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D-Optimal Design for Duality Quadratic Polynomial Regression Models in Circle Region
DOI: 10.12677/AAM.2019.87143, PDF, HTML, XML, 下载: 1,016  浏览: 3,315

Abstract: For duality quadratic polynomial regression models in circle region, D-op designs were given and proved according to the criterion “Symmetry + Uniform” with four vertexes of special square and central angle in Circle, and the least squares estimates were given, It is a very useful attempt to apply optimal design theory to polynomial regression models.

1. 引言

$y={f}^{\text{T}}\left(X\right)\beta +\epsilon$

2. 二元二次多项式回归模型的最优设计

$y={\beta }_{0}+{\beta }_{1}x+{\beta }_{2}y+{\beta }_{11}{x}^{2}+{\beta }_{22}{y}^{2}+\epsilon$ (1)

${f}^{\text{T}}\left(X\right)=\left(1,{x}^{2},{y}^{2},x,y\right)$

$X=\left[\begin{array}{ccccc}1& {a}^{2}& {b}^{2}& a& b\\ 1& {a}^{2}& {b}^{2}& -a& -b\\ 1& {b}^{2}& {a}^{2}& b& -a\\ 1& {b}^{2}& {a}^{2}& -b& a\\ 1& 0& 0& 0& 0\end{array}\right]$

$M=\left(\begin{array}{ccccc}1& 2\alpha & 2\alpha & 0& 0\\ 2\alpha & 2\left({a}^{4}+{b}^{4}\right)\alpha & 4{a}^{2}{b}^{2}\alpha & 0& 0\\ 2\alpha & 4{a}^{2}{b}^{2}\alpha & 2\left({a}^{4}+{b}^{4}\right)\alpha & 0& 0\\ 0& 0& 0& 2\alpha & 0\\ 0& 0& 0& 0& 2\alpha \end{array}\right)$

$|M|=16{\alpha }^{4}\left(1-4\alpha \right){\left(2{a}^{2}-1\right)}^{2}$

$\left\{\begin{array}{l}\frac{\partial |M|}{\partial \alpha }=0\\ \frac{\partial |M|}{\partial a}=0\end{array}$ ，则有效驻点是 $\left\{\begin{array}{l}\alpha =\frac{1}{5}\\ a=0\end{array}$$\left\{\begin{array}{l}\alpha =\frac{1}{5}\\ a=±\frac{\sqrt{2}}{2}\end{array}$

$d={f}^{\text{T}}\left(X\right){M}^{-1}\left({\xi }^{\ast }\right)f\left(X\right)\le 5$

${M}^{-1}\left({\xi }^{\ast }\right)=\left(\begin{array}{ccccc}5& -5& -5& 0& 0\\ -5& \frac{15}{2}& 5& 0& 0\\ -5& 5& \frac{15}{2}& 0& 0\\ 0& 0& 0& \frac{5}{2}& 0\\ 0& 0& 0& 0& \frac{5}{2}\end{array}\right)$

$d=5-\frac{15}{2}\left({x}^{2}+{y}^{2}\right)+\frac{15}{2}\left({x}^{4}+{y}^{4}\right)+10{x}^{2}{y}^{2}$

① 在 $\Re$ 的内部 ${x}^{2}+{y}^{2}<1$ ，由 $\left\{\begin{array}{l}\frac{\partial d}{\partial x}=30{x}^{3}+20x{y}^{2}-15x=0\\ \frac{\partial d}{\partial y}=30{y}^{3}+20{x}^{2}y-15y=0\end{array}$ ，则

$\left\{\begin{array}{l}x=0\\ y=0\\ d=5\end{array}$$\left\{\begin{array}{l}x=0\\ {y}^{2}=\frac{1}{2}\\ d=\frac{25}{8}\end{array}$$\left\{\begin{array}{l}{x}^{2}=\frac{1}{2}\\ y=0\\ d=\frac{25}{8}\end{array}$$\left\{\begin{array}{l}{x}^{2}=\frac{3}{10}\\ {y}^{2}=\frac{3}{10}\\ d=\frac{11}{4}\end{array}$

② 在 $\Re$ 的边界 ${x}^{2}+{y}^{2}=1$ ，有 $d=5-5{x}^{2}\left(1-{x}^{2}\right)\le 5$ ，当且仅当在 $x=0,±1$ 的时候取最大值。

3. 模型参数的最小二乘估计

$y={f}^{\text{T}}\left(X\right)\stackrel{\to }{\beta }+\epsilon$

$\stackrel{^}{\stackrel{\to }{\beta }}={\left({X}^{\text{T}}X\right)}^{-1}{X}^{\text{T}}Y=\frac{1}{5}{M}^{-1}{X}^{\text{T}}Y$ (2)

$\stackrel{^}{\stackrel{\to }{\beta }}=\left(\begin{array}{c}{\stackrel{^}{\beta }}_{0}\\ {\stackrel{^}{\beta }}_{11}\\ {\stackrel{^}{\beta }}_{22}\\ \begin{array}{c}{\stackrel{^}{\stackrel{\to }{\beta }}}_{1}\\ {\stackrel{^}{\stackrel{\to }{\beta }}}_{2}\end{array}\end{array}\right)=\left(\begin{array}{c}{y}_{5}\\ \frac{1}{2}\left({y}_{1}+{y}_{2}\right)-{y}_{5}\\ \frac{1}{2}\left({y}_{3}+{y}_{4}\right)-{y}_{5}\\ \begin{array}{l}\frac{1}{2}\left({y}_{1}-{y}_{2}\right)\\ \frac{1}{2}\left({y}_{3}-{y}_{4}\right)\end{array}\end{array}\right)$

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