Exact $D$-optimal designs for first-order trigonometric regression models on a partial circle

Abstract

Recently, various approximate design problems for low-degree trigonometric regression models on a partial circle have been solved. In this paper we consider approximate and exact optimal design problems for first-order trigonometric regression models without intercept on a partial circle. We investigate the intricate geometry of the non-convex exact trigonometric moment set and provide characterizations of its boundary. Building on these results we obtain a solution of the exact $D$-optimal design problem. It is shown that the structure of the optimal designs depends on both the length of the design interval and the number of observations.

Appendix

Proof of Theorem 4.1 (a)

Part (a) of Theorem 4.1 is an immediate consequence of Theorem 2.1, since the approximate optimal designs are already exact $n$-point designs when $n$ is even. $\square$

The following elementary lemma highlights a conclusion that can be drawn for three integers when their sum is odd and which may be false when the sum is even. It is needed several times in the proof of Theorem 4.1 (b).

Lemma 5.1

Suppose that $j_1, j_2, j_3$ are integers and $j_1+j_2+j_3$ is odd. If $j_2> |j_1-j_3|-1$, then $j_2\ge |j_1-j_3|+1$.

Proof

Suppose that $j_2$ is even. Then $j_1+j_3$ is odd, and so is $j_1-j_3$. Therefore $j_2\ne |j_1-j_3|$, and the assertion follows. The argument is similar when $j_2$ is odd. $\square$

Proof of Theorem 4.1 (b)

Let $n=2k+1$.

• (i) Let $0<a<\frac{1}{4}\pi$. Let $\xi _n$ be any $D$-optimal exact $n$-observation design, and let

  $$\begin{aligned} \xi _n^{*}:= \begin{Bmatrix}-a&a \\ k/n&(k+1)/n\end{Bmatrix} , \qquad \xi _n^{**}:= \begin{Bmatrix}-a&a \\ (k+1)/n&k/n\end{Bmatrix}. \end{aligned}$$

  Recall from Sect. 3 that $\mathbf{c}(\xi _n)$ is the moment point closest to the origin. Thus

  $$\begin{aligned} c_1^2(\xi _n)+c_2^2(\xi _n)\!=\! \Vert \mathbf{c}(\xi _n)\Vert ^2 \le \Vert \mathbf{c}(\xi _n^{*})\Vert ^2 =\Vert \mathbf{c}(\xi _n^{**})\Vert ^2 \!=\!\cos ^2(2a)+\frac{1}{n^2}\sin ^2(2a).\nonumber \\ \end{aligned}$$

  (5.1)

  Since $c_1(\xi _n)\ge \cos (2a)>0$, it follows that

  $$\begin{aligned} |c_2(\xi _n)|\le \frac{1}{n}\sin (2a). \end{aligned}$$

  (5.2)

  Therefore, the function $h$ defined by (3.3) and (3.2) satisfies

  $$\begin{aligned} h(c_2(\xi _n))=\frac{n-1}{n}\cos (2a)+\frac{1}{n} \sqrt{1-n^2c_2^2(\xi _n)}. \end{aligned}$$

  Hence, by Lemma 3.1,

  $$\begin{aligned} \Vert \mathbf{c}(\xi _n)\Vert ^2\ge h^2(c_2(\xi _n))+c_2^2(\xi _n) =\frac{1}{n^2}\psi (n^2c_2^2(\xi _n)), \end{aligned}$$

  (5.3)

  where

  $$\begin{aligned} \psi (u):=\left\{ (n-1)\cos (2a)+\sqrt{1-u} \right\} ^2+u, \quad 0\le u<1. \end{aligned}$$

  But

  $$\begin{aligned} \psi ^{\prime }(u)=-\frac{(n-1)\cos (2a)}{\sqrt{1-u}}<0, \quad 0\le u<1, \end{aligned}$$

  so that $\psi$ is strictly decreasing. Hence, by (5.2),

  $$\begin{aligned} \frac{1}{n^2}\psi (n^2c_2^2(\xi _n))\ge \frac{1}{n^2}\psi (\sin ^2(2a)) =\cos ^2(2a)+\frac{1}{n^2}\sin ^2(2a). \end{aligned}$$

  (5.4)

  Combining (5.1), (5.3) and (5.4), one obtains that none of the inequalities there can be strict. Equality in (5.3) implies that $c_1(\xi _n)=h(c_2(\xi _n))$, and equality in (5.4) implies that $|c_2(\xi _n)|=\frac{1}{n}\sin (2a)$. Thus, $c_1(\xi _n)=\cos (2a)$, and it follows that all the design points of $\xi _n$ belong to $\{-a, a\}$. The equation $|c_2(\xi _n)|=\frac{1}{n}\sin (2a)$ now yields that $\xi _n=\xi _n^*$ or $\xi _n=\xi _n^{**}$.

• (ii) If $a=\frac{1}{4}\pi$, the same arguments as in (i) show that $\xi _n^*$ and $\xi _n^{**}$ are $D$-optimal. However, in this case, the function $\psi$ is not strictly decreasing but constant, so that there may be more optimal designs. That all designs in the class given in Theorem 4.1 (b) (ii) are as good as $\xi _n^*$ is easily verified. That there are no other $D$-optimal designs can be shown using Lemma 3.1 (a). We omit the details.

• (iii) Let $\frac{1}{4}\pi <a<a_n^*$. It will first be shown that the moment set $\mathcal{C}_n$ does not contain the point $\mathbf{0}$. In view of (3.1), this implies that the moment vector of the optimal design is a boundary point of $\mathcal{C}_n$. Let

  $$\begin{aligned} \gamma :=\min \left\{ c_1(\xi _n) :\ c_2(\xi _n)=0\right\} \!, \end{aligned}$$

  and suppose the minimum is attained at the design $\zeta _n$ with $c_2(\zeta _n)=0$. We will show that $\gamma >0$. Obviously, $\mathbf{c}(\zeta _n)$ cannot be an interior point of $\mathcal{C}_n$, and so, by Lemma 3.2, $\zeta _n$ is of the form

  $$\begin{aligned} \zeta _n=\begin{Bmatrix}-a&\eta&a\\ j_1/n&j_2/n&j_3/n\end{Bmatrix}. \end{aligned}$$

  Note that $\zeta _n$ cannot be of the form (3.6) since $c_2(\zeta _n)=0$. We have

  $$\begin{aligned} \gamma&=c_1(\zeta _n)=\frac{j_1+j_3}{n}\cos (2a)+\frac{j_2}{n}\cos (2\eta ), \end{aligned}$$

  (5.5)
  $$\begin{aligned} 0&=c_2(\zeta _n)=\frac{-j_1+j_3}{n}\sin (2a)+\frac{j_2}{n}\sin (2\eta ). \end{aligned}$$

  (5.6)

  Reflecting $\zeta _n$ at the origin if necessary, we may assume that $j_1\ge j_3$. It follows from (5.6) that $0\le (-j_1+j_3)\sin (2a)+j_2$. Also, since $\frac{1}{4}\pi <a\le a_n^*$, $\sin (2a)>1-\cos ^2(2a)\ge 1-1/(4k^2)$. Consequently,

  $$\begin{aligned} j_2\!\ge \! (j_1-j_3)\sin (2a)\!\ge \! (j_1-j_3)\left(1-\frac{1}{4k^2} \right) \!\ge \! j_1-j_3-\frac{2k+1}{4k^2}> j_1-j_3-1. \end{aligned}$$

  Hence, by Lemma 5.1,

  $$\begin{aligned} j_2\ge j_1-j_3+1. \end{aligned}$$

  (5.7)

  As $j_1\ge j_3$, it now follows from (5.6) that $\eta \ge 0$. If $\eta \in [\frac{1}{4}\pi ,a)$, then $\sin (2\eta )\ge \sin (2a)$, so that, by (5.6), $0\ge (-j_1+j_3+j_2)\sin (2a)$, contradicting (5.7). Thus $\eta \in [0,\frac{1}{4}\pi )$, and so $\cos (2\eta )>0$. Therefore, by (5.5) and (5.6),

  $$\begin{aligned} n\gamma \!&= \! (j_1+j_3)\cos (2a)+j_2\sqrt{1-\sin ^2(2\eta )}\\ \quad&= (j_1+j_3)\cos (2a)+\sqrt{j_2^2-(j_1-j_3)^2\sin ^2(2a)}. \end{aligned}$$

  This expression is positive if and only if

  $$\begin{aligned} j_2^2>(j_1-j_3)^2+4j_1j_3\cos ^2(2a). \end{aligned}$$

  But this inequality follows from (5.7) and the fact that $4j_1j_3\cos ^2(2a)< 4k^2\cos ^2(2a_n^*)=1$. Hence $\gamma >0$. It is thus established that $\mathbf{0}\not \in \mathcal{C}_n$. Now let $\xi _n$ be $D$-optimal and

  $$\begin{aligned} \xi _n^*= \begin{Bmatrix}-a&0&a\\ k/n&1/n&k/n\end{Bmatrix}. \end{aligned}$$

  According to (3.1), $\mathbf{c}(\xi _n)$ has minimum norm in $\mathcal{C}_n$. In particular,

  $$\begin{aligned} \Vert \mathbf{c}(\xi _n)\Vert \le \Vert \mathbf{c}(\xi _n^*)\Vert = \frac{1+2k\cos (2a)}{n}<\frac{1}{n} \end{aligned}$$

  (5.8)

  and $\mathbf{c}(\xi _n)$ must be a boundary point of $\mathcal{C}_n$. Using (3.7) and (5.8), one may show that $\xi _n$ cannot have the form (3.6). It therefore follows by Lemma 3.2 that $\xi _n$ is of the form (3.5)

  $$\begin{aligned} \xi _n=\begin{Bmatrix}-a&\theta&a\\ k_1/n&k_2/n&k_3/n\end{Bmatrix}. \end{aligned}$$

  Observe that this implies that $\mathbf{c}(\xi _n)$ lies on the circle with center

  $$\begin{aligned} \overline{\mathbf{c}}=\frac{1}{n} \begin{pmatrix}(k_1+k_3)\cos (2a)\\ (-k_1+k_3)\sin (2a)\end{pmatrix} \end{aligned}$$

  and radius $k_2/n$. If $k_2\le |k_1-k_3|-1$, then

  $$\begin{aligned} \Vert \mathbf{c}(\xi _n)\Vert&\ge \Vert \overline{\mathbf{c}}\Vert -\frac{k_2}{n} = \frac{1}{n}\sqrt{(k_1-k_3)^2+4k_1k_3\cos ^2(2a)}-\frac{k_2}{n}\\&\ge \frac{1}{n}|k_1-k_3| -\frac{1}{n}(|k_1-k_3|-1) = \frac{1}{n}. \end{aligned}$$

  This contradicts (5.8). Thus $k_2>|k_1-k_3|-1$, and so, by Lemma 5.1, $k_2\ge |k_1-k_3|+1$. In particular, since $k_2\ge 1$, $k_1k_3=k_1(2k+1-k_1-k_2)\le k^2$, and $k_1k_3=k^2$ if and only if $k_1=k_3=k$ and $k_2=1$. It follows that

  $$\begin{aligned} \Vert \mathbf{c}(\xi _n)\Vert&\ge \frac{k_2}{n}-\Vert \overline{\mathbf{c}}\Vert = \frac{k_2}{n}-\frac{1}{n}\sqrt{(k_1-k_3)^2+4k_1k_3\cos ^2(2a)}\end{aligned}$$

  (5.9)
  $$\begin{aligned}&\ge \frac{|k_1-k_3|+1}{n}-\frac{1}{n}\sqrt{(k_1-k_3)^2+(2k\cos (2a))^2} =:A. \end{aligned}$$

  (5.10)

  If $\beta <0$, then $\sqrt{\alpha ^2+\beta ^2}\le |\alpha |-\beta$ with equality if and only if $\alpha =0$. Using this inequality and (5.8), one obtains that

  $$\begin{aligned} A\ge \frac{|k_1-k_3|+1}{n} -\frac{1}{n} \left\{ |k_1-k_3|-2k\cos (2a)\right\} =\frac{1+2k\cos (2a)}{n} \ge \Vert \mathbf{c}(\xi _n)\Vert .\nonumber \\ \end{aligned}$$

  (5.11)

  Consequently, all inequalities in (5.9), (5.10) and (5.11) must be equalities, so that $k_1=k_3=k$ and $k_2=1$. Moreover, in view of (5.9), $\Vert \overline{\mathbf{c}}\Vert + \Vert \mathbf{c}(\xi _n)\Vert =k_2/n=\Vert \overline{\mathbf{c}}-\mathbf{c}(\xi _n)\Vert$, which implies that $\overline{\mathbf{c}}$ and $\mathbf{c}(\xi _n)$ are proportional. Thus $\theta =0$.

• (iv) If $a_n^*\le a \le \pi$ and

  $$\begin{aligned} \xi _n = \begin{Bmatrix}-a_n^*+\theta&\theta&a_n^*+\theta \\ k/n&1/n&k/n\end{Bmatrix} \end{aligned}$$

  for some $\theta \in [-a+a_n^*, a-a_n^*]$, then $\Vert \mathbf{c}(\xi _n)\Vert =0$, so that, by (3.1), $\xi _n$ is $D$-optimal. $\square$