Corrigendum to "Robust smoothing of gridded data in one and higher dimensions with missing values" [Comput. Statist. Data Anal. 54 (2010) 1167-1178]
Corrigendum to "Robust smoothing of gridded data in one and higher dimensions with missing values" [Comput. Statist. Data Anal. 54 (2010) 1167-1178]
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DOI：
10.1016/j.csda.2011.12.001
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发表时间：
2012-06
期刊：
影响因子：
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通讯作者：
L. L. Tarnec-L.;Damien Garcia
中科院分区：
文献类型：
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作者：
L. L. Tarnec-L.;Damien Garcia
在第1170页，Eq. (14)收敛，对于任何初始条件，如果矩阵A是正定的。在最初的论文中，断言D是非奇异的。这显然是错误的，因为一个特征值是零（见方程。（八））。因此，A的正定性仍有待证明。我们假设非负权重wi不完全为零。根据定义，A= sDTD+ W并且s> 0。因为对于任何X，我们有XT（DTD）X=<$DX <$2 ≥ 0和XTWX=<$n i= 1 wix 2 i≥ 0，所以有XTAX≥ 0。因为A是对称的，所以A是半正定的。现在，设X是一个向量，使得XTAX= 0;那么（1）DX= 0和（2）XTWX= 0。(1)从等式(8)，n× n矩阵D有n个不同的特征值，其中一个为零。因此，D的核是一维的。因为很明显任何常向量都属于这个核，所以后者由常向量的集合组成。因此，由于DX= 0，我们推断X是常数。(2)我们写XTWX= n
On page 1170, Eq.(14) converges, for any initial conditions, if the matrix A is positive definite. In the original paper, it was asserted that D is nonsingular. This is obviously wrong since one eigenvalue is zero (see Eq.(8)). Thus the positive definiteness of A still remains to be proved. We assume that the non negative weights wi are not identically zero. By definition, A= sDTD+ W and s> 0. Since, for any X, we have XT (DTD) X=∥ DX∥ 2≥ 0 and XTWX= n i= 1 wix2 i≥ 0, one has XTAX≥ 0. Since A is symmetric, A is positive semidefinite.Now, let X be a vector such that XTAX= 0; then (1) DX= 0 and (2) XTWX= 0.(1) From Eq.(8), the n× n matrix D has n distinct eigenvalues, one of them being zero. Therefore, the kernel of D is of dimension 1. Since it is clear that any constant vector belongs to this kernel, the latter consists of the set of the constant vectors. Therefore, since DX= 0, we deduce that X is constant.(2) We write XTWX= n