异方差情况下二次回归参数的ls估计和Aitken估计符合的充分条件

IF 0.1
M. Savkina
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引用次数: 0

摘要

本文研究了异方差独立偏差情况下函数为$f(x) = ax^2+bx+c$的回归模型,其中$a$, $b$和$c$为未知参数。函数$f(x)$的近似值(观测值)记录在线段的等距点上。本文所证明的定理给出了在奇数观测点和双对称协方差矩阵下,参数$a$的艾特肯估计与其LS估计重合的方差的充分条件。在这种情况下,$b$和$c$的艾特肯估计和ls估计将不一致。这个定理的证明包括以下步骤。首先,对原多项式系统进行简化,得到二阶多项式系统。两个系统的变量都是未知的偏差方差,原始系统的每个解都给出了一组偏差方差,在这些偏差方差处,Aitken和LS参数的估计重合。在下一步中,原始系统多项式的求解被简化为求解一个有三个未知数的方程,所有其他的未知数都以某种方式通过这三个来表示。最后证明了这三个未知量存在正不等值,这就是所得方程的解。所有其他未知数代入它们的表达式时这些值都是正的。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
SUFFICIENT CONDITION FOR COINCIDENCE OF THE LS AND AITKEN ESTIMATIONS OF PARAMETER OF QUADRATIC REGRESSION IN CASE HETEROSCEDASTIC DEVIATIONS
In the paper in case heteroscedastic independent deviations a regression model whose function has the form $f(x) = ax^2+bx+c$, where $a$, $b$ and $c$ are unknown parameters, is studied. Approximate values (observations) of functions $f(x)$ are registered at equidistant points of a line segment. The theorem which is proved at the paper gives a sufficient condition on the variance of the deviations at which the Aitken estimation of parameter $a$ coincides with its estimation of the LS in the case of odd number of observation points and bisymmetric covariance matrix. Under this condition, the Aitken and LS estimations of $b$ and $c$ will not coincide. The proof of the theorem consists of the following steps. First, the original system of polynomials is simplified: we get the system polynomials of the second degree. The variables of both systems are unknown variances of deviations, each of the solutions of the original system gives a set variances of deviations at which the estimations of Aitken and LS parameter a coincide. In the next step the solving of the original system polynomials is reduced to solving an equation with three unknowns, and all other unknowns are expressed in some way through these three. At last it is proved that there are positive unequal values of these three unknowns, which will be the solution of the obtained equation. And all other unknowns when substituting in their expression these values will be positive.
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