Optimal estimator for uncertainty-based measurement quality control

Abstract

This paper presents an optimal estimator for uncertainty-based measurement quality control. The optimal estimator is developed based on an acceptance probability approach under a risk balance criterion. It yields a balance between the false acceptance and false rejection when the measurement quality index is equal to unity. This paper also presents a minimum mean absolute percentage error (MAPE) estimator and a minimum mean squared percentage error estimator based on the frequentist decision-theoretic approach. The mathematical formulations for computing the MAPE, relative bias error (RBE), relative precision error (RPE), and root mean squared percentage error (RMSPE) of the three presented estimators and five existing estimators are derived. The performance of these estimators are compared and evaluated in terms of the false acceptance/rejection probability, MAPE, RBE, RPE, and RMSPE. Among the eight estimators considered, the presented optimal estimator is the most meaningful and the best estimator from the measurement quality control perspective.

Appendices

Appendix 1: Solution to the integrals in Eq. (11)

Equation (11) is rewritten here

$$ {MAPE} = \int\limits_{0}^{a} {\left( {1 - \frac{x}{a}} \right)} f(x,k)\,{\text{d}}x + \int\limits_{a}^{\infty } {\left( {\frac{x}{a} - 1} \right)} \,f(x,k)\,{\text{d}}x $$

(31)

Rearranging the integrals in Eq. (31) yields

$$ \begin{aligned} {MAPE} = & \int\limits_{0}^{a} {f(x,k){\text{d}}x - } \frac{1}{a}\int\limits_{0}^{a} {x\,f(x,k){\text{d}}x} + \frac{1}{a}\int\limits_{a}^{\infty } {x\,f(x,k){\text{d}}x - } \int\limits_{a}^{\infty } {f(x,k){\text{d}}x} \\ = & {\rm P}\left( {\frac{k}{2},\frac{{a^{2} }}{2}} \right) - \frac{1}{a}\int\limits_{0}^{a} {x\,f(x,k){\text{d}}x} + \frac{1}{a}\int\limits_{0}^{\infty } {x\,f(x,k){\text{d}}x - \frac{1}{a}\int\limits_{0}^{a} {x\,f(x,k){\text{d}}x - \left[ {1 - {\rm P}\left( {\frac{k}{2},\frac{{a^{2} }}{2}} \right)} \right]} } \\ \end{aligned} $$

(32)

where ${\rm P}(\frac{k}{2},\frac{{a^{2} }}{2})$ is the regularized Gamma function with k degrees of freedom. Note that $\int\limits_{0}^{\infty } {x\,f(x,k){\text{d}}x} = c_{4} \sqrt {n - 1}$, which is the mean of the Chi distribution with k degrees of freedom. After some algebraic manipulation, Eq. (32) can be written as

$$ {MAPE} = 2 {\rm P}\left( {\frac{k}{2},\frac{{a^{2} }}{2}} \right) - 1 + c_{4} C - \frac{{2{\kern 1pt} C}}{{\sqrt {n - 1} }}\int\limits_{0}^{a} {x\,f(x,k){\text{d}}x} $$

(33)

The probability density function f(x, k) is written as

$$ f(x,k) = \frac{{2^{{1 - \frac{k}{2}}} x^{k - 1} {\text{e}}^{{ - \frac{{x^{2} }}{2}}} }}{{\Gamma \left( \frac{k}{2} \right)}} $$

(34)

where Г(.) stands for the Gamma function. xf(k, x) is then written as

$$ xf(x,k) = \frac{{2^{{1 - \frac{k}{2}}} x^{k} {\text{e}}^{{ - \frac{{x^{2} }}{2}}} }}{{\Gamma \left( \frac{k}{2} \right)}} $$

(35)

On the other hand, f(x, k + 1) can be written as

$$ f(x,k + 1) = \frac{{2^{{1 - \frac{k + 1}{2}}} x^{k} {\text{e}}^{{ - \frac{{x^{2} }}{2}}} }}{{\Gamma \left( {\frac{k + 1}{2}} \right)}} $$

(36)

Substituting Eq. (36) into Eq. (35) and after some algebraic manipulation, xf(k, x) becomes

$$ xf(x,k) = \frac{{\sqrt 2 \,\Gamma \left( {\frac{k + 1}{2}} \right)}}{{\Gamma \left( \frac{k}{2} \right)}}f(x,k + 1) $$

(37)

The fourth term in Eq. (33) then becomes

$$ \frac{{2{\kern 1pt} C}}{{\sqrt {n - 1} }}\int\limits_{0}^{a} {x\,f(x,k){\text{d}}x} = \frac{{2{\kern 1pt} C}}{{\sqrt {n - 1} }}\frac{{\sqrt 2 \,\Gamma \left( {\frac{k + 1}{2}} \right)}}{{\Gamma \left( \frac{k}{2} \right)}}\int\limits_{0}^{a} {f(x,k + 1){\text{d}}x} = 2\,c_{4} C\,{\rm P}\left( {\frac{k + 1}{2},\frac{{a^{2} }}{2}} \right) $$

(38)

where ${\rm P}(\frac{k + 1}{2},\frac{{a^{2} }}{2})$ is the regularized Gamma function with k + 1 degrees of freedom. Substituting Eq. (38) into Eq. (33) yields Eq. (12).

Appendix 2: Solution to the integral in Eq. (20)

Equation (20) is rewritten here

$$ {{MSPE}} = \int\limits_{0}^{\infty } {\left( {1 - \frac{{C{\kern 1pt} x}}{{\sqrt {n - 1} }}} \right)^{2} } \,f(x,k)\,{\text{d}}x $$

(39)

Equation (39) can be rewritten as

$$ \begin{aligned} {{MSPE}} = & \int\limits_{0}^{\infty } {\left( {1 - \frac{{2{\kern 1pt} C}}{{\sqrt {n - 1} }}x + \frac{{C^{2} {\kern 1pt} x^{2} }}{n - 1}} \right)} \,f(x,k)\,{\text{d}}x \\ = & \int\limits_{0}^{\infty } {\,f(x,k)\,{\text{d}}x} \, - \frac{{2{\kern 1pt} C}}{{\sqrt {n - 1} }}\int\limits_{0}^{\infty } x \,f(x,k)\,{\text{d}}x + \frac{{C^{2} }}{n - 1}\int\limits_{0}^{\infty } {x^{2} } \,f(x,k)\,{\text{d}}x \\ \end{aligned} $$

(40)

The first integral in Eq. (40) is equal to 1. The second term in Eq. (40) can be written as (replacing a with ∞ in Eq. (38))

$$ \frac{{2{\kern 1pt} C}}{{\sqrt {n - 1} }}\int\limits_{0}^{\infty } {x\,f(x,k){\text{d}}x} = \frac{2\,C}{{\sqrt {n - 1} }}\frac{{\sqrt 2 \,\Gamma \left( {\frac{k + 1}{2}} \right)}}{{\Gamma \left( \frac{k}{2} \right)}}\int\limits_{0}^{\infty } {f(x,k + 1){\text{d}}x} = 2\,c_{4} C $$

(41)

The integral of x ² f(k, x) in Eq. (40) can be resolved in a similar way as that for solving the integral of xf(k, x). First, write x ² f(k, x) as

$$ x^{2} f(x,k) = \frac{{2^{{1 - \frac{k}{2}}} x^{k + 1} {\text{e}}^{{ - \frac{{x^{2} }}{2}}} }}{{\Gamma \left( \frac{k}{2} \right)}} $$

(42)

Second, write f(x, k + 2) as

$$ f(x,k + 2) = \frac{{2^{{1 - \frac{k + 2}{2}}} x^{k + 1} {\text{e}}^{{ - \frac{{x^{2} }}{2}}} }}{{\Gamma \left( {\frac{k + 2}{2}} \right)}} $$

(43)

Then, substitute Eq. (43) into Eq. (42). After some algebraic manipulation, x ² f(k, x) becomes

$$ x^{2} f(x,k) = \frac{{2\,\Gamma \left( {\frac{k + 2}{2}} \right)}}{{\Gamma \left( \frac{k}{2} \right)}}f(x,k + 2) $$

(44)

The third term in Eq. (40) then becomes

$$ \begin{aligned} \frac{{C^{2} }}{n - 1}\int\limits_{0}^{\infty } {x\,^{2} f(x,k){\text{d}}x} = & \frac{{2\,C^{2} }}{n - 1}\frac{{\,\Gamma \left( {\frac{k + 2}{2}} \right)}}{{\Gamma \left( \frac{k}{2} \right)}}\int\limits_{0}^{\infty } {f(x,k + 2){\text{d}}x} \\ = & \frac{{2\,C^{2} }}{n - 1}\frac{{\,\frac{k}{2}\Gamma \left( \frac{k}{2} \right)}}{{\Gamma \left( \frac{k}{2} \right)}} = C^{2} \\ \end{aligned} $$

(45)

Substituting Eqs. (41) and (45) into Eq. (40) yields ${{MSPE}} = 1 - 2{\kern 1pt} c_{4} C + C^{2}$.