Impact of measurement errors on the performance and distributional properties of the multivariate capability index $\mathbf{NMC }_\mathbf{pm }$

Abstract

Current industrial processes are sophisticated enough to be tied to only one quality variable to describe the process result. Instead, many process variables need to be analyze together to assess the process performance. In particular, multivariate process capability analysis (MPCIs) has been the focus of study during the last few decades, during which many authors proposed alternatives to build the indices. These measures are extremely attractive to people in charge of industrial processes, because they provide a single measure that summarizes the whole process performance regarding its specifications. In most practical applications, these indices are estimated from sampling information collected by measuring the variables of interest on the process outcome. This activity introduces an additional source of variation to data, that needs to be considered, regarding its effect on the properties of the indices. Unfortunately, this problem has received scarce attention, at least in the multivariate domain. In this paper, we study how the presence of measurement errors affects the properties of one of the MPCIs recommended in previous researches. The results indicate that even little measurement errors can induce distortions on the index value, leading to wrong conclusions about the process performance.

Appendix: Proof of positive definiteness of \( \left( {\varvec{{\varSigma }}}^{-\mathbf{1}} {\varvec{{\varSigma }}}_{{\varvec{V}}} \left( {\varvec{{\varSigma }}}+{\varvec{{\varSigma }}}_{{\varvec{V}}}\right) ^{-\mathbf{1 }}\right) \)

After Eq. (13) of Sect. 4, it was stated that the quadratic form:

$$\begin{aligned} \left( {\varvec{\mu }}-{{\varvec{T}}} \right) ^{'} \left( {\varvec{{\varSigma }}}^{-\mathbf{1}} {\varvec{{\varSigma }}}_{{\varvec{V}}} \left( {\varvec{{\varSigma }}}+{\varvec{{\varSigma }}}_{{\varvec{V}}}\right) ^{-\mathbf{1}}\right) \left( {\varvec{\mu }}-{{\varvec{T}}} \right) \end{aligned}$$

(21)

is definite positive, which holds if the matrix

$$\begin{aligned} {\varvec{{\varSigma }}}^{-\mathbf{1}} {\varvec{{\varSigma }}}_{{\varvec{V}}} \left( {\varvec{{\varSigma }}}+{\varvec{{\varSigma }}}_{{\varvec{V}}}\right) ^{-\mathbf{1}} \end{aligned}$$

(22)

is positive definite.

In first place, given the structure assumed for matrix ${\varvec{{\varSigma }}}_{{\varvec{V}}}$ in this work, then the matrix in (22) can be written as:

$$\begin{aligned} {\varvec{{\varSigma }}}^{-{1}} {\varvec{{\varSigma }}}_{{\varvec{V}}} \left( {\varvec{{\varSigma }}}+{\varvec{{\varSigma }}}_{{\varvec{V}}}\right) ^{-{1}}={\varvec{\sigma }}_{{\varvec{G}}}^\mathbf{2 } {\varvec{{\varSigma }}}^{-\mathbf{1}} \left( {\varvec{{\varSigma }}}+{\varvec{{\varSigma }}}_{{\varvec{V}}} \right) ^{-\mathbf{1}}={\varvec{\sigma }}_{{\varvec{G}}}^\mathbf{2 } \left( {\varvec{{\varSigma }}}_{{\varvec{Y}}} {\varvec{{\varSigma }}}\right) ^{-\mathbf{1}} \end{aligned}$$

(23)

Therefore, it is necessary to prove that all the eigenvalues of the matrix ${\varvec{{\varSigma }}}_{{\varvec{Y}}}\varvec{{\varSigma }}$ are positive, in which case those of its inverse will be positive as well. Assuming that ${\varvec{{\varSigma }}}$ is a positive definite matrix, then all its eigenvalues, named ${{\lambda }}_{1}, {{\lambda }}_{{2}},\ldots ,{{\lambda }}_{{p}}$, are positive. On the other hand, matrix ${\varvec{{\varSigma }}}_{{\varvec{Y}}}$ is also positive definite. In fact, let ${\varvec{{\varSigma }}}={\varvec{U} \varvec{DU}}^{'}$ be the spectral decomposition of $\varvec{{\varSigma }}$, where ${\varvec{U}}=\left( {\varvec{u}}_{\varvec{1}}, {\varvec{u}}_{\varvec{2}},\ldots , {\varvec{u}}_{\varvec{p}}\right)$ and ${\varvec{u}}_{\varvec{i}}$ is the ith eigenvector associated with the ith eigenvalue ${{\lambda }}_{{i}}$. Being ${{\sigma }}_{{G}}^{{2}}>{{0}}$, let rewrite ${{\lambda }}_{{i}}$ as ${{\lambda }}_{{i}}^{{*}}-{{\sigma }}_{{G}}^{{2}}$. Then

$$\begin{aligned} \begin{aligned} \left( {\varvec{{\varSigma }}}-\left( {{\lambda }}_{{i}}^{*}-\sigma _G^2\right) {\varvec{I}}\right) {\varvec{u}}_{{\varvec{i}}}&=0 \ \Rightarrow \ \left( {\varvec{{\varSigma }}}+{{\sigma }}_{{G}}^{{2}}\ {\varvec{I}}\ -{{\lambda }}_{{i}}^{*}\ {\varvec{I}}\right) {\varvec{u}}_{{\varvec{i}}}=0 \\&\Rightarrow \left( {\varvec{{\varSigma }}}_{{\varvec{Y}}} -{{\lambda }}_{{i}}^{*}\ {\varvec{I}}\right) {\varvec{u}}_{{\varvec{i}}}=0 \end{aligned} \end{aligned}$$

(24)

Hence, ${{\lambda }}_{{i}}^{*}$ are the eigenvalues of ${\varvec{{\varSigma }}}_{{\varvec{Y}}}$. Since they are obtained from the eigenvalues of the matrix ${\varvec{{\varSigma }}}$ as ${{\lambda }}_{{i}}^{*}={{\lambda }}_{{i}}+{{\sigma }}_{{G}}^{{2}}$, they result positive. Moreover, from (24) it also result that ${\varvec{{\varSigma }}}$ and ${\varvec{{\varSigma }}}_{{\varvec{Y}}}$ have the same eigenvectors, ${\varvec{u}}_i$. In summary, ${\varvec{{\varSigma }}}$ and ${\varvec{{\varSigma }}}_{{\varvec{Y}}}$ are both positive definite and they have common eigenvectors.

It remains to see if the product between themselves is also definite positive. In general, given two matrices ${\varvec{A}}$ and ${\varvec{B}}$ there is no relationship between the eigenvalues of A B and those of ${\varvec{A}}$ and ${\varvec{B}}$, unless both matrices have the same eigenvectors. In such case, being ${{\lambda }}_{{i}}$ and ${{\lambda }}_{{i}}^{*}$ eigenvalues of ${\varvec{{\varSigma }}}$ and ${\varvec{{\varSigma }}}_{\varvec{Y}}$, respectively, both associated to the same eigenvector ${\varvec{u}}_{{\varvec{i}}}$, then ${{\lambda }}_{{i}} {{\lambda }}_{{i}}^{*}$ is eigenvalue of ${\varvec{{\varSigma }}}{\varvec{{\varSigma }}}_{{\varvec{Y}}}$. In fact:

$$\begin{aligned} {\varvec{{\varSigma }}}_{{\varvec{Y}}} {\varvec{{\varSigma }}} {{\varvec{u}}}_{{\varvec{i}}}={\varvec{{\varSigma }}}_{{\varvec{Y}}} \left( {{\lambda }}_{{i}} {\varvec{u}}_{{\varvec{i}}}\right) ={{\lambda }}_{{i}} \left( {\varvec{{\varSigma }}}_{{\varvec{Y}}} {{\varvec{u}}}_{{\varvec{i}}}\right) = {{\lambda }}_{{i}} {{\lambda }}_{{i}}^{*} {\varvec{u}}_{{\varvec{i}}} \end{aligned}$$

(25)

This implies that the eigenvalues of ${\varvec{{\varSigma }}}_{{\varvec{Y}}} {\varvec{{\varSigma }}}$ can be obtained as the product of those of ${\varvec{{\varSigma }}}_{{\varvec{Y}}}$ and ${\varvec{{\varSigma }}}$. Therefore, they are all positive, and hence so are those of the inverse of ${\varvec{{\varSigma }}}_{{\varvec{Y}}} {\varvec{{\varSigma }}}$. From all this matrix in (22) is positive definite.