Some explicit solutions of c-optimal design problems for polynomial regression with no intercept

Abstract

In this paper, we consider the optimal design problem for extrapolation and estimation of the slope at a given point, say z, in a polynomial regression with no intercept. We provide explicit solutions of these problems in many cases and characterize those values of z, where this is not possible.

Appendix: Proof of Lemma 2

The proof of this result can be found in the PhD thesis of Sahm (1998). As it is difficult to get access to this thesis, we repeat his arguments here for the sake of completeness. We begin with two statements, from which the proof of Lemma 2 will easily follow.

Lemma 4

Let $Q(x)=\prod _{i=1}^n(x-q_i)$ and $P(x)=\prod _{i=1}^m(x-p_i)$ denote two polynomials of degree n and m, respectively, where $m=n-1$ or $m=n$. Assume that $P\ne Q$ and set $q_{n+1}=-\infty$. If

$$\begin{aligned} q_i\ge p_i\ge q_{i+1}, \quad i=1,\ldots ,m, \end{aligned}$$

(25)

then we have $R(x) :=(P(x) /Q(x) )'<0,$ whenever the polynomial R(x) is defined.

Proof

In the case $n=1$, the result is nearly obvious: if $m=n-1=0$ we have $R(x) =-Q^{-2}(x) <0$ and for $m=n=1$

$$\begin{aligned} R(x)=\frac{p_1-q_1}{(x-q_1)^2}<0\quad \hbox {for all}\ x\ne q_1. \end{aligned}$$

Now we turn to the case $n>1$ and assume that the pair (P, Q) form a counterexample of minimal degree. This implies in particular that P(x) and Q(x) cannot have a root in common (otherwise their degree would not be minimal). Furthermore, all roots of P(x) and Q(x) must be simple. As the pair (P, Q) is a counterexample, there exists some $z\in \mathbb {R}$ where $R(z)\ge 0.$

The idea is following. Move the polynomial P up (or down) without changing the property $R(z)\ge 0$ until one of the zeros of P(x) and Q(x) coincide. Then divide the polynomials by this factor and produce a counterexample of smaller degree, which contradicts the assumption of minimality. For this purpose, we have to consider the two cases $Q'(z)\ge 0$ and $Q'(z)< 0$ separately.

We restrict ourselves to the case $Q'(z)\ge 0$ and mention that the case $Q'(z)< 0$ can be obtained by exactly the same arguments. To be precise, define

$$\begin{aligned} \epsilon :=\sup \{\delta >0 |\,\text {the roots of the polynomials}\;P(x) -\delta \;\text {and}\;Q(x)\;\text {interlace}\}. \end{aligned}$$

This set is not empty due to the continuity of the roots of the polynomial $P(x) -\delta$ with respect to $\delta$. Now let $\bar{P} (x) =P (x) -\epsilon .$ Then it is clear from the definition of $\epsilon$ that $\bar{P}(x)$ has m zeros $\bar{p}_1, \ldots , \bar{p}_m,$ which interlace with those of Q(x), and $\bar{P}(x)$ and Q(x) have at least one zero in common. Furthermore, since $\bar{P}'(x)=P'(x)$ we have

$$\begin{aligned} \bar{R}(z)=(\bar{P}/Q)'(z)=R(z)+\frac{\epsilon Q'(z)}{Q^2(z)}\ge 0; \end{aligned}$$

Before we divide $\bar{P}(x)$ and Q(x) by the factors that they have in common, note that $\bar{P}$ cannot equal Q because in this case we would have $P(x)=Q(x)+\epsilon$, which contradicts the interlacing property for $n>1.$ Now let $\tilde{P}(x)$ and $\tilde{Q}(x)$ denote the polynomials obtained by dividing $\bar{P}(x)$ and Q(x) by their greatest common denominator. These polynomials still have the interlacing property, are of degree smaller than n and are not equal. For the corresponding ratio, we find

$$\begin{aligned} \hat{R}(z)=(\tilde{P}/\tilde{Q})'(z)=(\bar{P}/Q)'(z)=\bar{R}(z)\ge 0 \end{aligned}$$

which contradicts the assumption that the pair (P, Q) forms a counterexample of minimal degree. $\square$

Theorem 7

Consider two polynomials $Q(x)=\prod _{i=1}^n(x-q_i),$ and $P(x)=\prod _{i=1}^m(x-p_i),$ of degree n and m, where $m\le n\le m+1$, the roots $q_1>\cdots > q_n,$ $p_1>\cdots > p_m,$ fulfill condition (25) and $P\ne Q$. Then the zeros of $q_1'>\cdots >q'_{n-1}$ and $p_1'>\cdots >p'_{m-1},$ the derivatives $Q'(x)$ and $P'(x)$ satisfy

$$\begin{aligned} q_i'>p_i'>q'_{i+1}\quad \hbox {for}\ i=1,\ldots ,m-1 \end{aligned}$$

(Here $q_n'$ id defined as $q_n'=-\infty$.) In other words, if the polynomials $P\ne Q$ have only simple roots, which interlace, then the roots of their derivatives strictly interlace.

Proof

We first consider the case where the polynomial P(x) has degree $m=n-1$. From Lemma 4 for $i=1,\ldots ,n-1$

$$\begin{aligned} 0>R(q'_i)=\left( \frac{P}{Q}\right) '(q'_i)=\frac{P'(q_i')Q(q_i')-P(q_i')Q'(q_i')}{Q(q_i')^2}= \frac{P'(q_i')}{Q(q_i')}. \end{aligned}$$

Since the denominator alternates in sign (the roots of $Q'$ interlace with those of Q) and the leading coefficient is 1, it follows that $\text {sign}( Q(q_i'))=(-1)^i.$ This implies that the numerator must also alternate in sign, i.e.,

$$\begin{aligned} \text {sign} ( P'(q_i')) = (-1)\text {sign} ( P'(q_{i+1}')) =(-1)^{i-1}\quad (i=1,\ldots ,n-2). \end{aligned}$$

This means that between any pair of consecutive roots $q_i',$ $q_{i+1}'$ of $Q'(x)$ there is a root of $P'(x)$, which proves Theorem 7 for $n=m+1.$

If both polynomials P(x) and Q(x) have the same degree m, note that

$$\begin{aligned} \lim _{x\rightarrow -\infty } \text {sign} (P'(x) )= (-1)^{m-1}=(-1)P'(q_{m-1}'). \end{aligned}$$

Consequently, $P'(x)$ also has a root in the interval $(-\infty ,q_{m-1}'),$ which completes the proof of Theorem 7. $\square$

Proof of Lemma 4

Without loss of generality, we assume that the leading coefficients of the polynomial $P_1$ and $P_2$ are equal to 1. With the notation $P(x)=P_2(x),$ $Q(x)=P_1(x)$, $p_i=t_{(n-i+1,2)},$ $q_i=t_{(n-i+1,1)}$ ($i=1\ldots ,n$) the proposition of Lemma 2 follows from Theorem 7.$\square$