Estimators of parameters of a mixture of three multinomial distributions based on simple majority results

Abstract

For assessing the precision of measurement systems that classify items dichotomically with the possibility of repeated ratings, the maximum likelihood method is commonly used to evaluate misclassification probabilities. However, a computationally simpler and more intuitive approach is the method of simple majority. In this approach, each item is labelled as conforming if the majority of repeated classification outcomes are conforming. A previous study has indicated that this technique yields estimators that have a lower mean squared error than but the same asymptotic properties as the corresponding maximum likelihood estimators. However, there are circumstances in which the use of measurement systems with a wider scale of responses is necessary. In this paper, we propose estimators based on simple majority results for evaluating the classification errors of measurement systems that rate items in a trichotomous domain. We investigate their properties and compare their performance with that of maximum likelihood estimators. We focus on the context in which the true quality states of the objects cannot be determined. The simple majority procedure is modelled using a mixture of three multinomial distributions. The proposed estimators are shown to be a competitive alternative because they offer closed-form expressions and demonstrate a performance similar to that of maximum likelihood estimators.

Appendix

Proof of Lemma:

The condition “$W_0^i> W_1^i \text{ and } W_0^i > W_2^i$” in the definition of $F_i$ is equivalent to the event $``2 W_1^i + W_2^i< r \text{ and } W_1^i + 2 W_2^i < r''$, as $W_0^i + W_1^i + W_2^i = r$. Therefore,

$$\begin{aligned} \begin{array}{lll} \big [ F_i = 0 \big ] &{} = &{} \big [ 2 W_1^i + W_2^i< r,\ W_1^i + 2 W_2^i < r \big ] \bigcup \Big [ U_1 = 0,\ W_0^i = W_2^i> W_1^i\Big ] \\ &{} &{} \bigcup \Big [ U_2 = 0,\ W_0^i = W_1^i > W_2^i \Big ] \bigcup \Big [ U_4 = 0,\ \displaystyle { W_0^i = W_1^i = W_2^i } \Big ] ,\ \end{array} \end{aligned}$$

and according to Fig. 3, $\Big [ W_1^i + W_2^i < \displaystyle { \frac{r}{2}} \Big ] \subset \big [ F_i = 0 \big ]$; consequently,

Fig. 3

Regions of “integration”

$$\begin{aligned} P \Big ( W_1^i + W_2^i < \displaystyle { \frac{r}{2}}\ \Big |\ Y_i =0 \Big ) \le P\big ( F_i = 0 \ \big |\ Y_i = 0 \big ) . \end{aligned}$$

However, $P \Big ( W_1^i + W_2^i < \displaystyle { \frac{r}{2}}\ \Big |\ Y_i =0 \Big ) = P \Big ( W_0^i > \displaystyle { \frac{r}{2} }\ \Big |\ Y_i =0 \Big )$, and $W_0^i \Big |\ Y_i = 0$ follows a binomial distribution with a number of trials r and a probability of “success” of $1 - \pi _0^1 - \pi _0^2$. Therefore, for sufficiently large r, we obtain the following approximation:

$$\begin{aligned} \begin{array}{lll} P\Big ( W_0^i > \displaystyle { \frac{r}{2}} \Big |\ Y_i = 0 \Big ) &{} \approx &{} P \Bigg ( Z \ge \displaystyle { \frac{ \displaystyle {\lfloor {r/2} \rfloor + 1 } - r \cdot \Big [ 1- \pi _0^1 - \pi _0^2 \Big ]}{ \sqrt{ r \cdot \Big [ \pi _0^1 + \pi _0^2 \Big ] \cdot \Big [ 1 - \pi _0^1 - \pi _0^2 \Big ] }}} \Bigg ) \\ &{} = &{} 1- \Phi \Bigg (\displaystyle { \frac{ \displaystyle {\lfloor {r/2} \rfloor + 1 } - r \cdot \Big [ 1- \pi _0^1 - \pi _0^2 \Big ]}{ \sqrt{ r \cdot \Big [ \pi _0^1 + \pi _0^2 \Big ] \cdot \Big [ 1 - \pi _0^1 - \pi _0^2 \Big ] }}} \Bigg ) , \end{array} \end{aligned}$$

where Z follows a standard normal distribution and $\Phi$ is its cumulative distribution function. The restrictions expressed in (3) imply that

$$\begin{aligned} \displaystyle { \frac{ \displaystyle {\lfloor {r/2} \rfloor + 1 } - r \cdot \Big [ 1- \pi _0^1 - \pi _0^2 \Big ]}{ \sqrt{ r \cdot \Big [ \pi _0^1 + \pi _0^2 \Big ] \cdot \Big [ 1 - \pi _0^1 - \pi _0^2 \Big ] }}} \longrightarrow - \infty \ \text{ as } r \rightarrow \infty . \end{aligned}$$

Hence, $\displaystyle { \lim _{r \rightarrow \infty } }\ P \Big ( W_1^i + W_2^i < \displaystyle { \frac{r}{2}}\ \Big |\ Y_i =0 \Big ) = 1$; thus,

$$\begin{aligned} \displaystyle { \lim _{r \rightarrow \infty }}\ P\big ( F_i = k \ \big | \ Y_i =k \big ) = 1\ \text{ for } k=0 . \end{aligned}$$

The proofs are analogous for $k=1$ and $k =2$. This completes the demonstration of the assertion presented in (8). The equality given in (9) is an immediate consequence of (8), and from

$$\begin{aligned} \begin{array}{lll} P \big ( F_i = 0 \big ) &{} = &{} \displaystyle { \sum _{y=0}^2 P\big ( F_i = 0\ \big |\ Y_i = y \big )\cdot P\big ( Y_i =y \big )} \\ &{} = &{} \displaystyle { \theta _0 \cdot P\big ( F_i = 0\ \big |\ Y_i = 0 \big ) + \sum _{y=1}^2 \theta _y \cdot P\big ( F_i = 0\ \big |\ Y_i = y \big )} , \end{array} \end{aligned}$$

it follows that $\displaystyle { \lim _{r \rightarrow \infty } }\ P \big ( F_i = k \big ) = \theta _k$ for $k=0$. Similarly, we can demonstrate that this fact is valid for $k=1$ and $k=2$, and thus, equality (10) is proven. $\square$

Proof of Theorem 1:

Note that

$$\begin{aligned} \text{ E } \big [ F_i( F_i -2) \big ] = \sum _{f=0}^2 \text{ E } \big [ F_i( F_i -2) \big | F_i =f \big ]\cdot P\big ( F_i = f \big ) = - P\big ( F_i =1 \big ) . \end{aligned}$$

Therefore, $\text{ E } \big [ \widehat{\theta }_1 \big ] = \displaystyle {- \frac{1}{n} } \displaystyle {\sum _{i= 1 }^{n}} \text{ E } \big [ F_i ( F_i -2) \big ] = P\big ( F_i =1 \big )$, and by the limit given in (10), we can conclude that $\displaystyle { \lim _{r \rightarrow \infty } }\, \text{ E } \big [ \widehat{\theta }_k \big ] = \theta _k$ for $k=1$. This equality can also be established for $k=0$ and $k=2$ in a similar manner. $\square$

Proof of Theorem 2:

The variance of $\widehat{\theta }_1$ can be written as

$$\begin{aligned} \text{ Var }\big [\widehat{\theta }_1 \big ] =\displaystyle { \frac{1}{n^2} }\displaystyle { \sum _{i=1}^n } \text{ Var }\big [ F_i ( F_i -2) \big ] . \end{aligned}$$

Moreover,

$$\begin{aligned} \text{ E } \big [ \big ( F_i( F_i -2)\big )^2 \big ] = \displaystyle { \sum _{f=0}^2 } \text{ E } \big [\big ( F_i( F_i -2)\big )^2 \big | F_i =f \big ] \cdot P\big ( F_i =f \big ) = P\big ( F_i = 1 \big ) . \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{array}{lll} \text{ Var }\big [ F_i ( F_i -2) \big ] &{} = &{} \text{ E } \Big [ \big ( F_i( F_i -2)\big )^2 \Big ] - \Big ( \text{ E } \big [ F_i( F_i -2) \big ] \Big )^2 \\ &{} = &{} P\big (F_i =1 \big ) \big [ 1 - P\big (F_i =1 \big ) \big ] , \end{array} \end{aligned}$$

and thus, $\begin{array}{lll} \text{ Var }\big [\widehat{\theta }_1 \big ]= & {} \displaystyle { \frac{1}{n} \cdot P\big (F_i =1 \big ) \big [ 1 - P\big (F_i =1 \big ) \big ]}. \end{array}$ By Chebyshev’s inequality, for all real $\epsilon >0$,

$$\begin{aligned} P\big ( \big | \widehat{\theta }_1 - \theta _1 \big | \ge \epsilon \big ) \le \displaystyle { \frac{ \text{ E } \big [ \big ( \widehat{\theta }_1 - \theta _1 \big )^2 \big ] }{\epsilon ^2} = \frac{\text{ Var }\big [\widehat{\theta }_1 \big ] + \big ( \text{ E } \big [ \widehat{\theta }_1 \big ] - \theta _1 \big )^2 }{ \epsilon ^2 }} . \end{aligned}$$

However, $\text{ Var }\big [\widehat{\theta }_1 \big ] \rightarrow 0$ as $n \rightarrow \infty$, and $\text{ E } \big [ \widehat{\theta }_1 \big ] \rightarrow \theta _1$ as $r\rightarrow \infty$. Consequently, $\widehat{\theta }_1$ converges in probability to $\theta _1$ as $n \rightarrow \infty$ and $r\rightarrow \infty$, that is, $\widehat{\theta }_1$ is consistent. The same reasoning and conclusion are valid for the statistics $\widehat{\theta }_0$ and $\widehat{\theta }_2$. $\square$

Proof of Theorem 3:

To prove that $\widehat{\pi }_0^1$ satisfies the conditions described, we note that the sequence $\displaystyle { \Bigg \{ \frac{(F_i-1) (F_i - 2)}{2} \Bigg \}_{r \ge 1}}$ converges in probability$\big (\overset{P}{\longrightarrow }\big )$ to the variable ${\displaystyle { \frac{(Y_i -1)(Y_i -2)}{2}}}$, that is, for all real $\epsilon > 0$,

$$\begin{aligned} \displaystyle { \lim _{r \rightarrow \infty } \displaystyle { P \Bigg [ \Bigg | \frac{(F_i-1) (F_i - 2)}{2} - \frac{(Y_i -1)(Y_i -2)}{2} \Bigg | \ge \epsilon \Bigg ]}} = 0 . \end{aligned}$$

(14)

In fact, if we define $I_{\big [F_i = 0\big ]} = \displaystyle { \frac{(F_i-1) (F_i - 2)}{2}}$ and $I_{\big [Y_i = 0\big ]} = \displaystyle { \frac{(Y_i -1)(Y_i -2)}{2}}$, then we can write

$$\begin{aligned}&\displaystyle { \displaystyle { P \Bigg [ \Bigg | \frac{(F_i-1) (F_i - 2)}{2} - \frac{(Y_i -1)(Y_i -2)}{2} \Bigg | \ge \epsilon \Bigg ]}} \\= & {} \displaystyle { \sum _{y=0}^{2} \sum _{f=0}^{2} } \displaystyle { \Big \{ \displaystyle { P \Big [ \Big | I_{\big [F_i = 0\big ]} - I_{\big [Y_i = 0\big ]} \Big | \ge \epsilon \ \Big | Y_i =y,\ F_i = f \Big ]}} \\&\times P\big ( F_i = f \big | Y_i =y \big ) \cdot P\big ( Y_i =y \big ) \Big \} \\= & {} \displaystyle {\sum _{(y,f) \in A} } P\big ( 1 \ge \epsilon \big | Y_i =y,\ F_i = f \big ) \cdot P\big ( F_i = f \big | Y_i =y \big )\cdot P\big ( Y_i =y \big ) , \end{aligned}$$

where $A = \{ (0,1),\ (0,2),\ (1,0),\ (2,0) \}$. Therefore, by applying (9), we obtain the equality given in (14). In addition, the sequence $\displaystyle { \Bigg \{ \frac{ W_1^i}{r} \cdot \frac{(F_i-1) (F_i - 2)}{2}} \Bigg \}_{ r \ge 1}$ converges in mean square to the variable $\displaystyle { \pi _{0}^{1}} \cdot \frac{(Y_i -1)(Y_i -2)}{2}$, that is,

$$\begin{aligned} \displaystyle {\lim _{r \rightarrow \infty } \text{ E } \Bigg [ \Bigg \{ \frac{ W_1^i}{r} \cdot I_{\big [F_i = 0\big ]} - \pi _{0}^{1} \cdot I_{\big [Y_i = 0\big ]} \Bigg \}^2 \Bigg ]} = 0. \end{aligned}$$

(15)

Indeed, if we let

$$\begin{aligned} \displaystyle { \text{ E }_{y,f} = \text{ E } \Bigg [ \Bigg \{ \frac{ W_1^i}{r} \cdot I_{\big [F_i = 0\big ]} - \pi _{0}^{1} \cdot I_{\big [Y_i = 0\big ]} \Bigg \}^2 \Bigg | Y_i =y,F_i = f \Bigg ] } , \end{aligned}$$

then we can write $\displaystyle { \text{ E } \Bigg [ \Bigg \{ \frac{ W_1^i}{r} \cdot I_{\big [F_i = 0\big ]} - \pi _{0}^{1} \cdot I_{\big [Y_i = 0\big ]} \Bigg \}^2 \Bigg ]}$

$$\begin{aligned} = \displaystyle { \sum _{y=0}^{2} \sum _{f=0}^{2} } \text{ E }_{y,f} \cdot P\big ( Y_i =y,\ F_i = f \big )= \text{ E }_{0,0} \cdot P\big (F_i = 0 \big | Y_i =0 \big )\cdot P\big ( Y_i = 0 \big ) + \mathfrak {R} , \end{aligned}$$

where $\mathfrak {R} = \displaystyle { \sum _{(y,f) \in A} \text{ E }_{y,f} \cdot P\big (F_i = f \big | Y_i =y \big )\cdot P\big ( Y_i = y \big )}$ and A is the set defined previously. Because the expected values that appear in $\mathfrak {R}$ are bounded, the limits given in (9) imply that $\displaystyle { \lim _{r \rightarrow \infty } } \mathfrak {R} = 0$. For this reason, we now focus on the asymptotic behaviour of $\text{ E }_{0,0}$.

Note that

$$\begin{aligned} P \big ( W_1^i = w_1 \big | Y_i = 0, F_i =0 \big ) = \displaystyle { \frac{ P \big ( W_1^i = w_1, Y_i = 0, F_i =0 \big )}{ P \big ( Y_i =0, F_i =0 \big ) }} \end{aligned}$$

$$\begin{aligned} \begin{array}{lll} &{} \le &{} \displaystyle { \frac{ P \big ( W_1^i = w_1, Y_i = 0 \big )}{ P \big ( Y_i =0, F_i =0 \big ) }} = \displaystyle { \frac{ P \big ( W_1^i = w_1, Y_i = 0 \big )}{ P \big ( Y_i =0 \big ) }} \cdot \displaystyle { \frac{ P \big ( Y_i = 0 \big )}{ P \big ( Y_i =0, F_i =0 \big ) }} \\ &{} = &{} P \big ( W_1^i = w_1 \big | Y_i = 0 \big ) \cdot \displaystyle {\frac{1}{ P \big ( F_i = 0 \big | Y_i = 0 \big ) }} . \end{array} \end{aligned}$$

Thus, $\text{ E }_{0,0} = \text{ E } \Bigg [ \Bigg \{ \displaystyle { \frac{ W_1^i}{r} } - \pi _{0}^{1} \Bigg \}^2 \Bigg | Y_i =0,\ F_i = 0 \Bigg ]$

$$\begin{aligned} \le \text{ E } \Bigg [ \Bigg \{ \displaystyle { \frac{ W_1^i}{r} } - \pi _{0}^{1} \Bigg \}^2 \Bigg | Y_i =0 \Bigg ] \cdot \displaystyle {\frac{1}{ P \big ( F_i = 0 \big | Y_i = 0 \big ) }} . \end{aligned}$$

(16)

Furthermore, as $W_1^i \big |\, Y_i = 0$ follows a binomial distribution with a number of runs r and a probability of “success” $\pi _0^1$, we conclude that $\displaystyle { \text{ E } \Bigg [ \frac{ W_1^i}{r} \ \Bigg |\ Y_i = 0 \Bigg ] = \pi _0^1}$ and

$$\begin{aligned} \text{ E } \Bigg [ \Bigg \{ \displaystyle { \frac{ W_1^i}{r} } - \pi _{0}^{1} \Bigg \}^2 \Bigg | Y_i =0 \Bigg ] = \displaystyle { \text{ Var } \Bigg [ \frac{ W_1^i}{r} \ \Bigg |\ Y_i = 0 \Bigg ] = \frac{1}{r} \cdot \pi _0^1 \cdot \Big [1 - \pi _0^1 \Big ] } . \end{aligned}$$

From the last equality, inequality (16) and result (10), it follows that $\displaystyle {\lim _{r \rightarrow \infty } \text{ E }_{0,0} } = 0$. This proves (15) and ensures the abovementioned convergence.

Markov’s inequality guarantees that

$$\begin{aligned} \displaystyle {\displaystyle {\frac{ W_1^i }{r} \cdot \frac{(F_i-1) (F_i - 2)}{2} \ \ \overset{P}{\longrightarrow } \ \ \displaystyle { \displaystyle { \pi _{0}^{1}} \cdot \frac{(Y_i -1)(Y_i -2)}{2} }}} . \end{aligned}$$

Thus, as a consequence of the probability limits established above, we have

$$\begin{aligned} \displaystyle {\frac{1}{2} \sum _{i=1}^n (F_i-1) (F_i - 2) } \ \ \overset{P}{\longrightarrow }\ \ {\displaystyle \frac{1}{2} \sum _{i=1}^n (Y_i -1)(Y_i -2) } \end{aligned}$$

(17)

and

$$\begin{aligned} \displaystyle {\displaystyle {\frac{1}{2r}} \sum _{i=0}^n W_1^i} \cdot (F_i-1) (F_i - 2) \ \ \overset{P}{\longrightarrow } \ \ \displaystyle { \frac{\pi _{0}^{1} }{2} \sum _{i=1}^n (Y_i -1)(Y_i -2) } , \end{aligned}$$

which imply that

$$\begin{aligned} \displaystyle { \widehat{\pi }_{0}^{1} } = \frac{ \displaystyle { \frac{1}{2r} \sum _{i=0}^n } W_1^i \cdot (F_i-1) (F_i - 2) }{\displaystyle { \frac{1}{2} \sum _{i=1}^n} (F_i-1) (F_i - 2)} \ \ \overset{P}{\longrightarrow }\ \ \frac{ \displaystyle { \frac{\pi _{0}^{1} }{2} \sum _{i=1}^n (Y_i -1)(Y_i -2) } }{ \displaystyle { \frac{1}{2} \sum _{i=1}^n (Y_i -1)(Y_i -2) } } = \pi _{0}^{1}\ , \end{aligned}$$

i.e, $\widehat{\pi }_{0}^{1}$ is a consistent estimator. Additionally, as this statistic is bounded,

$$\begin{aligned} \displaystyle { \lim _{r \rightarrow \infty } \text{ E } \Big [ \widehat{\pi }_{0}^{1} \Big ] = \pi _{0}^{1}} , \end{aligned}$$

that is, $\widehat{\pi }_{0}^{1}$ is an asymptotically unbiased estimator. By the same reasoning, it can be shown that the other statistics $\widehat{\pi }_k^{k'}$ are consistent and asymptotically unbiased. $\square$

Proof of Remark 2:

The convergence of the first two components follows directly from (10). To prove $\displaystyle { \lim _{r \rightarrow \infty } \pi _{k,r}^{k'}} = \pi _k^{k'}$, note that Bayes’ Theorem and limits (8)–(10) together imply that

$$\begin{aligned} \displaystyle { \lim _{r \rightarrow \infty } P\big ( Y_i = y\ \big | F_i = f \big )} \ = \ \left\{ \begin{array}{rrr} 1 &{} \text{ if } &{} f =y , \\ 0 &{} \text{ if } &{} f \ne y . \end{array} \right. \end{aligned}$$

(18)

Moreover, we can write

$$\begin{aligned} \begin{array}{lll} \pi _{0,r}^1 &{} = &{} P \big ( X_{ij} = 1 \big | F_i = 0 \big ) = \displaystyle { \frac{ P \big ( X_{ij} = 1, F_i =0 \big )}{P \big ( F_i = 0 \big )}} \\ &{} = &{} \displaystyle { \frac{1}{P \big ( F_i = 0 \big )} \sum _{y =0}^2 P \big ( X_{ij} = 1, F_i =0, Y_i = y \big )} \\ &{}\le &{} \displaystyle { \frac{ P \big ( X_{ij} = 1, Y_i = 0 \big ) + P \big ( F_i =0, Y_i = 1 \big )+ P \big ( F_i =0, Y_i = 2 \big ) }{P \big ( F_i = 0 \big )}} \\ &{} = &{} \displaystyle { \frac{ P \big ( X_{ij} = 1, Y_i = 0 \big )}{ P \big ( Y_i = 0 \big ) } \frac{ P\big (Y_i = 0 \big )}{P\big (F_i = 0 \big )} + P \big ( Y_i =1 \big | F_i = 0 \big ) + P \big ( Y_i = 2 \big | F_i = 0 \big ) } \\ &{} = &{} \displaystyle { \pi _0^1 \cdot \frac{ P\big (Y_i = 0 \big )}{P\big (F_i = 0 \big )} + P \big ( Y_i =1 \big | F_i = 0 \big ) + P \big ( Y_i = 2 \big | F_i = 0 \big )} , \end{array} \end{aligned}$$

and through a similar strategy, we obtain

$$\begin{aligned} \displaystyle { \pi _0^1 \le \pi _{0,r}^1 \cdot \frac{ P\big (F_i = 0 \big )}{P\big (Y_i = 0 \big ) } + P \big ( F_i =1 \big | Y_i = 0 \big ) + P \big ( F_i = 2 \big | Y_i = 0 \big ) } . \end{aligned}$$

From the above inequality and limits (8)–(10) and (18), it can be found that $\displaystyle { \lim _{r \rightarrow \infty } \pi _{0,r}^1 = \pi _0^1}$. Using the same arguments, this result can be proven for the other values of k and $k'$. $\square$