}_{h}^{*} \right) } \end{aligned}$$
(69)
for \(g,h=1, \ldots , k\).
Appendix 3
The sample estimates of the eight boundaries of the stanines (Eq. 28) can be written using the generalized exp-log notation (Eq. 5) as follows:
$$\begin{aligned} {{\mathbf{St}}}_{{\varvec{b}}}={\mathbf{A}}_{5}\cdot \mathrm {exp}\left( {\mathbf{A}}_{4}\cdot \mathrm {log}\left( {\mathbf{A}}_{3}\cdot \mathrm {exp}\left( {\mathbf{A}}_{2}\cdot \mathrm {log}\left( {\mathbf{A}}_{1}\cdot {\hat{\mathbf{m}}} \right) \right) \right) \right) . \end{aligned}$$
(70)
Let \({\mathbf{A}}_{1}\) be the \(4 \times k\) matrix,
$$\begin{aligned} {\mathbf{A}}_{1}=\left[ {\begin{array}{llll} {\mathbf{r}} &{} {\mathbf{r}}^{\left( 2 \right) } &{} {\mathbf {1}}_{k} &{} {\mathbf {1}}_{k}\\ \end{array} } \right] ^{\prime }, \end{aligned}$$
(71)
let \({{\mathbf{A}}}_{2}\) be the \(5 \times 4\) matrix
$$\begin{aligned} {{\mathbf{A}}}_{2}=\left[ {\begin{array}{cccc} 2 &{} 0 &{} 0 &{} -1 \\ 0 &{} 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 1 \\ 1 &{} 0 &{} 0 &{} -1 \\ 0 &{} 0 &{} 1 &{} -1 \\ \end{array} } \right] , \end{aligned}$$
(72)
let \({{\mathbf{A}}}_{3}\) be the \(3 \times 5\) matrix
$$\begin{aligned} {\mathbf{A}}_{3}=\left[ {\begin{array}{ccccc} -1 &{} 1 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 &{} 0 &{} -1 \\ 0 &{} 0 &{} 0 &{} 1 &{} 0 \\ \end{array} } \right] , \end{aligned}$$
(73)
let \({{\mathbf{A}}}_{4}\) be the \(2 \times 3\) matrix,
$$\begin{aligned} {{\mathbf{A}}}_{4}=\left[ {\begin{array}{ccc} {\frac{1}{2}} &{} {-\frac{1}{2}} &{} 0 \\ 0 &{} 0 &{} 1 \end{array} } \right] , \end{aligned}$$
(74)
and let \({{\mathbf{A}}}_{5}\) be the \(8 \times 2\) matrix
$$\begin{aligned} {{\mathbf{A}}}_{5}=\left[ {\begin{array}{cc} {\mathbf{f}} &{} {\mathbf {1}}_{8}\\ \end{array} } \right] . \end{aligned}$$
(75)
It follows that \({\mathbf{g}}_{0}={\hat{\mathbf{m}}}\), \({\mathbf{g}}_{1}\) is the \(4 \times 1\) vector
$$\begin{aligned} {\mathbf{g}}_{1}=\mathrm {log}\left( {\mathbf{A}}_{1}\cdot {\mathbf{g}}_{0} \right) =\log \left( \left[ {\begin{array}{llll} {\mathbf{r}} &{} {\mathbf{r}}^{\left( 2 \right) } &{} {\mathbf {1}}_{k} &{} {\mathbf {1}}_{k}\\ \end{array} } \right] ^{\prime }\cdot {\hat{\mathbf{m}}} \right) =\log \left( \left[ {\begin{array}{c} \sum X_{i} \\ \sum X_{i}^{2} \\ N\\ N\\ \end{array} } \right] \right) , \end{aligned}$$
(76)
\({\mathbf{g}}_{2}\) is the \(5 \times 1\) vector
$$\begin{aligned} {\mathbf{g}}_{2}=\mathrm {exp}({\mathbf{A}}_{2}\cdot {\mathbf{g}}_{1})=\mathrm {exp}\left( \left[ {\begin{array}{cccc} 2 &{} 0 &{} 0 &{} -1\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 1\\ 1 &{} 0 &{} 0 &{} -1\\ 0 &{} 0 &{} 1 &{} -1\\ \end{array} } \right] \cdot \log \left( \left[ {\begin{array}{c} \sum X_{i} \\ \sum X_{i}^{2} \\ N\\ N\\ \end{array} } \right] \right) \right) =\left[ {\begin{array}{c} \frac{\left( \sum X_{i} \right) ^{2}}{N}\\ \sum X_{i}^{2} \\ N\\ {\bar{X}}\\ \mathrm {1}\\ \end{array} } \right] , \end{aligned}$$
(77)
\({\mathbf{g}}_{3}\) is the \(3 \times 1\) vector
$$\begin{aligned} {\mathbf{g}}_{3}= & {} \mathrm {log}\left( {\mathbf{A}}_{3}\cdot {\mathbf{g}}_{2} \right) =\mathrm {log}\left( \left[ {\begin{array}{ccccc} -1 &{} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 1 &{} 0 &{} -1\\ 0 &{} 0 &{} 0 &{} 1 &{} 0\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} \frac{\left( \sum X_{i} \right) ^{2}}{N}\\ \sum X_{i}^{2} \\ N\\ {\bar{X}}\\ \mathrm {1}\\ \end{array} } \right] \right) , \nonumber \\= & {} \mathrm {log}\left( \left[ {\begin{array}{c} SS\\ N-1\\ {\bar{X}}\\ \end{array} } \right] \right) , \end{aligned}$$
(78)
\({\mathbf{g}}_{4}\) is the \(2 \times 1\) vector
$$\begin{aligned} {\mathbf{g}}_{4}=\mathrm {exp}({{\mathbf{A}}}_{4}\cdot {\mathbf{g}}_{3})= \exp \left( \left[ {\begin{array}{ccc} {\frac{1}{2}} &{} {-\frac{1}{2}} &{} 0 \\ 0 &{} 0 &{} 1 \\ \end{array} } \right] \cdot \mathrm {log}\left( \left[ {\begin{array}{c} SS\\ N-1\\ {\bar{X}}\\ \end{array} } \right] \right) \right) =\left[ {\begin{array}{c} s_{X}\\ {\bar{X}}\\ \end{array} } \right] , \end{aligned}$$
(79)
and \({\mathbf{g}}_{5}\) is the \(8 \times 1\) vector
$$\begin{aligned} {\mathbf{g}}_{5}={{\mathbf{A}}}_{5}\cdot {\mathbf{g}}_{4} = \left[ {\begin{array}{ll} {\mathbf{f}} &{} {\mathbf {1}}_{8}\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} s_{X}\\ {\bar{X}}\\ \end{array} } \right] ={\mathbf {g}}({\hat{\mathbf {m}}})={{\mathbf{St}}}_{{\mathbf{b}}}. \end{aligned}$$
(80)
Next, \({\mathbf{G}}_{0}={\mathbf{I}}_{{\varvec{k}}}\). \({\mathbf{G}}_{1}\) is the \(4 \times k\) matrix,
$$\begin{aligned} {{\mathbf{G}}_{1}{\mathbf{=D}}}^{-1}\left[ {\mathbf{A}}_{1}{{\mathbf{\cdot g}}}_{0} \right] \cdot {\mathbf{A}}_{1}\cdot {\mathbf{G}}_{0}={\mathbf{D}}^{-1}\left[ {\begin{array}{c} \sum X_{i} \\ \sum X_{i}^{2} \\ N\\ N\\ \end{array} } \right] \cdot \left[ {\begin{array}{llll} {\mathbf{r}} &{} {\mathbf{r}}^{\left( 2 \right) } &{} {\mathbf {1}}_{k} &{} {\mathbf {1}}_{k}\\ \end{array} } \right] ^{\prime }=\left[ {\begin{array}{c} \frac{{\mathbf{r}}^{{\prime }}}{\sum X_{i} }\\ \frac{{\mathbf{r}}^{\left( 2 \right) {\prime }}}{\sum X_{i}^{2} }\\ \frac{{\mathbf {1}}_{k}^{{\prime }}}{N}\\ \frac{{\mathbf {1}}_{k}^{{\prime }}}{N}\\ \end{array} } \right] . \end{aligned}$$
(81)
Then \({\mathbf{G}}_{2}\) is the \(5 \times k\) matrix
$$\begin{aligned} {\mathbf{G}}_{2}= & {} {\mathbf{D}}\left[ \exp \left( {\mathbf{A}}_{2}\cdot {\mathbf{g}}_{\mathrm {1}} \right) \right] \cdot {\mathbf{A}}_{\mathrm {2}} \cdot {\mathbf{G}}_{1} \nonumber \\= & {} {\mathbf{D}}\left( \left[ {\begin{array}{c} \frac{\left( \sum X_{i} \right) ^{2}}{N}\\ \sum X_{i}^{2} \\ N\\ {\bar{X}}\\ \mathrm {1}\\ \end{array} } \right] \right) \left[ {\begin{array}{cccc} 2 &{} 0 &{} 0 &{} -1\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 1\\ 1 &{} 0 &{} 0 &{} -1\\ 0 &{} 0 &{} 1 &{} -1\\ \end{array} } \right] \left[ {\begin{array}{c} \frac{{\mathbf{r}}^{{\prime }}}{\sum X_{i} }\\ \frac{{\mathbf{r}}^{\left( 2 \right) {\prime }}}{\sum X_{i}^{2} }\\ \frac{{\mathbf {1}}_{k}^{{\prime }}}{N}\\ \frac{{\mathbf {1}}_{k}^{{\prime }}}{N}\\ \end{array} } \right] {\mathbf{=}}\left[ {\begin{array}{c} 2{\mathbf{r}}^{{\prime }}{\bar{X}}-{\bar{X}}^{2}\\ {\mathbf{r}}^{(2)\prime } \\ {\mathbf {1}}_{k}^{{\prime }}\\ \frac{{\mathbf{d}}^{{\prime }}}{N}\\ {\mathbf {0}}_{k}^{{\prime }}\\ \end{array} } \right] , \end{aligned}$$
(82)
\({\mathbf{G}}_{3}\) is the \(3 \times k\) matrix
$$\begin{aligned} {\mathbf{G}}_{3}= & {} {\mathbf{D}}^{-1}\left[ {\mathbf{A}}_{3}\cdot {\mathbf{g}}_{\mathrm {2}} \right] \cdot {\mathbf{A}}_{\mathrm {3}} \cdot {\mathbf{G}}_{2} \nonumber \\= & {} {\mathbf{D}}^{-1}\left( \left[ {\begin{array}{c} SS\\ N-1\\ {\bar{X}}\\ \end{array} } \right] \right) \cdot \left[ {\begin{array}{ccccc} -1 &{} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 1 &{} 0 &{} -1\\ 0 &{} 0 &{} 0 &{} 1 &{} 0 \\ \end{array} } \right] \cdot \left[ {\begin{array}{c} 2{\mathbf{r}}^{{\prime }}{\bar{X}}-{\bar{X}}^{2}\\ {\mathbf{r}}^{(2) \prime }\\ {\mathbf {1}}_{k}^{{\prime }}\\ \frac{{\mathbf{d}}^{{\prime }}}{N}\\ {\mathbf {0}}_{k}^{{\prime }}\\ \end{array} } \right] =\left[ {\begin{array}{c} {\mathbf{d}}^{*{\prime }}\\ {\mathbf{e}}^{\prime } \\ \frac{{\mathbf{d}}^{{\prime }}}{\sum X_{i} }\\ \end{array} } \right] , \end{aligned}$$
(83)
where \({\mathbf{d}}^{*}=\frac{{\mathbf{d}}^{\mathbf {(2)}}}{SS}\). Then \({\mathbf{G}}_{4}\) is the \(2 \times k\) matrix
$$\begin{aligned} {\mathbf{G}}_{4}= & {} {\mathbf{D}}\left[ \exp \left( {\mathbf{A}}_{4}\cdot {\mathbf{g}}_{\mathrm {3}} \right) \right] \cdot {\mathbf{A}}_{\mathrm {4}} \cdot {\mathbf{G}}_{3} \nonumber \\= & {} \left[ {\begin{array}{cc} s_{X} &{} 0\\ 0 &{} {\bar{X}}\\ \end{array} } \right] \cdot \left[ {\begin{array}{ccc} {\frac{1}{2}} &{} {-\frac{1}{2}} &{} 0 \\ 0 &{} 0 &{} 1 \\ \end{array} } \right] {\cdot }\left[ {\begin{array}{c} {\mathbf{d}}^{*{\prime }}\\ {\mathbf{e}}^{\prime } \\ \frac{{\mathbf{d}}^{{\prime }}}{\sum X_{i} }\\ \end{array} } \right] =\left[ {\begin{array}{c} \frac{s_{X}}{2}\cdot \left( {\mathbf{d}}^{*{\prime }}-{\mathbf{e}}^{\prime } \right) \\ \frac{{\mathbf{d}}^{{\prime }}}{N}\\ \end{array} } \right] , \end{aligned}$$
(84)
and \({\mathbf{G}}_{5}\) is the \(8 \times k\) matrix
$$\begin{aligned} {\mathbf{G}}_{5}={\mathbf{G}}={\mathbf{A}}_{5}\cdot {\mathbf{G}}_{4}=\left[ {\begin{array}{ll} {\mathbf{f}} &{} {\mathbf {1}}_{8}\\ \end{array} } \right] \left[ {\begin{array}{c} \frac{s_{X}}{2}\cdot \left( {\mathbf{d}}^{*{\prime }}-{\mathbf{e}}^{\prime } \right) \\ \frac{{\mathbf{d}}^{{\prime }}}{{\varvec{N}}}\\ \end{array} } \right] =\frac{{{\mathbf{f}}\cdot s}_{X}}{2}\cdot \left( {\mathbf{d}}^{*{\prime }}-{\mathbf{e}}^{\prime } \right) +\frac{{{\mathbf {1}}_{8}{\mathbf{\cdot d}}}^{{\prime }}}{N}. \end{aligned}$$
(85)
The stanine boundaries are not homogeneous of order 0, because \({\mathbf{St}}_{{\varvec{b}}}\) is obtained using \(s_{X}\). As a result, the asymptotic variance is approximated by \({\mathbf{V}}_{{\mathbf{St}}}\approx {\mathbf{G}}{\mathbf{V}}_{{\hat{\mathbf{m}}}}{\mathbf{G}}^{\prime }\) (cf. Eq. 2). Some tedious algebra shows that the sample estimates of the elements of the asymptotic covariance matrix of \({\mathbf{V}}_{{\mathbf{St}}}\) are
$$\begin{aligned} V_{{\varvec{St}}_{gh}}\approx {\sum }_{j=1}^k {\sum }_{i=1}^k {t \cdot \left[ \frac{{f_{g}{\mathbf{\cdot }}s\!}_{X}}{2}\cdot \left( d_{i}^{*}-e \right) +\frac{d_{i}}{N} \right] } \cdot \left[ \frac{{f_{_{h}}{\mathbf{\cdot }}s\!}_{X}}{2}\cdot \left( d_{j}^{*}-e \right) +\frac{d_{j}}{N} \right] , \end{aligned}$$
(86)
where \(t={{\delta }_{ij}{\hat{\mathrm{m}}}}_{j} -\frac{{\hat{\mathrm{m}}}_{i} {\hat{\mathrm{m}}}_{j}}{N}\). For large N, Eq. 86 reduces to
$$\begin{aligned} V_{{\varvec{St}}_{gh}}\approx {\sum }_{i=1}^k {{\hat{\mathrm{m}}}_{i} \cdot \left[ \frac{{f_{g} \cdot s\!}_{X}}{2}\cdot \left( d_{i}^{*}-\frac{1}{N} \right) +\frac{d_{i}}{N} \right] \cdot \left[ \frac{{{f}_{_{h}} \cdot s\!}_{X}}{2}\cdot \left( d_{i}^{*}-\frac{1}{N} \right) +\frac{d_{i}}{N} \right] }. \end{aligned}$$
(87)
Appendix 4
The sample estimate of the k standardized scores corresponding to r, collected in a \(k \times 1\) vector z (Eq. 31), can be written as
$$\begin{aligned} {\mathbf{z}}={\mathbf{A}}_{7}\cdot \mathrm {exp}\left( {\mathbf{A}}_{6}\cdot \mathrm {log}\left( {\mathbf{A}}_{5}\cdot \mathrm {exp}\left( {\mathbf{A}}_{4}\cdot \mathrm {log}\left( {\mathbf{A}}_{3}\cdot \mathrm {exp}\left( {\mathbf{A}}_{2}\cdot \mathrm {log}\left( {\mathbf{A}}_{1}\cdot {\hat{\mathbf{m}}} \right) \right) \right) \right) \right) \right) . \end{aligned}$$
(88)
Let \({\mathbf{A}}_{1}\) be the \((k+2) \times k\) matrix
$$\begin{aligned} {\mathbf{A}}_{1}=\left[ {\begin{array}{lll} {\mathbf{I}}_{k} &{} {\mathbf{1}}_{k} &{} {\mathbf{1}}_{k}\\ \end{array} } \right] ^{\prime }. \end{aligned}$$
(89)
Let \(\oplus \) indicate the direct sum, for example \({\mathbf{X}}\oplus {\mathbf{Y}}=\left[ {\begin{array}{cc} {\mathbf{X}} &{} 0\\ 0 &{} {\mathbf{Y}}\\ \end{array} } \right] \). Then \({\mathbf{A}}_{2}\) is the \((k+1) \times (k+2)\) matrix
$$\begin{aligned} {\mathbf{A}}_{2}={\mathbf{I}}_{k}\oplus \left[ {\begin{array}{ll} 1 &{} -1\\ \end{array} } \right] , \end{aligned}$$
(90)
\({\mathbf{A}}_{3}\) is the \((k+4) \times (k+1)\) matrix
$$\begin{aligned} {\mathbf{A}}_{3}=\left[ {\begin{array}{llll} {\mathbf{r}} &{} {\mathbf{r}}^{\left( 2 \right) } &{} {\mathbf{1}}_{k} &{} {\mathbf{1}}_{k} \\ \end{array} } \right] ^{\prime } \oplus {\mathbf{r}}, \end{aligned}$$
(91)
\({{\mathbf{A}}}_{4}\) is the \((5+k) \times (4+k)\) matrix
$$\begin{aligned} {\mathbf{A}}_{4}=\left[ {\begin{array}{cccc} 1 &{} 0 &{} 0 &{} -1 \\ 2 &{} 0 &{} 0 &{} -1 \\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 1 \\ 0 &{} 0 &{}1 &{} -1\\ \end{array} } \right] \oplus {\mathbf{I}}_{k}, \end{aligned}$$
(92)
\({{\mathbf{A}}}_{5}\) is the \((k+3) \times (k+5)\) matrix
$$\begin{aligned} {{\mathbf{A}}}_{5}=1\oplus \left[ {\begin{array}{ll} -1 &{} 1\\ \end{array} } \right] \oplus \left[ {\begin{array}{ll} 1 &{} -1\\ \end{array} } \right] \oplus {\mathbf{I}}_{k}, \end{aligned}$$
(93)
\({\mathbf{A}}_{6}\) is the \((1+k) \times (3+k)\) matrix
$$\begin{aligned} {\mathbf{A}}_{6}=\left[ {\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 1 &{} -\frac{1}{{{2}}} &{} \frac{1}{{2}} &{} {\mathbf {0}}_{k}^{\prime }\\ {\mathbf {0}}_{k} &{} -\frac{1}{2}\cdot {\mathbf {1}}_{k} &{} \frac{1}{{{2}}}\cdot {\mathbf {1}}_{k}&{} {\mathbf{I}}_{\mathrm {k}}\\ \end{array} } \right] , \end{aligned}$$
(94)
and \({\mathbf{A}}_{7}\) is the \(k \times (k+1)\) matrix
$$\begin{aligned} {\mathbf{A}}_{7}=\left[ {\begin{array}{cc} {-{\mathbf {1}}}_{k} &{} {\mathbf{I}}_{k}\\ \end{array} } \right] . \end{aligned}$$
(95)
It follows that \({\mathbf{g}}_{0}={\hat{\mathbf{m}}}\), the \((k+2) \times 1\) vector \({\mathbf{g}}_{1}\) equals
$$\begin{aligned} {\mathbf{g}}_{1}=\log \left( {\mathbf{A}}_{1}\cdot {\hat{\mathbf{m}}} \right) =\log \left( \left[ {\begin{array}{ccc} {\mathbf{I}}_{k} &{} {\mathbf{1}}_{k} &{} {\mathbf{1}}_{k}\\ \end{array} } \right] ^{\prime } \cdot {\hat{\mathbf{m}}} \right) =\log {\left( \left[ {\begin{array}{ccc} {\hat{\mathbf{m}}}^{\prime } &{} N &{} N\\ \end{array} } \right] ^{\prime } \right) ,} \end{aligned}$$
(96)
the \((k+1) \times 1\) vector \({\mathbf{g}}_{2}\) equals
$$\begin{aligned} {\mathbf{g}}_{2}=\mathrm {exp}\left( {\mathbf{A}}_{2}\cdot {\mathbf{g}}_{1} \right) =\mathrm {exp}\left( {\mathbf{I}}_{k}\oplus \left[ {\begin{array}{cc} 1 &{} -1\\ \end{array} } \right] \cdot \log \left( \left[ {\begin{array}{cccc} {\hat{\mathbf{m}}}^{\prime } &{} N &{} N\\ \end{array} } \right] ^{\prime } \right) \right) =\left[ {\begin{array}{cc} {\hat{\mathbf{m}}}^{\prime } &{} 1\\ \end{array} } \right] ^{\prime }, \end{aligned}$$
(97)
the \((k+4) \times 1\) vector \({\mathbf{g}}_{3}\) equals
$$\begin{aligned} {\mathbf{g}}_{3}=\mathrm {log}\left( {\mathbf{A}}_{3}\cdot {\mathbf{g}}_{2} \right) =\mathrm {log}\left( \left( \left[ {\begin{array}{cccc} {\mathbf{r}} &{} {\mathbf{r}}^{(2)} &{} {\mathbf{1}}_{k} &{} {\mathbf{1}}_{k}\\ \end{array} } \right] '\oplus {\mathbf{r}} \right) \cdot \left[ {\begin{array}{cc} {\hat{\mathbf{m}}}^{\prime } &{} 1\\ \end{array} } \right] ^{\prime } \right) =\mathrm {log}\left[ {\begin{array}{c} \sum X_{i} \\ \sum X_{i}^{2} \\ N\\ N\\ {\mathbf{r}}\\ \end{array} } \right] , \end{aligned}$$
(98)
the \((k+5) \times 1\) vector \({\mathbf{g}}_{4}\) equals
$$\begin{aligned} {\mathbf{g}}_{4}= & {} \exp \left( {{\mathbf{A}}}_{4}\cdot {\mathbf{g}}_{3} \right) \nonumber \\= & {} \mathrm {exp}\left( \left( \left[ {\begin{array}{cccc} 1 &{} 0 &{} 0 &{} -1\\ 2 &{} 0 &{} 0 &{} -1\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 1\\ 0 &{} 0 &{} 1 &{} -1\\ \end{array} } \right] \oplus {\mathbf{I}}_{k} \right) \cdot \mathrm {log}\left[ {\begin{array}{c} \sum X_{i} \\ \sum X_{i}^{2} \\ N\\ N\\ {\mathbf{r}}\\ \end{array} } \right] \right) =\left[ {\begin{array}{c} {\bar{X}}\\ \frac{\left( \sum X_{i} \right) ^{2}}{N}\\ \sum X_{i}^{2} \\ N\\ 1\\ {\mathbf{r}}\\ \end{array} } \right] , \end{aligned}$$
(99)
the \((k+3) \times 1\) vector \({\mathbf{g}}_{5}\) equals
$$\begin{aligned} {\mathbf{g}}_{5}= & {} \log \left( {\mathbf{A}}_{5}\cdot {\mathbf{g}}_{4} \right) \nonumber \\= & {} \mathrm {log}\left( \left[ 1\oplus \left[ {\begin{array}{cc} -1 &{} 1\\ \end{array} } \right] \oplus \left[ {\begin{array}{cc} 1 &{} -1\\ \end{array} } \right] \oplus {\mathbf{I}}_{k} \right] \cdot \left[ {\begin{array}{c} {\bar{X}}\\ \frac{\left( \sum X_{i} \right) ^{2}}{N}\\ \sum X_{i}^{2} \\ N\\ 1\\ {\mathbf{r}}\\ \end{array} } \right] \right) =\mathrm {log}\left[ {\begin{array}{c} {\bar{X}}\\ SS\\ N-1\\ {\mathbf{r}}\\ \end{array} } \right] , \end{aligned}$$
(100)
the \((k+1) \times 1\) vector \({\mathbf{g}}_{6}\) equals
$$\begin{aligned} {\mathbf{g}}_{6}= & {} \exp \left( {{\mathbf{A}}}_{6}\cdot {\mathbf{g}}_{5} \right) \nonumber \\= & {} \mathrm {exp}\left( \left[ {\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 1 &{} -\frac{1}{{{2}}} &{} \frac{1}{{{2}}} &{} {\mathbf {0}}_{k}^{\prime }\\ {\mathbf {0}}_{k} &{} -\frac{1}{2}\cdot {\mathbf {1}}_{k} &{} \frac{1}{{{2}}}\cdot {\mathbf {1}}_{k}&{} {\mathbf{I}}_{\mathrm {k}}\\ \end{array} } \right] \cdot \mathrm {log}\left[ {\begin{array}{c} {\bar{X}}\\ SS\\ N-1\\ {\mathbf{r}}\\ \end{array} } \right] \right) =\left[ {\begin{array}{c} \frac{{\bar{X}}}{s_{X}}\\ \frac{{\mathbf{r}}}{s_{X}}\\ \end{array} } \right] , \end{aligned}$$
(101)
and the \(k \times 1\) vector \({\mathbf{g}}_{7}\) equals
$$\begin{aligned} {\mathbf{g}}_{{\mathbf {7}}}={\mathbf {g}}({\hat{\mathbf {m}}})={{\mathbf{A}}}_{7}\cdot {\mathbf{g}}_{6}=\left[ {\begin{array}{cc} {-{\mathbf {1}}}_{k} &{} {\mathbf{I}}_{k}\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} \frac{{\bar{X}}}{s_{X}}\\ \frac{{\mathbf{r}}}{s_{X}}\\ \end{array} } \right] =\left[ \frac{{\mathbf{d}}}{s_{X}} \right] ={\mathbf{z}}. \end{aligned}$$
(102)
Next, \({\mathbf{G}}_{0}={\mathbf{I}}_{k}\), the \((k+2) \times k\) matrix \({\mathbf{G}}_{1}\) equals
$$\begin{aligned} {\mathbf{G}}_{1}= & {} {\mathbf{D}}^{-1}\left[ {\mathbf{A}}_{1} \cdot {\mathbf{g}}_{0} \right] \cdot {\mathbf{A}}_{1}\cdot {\mathbf{G}}_{0} \nonumber \\= & {} \mathbf{D}^{-1}{\left( \left[ {\begin{array}{ccc} {\hat{\mathbf{m}}}^{{\prime }} &{} N &{} N\\ \end{array} } \right] ^{\prime } \right) {\cdot \left[ {\begin{array}{ccc} {\mathbf{I}}_{k} &{} {\mathbf{1}}_{k} &{} {\mathbf{1}}_{k}\\ \end{array} } \right] }^{\prime }=\left[ {\begin{array}{c} {\mathbf{D}}^{-1}({\hat{\mathbf{m}}})\\ 1_{k}^{\prime } / N\\ 1_{k}^{\prime } / N\\ \end{array} } \right] }, \end{aligned}$$
(103)
the \((k+1) \times k\) matrix \({\mathbf{G}}_{2}\) equals
$$\begin{aligned} {\mathbf{G}}_{2}= & {} {\mathbf{D}}\left[ \mathrm {exp}({\mathbf{A}}_{2}\cdot {\mathbf{g}}_{\mathrm {1}}) \right] \cdot {\mathbf{A}}_{\mathrm {2}} \cdot {\mathbf{G}}_{1} \nonumber \\= & {} {\mathbf{D}}\left( \left[ {\begin{array}{cc} {\hat{\mathbf{m}}} &{} 1\\ \end{array} } \right] ^{\prime } \right) \cdot \left[ {\mathbf{I}}_{k}\oplus \left[ {\begin{array}{cc} 1 &{} -1\\ \end{array} } \right] \right] \cdot \left[ {\begin{array}{c} {\mathbf{D}}^{-1}({\hat{\mathbf{m}}})\\ 1_{k}^{\prime } / N\\ 1_{k}^{\prime } / N\\ \end{array} } \right] =\left[ {\begin{array}{c} {\mathbf{I}}_{k}\\ {\mathbf {0}}_{k}^{\prime }\\ \end{array} } \right] , \end{aligned}$$
(104)
the \((k+4) \times k\) matrix \({\mathbf{G}}_{3}\) equals
$$\begin{aligned} {\mathbf{G}}_{3}={\mathbf{D}}^{-1}\left( \left[ {\begin{array}{c} \sum X_{i} \\ \sum X_{i}^{2} \\ N\\ N\\ {\mathbf{r}}\\ \end{array} } \right] \right) \cdot \left[ \left[ {\begin{array}{cccc} {\mathbf{r}} &{} {\mathbf{r}}^{\left( 2 \right) } &{} {\mathbf{1}}_{k} &{} {\mathbf{1}}_{k}\\ \end{array} } \right] ^{\prime }\oplus {\mathbf{r}} \right] \cdot \left[ {\begin{array}{c} {\mathbf{I}}_{k}\\ {\mathbf {0}}_{k}^{\prime }\\ \end{array} } \right] =\left[ {\begin{array}{c} {\mathbf{r}}^{{\prime }} / \sum X_{i} \\ {\mathbf{r}}^{(2) {\prime }} / \sum X_{i}^{2} \\ {\mathbf {1}}_{k}^{\prime } / N\\ {\mathbf {1}}_{k}^{\prime } / N\\ {\mathbf {0}}_{k\times k}\\ \end{array} } \right] , \end{aligned}$$
(105)
the \((k+5) \times k\) matrix \({\mathbf{G}}_{4}\) equals
$$\begin{aligned} {\mathbf{G}}_{4}= & {} {\mathbf{D}}\left[ \exp \left( {\mathbf{A}}_{4}\cdot {\mathbf{g}}_{\mathrm {3}} \right) \right] \cdot {\mathbf{A}}_{\mathrm {4}} \cdot {\mathbf{G}}_{3} \nonumber \\= & {} {\mathbf{D}}\left( \left[ {\begin{array}{c} {\bar{X}}\\ \frac{\left( \sum X_{i} \right) ^{2}}{N}\\ \sum X_{i}^{2} \\ N\\ 1\\ {\mathbf{r}}\\ \end{array} } \right] \right) \cdot \left[ \left[ {\begin{array}{cccc} 1 &{} 0 &{} 0 &{} -1 \\ 2 &{} 0 &{} 0 &{} -1 \\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} 1 &{} -1 \\ \end{array} } \right] \oplus {\mathbf{I}}_{k} \right] \cdot \left[ {\begin{array}{c} {\mathbf{r}}^{{\prime }} / \sum X_{i} \\ {\mathbf{r}}^{\left( 2 \right) {\prime }} / \sum X_{i}^{2} \\ {\mathbf {1}}_{k}^{\prime } / N\\ {\mathbf {1}}_{k}^{\prime } / N\\ {\mathbf {0}}_{k\times k}\\ \end{array} } \right] =\left[ {\begin{array}{cc} \frac{{\mathbf{d}}^{{\prime }}}{N}\\ {\mathbf{r}}^{{\prime }}2{\bar{X}}-{\bar{X}}^{2}\\ {\mathbf{r}}^{\left( 2 \right) {\prime }}\\ {\mathbf {1}}_{k}^{{\prime }}\\ {\mathbf {0}}_{(k+1)\times k}\\ \end{array} } \right] ,\nonumber \\ \end{aligned}$$
(106)
the \((k+3) \times k\) matrix \({\mathbf{G}}_{5}\) equals
$$\begin{aligned} {\mathbf{G}}_{5}= & {} {\mathbf{D}}^{-1}\left[ {\mathbf{A}}_{5}\cdot {\mathbf{g}}_{\mathrm {4}} \right] \cdot {\mathbf{A}}_{\mathrm {5}} \cdot {\mathbf{G}}_{4} \nonumber \\= & {} {\mathbf{D}}^{-1}\left[ {\begin{array}{c} {\bar{X}}\\ SS\\ N-1\\ {\mathbf{r}}\\ \end{array} } \right] \cdot \left[ 1\oplus \left[ {\begin{array}{cc} -1 &{} 1\\ \end{array} } \right] \oplus \left[ {\begin{array}{cc} 1 &{} -1\\ \end{array} } \right] \oplus {\mathbf{I}}_{k} \right] \cdot \left[ {\begin{array}{c} \frac{{\mathbf{d}}^{{\prime }}}{N}\\ {\mathbf{r}}^{{\prime }}2{\bar{X}}-{\bar{X}}^{2}\\ {\mathbf{r}}^{\left( 2 \right) {\prime }}\\ {\mathbf {1}}_{k}^{{\prime }}\\ {\mathbf {0}}_{(k+1)\times k}\\ \end{array} } \right] \nonumber \\= & {} \left[ {\begin{array}{c} \frac{{\mathbf{r}}^{{\prime }}}{\sum X_{i} }-\frac{\mathrm {1}}{N}\\ {\mathbf{d}}^{*{\prime }}\\ {\mathbf{e}}^{\prime }\\ {\mathbf {0}}_{k\times k}\\ \end{array} } \right] , \end{aligned}$$
(107)
and the \(\left( k+1 \right) \times k\) matrix \({\mathbf{G}}_{6}\) equals
$$\begin{aligned} {\mathbf{G}}_{6}= & {} {\mathbf{D}}\left[ \exp \left( {\mathbf{A}}_{6}\cdot {\mathbf{g}}_{\mathrm {5}} \right) \right] \cdot {\mathbf{A}}_{\mathrm {6}} \cdot {\mathbf{G}}_{5} \nonumber \\= & {} {\mathbf{D}}\left( \left[ {\begin{array}{c} \frac{{\bar{X}}}{s_{X}}\\ \frac{{\mathbf{r}}}{s_{X}}\\ \end{array} } \right] \right) \cdot \left[ {\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 1 &{} -\frac{1}{{{2}}} &{} \frac{1}{{{2}}} &{} {\mathbf {0}}_{k}^{\prime }\\ {\mathbf {0}}_{k} &{} -\frac{1}{2}\cdot {\mathbf {1}}_{k} &{} \frac{1}{{{2}}}\cdot {\mathbf {1}}_{k}&{} {\mathbf{I}}_{\mathrm {k}}\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} \frac{{\mathbf{r}}^{{\prime }}}{\sum X_{i} }-\frac{\mathrm {1}}{N}\\ {\mathbf{d}}^{*{\prime }}\\ {\mathbf{e}}^{\prime }\\ {\mathbf {0}}_{k\times k}\\ \end{array} } \right] \nonumber \\= & {} \left[ {\begin{array}{c} \frac{{\bar{X}}}{s_{X}}\cdot \left[ \frac{{\mathbf{r}}^{{\prime }}}{\sum X}-\frac{{\mathbf {1}}_{k}^{{\prime }}}{N}-0.5\left( {\mathbf{d}}^{*^{\prime }}-{\mathbf{e}}^{\prime } \right) \right] \\ \frac{-\mathrm {0.5{\mathbf {r}}}}{s_{X}} \cdot \left( {\mathbf{d}}^{*^{\prime }}-{\mathbf{e}}^{\prime } \right) \\ \end{array} } \right] . \end{aligned}$$
(108)
Finally, the \(k\times k\) matrix \({\mathbf{G}}_{7}\) equals
$$\begin{aligned} {\mathbf{G}} = {\mathbf{A}}_{\mathrm {7}} \cdot {\mathbf{G}}_{6}= & {} \left[ {\begin{array}{cc} {-{\mathbf {1}}}_{k} &{} {\mathbf{I}}_{k}\\ \end{array} } \right] \cdot \left[ {\begin{array}{c} \frac{{\bar{X}}}{s_{X}}\cdot \left[ \frac{{\mathbf{r}}^{{\prime }}}{\sum X}-\frac{{\mathbf {1}}_{k}^{{\prime }}}{N}-0.5\left( {\mathbf{d}}^{*^{\prime }}-{\mathbf{e}}^{\prime } \right) \right] \\ \frac{-\mathrm {0.5{\mathbf {r}}}}{s_{X}} \cdot \left( {\mathbf{d}}^{*^{\prime }}-{\mathbf{e}}^{\prime } \right) \\ \end{array} } \right] \nonumber \\= & {} \frac{{\mathbf {1}}_{k}}{s_{X}}\cdot \left\{ -{\bar{X}}\left[ \frac{{\mathbf{r}}^{{\prime }}}{\sum X }-\frac{{\mathbf {1}}_{k}^{{\prime }}}{N}-0.5\left( {\mathbf{d}}^{*^{\prime }}-{\mathbf{e}}^{\prime } \right) \right] -0.5{\mathbf{r}}\left( {\mathbf{d}}^{*^{\prime }}-{\mathbf{e}}^{\prime } \right) \right\} , \end{aligned}$$
(109)
with elements
$$\begin{aligned} G_{ij}=\frac{1}{s_{X}}\cdot \left\{ -{\bar{X}}\left[ \frac{r_{j}}{\sum X }-\frac{1}{N}-0.5\left( d_{j}^{*}-e \right) \right] -0.5r_{i}\left( d_{j}^{*}-e \right) \right\} . \end{aligned}$$
(110)
The Z-scores are not homogeneous of order 0, because \({\mathbf{z}}\) is obtained using \(s_{X}\). As a result, the asymptotic variance is approximated by \({\mathbf{V}}_{{\mathbf{z}}}\approx {\mathbf{G}}{\mathbf{V}}_{{\hat{\mathbf{m}}}}{\mathbf{G}}^{\prime }\) (cf. Eq. 2). Some tedious algebra shows that the sample estimate of the asymptotic covariance matrix of \({\mathbf{V}}_{{\mathbf{z}}}\) is
$$\begin{aligned} V_{Z_{gh}}\approx {\sum }_{j=1}^k {\sum }_{i=1}^k {t\left[ -{\bar{X}}\cdot \left( \frac{r_{i}}{\sum X }-\frac{\mathrm {1}}{N} -u_{i} \right) -r_{g}u_{i} \right] } \cdot \left[ -{\bar{X}}\left( \frac{r_{j}}{\sum X }-\frac{\mathrm {1}}{N} -u_{j} \right) -r_{h}u_{j} \right] ,\nonumber \\ \end{aligned}$$
(111)
where \(u_{i}= 0.5\left[ d_{i}^{*}-e \right] \). For large N,
$$\begin{aligned} V_{{Z}_{gh}}\approx {\sum }_i {\frac{{\hat{\mathrm{m}}}_{i}}{s_{X}^{2}}\left[ -{\bar{X}}\left( \frac{r_{i}}{\sum X }-\frac{\mathrm {1}}{N} -u_{i} \right) -r_{g}u_{i} \right] } \cdot \left[ -{\bar{X}}\left( \frac{r_{j}}{\sum X }-\frac{\mathrm {1}}{N} -u_{j} \right) -r_{h}u_{j} \right] .\nonumber \\ \end{aligned}$$
(112)
