Elfving’s theorem for R-optimality of experimental designs

Abstract

The present paper is devoted to the construction of R-optimal designs in multiresponse linear models. The R-optimality criterion introduced by Dette (J R Stat Soc Ser B 59:97–110, 1997) minimizes the volume of Bonferroni rectangular confidence region for the parameter estimation. A generalization of Elfving’s theorem is proved for the optimal designs with respect to R-optimality, which gives a geometric characterization of R-optimal designs. The geometric characterizations of the R-optimal designs are illustrated by four examples.

Appendix

1.1 Proof of Theorem 2

Here we present the proof by using similar arguments in Dette (1993) and invoking Theorem 1.

Let $\xi ^*=\bigg \{\begin{array}{c} {\varvec{x}}_v \\ w_v \end{array} \bigg \}_{v=1}^s$ denote an R-optimal design for the model (1). By Theorem 1 we have

$$\begin{aligned} \phi ({\varvec{x}},\xi ^*)= & {} \text{ tr }\left\{ {{\varvec{M}}}^{-1}(\xi ^*){{\varvec{F}}}^T({{\varvec{x}}}){\varvec{\Sigma }}^{-1} {{\varvec{F}}}({{\varvec{x}}}){{\varvec{M}}}^{-1}(\xi ^*) \right. \nonumber \\&\quad \left. \left( \sum _{i=1}^p\displaystyle \frac{{{\varvec{e}}}_i{{\varvec{e}}}_i^T}{{{\varvec{e}}}_i^T{{\varvec{M}}}^{-1}(\xi ^*){{\varvec{e}}}_i} \right) \right\} \leqslant p \end{aligned}$$

(18)

for all ${\varvec{x}}\in \mathcal {X}$, and

$$\begin{aligned} \phi ({\varvec{x}}_v,\xi ^*)=p \end{aligned}$$

(19)

for all $v=1,\ldots ,s$. Letting $\gamma _{i}^{-2}=p{\varvec{e}}_{i}^T{\varvec{M}}^{-1}(\xi ^*){\varvec{e}}_{i}, i=1, \ldots , p$ and ${\varvec{D}}={\varvec{M}}^{-1}(\xi ^*){\varvec{\Gamma }}$, it follows that

$$\begin{aligned}{\varvec{\Gamma }}={\varvec{M}}(\xi ^*){\varvec{D}}=\sum _{v=1}^{s}w_v{\varvec{F}}^T{({\varvec{x}}_v)}{\varvec{\Sigma }}^{-1/2}{\varvec{K}}_{v}, \end{aligned}$$

where ${\varvec{K}}_{v}={\varvec{\Sigma }}^{-1/2}{\varvec{F}}({\varvec{x}}_v){\varvec{D}}$, $v=1,\ldots ,s$. This proves the representation given in (a). Moreover, from these definitions and the fact in (19) we have

$$\begin{aligned} \parallel {\varvec{K}}_{v}\parallel ^2= & {} \text{ tr }\left\{ {\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}_v){\varvec{D}}\right\} \nonumber \\= & {} \text{ tr }\{{\varvec{\Gamma }}{\varvec{M}}^{-1}(\xi ^*){\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}_v){\varvec{M}}^{-1}(\xi ^*){\varvec{\Gamma }}\} \nonumber \\= & {} \text{ tr }\{{\varvec{M}}^{-1}(\xi ^*){\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}_v){\varvec{M}}^{-1}(\xi ^*){\varvec{\Gamma }}{\varvec{\Gamma }}\} \nonumber \\= & {} 1. \end{aligned}$$

(20)

From the inequality (18) and the Cauchy–Schwarz inequality we get

$$\begin{aligned} \left( \text{ tr }\{{\varvec{D}}^T{\varvec{F}}^T{({\varvec{x}})}{\varvec{\Sigma }}^{-1/2}{\varvec{K}}\}\right) ^2\leqslant \text{ tr }\{{\varvec{K}}^T{\varvec{K}}\}\text{ tr }\{{\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}){\varvec{\Sigma }}^{-1/2}{\varvec{\Sigma }}^{-1/2}{\varvec{F}}({\varvec{x}}){\varvec{D}}\}\leqslant 1 \end{aligned}$$

for all ${\varvec{x}}\in \mathcal {X}$, whenever the matrix ${\varvec{K}}$ satisfies the equation $\parallel {\varvec{K}}\parallel =1$. Observing (20) it is now easy to see that the point ${\varvec{\Gamma }}$ is a boundary point of the set $\mathcal {R}$ with supporting hyperplane ${\varvec{D}}$ which proves (b). Finally, the condition (c) follows readily from the definitions of ${\varvec{\Gamma }}$ and ${\varvec{D}}$.

To prove sufficiency, we let ${\varvec{D}}\in \mathbb {R}^{p\times p}$ denote a supporting hyperplane to the set $\mathcal {R}_{p,r}$ at the boundary point ${\varvec{\Gamma }}$. Thus we have for all ${\varvec{x}}\in \mathcal {X}$ and ${\varvec{K}}$ satisfying $\parallel {\varvec{K}}\parallel =1,$

$$\begin{aligned} \left| \text{ tr }\{{\varvec{D}}^T{\varvec{F}}^T{({\varvec{x}})}{\varvec{\Sigma }}^{-1/2}{\varvec{K}}\}\right| \leqslant 1. \end{aligned}$$

(21)

Defining ${\varvec{K}}({\varvec{x}})={\varvec{\Sigma }}^{-1/2}{\varvec{F}}({\varvec{x}}){\varvec{D}}/\sqrt{\text{ tr }\{{\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}){\varvec{D}}\}}$, we observe from (21) that

$$\begin{aligned} \text{ tr }\{{\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}){\varvec{D}}\}\leqslant 1 \quad {\text {for}}~ {\text {all}}\;\;{\varvec{x}}\in \mathcal {X}. \end{aligned}$$

(22)

Because ${\varvec{D}}$ is a supporting hyperplane to $\mathcal {R}_{p,r}$ at the boundary point ${\varvec{\Gamma }}$ we obtain from (21) (used at ${\varvec{x}}={\varvec{x}}_v$) and the representation (a)

$$\begin{aligned} 1=\text{ tr }\{{\varvec{D}}^T{\varvec{\Gamma }}\} =\sum _{v=1}^{s}w_v\text{ tr }\left\{ {\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1/2}{\varvec{K}}_{v}\right\} \leqslant 1 \end{aligned}$$

and this implies $\text{ tr }\left\{ {\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1/2}{\varvec{K}}_{v}\right\} =1, v=1,\ldots ,s$. By an application of the Cauchy–Schwarz inequality we now get for $v=1,\ldots ,s$

$$\begin{aligned} 1= & {} \left( \text{ tr }\left\{ {\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1/2}{\varvec{K}}_{v}\right\} \right) ^2 \nonumber \\\leqslant & {} \text{ tr }\{{\varvec{K}}_v^T{\varvec{K}}_v\}\text{ tr }\left\{ {\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1}{\varvec{F}}{({\varvec{x}}_v)}{\varvec{D}}\right\} \leqslant 1, \end{aligned}$$

(23)

where the last inequality results from (22). Therefore, we have ${\varvec{K}}_{v}=\lambda _v{\varvec{\Sigma }}^{-1/2}{\varvec{F}}({\varvec{x}}_v){\varvec{D}}$ for some $\lambda _v \in \mathbb {R}, v=1,\ldots ,s.$ From the normalizing conditions on the ${\varvec{K}}_{v}$ we thus obtain

$$\begin{aligned} 1=\parallel {\varvec{K}}_{v}\parallel ^2 =\lambda _v^2\text{ tr }\left\{ {\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}_v){\varvec{D}}\right\} =\lambda _v^2, \quad v=1,\ldots ,s. \end{aligned}$$

(24)

On the other hand, we have from the property that ${\varvec{\Gamma }}$ is a boundary point of $\mathcal {R}_{p,r}$ with supporting hyperplane ${\varvec{D}}$

$$\begin{aligned} 1= & {} \text{ tr }\{{\varvec{D}}^T{\varvec{\Gamma }}\}=\sum \limits _{v=1}^{s}w_v\text{ tr }\left\{ {\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1/2}{\varvec{K}}_{v}\right\} \\= & {} \sum \limits _{v=1}^{s}w_v\lambda _v\text{ tr }\left\{ {\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}_v){\varvec{D}}\right\} =\sum \limits _{v=1}^{s}w_v\lambda _v. \end{aligned}$$

It follows from (24) and noting $w_v> 0$ and $\sum _{v=1}^{s}w_v=1$ that $\lambda _v=1$ which implies ${\varvec{K}}_{v}={\varvec{\Sigma }}^{-1/2}{\varvec{F}}({\varvec{x}}_v){\varvec{D}}, v=1,\ldots ,s.$ From this representation we finally obtain

$$\begin{aligned} {\varvec{\Gamma }}=\sum _{v=1}^{s}w_v{\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1/2}{\varvec{K}}_{v} =\sum _{v=1}^{s}w_v{\varvec{F}}^T({\varvec{x}}_v){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}_v){\varvec{D}}={\varvec{M}}(\xi ^*){\varvec{D}}. \end{aligned}$$

Observing condition (c) it follows

$$\begin{aligned} \displaystyle \frac{1}{p}={\varvec{e}}_{i}^T{\varvec{D}}^T{\varvec{\Gamma }}{\varvec{e}}_{i} ={\varvec{e}}_{i}^T{\varvec{\Gamma }}^T{\varvec{M}}^{-1}(\xi ^*){\varvec{\Gamma }}{\varvec{e}}_{i} =\gamma _{i}^2{\varvec{e}}_{i}^T{\varvec{M}}^{-1}(\xi ^*){\varvec{e}}_{i} \end{aligned}$$

and the inequality (22) yields

$$\begin{aligned}&\text{ tr }\{{\varvec{D}}^T{\varvec{F}}^T({\varvec{x}}){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}){\varvec{D}}\} \\&\quad =\text{ tr }\{{\varvec{\Gamma }}^T{{\varvec{M}}}^{-1}(\xi ^*){\varvec{F}}^T({\varvec{x}}){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}){{\varvec{M}}}^{-1}(\xi ^*){\varvec{\Gamma }}\}\\&\quad =\text{ tr }\{{{\varvec{M}}}^{-1}(\xi ^*){\varvec{F}}^T({\varvec{x}}){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}){{\varvec{M}}}^{-1}(\xi ^*){\varvec{\Gamma }}{\varvec{\Gamma }}^T\}\\&\quad =p\,\text{ tr }\left\{ {{\varvec{M}}}^{-1}(\xi ^*){\varvec{F}}^T({\varvec{x}}){\varvec{\Sigma }}^{-1}{\varvec{F}}({\varvec{x}}){{\varvec{M}}}^{-1}(\xi ^*) \sum \limits _{i=1}^{p}\displaystyle \frac{{\varvec{e}}_{i}{\varvec{e}}_{i}^T}{{\varvec{e}}_{i}^T{M}^{-1}(\xi ^*){\varvec{e}}_{i}}\right\} \\&\quad \leqslant p \end{aligned}$$

for all ${\varvec{x}}\in \mathcal {X}$. By Theorem 1 it then immediately follows that the design $\xi ^*$ is R-optimal for the model (1), which completes the proof of Theorem 2.

1.2 Proof of the result in (15)

By $\mathcal {S}$ denote the set in the right side of (15), i.e.,

$$\begin{aligned} \mathcal {S}=\left\{ \left( \begin{array}{cc} h_1 &{}\quad h_2\\ h_3 &{}\quad h_4\end{array} \right) \bigg | h_1^2+h_2^2\leqslant 1,\sqrt{h_3^2+h_4^2}\leqslant 1-\sqrt{h_1^2+h_2^2}\right\} . \end{aligned}$$

We first show that $\mathcal {S}$ is convex and then show that $\mathcal {S}=\text {conv}(\mathcal {H})$ where $\mathcal {H}$ is defined by (14).

For any ${\varvec{U}}=\left( \begin{array}{cc} u_1 &{}\quad u_2\\ u_3 &{}\quad u_4\end{array} \right) \in \mathcal {S}$ and ${\varvec{V}}=\left( \begin{array}{cc} v_1 &{}\quad v_2\\ v_3 &{}\quad v_4\end{array} \right) \in \mathcal {S}$, we have

$$\begin{aligned}\sqrt{u_1^2+u_2^2}+\sqrt{u_3^2+u_4^2}\leqslant 1\end{aligned}$$

and

$$\begin{aligned}\sqrt{v_1^2+v_2^2}+\sqrt{v_3^2+v_4^2}\leqslant 1.\end{aligned}$$

Thus, for any $t\in [0,1]$, ${\varvec{W}}=\left( \begin{array}{cc} w_1 &{}\quad w_2\\ w_3 &{}\quad w_4\end{array} \right) =(1-t){\varvec{U}}+t{\varvec{V}}\in \mathcal {S}$, which follows from

$$\begin{aligned}&\sqrt{w_1^2+w_2^2}+\sqrt{w_3^2+w_4^2}\\&\quad \leqslant \left[ (1-t)\sqrt{u_1^2+u_2^2}+t\sqrt{v_1^2+v_2^2}\right] +\left[ (1-t)\sqrt{u_3^2+u_4^2}+t\sqrt{v_3^2+v_4^2}\right] \\&\quad =(1-t)\left[ \sqrt{u_1^2+u_2^2}+\sqrt{u_3^2+u_4^2}\right] +t\left[ \sqrt{v_1^2+v_2^2}+\sqrt{v_3^2+v_4^2}\right] \\&\quad \leqslant 1. \end{aligned}$$

Therefore, we conclude that $\mathcal {S}$ is a convex set. Next, we prove that $\text {conv}(\mathcal {H})=\mathcal {S}$.

Let us replace ${\varvec{\epsilon }}$ by ${\varvec{\epsilon }}=(\epsilon _{1}, \epsilon _{2})^T.$ Then

$$\begin{aligned} \mathcal {H}= & {} \left\{ \left( \begin{array}{cc} h_1 &{}\quad h_2\\ h_3 &{}\quad h_4\end{array}\right) =\left( \begin{array}{cc}(1-x)\epsilon _1 &{}\quad (1-x)\epsilon _2 \\ x^2\epsilon _1 &{}\quad x^2\epsilon _2\end{array} \right) \big | 0\leqslant x\leqslant 1, \; \epsilon _1^2+\epsilon _{2}^2=1\right\} \\= & {} \left\{ \left( \begin{array}{cc} h_1 &{}\quad h_2\\ h_3 &{}\quad h_4\end{array}\right) \big | h_1^2+h_2^2= (1-x)^2,\; h_3^2+h_4^2=x^4,\; 0\leqslant x\leqslant 1\right\} . \end{aligned}$$

Define

$$\begin{aligned}\mathcal {H}_1=\left\{ \left( \begin{array}{cc} h_1 &{}\quad h_2\\ h_3 &{}\quad h_4\end{array}\right) \big |h_1^2+h_2^2=1, h_3=h_4=0\right\} \end{aligned}$$

and

$$\begin{aligned} \mathcal {H}_2=\left\{ \left( \begin{array}{cc} h_1 &{}\quad h_2\\ h_3 &{}\quad h_4\end{array}\right) \big |h_3^2+h_4^2=1, h_1=h_2=0\right\} . \end{aligned}$$

It is to be noted that $\mathcal {H}_1$ and $\mathcal {H}_2$ are subsets of $\mathcal {H}$ corresponding to $x=0$ and $x=1$, respectively. Moreover, it is easy to check that

$$\begin{aligned}\text {conv}(\mathcal {H}_1)=\left\{ \left( \begin{array}{cc} h_1 &{}\quad h_2\\ h_3 &{}\quad h_4\end{array}\right) \big |h_1^2+h_2^2\leqslant 1, h_3=h_4=0\right\} \end{aligned}$$

and

$$\begin{aligned}\text {conv}(\mathcal {H}_2)=\left\{ \left( \begin{array}{cc} h_1 &{}\quad h_2\\ h_3 &{}\quad h_4\end{array}\right) \big |h_3^2+h_4^2\leqslant 1, h_1=h_2=0\right\} .\end{aligned}$$

In what follows, we shall prove both $\text {conv}(\mathcal {H})\subset \mathcal {S}$ and $\mathcal {S}\subset \text {conv}(\mathcal {H})$ hold, respectively. First, for any ${\varvec{H}}=\left( \begin{array}{cc} h_1 &{}\quad h_2\\ h_3 &{}\quad h_4\end{array}\right) \in \mathcal {H}$, we have

$$\begin{aligned}0\leqslant h_1^2+h_2^2\leqslant 1,\quad \sqrt{h_1^2+h_2^2}+\sqrt{h_3^2+h_4^2}=(1-x)+x^2\leqslant 1, \end{aligned}$$

which implies ${\varvec{H}}\in \mathcal {S}$. It means that $\mathcal {H}\subset \mathcal {S}$ and hence $\text {conv}(\mathcal {H})\subset \text {conv}(\mathcal {S})=\mathcal {S}$. Secondly, we will prove that

• (i) $\mathcal {S}\subset \text {conv}\left( \text {conv}(\mathcal {H}_1)\cup \text {conv}(\mathcal {H}_2)\right)$, and

• (ii) $\text {conv}\left( \text {conv}(\mathcal {H}_1)\cup \text {conv}(\mathcal {H}_2)\right) \subset \text {conv}(\mathcal {H})$.

Let ${\varvec{H}}=\left( \begin{array}{cc} h_1 &{}\quad h_2\\ h_3 &{}\quad h_4\end{array}\right)$ be any element of $\mathcal {S}$. It is obvious that if ${\varvec{H}}$ is a null matrix then ${\varvec{H}}\in \mathcal {H}$. For the non-null ${\varvec{H}}$, we let

$$\begin{aligned}z=\displaystyle \frac{\sqrt{h_1^2+h_2^2}}{\sqrt{h_1^2+h_2^2}+\sqrt{h_3^2+h_4^2}},\end{aligned}$$

and then we can rewrite ${\varvec{H}}$ as

$$\begin{aligned} {\varvec{H}}= & {} z\left( \begin{array}{cc}h_1/z &{}\quad h_2/z\\ 0 &{}\quad 0\end{array}\right) +(1-z)\left( \begin{array}{cc}0 &{} 0 \\ h_3/(1-z) &{} h_4/(1-z)\end{array}\right) \\= & {} z{\varvec{U}}+(1-z){\varvec{V}},\quad \text{ say }. \end{aligned}$$

Note that

$$\begin{aligned}\Vert {\varvec{U}}\Vert =\displaystyle \frac{\sqrt{h_1^2+h_2^2}}{z}=\sqrt{h_1^2+h_2^2}+\sqrt{h_3^2+h_4^2}\leqslant 1\end{aligned}$$

and

$$\begin{aligned}\Vert {\varvec{V}}\Vert =\displaystyle \frac{\sqrt{h_3^2+h_4^2}}{1-z}=\sqrt{h_1^2+h_2^2}+\sqrt{h_3^2+h_4^2}\leqslant 1.\end{aligned}$$

Thus ${\varvec{U}}\in \text {conv}(\mathcal {H}_1)$ and ${\varvec{V}}\in \cup \text {conv}(\mathcal {H}_2)$. This means that ${\varvec{H}}\in \text {conv}\left( \text {conv}(\mathcal {H}_1) \text {conv}(\mathcal {H}_2)\right)$ and hence (i) holds.

As the proof of (ii), noting that $\mathcal {H}_1\subset \mathcal {H}$ and $\mathcal {H}_2\subset \mathcal {H}$, it immediately follows that $\text {conv}(\mathcal {H}_1)\subset \text {conv}(\mathcal {H})$ and $\text {conv}(\mathcal {H}_2)\subset \text {conv}(\mathcal {H}).$ We then have $(\text {conv}(\mathcal {H}_1)\cup \text {conv}(\mathcal {H}_2))\subset \text {conv}(\mathcal {H})$ which implies that $\text {conv}\left( \text {conv}(\mathcal {H}_1)\cup \text {conv}(\mathcal {H}_2)\right) \subset \text {conv}(\text {conv}(\mathcal {H}))=\text {conv}(\mathcal {H})$. This completes the proof of the result in (15).