Comparison of joint control schemes for multivariate normal i.i.d. output

Abstract

The performance of a product frequently relies on more than one quality characteristic. In such a setting, joint control schemes are used to determine whether or not we are in the presence of unfavorable disruptions in the location (${\varvec{\mu }}$) and spread (${\varvec{\varSigma }}$) of a vector of quality characteristics. A common joint scheme for multivariate output comprises two charts: one for ${\varvec{\mu }}$ based on a weighted Mahalanobis distance between the vector of sample means and the target mean vector; another one for ${\varvec{\varSigma }}$ depending on the ratio between the determinants of the sample covariance matrix and the target covariance matrix. Since we are well aware that there are plenty of quality control practitioners who are still reluctant to use sophisticated control statistics, this paper tackles Shewhart-type charts for ${\varvec{\mu }}$ and ${\varvec{\varSigma }}$ based on three pairs of control statistics depending on the nominal mean vector and covariance matrix, ${\varvec{\mu }}_0$ and ${\varvec{\varSigma }}_0$. We either capitalize on existing results or derive the joint probability density functions of these pairs of control statistics in order to assess the ability of the associated joint schemes to detect shifts in ${\varvec{\mu }}$ or ${\varvec{\varSigma }}$ for various out-of-control scenarios. A comparison study relying on extensive numerical and simulation results leads to the conclusion that none of the three joints schemes for ${\varvec{\mu }}$ and ${\varvec{\varSigma }}$ is uniformly better than the others. However, those results also suggest that the joint scheme with the control statistics $n \, ( \bar{\mathbf {X}}-{\varvec{\mu }}_0 )^\top \, {\varvec{\varSigma }}_0^{-1} \, ( \bar{\mathbf {X}}-{\varvec{\mu }}_0 )$ and $\hbox {det} \left( (n-1) \mathbf{S} \right) / \hbox {det} \left( {\varvec{\varSigma }}_0 \right)$ has the best overall average run length performance.

Appendices

Appendix A

Now, we prove Theorems 2 and 3, in this particular order.

Proof of Theorem 3

We assume that ${\varvec{\delta }}$ is arbitrary in part a) of this proof. In part b) it is necessary to assume that ${\varvec{\delta }} = \mathbf{0}$ to capitalize on an existing result.

a) Let: $~~\mathbf{U}_{i} = {\varvec{\varSigma }}^{-1/2} (\mathbf{X}_{i} - {\varvec{\mu }}_0)$, where $\mathbf{X}_{i} {\mathop {\sim }\limits ^{i.i.d.}} \mathcal{N}_p ( {\varvec{\mu }}_0 + {\varvec{\varSigma }}_0^{1/2} {\varvec{\delta }}/\sqrt{n}, \, {\varvec{\varSigma }})$; $\bar{\mathbf{U}} = \frac{1}{n} \sum _{i=1}^n \mathbf{U}_{i}$.  Then:

$$\begin{aligned} n \mathbf{S}^* = {\varvec{\varSigma }}^{1/2} \, \left( \sum _{i=1}^n \mathbf{U}_{i} \, \mathbf{U}_{i}^\top \right) \, {\varvec{\varSigma }}^{1/2}, \end{aligned}$$

(17)

where $\mathbf{U}_{i} {\mathop {\sim }\limits ^{i.i.d.}} \mathcal{N}_p ( {\varvec{\varSigma }}^{-1/2} {\varvec{\varSigma }}_0^{1/2} {\varvec{\delta }}/ \sqrt{n}, \, \mathbf{I})$;

$$\begin{aligned} T^{(3)}= & {} n \, (\bar{\mathbf{X}} - {\varvec{\mu }}_0)^\top \, (\mathbf{S}^*)^{-1}\, (\bar{\mathbf{X}} - {\varvec{\mu }}_0) \nonumber \\= & {} n (\sqrt{n} \, \bar{\mathbf{U}})^\top \, \left( \sum _{i=1}^n \mathbf{U}_{i} \, \mathbf{U}_{i}^\top \right) ^{-1} \, (\sqrt{n} \, \bar{\mathbf{U}}), \end{aligned}$$

(18)

with $\sqrt{n} \, \bar{\mathbf{U}} \sim \mathcal{N}_p( {\varvec{\varSigma }}^{-1/2} \, {\varvec{\varSigma }}_0^{1/2} {\varvec{\delta }}, \, \mathbf{I})$.

Let: $~\mathbf{U} = (\mathbf{U}_{1}, \ldots , \mathbf{U}_{n})^\top$ be an $n \times p$ random matrix whose rows we know are independent normal variates with common mean $({\varvec{\varSigma }}^{-1/2} {\varvec{\varSigma }}_0^{1/2} {\varvec{\delta }}/ \sqrt{n} \,)^\top$ and the same covariance matrix $\mathbf{I}$; $~\mathbf{H} = (\mathbf{h}_1, \ldots , \mathbf{h}_n)^\top$ be an orthogonal $n \times n$ matrix with $\mathbf{h}_1 = \mathbf{1}_n/\sqrt{n}$; $~\mathbf{Z} = (\mathbf{Z}_{1}, \ldots , \mathbf{Z}_{n})^\top = \mathbf{H} \, \mathbf{U}$ be a random matrix obtained from U by the $n \times n$ orthogonal transformation $\mathbf{H}$.  Then, according to Fujikoshi et al. (2010, p. 13, Theorem 1.2.6), $\mathbf{Z}_{1}, \ldots , \mathbf{Z}_{n}$ (i.e., the rows of $\mathbf{Z}$) are independent normal variates with the same covariance matrix $\mathbf{I}$, and $E(\mathbf{Z}) = \mathbf{H} \, E(\mathbf{U})$, that is, $\mathbf{Z}$ has the same properties as $\mathbf{U}$ except that the mean of $\mathbf{Z}$ is changed to $\mathbf{H} \, E(\mathbf{U})$. Furthermore:

$$\begin{aligned} \mathbf{Z}_{1}= & {} \sqrt{n} \, \bar{\mathbf{U}} \sim \mathcal{N}_p ( {\varvec{\varSigma }}^{-1/2} \, {\varvec{\varSigma }}_0^{1/2} {\varvec{\delta }}, \mathbf{I}); \\ \mathbf{Z}_{i}\sim & {} \mathcal{N}_p( \mathbf{0}, \mathbf{I}), \, i=2,\ldots ,n, \end{aligned}$$

where the zero vector mean follows from the orthogonality of $\mathbf{H}$ which translates into null inner products between the first row of $\mathbf{H}$ and any of its other rows, implying in turn that the sum of the entries of any of the remaining rows of $\mathbf{H}$ are also null.

b) Let us capitalize on the fact that $\sum _{i=1}^n \mathbf{U}_{i} \mathbf{U}_{i}^\top = \mathbf{U}^\top \mathbf{U} = (\mathbf{H} \mathbf{U})^\top (\mathbf{H}{} \mathbf{U}) = \mathbf{Z}^\top \mathbf{Z} = \sum _{i=1}^n \mathbf{Z}_{i} \mathbf{Z}_{i}^\top$, and consider: $\mathcal{A} = \sum _{i=1}^n \mathbf{Z}_{i} \mathbf{Z}_{i}^\top$; ${\varvec{\tau }} = \mathcal{A}^{-1/2} \mathbf{Z}_{1}$.

Firstly, by recalling result (13) and noting that $\mathcal{A} = {\varvec{\varSigma }}^{-1/2} (n \, \mathbf{S}^*) {\varvec{\varSigma }}^{-1/2}$, we can resort to Arnold (2006, p. 9185) to add that $\mathcal{A} \sim W_p \left( n, \mathbf{I}; {\varvec{\delta }}^* \, ({\varvec{\delta }}^\star )^\top \right)$, where ${\varvec{\delta }}^* = {\varvec{\varSigma }}^{-1/2} \, {\varvec{\varSigma }}_0^{1/2} \, {\varvec{\delta }}$. If we adopt the notation $etr(A) = \exp (tr(A))$, consider ${\varvec{\delta }} = \mathbf{0}$ and recall (Muirhead 1982, p. 85, Theorem 3.2.1), then we can write the marginal p.d.f. of the $p \times p$ matrix $\mathcal{A}$ as

$$\begin{aligned} f_{\mathcal{A}}(\mathcal{A}) = \frac{1}{2^{pn/2} \varGamma _p(n/2)} \, \left[ \hbox {det}(\mathcal{A}) \right] ^{(n-p-1)/2} \, etr\left( -\mathcal{A}/2 \right) , \end{aligned}$$

(19)

where $\varGamma _p (n/2) = \pi ^{p(p-1)/4} \prod _{i=1}^p \varGamma [(n-i+1)/2]$ according to Muirhead (1982, p. 100).

Secondly, if ${\varvec{\delta }} = \mathbf{0}$, then $\mathbf{Z}_{1}, \ldots , \mathbf{Z}_{n}$ are independent $\mathcal{N}_p( \mathbf{0}, \mathbf{I})$ random vectors and in this case we can invoke Gupta and Nagar (2000, pp. 167–168, Theorem 5.2.3) or Muirhead (1982, p. 117, Exercise 3.15) and state that: the joint p.d.f. of $\mathcal{A}$ and ${\varvec{\tau }}$ is

$$\begin{aligned} f_{\mathcal{A}, {\varvec{\tau }}}(\mathcal{A}, {\varvec{\tau }}) = \frac{\left[ \hbox {det}(\mathcal{A}) \right] ^{(n-p-1)/2} \, etr\left( -\mathcal{A}/2 \right) }{2^{pn/2} \, \pi ^{p/2} \, \varGamma _p[(n-1)/2]} \times (1-{\varvec{\tau }}^\top {\varvec{\tau }})^{(n-p-2)/2}; \end{aligned}$$

the marginal p.d.f. of ${\varvec{\tau }}$ is

$$\begin{aligned} f_{{\varvec{\tau }}}({\varvec{\tau }}) = \frac{\varGamma (n/2)}{\pi ^{p/2} \, \varGamma [(n-p)/2]}\times (1-{\varvec{\tau }}^\top {\varvec{\tau }})^{(n-p-2)/2}, \end{aligned}$$

(20)

for ${\varvec{\tau }}^\top {\varvec{\tau }} < 1$.

Thirdly, combining results (19)–(20) and the fact that $\varGamma _p(n/2)/\varGamma _p((n-1)/2) = \varGamma (n/2)/\varGamma ((n-p)/2)$, we can write the joint distribution of $\mathcal{A}$ and ${\varvec{\tau }}$ as the product of the corresponding marginal p.d.f., $f_{\mathcal{A}, {\varvec{\tau }}}(\mathcal{A}, {\varvec{\tau }}) = f_\mathcal{A}(\mathcal{A}) \times f_{{\varvec{\tau }}}({\varvec{\tau }})$, that is to say that $\mathcal{A}$ and ${\varvec{\tau }}$ are independent provided that ${\varvec{\delta }} = \mathbf{0}$. (If ${\varvec{\delta }} \ne \mathbf{0}$ this is no longer the case.)

Finally, note that from (18) and (17) we get

$$\begin{aligned} T^{(3)}= & {} n \, (\bar{\mathbf{X}} - {\varvec{\mu }}_0)^\top \, (\mathbf{S}^*)^{-1}\, (\bar{\mathbf{X}} - {\varvec{\mu }}_0) = n \, {\varvec{\tau }}^\top {\varvec{\tau }} \\ U^{(3)}= & {} \frac{\hbox {det} ( n \mathbf{S}^* )}{\hbox {det}( {\varvec{\varSigma }}_0 )} = \frac{\hbox {det}({\varvec{\varSigma }})}{\hbox {det}({\varvec{\varSigma }}_0)} \times \hbox {det}(\mathcal{A}), \end{aligned}$$

hence these two measurable functions of ${\varvec{\tau }}$ and $\mathcal{A}$ (respectively) are independent control statistics by the disjoint blocks theorem, as long as ${\varvec{\delta }} = \mathbf{0}$. Moreover, by taking advantage of results (12) and (14), $\eta ^{(3)} (\mathbf{0}, \, {\varvec{\varSigma }})$ follows in a straightforward manner, thus we conclude the proof of Theorem 3. $\square$

Proof of Theorem 2

Firstly, recall that

$$\begin{aligned} n \mathbf{S}^* = (n-1) \mathbf{S} + n (\bar{\mathbf{X}} - {\varvec{\mu }}_0) \, (\bar{\mathbf{X}} - {\varvec{\mu }}_0)^\top . \end{aligned}$$

(21)

Secondly, note that if $\mathbf{A}$ is a $p \times p$ nonsingular matrix and $\mathbf{u}$ and $\mathbf{v}$ are two $p-$dimensional vectors, then

$$\begin{aligned} (\mathbf{A} + \mathbf{u} \, \mathbf{u}^\top )^{-1} = \mathbf{A}^{-1} - \frac{\mathbf{A}^{-1} \, \mathbf{u} \, \mathbf{u}^\top \, \mathbf{A}^{-1}}{1+ \mathbf{u}^\top \, \mathbf{A}^{-1} \, \mathbf{u}} \end{aligned}$$

(Fujikoshi et al. 2010, A.1.3, p. 496) and

$$\begin{aligned} \mathbf{u}^\top (\mathbf{A} + \mathbf{u}{} \mathbf{u}^\top )^{-1} \mathbf{u} = \frac{\mathbf{u}^\top \mathbf{A}^{-1} \mathbf{u}}{1+ \mathbf{u}^\top \mathbf{A}^{-1} \mathbf{u}} . \end{aligned}$$

(22)

Thus, by considering $\mathbf{A} = (n-1) \mathbf{S}$ and $\mathbf{u} = \sqrt{n} (\bar{\mathbf{X}} - {\varvec{\mu }}_0)$, we get

$$\begin{aligned} T^{(3)} = n^2 \, ( \bar{\mathbf {X}}-{\varvec{\mu }}_0 )^\top \, (n \mathbf{S}^*)^{-1} \, ( \bar{\mathbf {X}}-{\varvec{\mu }}_0 ) = \frac{n \, T^{(2)}}{n-1+ T^{(2)}}. \end{aligned}$$

(23)

Thirdly, we need to invoke Sylvester’s determinant theorem (Sylvester 1851; Akritas et al. 1996), namely one of its consequences: for the case of column vector $\mathbf{c}$ and row vector $\mathbf{r}$, each with p components, we have $\hbox {det}(\mathbf{I} + \mathbf{c} \, \mathbf{r}) = 1 + \mathbf{r} \, \mathbf{c}$ and, if $\mathbf{A}$ is positive definite,

$$\begin{aligned} \hbox {det}( \mathbf{A} + \mathbf{u}{} \mathbf{u}^\top ) = \hbox {det}(\mathbf{A}) \; (1+\mathbf{u}^\top \mathbf{A}^{-1} \mathbf{u} ) . \end{aligned}$$

(24)

Hence, using (21),

$$\begin{aligned} U^{(3)} = \frac{\hbox {det} ( n \mathbf{S}^* )}{\hbox {det} ( {\varvec{\varSigma }}_0 )} = \left( 1 + \frac{T^{(2)}}{n-1} \right) U^{(2)}. \end{aligned}$$

Finally, by recalling Rohatgi (1976, Theorem 6, p. 135) or Karr (1993, Theorem 2.43, p. 62), we can obtain the p.d.f. of a one-to-one transformation such as $(T^{(2)}, \, U^{(2)})$ of the random vector $(T^{(3)}, \, U^{(3)}) = \underline{g}^{-1}(T^{(2)}, \, U^{(2)})$:

$$\begin{aligned} f_{T^{(2)}, \, U^{(2)}}(x,y)= & {} f_{T^{(3)}, \, U^{(3)}} \left[ \underline{g}^{-1}(x,y) \right] \times \left| J \left[ \underline{g}^{-1}(x,y) \right] \right| \\= & {} f_{T^{(3)}, \, U^{(3)}} \left[ \frac{n x}{n-1 + x}, \, \left( 1 + \frac{x}{n-1} \right) y \right] \times \frac{n}{n-1 + x}. \end{aligned}$$

After obtaining the joint p.d.f. of $T^{(3)}$ and $U^{(3)}$ from the joint c.d.f. in (15), we get the joint c.d.f. $T^{(2)}$ and $U^{(2)}$:

$$\begin{aligned} F_{{T}^{(2)}, \, U^{(2)}}(x,y)= & {} \int _{0}^x \frac{n-1}{(n-1+t)^2} \times f_{beta(p/2,(n-p)/2)} \left( \frac{t}{n-1+t} \right) \\&\times P\left[ \prod _{i=1}^p \chi _{n-i+1}^2 \le \frac{\hbox {det} \left( {\varvec{\varSigma }}_0 \right) }{\hbox {det} \left( {\varvec{\varSigma }} \right) } \times \left( 1+\frac{t}{n-1} \right) y \right] \, \hbox {d}t. \end{aligned}$$

$\square$

Appendix B

Let us remind the reader that the results in this appendix play a pivotal role in the obtention of control limits and out-of-control ARL values.

We can add the following auxiliary results referring to the product of independent central chi-square distributions $Q_p \equiv Q_p^{(n)} = \prod _{i=1}^p \chi _{n-i}^2$, for $p=2,3$, by capitalizing on the fact that $~2 \sqrt{Q_2} \sim \chi _{2(n-2)}^2$ (Anderson 1958, p. 172):

$$\begin{aligned} F_{Q_2} (q)= & {} F_{\chi _{2 (n-2)}^2} (2 \sqrt{q}) \\ F_{Q_2}^{-1}(z)= & {} \frac{1}{4} \left[ F_{\chi _{2 (n-2)}^2}^{-1}(z) \right] ^2, \, z \in (0,1); \\ Q_3= & {} Q_2 \times \chi _{n-3}^2 \\ F_{Q_3} (q)= & {} \int _{0}^{+\infty } F_{\chi _{2 (n-2)}^2} \left( 2 \sqrt{q/y} \right) \times f_{\chi _{n-3}^2} (y) \, \hbox {d}y \\ F_{Q_3}^{-1}(z)&\text{ is } \text{ the } \text{ root } q \text{ of } F_{Q_3} (q) = z, \, z \in (0,1). \end{aligned}$$

As for $p=4$, we have to take advantage of another result stated and proved by Wells et al. (1962). Let $I_n(x)$ and $K_\nu (x)$ represent the modified Bessel functions of the first and second kinds defined by

$$\begin{aligned} I_n(x)= & {} \sum _{m=0}^{+\infty } \frac{x^{2m+n}}{2^{2m+n} \, m! \, \varGamma (m+n+1)} \\ K_\nu (x)= & {} \frac{\pi \, [ I_{-\nu } (x) - I_{\nu } (x)]}{2 \sin (\nu \pi )}. \end{aligned}$$

Moreover, let $Y_1$ and $Y_2$ be two independent chi-square r.v. with $k_1$ and $k_2$ degrees of freedom, respectively. Then the p.d.f. of $W = Y_1 \, Y_2$ is equal to

$$\begin{aligned} f_W(w) = \frac{ w^{\frac{k_1}{4}+\frac{k_2}{4}-1} \, K_{\frac{k_1}{2}-\frac{k_2}{2}} (\sqrt{w})}{ 2^{\frac{k_1}{2}+\frac{k_2}{2}-1} \, \varGamma (\frac{k_1}{2}) \, \varGamma (\frac{k_2}{2})}. \end{aligned}$$

Finally, since $4 \sqrt{Q_4} \equiv W$ when $k_1=2(n-2)$ and $k_2 = 2(n-4)$, we get:

$$\begin{aligned} f_{4 \sqrt{Q_4}}(w)= & {} \frac{ w^{n-4} \, K_2 \left( \sqrt{w} \right) }{ 2^{2n-7} \, (n-3)! \, (n-5)! }\\ F_{Q_4} (q)= & {} P \left( \frac{Y_1^2}{4} \, \frac{Y_2^2}{4} \le q \right) = P \left( Y_1 \, Y_2 \le 4 \sqrt{q} \right) = \int _0^{4 \sqrt{q}} f_{4 \sqrt{Q_4}}(w) \, \hbox {d}w \\ F_{Q_4}^{-1}(z)&\text{ is } \text{ the } \text{ root } q \text{ of } F_{Q_4} (q) = z, \, z \in (0,1). \end{aligned}$$

We ought to add that, by taking advantage of Theorem 2 and condition (16), we can conclude that $\beta ^{(2)}$ is the root of

$$\begin{aligned} \int _{0}^{x} \frac{n-1}{(n-1+t)^2}\times & {} f_{beta(p/2,(n-p)/2)} \left( \frac{t}{n-1+t} \right) \\\times & {} F_{Q_p^{(n+1)}}\left[ \left( 1+\frac{t}{n-1} \right) y \right] \, \hbox {d}t = 1 - (ARL^\star )^{-1}, \end{aligned}$$

where $x = \frac{(n-1)p}{n-p} \times F_{F_{p,n-p}}^{-1} (1- \beta ^{(2)})$ and $y = F_{Q_p^{(n)}}^{-1}(1-\beta ^{(2)})$. Once $\beta ^{(2)}$ is obtained numerically, the upper control limits of the individual charts of scheme 2 soon follow.