[发明专利]一种确定累积法建模最小阶次的方法

摘要

本发明公开了一种确定累积法建模最小阶次的方法，包括设累积回归模型为：式中为模型参数估计值；取K＝N+2，根据样本数据构建K阶累积和及累积广义均值；根据K阶累积和，结合最小二乘原理，构建K(>N+1)阶累积方程组，并转化成矩阵形式；利用最小二乘原理，将累积正规方程中的系数作为数据源输入最小二乘法求解，并得出此时模型的残余标准差；基于F分部检验，对K＝N+2时的模型残余标准差进行单侧检验，相较于K＝N+1时的残余标准差是否有显著减小。本发明可以在提高建模精度的同时，减小算法的复杂度，有效解决了累积法建模中累积阶次K的选取问题。

主权项

一种确定累积法建模最小阶次的方法，其特征是：包括下列步骤：步骤1，设累积回归模型为：式中为模型参数估计值，(X_j1,X_j2…X_jn),j＝1,2…m为n个样本观察值，为预测值；步骤2，取K>N+1，根据样本数据构建K阶累积和及累积广义均值$\underset{j=1}{\overset{m}{\Σ}}{{}^{\left(k\right)}x}_j=\underset{j=1}{\overset{m}{\Σ}}\left(\begin{array}{c}m-j+k-1\\ k-1\end{array}\right)x_j=\left(\begin{array}{c}m-j+k-1\\ k-1\end{array}\right)x_j$$\stackrel{\‾}{T_k}\left(x_n\right)=\underset{j=1}{\overset{m}{\Σ}}{{}^{\left(k\right)}x}_{jn}/\left(\begin{array}{c}m+k-1\\ k\end{array}\right)$步骤3，根据K阶累积和，结合最小二乘原理，构建K(>N+1)阶累积方程组，使方程数大于未知量数：$\left\{\begin{array}{c}\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(1\right)}Y}_j={\mathrm{\β}}_0\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{}^{\left(1\right)}+{\mathrm{\β}}_1\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(1\right)}x}_{j1}+\mathrm{...}+{\mathrm{\β}}_n\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(1\right)}x}_{jn}\\ \underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(2\right)}Y}_j={\mathrm{\β}}_0\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{}^{\left(2\right)}+{\mathrm{\β}}_1\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(2\right)}x}_{j1}+\mathrm{...}+{\mathrm{\β}}_n\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(2\right)}x}_{jn}\\ \mathrm{.........}\\ \underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(n\right)}Y}_j={\mathrm{\β}}_0\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{}^{\left(n\right)}+{\mathrm{\β}}_1\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(n\right)}x}_{j1}+\mathrm{...}+{\mathrm{\β}}_n\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(n\right)}x}_{jn}\\ \underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(n+1\right)}Y}_j={\mathrm{\β}}_0\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{}^{\left(n+1\right)}+{\mathrm{\β}}_1\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(n+1\right)}x}_{j1}+\mathrm{...}+{\mathrm{\β}}_n\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{{}^{\left(n+1\right)}x}_{jn}\\ \mathrm{.........}\end{array}\right.$步骤4，根据累积广义均值将累积方程组转化成矩阵形式：$\left[\begin{array}{cccc}1& {\stackrel{\‾}{T}}_1\left({\stackrel{\‾}{x}}_1\right)& ...& {\stackrel{\‾}{T}}_1\left({\stackrel{\‾}{x}}_n\right)\\ 1& {\stackrel{\‾}{T}}_2\left({\stackrel{\‾}{x}}_1\right)& ...& {\stackrel{\‾}{T}}_2\left({\stackrel{\‾}{x}}_n\right)\\ \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}\\ \\ \end{array}& \begin{array}{c}.\\ .\\ .\end{array}\\ 1& {\stackrel{\‾}{T}}_n\left({\stackrel{\‾}{x}}_1\right)& ...& {\stackrel{\‾}{T}}_n\left({\stackrel{\‾}{x}}_n\right)\\ 1& {\stackrel{\‾}{T}}_{n+1}\left({\stackrel{\‾}{x}}_1\right)& ...& {\stackrel{\‾}{T}}_{n+1}\left({\stackrel{\‾}{x}}_n\right)\\ \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}\\ \\ \end{array}& \begin{array}{c}.\\ .\\ .\end{array}\end{array}\right]\left[\begin{array}{c}{\mathrm{\β}}_0\\ {\mathrm{\β}}_1\\ \begin{array}{c}.\\ .\\ .\end{array}\\ {\mathrm{\β}}_n\end{array}\right]=\left[\begin{array}{c}{\stackrel{\‾}{T}}_1\left(Y\right)\\ {\stackrel{\‾}{T}}_2\left(Y\right)\\ \begin{array}{c}.\\ .\\ .\end{array}\\ {\stackrel{\‾}{T}}_n\left(Y\right)\\ {\stackrel{\‾}{T}}_{n+1}\left(Y\right)\\ \begin{array}{c}.\\ .\\ .\end{array}\end{array}\right]$步骤5，利用最小二乘原理，将累积正规方程中的系数作为数据源输入进行最小二乘法求解，$\left[\begin{array}{cccc}\underset{i=1}{\overset{n}{\mathrm{\Σ}}}1& \underset{i=1}{\overset{n}{\mathrm{\Σ}}}{x}_{1i}& ...& \underset{i=1}{\overset{n}{\mathrm{\Σ}}}{x}_{ni}\\ \underset{i=1}{\overset{n}{\mathrm{\Σ}}}{x}_{1i}& \underset{i=1}{\overset{n}{\mathrm{\Σ}}}{x_{1i}}^2& ...& \underset{i=1}{\overset{n}{\mathrm{\Σ}}}{x}_{1i}{x}_{ni}\\ \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}\\ \\ \end{array}& \begin{array}{c}.\\ .\\ .\end{array}\\ \underset{i=1}{\overset{n}{\mathrm{\Σ}}}{x}_{ni}& \underset{i=1}{\overset{n}{\mathrm{\Σ}}}{x}_{ni}{x}_{1i}& ...& \underset{i=1}{\overset{n}{\mathrm{\Σ}}}{x_{ni}}^2\end{array}\right]\left[\begin{array}{c}{\mathrm{\β}}_0\\ {\mathrm{\β}}_1\\ \begin{array}{c}.\\ .\\ .\end{array}\\ {\mathrm{\β}}_n\end{array}\right]=\left[\begin{array}{c}\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{y}_i\\ \underset{i=1}{\overset{n}{\mathrm{\Σ}}}{x}_{1i}{y}_i\\ \begin{array}{c}.\\ .\\ .\end{array}\\ \underset{i=1}{\overset{n}{\mathrm{\Σ}}}{x}_{ni}{y}_i\end{array}\right]$步骤6，基于F分部检验，对K>N+1时的模型残余标准差进行检验，相较于K＝N+1时的残余标准差是否有显著减小。若没有，则增加阶次K直至残余标准差有显著减小，此时的K即为建模的最小阶次。