Box–Cox elliptical distributions with application

Abstract

We propose and study the class of Box–Cox elliptical distributions. It provides alternative distributions for modeling multivariate positive, marginally skewed and possibly heavy-tailed data. This new class of distributions has as a special case the class of log-elliptical distributions, and reduces to the Box–Cox symmetric class of distributions in the univariate setting. The parameters are interpretable in terms of quantiles and relative dispersions of the marginal distributions and of associations between pairs of variables. The relation between the scale parameters and quantiles makes the Box–Cox elliptical distributions attractive for regression modeling purposes. Applications to data on vitamin intake are presented and discussed.

Appendices

Appendix A: Proof of the Theorem 1

The conditional PDF of $W_k|{\varvec{W}}_{-k}$, $k=1,\ldots ,p$, is given by

$$\begin{aligned} f_{W_k|{\varvec{W}}_{-k}}(w_k) = \frac{g(({\varvec{w}}-{\varvec{\mu }})'{\varvec{\varSigma }}^{-1}({\varvec{w}}-{\varvec{\mu }}))}{\int _{a_k}^{b_k} g(({\varvec{w}}-{\varvec{\mu }})'{\varvec{\varSigma }}^{-1}({\varvec{w}}-{\varvec{\mu }}))\,\mathrm{{d}}w_k},\quad w_k\in (a_k,b_k). \end{aligned}$$

From the identity $({\varvec{w}}-{\varvec{\mu }})'{\varvec{\varSigma }}^{-1}({\varvec{w}}-{\varvec{\mu }}) = [(w_k-\mu _{k.-k})/\sigma _{k.-k}]^2+q({\varvec{w}}_{-k})$, we get the result.

Appendix B: Proof of the Theorem 2

If ${\varvec{W}}=T_{{\varvec{\lambda }},{\varvec{\mu }}}({\varvec{Y}})\sim {\text{ TE }l}_p({\varvec{\xi }},{\varvec{\varSigma }};R({\varvec{\lambda }});g)$, then its PDF is given by

$$\begin{aligned} f_{{\varvec{W}}}({\varvec{w}})=\dfrac{g(({\varvec{w}}-{\varvec{\xi }})'{\varvec{\varSigma }}^{-1}({\varvec{w}}-{\varvec{\xi }}))}{\int _{R({\varvec{\lambda }})} g(({\varvec{w}}-{\varvec{\xi }})'{\varvec{\varSigma }}^{-1}({\varvec{w}}-{\varvec{\xi }}))\,\mathrm{d}{\varvec{w}}},\quad {\varvec{w}}\in R({\varvec{\lambda }}). \end{aligned}$$

(16)

Let $V:R({\varvec{\lambda }})\rightarrow R({\varvec{\lambda }})$ be the transformation defined as $V({\varvec{w}})={\varvec{D}}_{{\varvec{\alpha }}}^{-1}({\varvec{w}}-{\varvec{\xi }})$, and let ${\varvec{U}}=V({\varvec{W}})$, with Jacobian $J({\varvec{w}}\rightarrow {\varvec{u}})=\prod _{k=1}^{p}(1+\lambda _k \xi _k)$. The PDF of ${\varvec{U}}$ is

$$\begin{aligned} f_{{\varvec{U}}}({\varvec{u}})&= \dfrac{g({\varvec{u}}'({\varvec{D}}_{{\varvec{\alpha }}}^{-1}{\varvec{\varSigma }}{\varvec{D}}_{{\varvec{\alpha }}}^{-1})^{-1}{\varvec{u}})}{\int _{R({\varvec{\lambda }})} g({\varvec{u}}'({\varvec{D}}_{{\varvec{\alpha }}}^{-1}{\varvec{\varSigma }}{\varvec{D}}_{{\varvec{\alpha }}}^{-1})^{-1}{\varvec{u}})\,\mathrm{d}{\varvec{u}}},\quad {\varvec{u}}\in R({\varvec{\lambda }}). \end{aligned}$$

Hence, ${\varvec{U}}\sim {\text{ TE }l}_p({\varvec{0}},{\varvec{D}}_{{\varvec{\alpha }}}^{-1}{\varvec{\varSigma }}{\varvec{D}}_{{\varvec{\alpha }}}^{-1};R({\varvec{\lambda }});g)$. Because ${\varvec{U}}=V(T_{{\varvec{\lambda }},{\varvec{\mu }}}({\varvec{Y}}))=T_{{\varvec{\lambda }},{\varvec{\delta }}}({\varvec{Y}})$, where ${\varvec{\delta }}=T_{{\varvec{\lambda }},{\varvec{\mu }}}^{-1}({\varvec{\xi }})$, then from Definition 4 we have ${\varvec{Y}}=T_{{\varvec{\lambda }},{\varvec{\delta }}}^{-1}({\varvec{U}})\sim \text{ BCE }\ell _p({\varvec{\delta }},{\varvec{\lambda }},{\varvec{D}}_{{\varvec{\alpha }}}^{-1}{\varvec{\varSigma }}{\varvec{D}}_{{\varvec{\alpha }}}^{-1};g)$.

On the other hand, if ${\varvec{Y}}\sim \text{ BCE }\ell _p({\varvec{\delta }},{\varvec{\lambda }},{\varvec{D}}_{{\varvec{\alpha }}}^{-1}{\varvec{\varSigma }}{\varvec{D}}_{{\varvec{\alpha }}}^{-1};g)$, then its PDF is

$$\begin{aligned} f_{{\varvec{Y}}}({\varvec{y}}) = \dfrac{g({\varvec{w}}'({\varvec{D}}_{{\varvec{\alpha }}}^{-1}{\varvec{\varSigma }}{\varvec{D}}_{{\varvec{\alpha }}}^{-1})^{-1}{\varvec{w}})\prod _{k=1}^{p}\frac{y_k^{\lambda _k-1}}{{\delta _k}^{\lambda _k}}}{\int _{R({\varvec{\lambda }})}g({\varvec{w}}'({\varvec{D}}_{{\varvec{\alpha }}}^{-1}{\varvec{\varSigma }}{\varvec{D}}_{{\varvec{\alpha }}}^{-1})^{-1}{\varvec{w}})\,\mathrm{d}{\varvec{w}}},\quad {\varvec{w}}=T_{{\varvec{\lambda }},{\varvec{\delta }}}({\varvec{y}}),\quad {\varvec{y}}\in \mathbb {R}_{+}^p. \end{aligned}$$

(17)

Now, from the transformation ${\varvec{W}}=T_{{\varvec{\lambda }},{\varvec{\mu }}}({\varvec{Y}})$, with Jacobian $J({\varvec{y}}\rightarrow {\varvec{w}})=\prod _{k=1}^{p}\mu _k(1+\lambda _k w_k)^{1/\lambda _k-1}$, in the PDF (17) we arrive at PDF (16).

Appendix C: Proof of Theorem 3

• 1. From ${\varvec{T}}={\varvec{D_\alpha Y}}$, with Jacobian $J({\varvec{y}}\rightarrow {\varvec{t}})=\prod _{k=1}^{p}\alpha _k^{-1}$, in (6), we get the PDF of ${\varvec{T}}$ as

  $$\begin{aligned} f_{{\varvec{T}}}({\varvec{t}}) = \dfrac{g({\varvec{w}}'{\varvec{\varSigma }}^{-1}{\varvec{w}})\prod _{k=1}^{p}\frac{t_k^{\lambda _k-1}}{(\alpha _k\mu _k)^{\lambda _k}}}{\int _{R({\varvec{\lambda }})}g({\varvec{w}}'{\varvec{\varSigma }}^{-1}{\varvec{w}})\,\mathrm{d}{\varvec{w}}},\quad {\varvec{t}}\in \mathbb {R}_{+}^p, \end{aligned}$$

  where ${\varvec{w}}{=}T_{{\varvec{\lambda }},{\varvec{\mu }}}({\varvec{D}}_{{\varvec{\alpha }}}^{-1}{\varvec{t}}){=}T_{{\varvec{\lambda }},{\varvec{D_\alpha \mu }}}({\varvec{t}})$. Hence, ${\varvec{T}}={\varvec{D_{\alpha }Y}}\sim \text{ BCE }\ell _p({\varvec{D_\alpha \mu }},{\varvec{\lambda }},{\varvec{\varSigma }};g)$.

• 2. Note that the PDF of ${\varvec{Y}}$, given in (6), can be expressed as

  $$\begin{aligned} f_{{\varvec{Y}}}({\varvec{y}})=\dfrac{g({\varvec{v}}'({\varvec{D_\beta \varSigma D_\beta }})^{-1}{\varvec{v}})\prod _{k=1}^{p}\frac{|\beta _k|y_k^{\lambda _k-1}}{\mu _k^{\lambda _k}}}{\int _{R({\varvec{D}}_{{\varvec{\beta }}}^{-1}{\varvec{\lambda }})}g({\varvec{v}}'({\varvec{D_\beta \varSigma D_\beta }})^{-1}{\varvec{v}})\,\mathrm{d}{\varvec{v}}},\quad {\varvec{y}}\in \mathbb {R}_{+}^p, \end{aligned}$$

  where ${\varvec{v}}={\varvec{D_\beta }}T_{{\varvec{\lambda }},{\varvec{\mu }}}({\varvec{y}})$ has its kth component given by

  $$\begin{aligned} v_k= {\left\{ \begin{array}{ll} \dfrac{[(y_k/\mu _k)^{\beta _k}]^{\lambda _k/\beta _k}-1}{\lambda _k/\beta _k}, &{}\quad \lambda _k\ne 0,\\ \quad \,\,\,\,\log (y_k/\mu _k)^{\beta _k}, &{}\quad \lambda _k=0, \end{array}\right. } \end{aligned}$$

  for $k=1,\ldots ,p$. From $U_k=(Y_k/\mu _k)^{\beta _k}$, $k=1,\ldots ,p$, with Jacobian $J({\varvec{y}}\rightarrow {\varvec{u}})=\prod _{k=1}^{p}\mu _k \beta _k^{-1}u_k^{1/\beta _k-1}$, we arrive at the desired result.

• 3. Plugging ${\varvec{\lambda }}={\varvec{1}}$ in (6) we have that the PDF of ${\varvec{Y}}$ is

  $$\begin{aligned} f_{{\varvec{Y}}}({\varvec{y}})=\dfrac{g(({\varvec{y}}-{\varvec{\mu }})'({\varvec{D_{\mu }\varSigma D_{\mu }}})^{-1}({\varvec{y}}-{\varvec{\mu }}))\prod _{k=1}^{p}\frac{1}{\mu _k}}{\int _{R({\varvec{1}})}g({\varvec{w}}'{\varvec{\varSigma }}^{-1}{\varvec{w}})\,\mathrm{d}{\varvec{w}}},\quad {\varvec{y}}\in \mathbb {R}_{+}^p. \end{aligned}$$

  From the change of variables ${\varvec{w}}={\varvec{D}}_{{\varvec{\mu }}}^{-1}({\varvec{y}}-{\varvec{\mu }})$ we arrive at the desired result.

Appendix D: Proof of Theorem 4

Plugging ${\varvec{\varSigma }}_{12}={\varvec{0}}$ in (8), and then making the change of variables ${\varvec{s}}=T({\varvec{w}}_2)={\varvec{\varSigma }}_{22}^{-1/2}{\varvec{w}}_2$, the marginal PDF of ${\varvec{Y}}_1$ is

$$\begin{aligned} f_{{\varvec{Y}}_1}({\varvec{y}}_1)&=\dfrac{\bigr \{\int _{R({\varvec{\lambda _2}})}g({\varvec{w}}_1'{\varvec{\varSigma }}_{11}^{-1}{\varvec{w}}_1+{\varvec{w}}_2'{\varvec{\varSigma }}_{22}^{-1}{\varvec{w}}_2)\,\mathrm{d}{\varvec{w}}_2\bigl \}\prod _{k=1}^{r}\frac{y_k^{\lambda _k-1}}{\mu _k^{\lambda _k}}}{\int _{R({\varvec{\lambda }}_1)}\bigl \{\int _{R({\varvec{\lambda }}_2)}g({\varvec{w}}_1'{\varvec{\varSigma }}_{11}^{-1}{\varvec{w}}_1+{\varvec{w}}_2'{\varvec{\varSigma }}_{22}^{-1}{\varvec{w}}_2)\,\mathrm{d}{\varvec{w}}_2\bigl \}\,\mathrm{d}{\varvec{w}}_1}\\&=\dfrac{g_1({\varvec{w}}_1'{\varvec{\varSigma }}_{11}^{-1}{\varvec{w}}_1)\prod _{k=1}^{r}\frac{y_k^{\lambda _k-1}}{\mu _k^{\lambda _k}}}{\int _{R({\varvec{\lambda }}_1)}g_1({\varvec{w}}_1'{\varvec{\varSigma }}_{11}^{-1}{\varvec{w}}_1)\,\mathrm{d}{\varvec{w}}_1},\quad {\varvec{w}}_1=T_{{\varvec{\lambda }}_1,{\varvec{\mu }}_1}({\varvec{y}}_1),\quad {\varvec{y}}_1\in \mathbb {R}_{+}^r. \end{aligned}$$

Note that $g_{1}(u) = \int _{T(R({\varvec{\lambda _2}}))}g(u + {\varvec{s}}'{\varvec{s}})\,\mathrm{d}{\varvec{s}} \le \int _{\mathbb {R}^{p-r}}g(u+{\varvec{s}}'{\varvec{s}})\,\mathrm{{d}}{\varvec{s}}=h_1(u)$, $u\ge 0$, where $h_1$ is such that $\int _{0}^{\infty }t^{r-1}h_1(t^2)\,\text {d}t<\infty$ (Fang et al. 1990, Sec. 2.2). This completes the proof.

Appendix E: Proof of Theorem 5

Because ${\varvec{Y}}\sim \text{ LE }\ell _p({\varvec{\mu }},{\varvec{\varSigma }};g)$, with $g(u)\propto \int _{0}^{\infty }t^{p/2}\exp (-ut/2)\,\mathrm{d}H(t)$, $u\ge 0$, then ${\varvec{X}}=T_{{\varvec{0}},{\varvec{\mu }}}({\varvec{Y}})\sim \text{ E }\ell _p({\varvec{0}},{\varvec{\varSigma }};g)$. Thus, ${\varvec{X}}_1=T_{{\varvec{0}},{\varvec{\mu }}_1}({\varvec{Y}}_1)\sim \text{ E }\ell _p({\varvec{0}},{\varvec{\varSigma }}_{11};g)$ (Kano 1994). Hence, ${\varvec{Y}}_1\sim \text{ LE }\ell _r({\varvec{\mu }}_1,{\varvec{\varSigma }}_{11};g)$.

Appendix F: Proof of Theorem 6

The conditional PDF of ${\varvec{Y}}_1|{\varvec{Y}}_2$ is given by

$$\begin{aligned} f_{{\varvec{Y}}_1|{\varvec{Y}}_2}({\varvec{y}}_1)=\frac{g({\varvec{w}}'{\varvec{\varSigma }}^{-1}{\varvec{w}})\prod _{k=1}^{r}\frac{y_k^{\lambda _k-1}}{\mu _k^{\lambda _k}}}{\int _{R({\varvec{\lambda }}_1)}g({\varvec{w}}'{\varvec{\varSigma }}^{-1}{\varvec{w}})\,\mathrm{{d}}{\varvec{{\varvec{w}}_1}}},\quad {\varvec{w}}_1=T_{{\varvec{\lambda }}_1,{\varvec{\mu }}_1}({\varvec{y}}_1),\quad {\varvec{y}}_1\in \mathbb {R}_{+}^r. \end{aligned}$$

(18)

Because ${\varvec{w}}'{\varvec{\varSigma }}^{-1}{\varvec{w}}={\varvec{u}}_1' \bigl ({\varvec{D}}_{{\varvec{\alpha }}({\varvec{w}}_2)}^{-1}{\varvec{\varSigma }}_{11\cdot 2}{\varvec{D}}_{{\varvec{\alpha }}({\varvec{w}}_2)}^{-1}\bigr )^{-1}{\varvec{u}}_1+q({\varvec{w}}_2)$, where ${\varvec{u}}_1={\varvec{D}}_{{\varvec{\alpha }}({\varvec{w}}_2)}^{-1} ({\varvec{w}}_1-{\varvec{\mu }}_1({\varvec{w}}_2))=T_{{\varvec{\lambda }}_1,{\varvec{\delta }}_1}({\varvec{y}}_1)$, (18) can be expressed as

$$\begin{aligned} f_{{\varvec{Y}}_1|{\varvec{Y}}_2}({\varvec{y}}_1) = \frac{g_{q({\varvec{w}}_2)}\bigl ({\varvec{u}}_1'\bigl ({\varvec{D}}_{{\varvec{\alpha }}({\varvec{w}}_2)}^{-1}{\varvec{\varSigma }}_{11\cdot 2}{\varvec{D}}_{{\varvec{\alpha }}({\varvec{w}}_2)}^{-1}\bigr )^{-1}{\varvec{u}}_1\bigr )\prod _{k=1}^{r}\frac{y_k^{\lambda _k-1}}{\mu _k^{\lambda _k}(1+\lambda _k\mu _{1k}({\varvec{w}}_2))}}{\int _{R({\varvec{\lambda }}_1)}g_{q({\varvec{w}}_2)}\bigl ({\varvec{u}}_1'\bigl ({\varvec{D}}_{{\varvec{\alpha }}({\varvec{w}}_2)}^{-1}{\varvec{\varSigma }}_{11\cdot 2}{\varvec{D}}_{{\varvec{\alpha }}({\varvec{w}}_2)}^{-1}\bigr )^{-1}{\varvec{u}}_1\bigr )\,\mathrm{{d}}{{\varvec{u}}_1}},\quad {\varvec{y}}_1\in \mathbb {R}_{+}^p, \end{aligned}$$

where ${\varvec{u}}_1=T_{{\varvec{\lambda }}_1,{\varvec{\delta }}_1}({\varvec{y}}_1)$. Since $\prod _{k=1}^{r}\mu _k^{\lambda _k}(1+\lambda _k\mu _{1k}({\varvec{w}}_2)) = \prod _{k=1}^{r}{\delta _k}^{\lambda _k}$ the proof is complete.

Appendix G: Proof of Theorem 7

${\varvec{Y}}_1$ and ${\varvec{Y}}_2$ are independent if, and only if, the PDF of ${\varvec{Y}}\sim \text{ BCE }\ell _p({\varvec{\mu }},{\varvec{\lambda }},{\varvec{\varSigma }};g)$ given in (6) is such that $f_{{\varvec{Y}}}({\varvec{y}})=f_{{\varvec{Y}}_1}({\varvec{y}}_1)f_{{\varvec{Y}}_2}({\varvec{y}}_2)$. This condition is satisfied if, and only if, ${\varvec{\varSigma }}_{12}={\varvec{0}}$ and the DGF g satisfies the functional equation $g(u+v)=g(u)g(v)$, with $u\ge 0$  and  $v\ge 0$, for which $g(u)=\exp (-ku)$, for some $k\ge 0$, is a solution (Gupta et al. 2013, Sec. 1.3). From $\int _{0}^{\infty }t^{p-1}\exp (-kt^2)\,\text {d}t=2^{p/2-1}\varGamma (p/2)$, we find that $k=1/2$. Hence, ${\varvec{Y}}_1$ and ${\varvec{Y}}_2$ are independent if, and only if, ${\varvec{\varSigma }}_{12}={\varvec{0}}$ and ${\varvec{Y}}\sim \text{ BCN }_p({\varvec{\mu }},{\varvec{\lambda }},{\varvec{\varSigma }})$.

Appendix H: Proof of Theorem 8

From (6) we have

$$\begin{aligned} {\text{ E }}\biggl (\prod _{k=1}^{p}Y_k^{h_k}\biggr ) = \dfrac{\int _{\mathbb {R}_{+}^p} g({\varvec{w}}'{\varvec{\varSigma }}^{-1}{\varvec{w}}) \prod _{k=1}^{p}\frac{y_k^{\lambda _k+h_k-1}}{\mu _k^{\lambda _k}}\,\mathrm{d}{\varvec{y}}}{\int _{R({\varvec{\lambda }})}g({\varvec{w}}'{\varvec{\varSigma }}^{-1}{\varvec{w}})\,\mathrm{d}{\varvec{w}}}, \end{aligned}$$

where ${\varvec{w}}=T_{{\varvec{\lambda }},{\varvec{\mu }}}({\varvec{y}})$. By making the change of variables ${\varvec{u}} ={\varvec{D}}_{{\varvec{\mu }}}^{-1}{\varvec{y}}$ we arrive at the desired result.

Appendix I: Marginal PDF of $Y_k$

The function $g_{{{\varvec{\varUpsilon }}_k}}$ given in (10) can be defined in $\mathbb {R}$. Hence, we can define a random variable $U_k\in \mathbb {R}$ from the PDF

$$\begin{aligned} f_{U_k}(u_k)=c_k g_{{{\varvec{\varUpsilon }}_k}}(u_k),\quad u_k\in \mathbb {R}, \end{aligned}$$

where $c_k^{-1} = \int _{-\infty }^{\infty }g_{{{\varvec{\varUpsilon }}_k}}(t)\,\mathrm{d}t$. The CDF of $U_k$ is given by (11). We now define $S_k\in I(\lambda _k\sqrt{\sigma _{kk}})$ as a random variable $U_k$ truncated on $I(\lambda _k\sqrt{\sigma _{kk}})$. The PDF of $S_k$ is given by

$$\begin{aligned} f_{S_k}(s_k) = \frac{g_{{{\varvec{\varUpsilon }}_k}}(s_k)}{\int _{I(\lambda _k\sqrt{\sigma _{kk}})}g_{{{\varvec{\varUpsilon }}_k}}(s_k)\,\mathrm{d}s_k},\quad s_k\in I(\lambda _k\sqrt{\sigma _{kk}}). \end{aligned}$$

(19)

From the transformation $S_k=\sigma _{kk}^{-1/2}T_{\lambda _k,\mu _k}(Y_k)$, with Jacobian $J(s_k\rightarrow y_k)=\sigma _{kk}^{-1/2}\mu _{k}^{-\lambda _k}y_k^{\lambda _k-1}$, we arrive at the PDF of $Y_k$ given in (9).

Appendix J: Proof of Theorem 9

Because ${\varvec{Y}}\sim \text{ BCE }\ell _p({\varvec{\mu }},{\varvec{\lambda }},{\varvec{\varSigma }};g)$, the PDF of $Y_k$, $k=1,\ldots ,p$, is given by (9), where $S_k=\sigma _{kk}^{-1/2}T_{\lambda _k,\mu _k}(Y_k)$ has PDF given in (19). The CDF of $S_k$ is

$$\begin{aligned} F_{S_k}(s_k) = {\left\{ \begin{array}{ll} \,\,\,\,\dfrac{F_{U_k}(s_k)-F_{U_k}(-1/{\lambda _k\sqrt{\sigma _{kk}})}}{1-F_{U_k}(-1/{\lambda _k\sqrt{\sigma _{kk}})}}, &{}\quad \lambda _k > 0,\\ \dfrac{1 - F_{U_k}(-1/{\lambda _k\sqrt{\sigma _{kk}})}+F_{U_k}(s_k)}{F_{U_k}(-1/{\lambda _k\sqrt{\sigma _{kk}})}}, &{}\quad \lambda _k<0,\\ \quad \quad \quad \quad \,\, F_{U_k}(s_k), &{}\quad \lambda _k = 0, \end{array}\right. } \end{aligned}$$

(20)

from which we have that the $\alpha$-quantile $y_{k,\alpha }$ of $Y_k$, $\alpha \in (0,1)$, is such that ${\text{ P }}(Y_k\le y_{k,\alpha })=\alpha$, or equivalently ${\text{ P }}[S_k \le \sigma _{kk}^{-1/2}T_{\lambda _k,\mu _k}(y_{k,\alpha })]=\alpha$. Hence,

$$\begin{aligned} y_{k,\alpha } = {\left\{ \begin{array}{ll} \mu _k (1+\lambda _k\sqrt{\sigma _{kk}} s_{k,\alpha })^{1/\lambda _k}, &{}\quad \lambda _k\ne 0,\\ \quad \,\,\mu _k\exp (\sqrt{\sigma _{kk}}s_{k,\alpha }), &{}\quad \lambda _k=0, \end{array}\right. } \end{aligned}$$

where $s_{k,\alpha }$ is such that $F_{S_k}(s_{k,\alpha })=\alpha$, with $F_{S_k}$ given in (20). Therefore, $s_{k,\alpha }$ is given by

$$\begin{aligned} s_{k,\alpha } = {\left\{ \begin{array}{ll} F_{U_k}^{-1}(\alpha + (1-\alpha )F_{U_k}(-1/{\lambda _k\sqrt{\sigma _{kk}}})), &{} \quad \lambda _k>0, \\ F_{U_k}^{-1}((1+\alpha )F_{U_k}(-1/{\lambda _k\sqrt{\sigma _{kk}}}) - 1), &{}\quad \lambda _k<0, \\ \quad \quad \quad \quad \quad \quad F_{U_k}^{-1}(\alpha ), &{}\quad \lambda _k = 0, \end{array}\right. } \end{aligned}$$

where $F_{U_k}$ is the CDF given in (11).