Standardized maximin D- and c-optimal designs for the Poisson–Gamma model

Abstract

The Poisson–Gamma model is obtained as a generalization of the Poisson model, when Gamma distributed block effects are assumed for Poisson count data. We show that optimal designs for estimating linear combinations of the model parameters coincide for the case of known and unknown parameters of the Gamma distribution. To obtain robust designs regarding parameter misspecification we determine standardized maximin D-optimal designs for a binary and a continuous design region. For standardized maximin c-optimality we show that the optimal designs for the Poisson–Gamma and Poisson model are equal and derive optimal designs for both models.

Appendix A

Proof of Theorem 4.1

For the criterion functions we have:

$$\begin{aligned} \phi \bigl (\varvec{A}^T\varvec{M}(\zeta ;\varvec{\beta },a)^-\varvec{A}\bigr )&=\phi \Biggl (\begin{pmatrix}\varvec{\tilde{A}}^T&\quad \varvec{0}_s \end{pmatrix}\cdot \begin{pmatrix}\varvec{M}(\zeta ;\varvec{\beta })^- &{}\quad \varvec{0}_p\\ \varvec{0}_p^T &{}\quad M(\zeta ;a)^- \end{pmatrix}\cdot \begin{pmatrix}\varvec{\tilde{A}}\\ \varvec{0}_s^T \end{pmatrix}\Biggr )\\&=\phi \bigl (\varvec{\tilde{A}}^T\varvec{M}(\zeta ;\varvec{\beta })^-\varvec{\tilde{A}}\bigr ). \end{aligned}$$

From the equality of the criterion functions the theorem follows. $\square$

Proof of Theorem 5.2

Let $\lambda _j=\lambda _j(\beta _1)=\exp (\beta _0+\beta _1x_j)$ and let $\delta _j=\delta _j(\beta _1)=1+\frac{m}{b}\cdot \lambda _j(\beta _1)$ for $j=1,2$, where $x_1=0$ and $x_2=1$. To simplify the notation, the squared efficiency ${{\,\textrm{eff}\,}}_D(\xi ;\varvec{\beta })^2=\det \bigl (\varvec{M}(\xi ;\varvec{\beta })\bigr )/\det \bigl (\varvec{M}(\xi _{\varvec{\beta }}^{*};\varvec{\beta })\bigr )$ is considered as criterion function, where $\xi _{\varvec{\beta }}^{*}$ is the locally D-optimal design for $\varvec{\beta }$. With Eq. (9) the determinant of the information matrix for a design $\xi$ with support points $x_1$ and $x_2$ and corresponding weights $w_1$ and $w_2$ is given by:

$$\begin{aligned}&\det \bigl (\varvec{M}(\xi ;\varvec{\beta })\bigr )=\left( \frac{a}{b}\right) ^2\cdot \frac{\det \bigl (\varvec{M}_{{{\,\textrm{Po}\,}}}(\xi ;\varvec{\beta })\bigr )}{1+\frac{m}{b}\cdot \varvec{e}_{1,2}^T\varvec{M}_{{{\,\textrm{Po}\,}}}(\xi ;\varvec{\beta })\varvec{e}_{1,2}}\nonumber \\&=\left( \frac{a}{b}\right) ^2\cdot \frac{w_1w_2\lambda _1\lambda _2\cdot (x_2-x_1)^2}{1+\frac{m}{b}\cdot \left( w_1\lambda _1+w_2\lambda _2\right) }=\left( \frac{a}{b}\right) ^2\cdot \frac{w_1w_2\lambda _1\lambda _2\cdot (x_2-x_1)^2}{w_1\delta _1+w_2\delta _2}. \end{aligned}$$

(A1)

The locally D-optimal design $\xi _{\varvec{\beta }}^{*}$ given in Eq. (15) has support points $x_1^{*}=0$ and $x_2^{*}=1$ and corresponding D-optimal weights $w_{1,\varvec{\beta }}^{*}=\sqrt{\delta _2}/\left( \sqrt{\delta _1}+\sqrt{\delta _2}\right)$ and $w_{2,\varvec{\beta }}^{*}=\sqrt{\delta _1}/\left( \sqrt{\delta _1}+\sqrt{\delta _2}\right)$. The determinant of the information matrix of $\xi _{\varvec{\beta }}^{*}$ is given by:

$$\begin{aligned} \det \bigl (\varvec{M}(\xi _{\varvec{\beta }}^{*};\varvec{\beta })\bigr )=\left( \frac{a}{b}\right) ^2\cdot \frac{w_{1,\varvec{\beta }}^{*}w_{2,\varvec{\beta }}^{*}\lambda _1\lambda _2}{w_{1,\varvec{\beta }}^{*}\delta _1+w_{2,\varvec{\beta }}^{*}\delta _2}=\left( \frac{a}{b}\right) ^2\cdot \frac{\lambda _1\lambda _2}{\big (\sqrt{\delta _1}+\sqrt{\delta _2} \big )^2}. \end{aligned}$$

Hence, the criterion function for standardized maximin D-optimality for a design $\xi$ with support points $x_1=0$ and $x_2=1$ and corresponding weights $w_1=w$ and $w_2=1-w$ is given by

$$\begin{aligned} g(w,\beta _1):={{\,\textrm{eff}\,}}_D(\xi ;\varvec{\beta })^2&=\frac{\det \bigl (\varvec{M}(\xi ;\varvec{\beta })\bigr )}{\det \bigl (\varvec{M}(\xi _{\varvec{\beta }}^{*};\varvec{\beta })\bigr )}=\frac{w\cdot (1-w)\cdot \big (\sqrt{\delta _1}+\sqrt{\delta _2} \big )^2}{w\delta _1+(1-w)\delta _2} \end{aligned}$$

(A2)

$$\begin{aligned}&=\frac{w\cdot (1-w)}{w\cdot (w_{2,\varvec{\beta }}^{*})^2+(1-w)\cdot (w_{1,\varvec{\beta }}^{*})^2}. \end{aligned}$$

(A3)

It is now shown that for arbitrary but fixed w the minima of the criterion function $g(w,\beta _1)$ with respect to $\beta _1$ on the interval $\left[ L,R\right]$ are in the set of boundary points $\left\{ L,R\right\}$. We have $\partial \delta _1/\partial \beta _1=0$ and $\partial \delta _2/\partial \beta _1=(m/b)\cdot \lambda _2$. Deriving the criterion function $g(w,\beta _1)$ in Eq. (A2) with respect to $\beta _1$ yields:

$$\begin{aligned} \frac{\partial g(w,\beta _1)}{\partial \beta _1} =w(1-w)\left( \frac{\big (\sqrt{\delta _1}+\sqrt{\delta _2} \big )\cdot \frac{m}{b}\cdot \lambda _2}{\sqrt{\delta _2}\big (w\delta _1+(1-w)\delta _2\big )}-\frac{\big (\sqrt{\delta _1}+\sqrt{\delta _2} \big )^2(1-w)\cdot \frac{m}{b}\cdot \lambda _2}{\big (w\delta _1+(1-w)\delta _2\big )^2}\right) . \end{aligned}$$

The equation $\partial g(w,\beta _1)/\partial \beta _1=0$ holds if and only if $0=w\delta _1-(1-w)\cdot \sqrt{\delta _1}\cdot \sqrt{\delta _2}$. We obtain $\sqrt{\delta _2}=(w/(1-w))\cdot \sqrt{\delta _1}$ and hence $\delta _2=(w/(1-w))^2\cdot \delta _1$. Solving this equation for $\beta _1$, we obtain for $(w/(1-w))^2>\left( 1+\frac{m}{b}\cdot \exp (\beta _0)\right) ^{-1}$ the unique solution

$$\begin{aligned} \beta _1^{*}=\ln \left( \left( \left( \frac{w}{1-w}\right) ^2\cdot \left( 1+\frac{m}{b}\cdot \exp (\beta _0)\right) -1\right) \cdot \frac{b}{m}\right) -\beta _0. \end{aligned}$$

For $\beta _1=\beta _1^{*}$ we have the relation $\sqrt{\delta _2}=(w/(1-w))\cdot \sqrt{\delta _1}$. Solving for w yields $w=\sqrt{\delta _2}/(\sqrt{\delta _1}+\sqrt{\delta _2})$. Thus w is equal to the optimal weight $w_{1,\varvec{\beta }}^{*}$ of the locally D-optimal design $\xi _{\varvec{\beta }}^{*}$ for $\varvec{\beta }=(\beta _0,\beta _1^{*})^T$ and thus $g(w,\beta _1^{*})=1$ holds. Since $g(w,\beta _1)\le 1$, there is a maximum at $\beta _1=\beta _1^{*}$ for fixed w. The derivative of $g(w,\beta _1)$ has no further zeros, thus there exist no further extrema in the open interval (L, R). Hence, $g(w,\beta _1)$ is unimodal with respect to $\beta _1$ and is minimal on the boundary of the interval $\left[ L,R\right]$.

For $(w/(1-w))^2\le (1+\frac{m}{b}\cdot \exp (\beta _0))^{-1}$, the derivative of the criterion function with respect to $\beta _1$ has no zero and is negative for fixed w for all $\beta _1\in \mathbb {R}$. Thus $g(w,\beta _1)$ is minimal in the interval $\left[ L,R\right]$ at R.

The criterion function attains its minima with respect to $\beta _1$ in the set $\left\{ L,R\right\}$. Dette et al. (2007) showed that for a standardized maximin optimal design $\xi ^{*}$ we have $\vert \mathcal {N}(\xi ^{*})\vert \ge 2$, which means that $\xi ^{*}$ attains its minimal efficiency at least at two different values of $\varvec{\beta }$. It follows that $g(w,L)=g(w,R)$ must hold for $\xi ^{*}$. With the representation of $g(w,\beta _1)$ in Eq. (A3), the optimal weight $w^{*}$ is thus the solution of the equation

$$\begin{aligned}&\frac{w^{*}\cdot (1-w^{*})}{w^{*}\cdot (1-w_{1,\varvec{\beta }_L}^{*})^2+(1-w^{*})\cdot (w_{1,\varvec{\beta }_L}^{*})^2}\\&=\frac{w^{*}\cdot (1-w^{*})}{w^{*}\cdot (1-w_{1,\varvec{\beta }_R}^{*})^2+(1-w^{*})\cdot (w_{1,\varvec{\beta }_R}^{*})^2}, \end{aligned}$$

where $w_{1,\varvec{\beta }_L}^{*}$ and $w_{1,\varvec{\beta }_R}^{*}$ are the weights of the locally D-optimal designs $\xi _{\varvec{\beta }_L}^{*}$ and $\xi _{\varvec{\beta }_R}^{*}$ for $\varvec{\beta }_L=(\beta _0,L)$ and $\varvec{\beta }_R=(\beta _0,R)$, respectively. Solving this equation for $w^{*}$, we obtain $w^{*}=\big (w_{1,\varvec{\beta }_L}^{*}+w_{1,\varvec{\beta }_R}^{*}\big )/2$. Hence, the standardized maximin D-optimal design is given by $\xi ^{*}=\frac{1}{2}\cdot (\xi _{\varvec{\beta }_L}^{*}+\xi _{\varvec{\beta }_R}^{*})$. $\square$

Lemma A.1

Let $\lambda _{\beta _1}(x)=\exp (\beta _0+\beta _1x)$, $\delta _{\beta _1}(x)=1+\frac{m}{b}\cdot \lambda _{\beta _1}(x)$ and $\delta (0)=1+\frac{m}{b}\cdot \exp (\beta _0)$. Further let

$$\begin{aligned} w_{L,R}(x):=\frac{\lambda _{R}(x)\cdot R^2\cdot \delta _{L}(x)-\lambda _{L}(x)\cdot L^2\cdot \delta _{R}(x)}{\lambda _{L}(x)\cdot L^2\cdot \big (\delta (0)-\delta _{R}(x)\big )-\lambda _{R}(x)\cdot R^2\cdot \big (\delta (0)-\delta _{L}(x)\big )}. \end{aligned}$$

(A4)

For $L<R<0$ and $\tilde{x}=2\cdot \ln (R/L)/(L-R)$, the function $w_{L,R}(x)$ has a unique zero $x_0>\tilde{x}$ in the interval $(0,\infty )$ and $w_{L,R}(\tilde{x})=1$ holds. Furthermore, $0<w_{L,R}(x)<1$ holds only for $x\in (\tilde{x},x_0)$. On the interval $(\tilde{x},x_0)$ the function $w_{L,R}(x)$ is strictly monotonically decreasing and has the following representation:

$$\begin{aligned} w_{L,R}(x)=\frac{1}{\displaystyle 1-\frac{\delta (0)}{\displaystyle 1+\frac{m}{b}\cdot \frac{L^2-R^2}{\displaystyle \frac{L^2}{\lambda _{R}(x)}-\frac{R^2}{\lambda _{L}(x)}}}}. \end{aligned}$$

(A5)

For $0<L<R$ the equation $w_{L,R}(x)=w_{-R,-L}(-x)$ holds.

Proof of Lemma A.1

Let $L<R<0$. The indices of $w_{L,R}(x)$ are omitted to simplify the notation.

First, it is shown that the numerator and denominator of w(x) have no common zeros. It is assumed that the numerator is equal to 0. Then the denominator simplifies to $\delta (0)\cdot \big (\lambda _{L}(x) L^2-\lambda _{R}(x) R^2\big )$. If the denominator would also be equal to 0, then $\lambda _{L}(x) L^2=\lambda _{R}(x) R^2$ must hold. Then for the numerator we would have:

$$\begin{aligned} 0=\lambda _{R}(x) R^2\delta _{L}(x)-\lambda _{L}(x) L^2 \delta _{R}(x)=\lambda _{R}(x) R^2\cdot \left( \delta _{L}(x)-\delta _{R}(x)\right) . \end{aligned}$$

It follows that $\delta _{L}(x)=\delta _{R}(x)$ and thus $L=R$, which is a contradiction.

For $\lambda _{R}(x) R^2 \delta _{L}(x)-\lambda _{L}(x) L^2 \delta _{R}(x)\ne 0$, the function w(x) can be rewritten as:

$$\begin{aligned} w(x)&=\frac{1}{\displaystyle 1+\frac{\lambda _{L}(x) L^2 \delta (0)-\lambda _{R}(x) R^2 \delta (0)}{\lambda _{R}(x) R^2 \delta _{L}(x)-\lambda _{L}(x) L^2 \delta _{R}(x)}}\nonumber \\&=\frac{1}{\displaystyle 1+\frac{\delta (0)\cdot \big (\lambda _{L}(x) L^2-\lambda _{R}(x) R^2\big )}{\lambda _{R}(x) R^2-\lambda _{L}(x) L^2+\frac{m}{b}\cdot \big (\lambda _{R}(x) R^2 \lambda _{L}(x)-\lambda _{L}(x) L^2 \lambda _{R}(x)\big )}}. \end{aligned}$$

(A6)

For $\lambda _{L}(x) L^2-\lambda _{R}(x) R^2\ne 0$ we obtain:

$$\begin{aligned} w(x)&=\frac{1}{\displaystyle 1-\frac{\delta (0)}{\displaystyle 1+\frac{m}{b}\cdot \frac{L^2-R^2}{\displaystyle \frac{L^2}{\lambda _{R}(x)}-\frac{R^2}{\lambda _{L}(x)}}}}. \end{aligned}$$

(A7)

In the following, the shape of the function w(x) is characterized. Let $\lambda (0)=\exp (\beta _0)$. For numerator and denominator of w(x) in Eq. (A4) the following limits hold:

$$\begin{aligned}&\lim _{x\searrow 0} \lambda _{R}(x) R^2 \delta _{L}(x)-\lambda _{L}(x) L^2 \delta _{R}(x)=\lambda (0) \delta (0)\cdot (R^2-L^2)<0,\\&\lim _{x\searrow 0} \lambda _{L}(x) L^2\cdot \big (\delta (0)-\delta _{R}(x)\big )-\lambda _{R}(x) R^2\cdot \big (\delta (0)-\delta _{L}(x)\big )=0. \end{aligned}$$

Hence, it follows that $\lim _{x\searrow 0} w(x)=-\infty$. With $L-R<0$ we have:

$$\begin{aligned} \lim _{x\rightarrow \infty } \frac{L^2}{\lambda _{R}(x)}-\frac{R^2}{\lambda _{L}(x)}=\lim _{x\rightarrow \infty } \exp (-\beta _0-Lx)\cdot \left( L^2\cdot \exp \bigl ((L-R) x\bigr )-R^2\right) =-\infty . \end{aligned}$$

We obtain the limit of w(x) for $x\rightarrow \infty$ using the representation in Eq. (A7):

$$\begin{aligned} \lim _{x\rightarrow \infty } w(x)=\frac{1}{1-\delta (0)}=-\frac{b}{m}\cdot \exp (-\beta _0)<0. \end{aligned}$$

(A8)

From Eq. (A6) it follows that $w(x)=1$ if and only if $\lambda _{L}(x) L^2-\lambda _{R}(x) R^2=0$. This equation has a unique solution, which is given by $\tilde{x}=2\cdot \ln (R/L)/(L-R)>0$.

Let g(x) be the numerator and h(x) be the denominator of w(x) in Eq. (A4). With the quotient rule we have $w'(x)=\left( g'(x)h(x)-g(x)h'(x)\right) /h(x)^2$. It can easily be shown that the numerator is given by

$$\begin{aligned} g'(x)h(x)-g(x)h'(x)&=\frac{m}{b}\cdot LR\lambda _{L}(x)\lambda _{R}(x)\delta (0)\cdot (L^2-R^2)\cdot \big (R\lambda _{R}(x)-L\lambda _{L}(x)\big )\\&=c(x)\cdot \big (R\lambda _{R}(x)-L\lambda _{L}(x)\big ) \end{aligned}$$

with $c(x)=\frac{m}{b}\cdot LR\lambda _{L}(x)\lambda _{R}(x)\delta (0)\cdot (L^2-R^2)>0$. Hence, $w'(x)=0$ if and only if $R\lambda _{R}(x)-L\lambda _{L}(x)=0$ holds. As unique solution we obtain $x=\ln (R/L)/(L-R)=\tilde{x}/2$. The zero of $w'(x)$ lies to the left of the point $x=\tilde{x}$, at which the function w(x) has the value 1.

Now, it is shown that $w(\tilde{x}/2)<0$ holds. With the representation of w(x) in Eq. (A7), it follows that $w(\tilde{x}/2)<0$ is equivalent to:

$$\begin{aligned}&\frac{\delta (0)}{\displaystyle 1+\frac{m}{b}\cdot \frac{L^2-R^2}{\displaystyle \frac{L^2}{\lambda _{R}(\tilde{x}/2)}-\frac{R^2}{\lambda _{L}(\tilde{x}/2)}}}>1 \;\;\Leftrightarrow \;\; \delta (0)>1+\frac{m}{b}\cdot \frac{L^2-R^2}{\displaystyle \frac{L^2}{\lambda _{R}(\tilde{x}/2)}-\frac{R^2}{\lambda _{L}(\tilde{x}/2)}}>0\\&\Leftrightarrow \;\; 1>\frac{L^2-R^2}{\displaystyle \frac{L^2}{\exp (R\cdot \tilde{x}/2)}-\frac{R^2}{\exp (L\cdot \tilde{x}/2)}}>-\frac{b}{m}\cdot \exp (-\beta _0). \end{aligned}$$

We have:

$$\begin{aligned} \frac{L^2-R^2}{\displaystyle \frac{L^2}{\exp (R\tilde{x}/2)}-\frac{R^2}{\exp (L\tilde{x}/2)}}&=\frac{L^2-R^2}{\displaystyle \frac{L^2}{\left( \frac{R}{L}\right) ^{\frac{R}{L-R}}}-\frac{R^2}{\left( \frac{R}{L}\right) ^{\frac{L}{L-R}}}}=\left( \frac{R}{L}\right) ^{\frac{R/L}{1-R/L}}\left( 1+\frac{R}{L}\right) . \end{aligned}$$

It can easily be shown that for the function $g(x)=x^{\frac{x}{1-x}}\cdot (1+x)$ we have $0<g(x)<1$ for $x\in (0,1)$. Since $0<R/L<1$, it follows that $w(\tilde{x}/2)<0$.

Let $n\in \mathbb {N}$ and $n\ge 2$. Now, we show that $w'(n\cdot \tilde{x}/2)<0$ holds. First, we consider the denominator. We have:

$$\begin{aligned}&h(n\cdot \tilde{x}/2)=h\Biggl (n\cdot \frac{\ln \bigl (\frac{R}{L}\bigr )}{L-R}\Biggr )\\&\quad =\frac{m}{b}\cdot e^{2\beta _0}\left[ \left( \frac{R}{L}\right) ^{\frac{n L}{L-R}} L^2\left[ 1-\left( \frac{R}{L}\right) ^{\frac{n R}{L-R}}\right] -\left( \frac{R}{L}\right) ^{\frac{n R}{L-R}} R^2\left[ 1-\left( \frac{R}{L}\right) ^{\frac{n L}{L-R}}\right] \right] \\&\quad =\frac{m}{b}\cdot e^{2\beta _0}\cdot \left( \frac{R}{L}\right) ^{\frac{n L}{L-R}}\cdot \left[ L^2\cdot \left( 1-\left( \frac{L}{R}\right) ^{n-2}\right) +(R^2-L^2)\cdot \left( \frac{R}{L}\right) ^{\frac{n\cdot R}{L-R}}\right] . \end{aligned}$$

Since $n\ge 2$, it follows that $1-(L/R)^{n-2}\le 0$. Furthermore, we have $R^2-L^2<0$. Thus it follows that $h(n\cdot \tilde{x}/2)<0$ and $h(n\cdot \tilde{x}/2)^2>0$. Now, it is shown that the numerator is negative:

$$\begin{aligned}&g'(n\cdot \tilde{x}/2)h(n\cdot \tilde{x}/2)-g(n\cdot \tilde{x}/2)h'(n\cdot \tilde{x}/2)\\&\quad =c(n\cdot \tilde{x}/2)\cdot e^{\beta _0}\cdot \left[ R\cdot \left( \frac{R}{L}\right) ^{\frac{n\cdot R}{L-R}}-L\cdot \left( \frac{R}{L}\right) ^{\frac{n\cdot L}{L-R}}\right] \\&\quad =c(n\cdot \tilde{x}/2)\cdot e^{\beta _0}\cdot R\cdot \left( \frac{R}{L}\right) ^{\frac{n\cdot R}{L-R}}\cdot \left[ 1-\left( \frac{R}{L}\right) ^{n-1}\right] . \end{aligned}$$

Since $R<0$ and $R/L<1$, the numerator of $w'(n\cdot \tilde{x}/2)$ is negative and it follows that $w'(n\cdot \tilde{x}/2)<0$ for $n\ge 2$. In particular, for $n=2$ we conclude that w(x) is decreasing at $x=\tilde{x}$.

From the results above we deduce that the function w(x) is strictly monotonically increasing on the interval $(0,\tilde{x}/2)$. At the single extremum at $\tilde{x}/2$ we have $w(\tilde{x}/2)<0$. There is no saddle point at $\tilde{x}/2$, since otherwise an $\hat{x}\in (\tilde{x}/2,\tilde{x})$ with $w(\hat{x})=1$ would exist because of $w'(\tilde{x})<0$. This results in a contradiction, since $w(x)=1$ only holds for $x=\tilde{x}$. Thus, w has a maximum at $\tilde{x}/2$.

Since $w'(\tilde{x})<0$ and $w(x)=1$ holds only for $x=\tilde{x}$, there must be exactly one pole $x_{\text {Pole}}$ in the interval $(\tilde{x}/2,\tilde{x})$, for which $\lim _{x\nearrow x_{\text {Pole}}} w(x)=-\infty$ and $\lim _{x\searrow x_{\text {Pole}}} w(x)=\infty$. There cannot be any poles in the interval $[\tilde{x},\infty )$. Suppose there is a pole $\hat{x}_{\text {Pole}}\in [\tilde{x},\infty )$. The case $\lim _{x\searrow \hat{x}_{\text {Pole}}} w(x)=-\infty$ leads to a contradiction, since $w'(n\cdot \tilde{x}/2)<0$ holds for $n\ge 2$. The case $\lim _{x\searrow \hat{x}_{\text {Pole}}} w(x)=\infty$ leads to a contradiction, since for $x\rightarrow \infty$ the function w(x) converges to a negative value and thus an $\hat{x}\ne \tilde{x}$ with $w(\hat{x})=1$ would exist.

Thus w(x) is strictly monotonically decreasing on the intervals $(\tilde{x}/2,x_{\text {Pole}})$ and $(x_{\text {Pole}},\infty )$. In particular, w(x) has a unique zero $x_0>\tilde{x}$ in the interval $(0,\infty )$. We have $0<w(x)<1$ only for $x\in (\tilde{x},x_0)$. On this interval w(x) is strictly monotonically decreasing and has the representation in Eq. (A7). The function w(x) is sketched in Fig. 5.

Fig. 5

Graph of $w_{L,R}(x)$ for $L<R<0$, where the segment $0\le w_{L,R}(x)\le 1$ is drawn in black

For $0<L<R$ we have $w(x)=w_{L,R}(x)=w_{-R,-L}(-x)$, since $\lambda _{\beta _1}(x)=\lambda _{-\beta _1}(-x)$ holds for all $\beta _1$. $\square$

Proof of Theorem 5.3

The case $R<0$ and $u=0$ is considered. The case $L>0$ and $v=0$ follows analogously.

Let $\lambda (z)=\lambda _{\beta _1}(z)=\exp (\beta _0+\beta _1z)$ and $\delta (z)=\delta _{\beta _1}(z)=1+\frac{m}{b}\cdot \lambda (z)$. The solution of Eq. (18) does not depend on the parameter $\beta _1$. We note that the condition $z^{*}\le \min \left\{ \vert L\vert ,\vert R\vert \right\} \cdot (v-u)=\vert R\vert \cdot v$ ensures that for all $\beta _1\in \left[ L,R\right]$ the point $-z^{*}/\beta _1$ lies in the design region $\mathscr {X}=\left[ 0,v\right]$. The locally D-optimal design $\xi _{\varvec{\beta }}^{*}$ given in Eq. (19) has the support points $x_1^{*}=0$ and $x_{2,\varvec{\beta }}^{*}=-z^{*}/\beta _1$ with weights $w_{1,\varvec{\beta }}^{*}=\sqrt{\smash [b]{\delta (x_{2,\varvec{\beta }}^{*})}}/\big (\smash [b]{\sqrt{\delta (x_1^{*})}}+\sqrt{\smash [b]{\delta (x_{2,\varvec{\beta }}^{*})}}\big )$ and $w_{2,\varvec{\beta }}^{*}=\sqrt{\smash [b]{\delta (x_1^{*})}}/\big (\sqrt{\smash [b]{\delta (x_1^{*})}}+\sqrt{\smash [b]{\delta (x_{2,\varvec{\beta }}^{*})}}\big )$. With Eq. (A1) the determinant of the information matrix for $\xi _{\varvec{\beta }}^{*}$ is given by:

$$\begin{aligned} \det \bigl (\varvec{M}(\xi _{\varvec{\beta }}^{*};\varvec{\beta })\bigr )&=\left( \frac{a}{b}\right) ^2\cdot \frac{w_{1,\varvec{\beta }}^{*}\cdot w_{2,\varvec{\beta }}^{*}\cdot \lambda (0)\cdot \lambda \bigl (-\frac{z^{*}}{\beta _1}\bigr )\cdot (x_{2,\varvec{\beta }}^{*}-x_1^{*})^2}{w_{1,\varvec{\beta }}^{*}\cdot \delta (0)+w_{2,\varvec{\beta }}^{*}\cdot \delta (-z^{*}/\beta _1)}\\&=\left( \frac{a}{b}\right) ^2\cdot \frac{\lambda (0)\cdot \lambda \bigl (-\frac{z^{*}}{\beta _1}\bigr )}{\big (\sqrt{\delta (0)}+\sqrt{\delta (-z^{*}/\beta _1)} \big )^2}\cdot \left( \frac{z^{*}}{\beta _1}\right) ^2. \end{aligned}$$

According to Eq. (A1), for a design $\xi$ with support points $x_1<x_2$ and corresponding weights w and $1-w$ we have:

$$\begin{aligned} \det \bigl (\varvec{M}(\xi ;\varvec{\beta })\bigr )=\left( \frac{a}{b}\right) ^2\cdot \frac{w(1-w)\lambda (x_1)\lambda (x_2)(x_2-x_1)^2}{w\delta (x_1)+(1-w)\delta (x_2)}. \end{aligned}$$

For simplicity of notation, we consider as criterion function the squared efficiency ${{\,\textrm{eff}\,}}_D(\xi ;\varvec{\beta })^2=\det \bigl (\varvec{M}(\xi ;\varvec{\beta })\bigr )/\det \bigl (\varvec{M}(\xi _{\varvec{\beta }}^{*};\varvec{\beta })\bigr )$. For the design $\xi$ the criterion function $g(w,\beta _1,x_1,x_2)={{\,\textrm{eff}\,}}_D(\xi ;\varvec{\beta })^2$ is given by

$$\begin{aligned} g(w,\beta _1,x_1,x_2)&=\frac{w(1-w)\lambda (x_1)\lambda (x_2)(x_2-x_1)^2}{w\delta (x_1)+(1-w)\delta (x_2)}\cdot \frac{\big (\sqrt{\delta (0)}+\sqrt{\delta \bigl (-\frac{z^{*}}{\beta _1}\bigr )} \big )^2\cdot \beta _1^2}{\lambda (0)\cdot \lambda \bigl (-\frac{z^{*}}{\beta _1}\bigr )\cdot (z^{*})^2}\\&=c_1(w,\beta _1,x_2)\cdot (x_2-x_1)^2\left( w\left( \frac{1}{\lambda (x_1)}+\frac{m}{b}\right) +(1-w)\cdot \frac{\delta (x_2)}{\lambda (x_1)}\right) ^{-1} \end{aligned}$$

with a function $c_1(w,\beta _1,x_2)$ not depending on $x_1$. Since $\beta _1<0$, the function $\lambda (x_1)$ is strictly monotonically decreasing in $x_1$. It follows that $g(w,\beta _1,x_1,x_2)$ is also strictly monotonically decreasing and is therefore maximized on the interval $\left[ 0,v\right]$ at $x_1=0$. Hence, the criterion function can be simplified to:

$$\begin{aligned}&\tilde{g}(w,\beta _1,x_2):=g(w,\beta _1,0,x_2)\\ {}&=\frac{w(1-w)\lambda (x_2) x_2^2}{w\delta (0)+(1-w)\delta (x_2)}\cdot \frac{\big (\sqrt{\delta (0)}+\sqrt{\delta \bigl (-\frac{z^{*}}{\beta _1}\bigr )} \big )^2\cdot \beta _1^2}{\lambda \bigl (-\frac{z^{*}}{\beta _1}\bigr )\cdot (z^{*})^2}. \end{aligned}$$

Now, we show that the function $\tilde{g}(w,\beta _1,x_2)$ is unimodal with respect to $\beta _1$ on the interval $\left[ L,R\right]$. The terms $\lambda (-z^{*}/\beta _1)=\exp (\beta _0-z^{*})$, $\delta (0)=1+(m/b)\cdot \exp (\beta _0)$ and $\delta (-z^{*}/\beta _1)=1+(m/b)\cdot \exp (\beta _0-z^{*})$ do not depend on the parameter $\beta _1$. With $\partial \delta (x_2)/\partial \beta _1=(m/b)\cdot x_2\cdot \lambda (x_2)$ we obtain:

$$\begin{aligned} \frac{\partial \tilde{g}(w,\beta _1,x_2)}{\partial \beta _1}&=\left[ (x_2\beta _1+2)\cdot \big (w\delta (0)+(1-w)\delta (x_2)\big )-\lambda (x_2)\beta _1(1-w)\cdot \frac{m}{b}\cdot x_2\right] \\&\;\;\;\;\cdot \frac{w(1-w)\cdot x_2^2\cdot \big (\sqrt{\delta (0)}+\sqrt{\delta \bigl (-\frac{z^{*}}{\beta _1}\bigr )} \big )^2\cdot \lambda \bigl (-\frac{z^{*}}{\beta _1}\bigr )\cdot \beta _1}{\big (w\delta (0)+(1-w)\delta (x_2)\big )^2\cdot \lambda \bigl (-\frac{z^{*}}{\beta _1}\bigr )\cdot (z^{*})^2}. \end{aligned}$$

A straightforward calculation shows that we have $\partial \tilde{g}(w,\beta _1,x_2)/\partial \beta _1=0$ if and only if $\beta _1=0$ or

$$\begin{aligned} \beta _1=\beta _1^{*}=-\frac{1}{x_2}\cdot \left( W\left( \frac{m}{b}\cdot \frac{2\cdot (1-w)}{w\cdot \delta (0)+1-w}\cdot \exp (\beta _0-2)\right) +2\right) , \end{aligned}$$

where W denotes the principal branch of the Lambert W function. We note that $\beta _1^{*}<0$. For $\beta _1=0$ we have $\tilde{g}(w,0,x_2)=0$ and for $\beta _1<0$ we have $\tilde{g}(w,\beta _1,x_2)>0$. Since $x_2>0$, we have $\lim _{\beta _1\rightarrow -\infty } \lambda (x_2)\beta _1^2=0$ and $\lim _{\beta _1\rightarrow -\infty } \delta (x_2)=1$ and hence $\lim _{\beta _1\rightarrow -\infty } \tilde{g}(w,\beta _1,x_2)=0$. Thus $\tilde{g}(w,\beta _1,x_2)$ has a unique maximum at $\beta _1=\beta _1^{*}$ and for $\beta _1<0$ there exist no further extrema. Hence, $\tilde{g}(w,\beta _1,x_2)$ is unimodal with respect to $\beta _1\le 0$ and attains its minima in the set $\left\{ L,R\right\}$.

Dette et al. (2007) showed that a standardized maximin optimal design $\xi ^{*}$ attains its minimal efficiency at least at two different values of $\varvec{\beta }$. It follows that $\tilde{g}(w,L,x_2)=\tilde{g}(w,R,x_2)$ must hold for $\xi ^{*}$. For $0<w<1$ this equation is equivalent to

$$\begin{aligned} \frac{\lambda _{L}(x_2) L^2 }{w\delta (0)+(1-w)\delta _{L}(x_2)}=\frac{\lambda _{R}(x_2) R^2 }{w\delta (0)+(1-w)\delta _{R}(x_2)}. \end{aligned}$$

Solving this equation for w, we obtain the solution

$$\begin{aligned} w(x_2)=\frac{\lambda _{R}(x_2) R^2\delta _{L}(x_2)-\lambda _{L}(x_2) L^2\delta _{R}(x_2)}{\lambda _{L}(x_2) L^2\cdot \big (\delta (0)-\delta _{R}(x_2)\big )-\lambda _{R}(x_2) R^2\cdot \big (\delta (0)-\delta _{L}(x_2)\big )}, \end{aligned}$$

provided that $x_2$ is chosen such that the denominator is not equal to zero. By Lemma A.1 the function $w(x_2)$ has values in the interval (0, 1) only for $\tilde{x}<x_2<x_0$, where $\tilde{x}=2\cdot \ln (R/L)/(L-R)$ and $x_0>\tilde{x}$ is the unique zero of $w(x_2)$ in the interval $(0,\infty )$. We have $w(x_2)=0$ if and only if $\lambda _{R}(x_2)\cdot R^2\cdot \delta _{L}(x_2)-\lambda _{L}(x_2)\cdot L^2\cdot \delta _{R}(x_2)=0$ holds. Rewriting this equation and dividing by $\exp (\beta _0)$ yields the equivalent condition $h(x)=0$, where h is given in Eq. (21). For $x_2\in (\tilde{x},x_0)$, according to Lemma A.1, the function $w(x_2)$ has the representation in Eq. (20).

The function $\tilde{g}(w(x_2),L,x_2)=\tilde{g}(w(x_2),R,x_2)$ must be maximized with respect to $x_2\in (\tilde{x},x_0)$ in order to obtain the standardized maximin D-optimal design. Equivalently, omitting terms not depending on $x_2$, we have to maximize:

$$\begin{aligned}&\frac{w(x_2)(1-w(x_2))\lambda _{L}(x_2)x_2^2}{w(x_2)\delta (0)+(1-w(x_2))\delta _{L}(x_2)}\\&=\frac{w(x_2)(1-w(x_2))x_2^2}{\frac{m}{b}\cdot w(x_2)(\exp (-Lx_2)-1)+\exp (-\beta _0-Lx_2)+\frac{m}{b}}. \end{aligned}$$

Since $w(\tilde{x})=1$ and $w(x_0)=0$, this function is equal to zero for $x_2=\tilde{x}$ and $x_2=x_0$. Hence, there exists a maximum at some $x^{*}\in (\tilde{x},x_0)$. If $x^{*}\in \mathscr {X}$, then the design with support points 0 and $x^{*}$ and corresponding weights $w(x^{*})$ and $1-w(x^{*})$ is standardized maximin D-optimal in the class of 2-point designs. $\square$

Proof of Theorem 5.6

The proof is analogous to that of Theorem 5.3, but has to be adjusted at some points. We consider the case $R<0$ and $u=0$. Let $z^{*}=2\cdot (1+W(e^{-1}))$. For $\varvec{\beta }=(\beta _0,\beta _1)^T$ the locally c-optimal design for $\varvec{c}=(0,1)^T$ in the Poisson model has the support points $x_1^{*}=0$ and $x_2^{*}=-z^{*}/\beta _1$ and weights $w_1^{*}=\sqrt{\exp (\beta _0-z^{*})}/\big (\sqrt{\exp (\beta _0)}+\sqrt{\exp (\beta _0-z^{*})}\big )$ and $w_2^{*}=1-w_1^{*}$ (cf. Ford et al. 1992; Schmidt 2019). This design is the limit of the locally D-optimal design for the Poisson–Gamma model given in Eq. (19) for $b\rightarrow 0$.

The function w(x) in Eq. (24) is the limit of the function in Eq. (A5) for $b\rightarrow 0$. The properties of w(x) result analogously to the proof of Lemma A.1, but w(x) does not have a zero in the interval $(0,\infty )$. In contrast to Eq. (A8), we have $\lim _{x\rightarrow \infty } w(x)=0$. Since w(x) is strictly monotonically decreasing on the interval $I=(\tilde{x},\infty )$, we have $0<w(x)<1$ for $x\in I$. The limit of the function $\tilde{g}(w(x),L,x)$ from the proof of Theorem 5.3 for $b\rightarrow 0$ must be maximized with respect to $x\in I$:

$$\begin{aligned} \lim _{b\rightarrow 0} \tilde{g}(w(x),L,x)=\frac{w(x)\cdot (1-w(x))\cdot x^2}{w(x)\cdot \left( \exp (-Lx)-1\right) +1}\cdot \frac{\big (1+\sqrt{\exp (-z^{*})} \big )^2\cdot L^2}{\exp (-z^{*})\cdot (z^{*})^2}. \end{aligned}$$

Since the second fraction does not depend on x, it can be omitted for the maximization. $\square$