Data-driven multiple criteria decision making for diagnosis of thyroid cancer

Abstract

In the era of the Internet and big data, data permeate the entire process of multiple criteria decision making (MCDM). Therefore, generation of rational solutions from current observations and historical data has become an important and interesting issue. To address this issue, this paper proposes a data-driven MCDM framework in the context of the evidential reasoning approach. Three challenges in the framework are met, including the transformation of observations into assessments, the learning of parameters and their constraints from historical data, and the generation of a data-driven solution. The proposed framework is then used to model the diagnosis of thyroid cancer and generate data-driven diagnostic results. The three challenges in the application are met to aid radiologists in improving the diagnostic accuracy of thyroid cancers. To examine whether the application of the proposed data-driven MCDM framework to the diagnosis of thyroid cancer can help improve diagnostic accuracy, we conduct a case study by using the examination reports of three radiologists from July 2015 to October 2017 in the ultrasonic department of a tertiary hospital located in Hefei, Anhui Province, China.

Appendix: Proof of Theorem 1

Theorem 1

Given the individual assessment B(e_i(a_l)) = {(P(Ω), 1)} and other individual assessments B(e_j(a_l)) = {(H_n, β_n,j(a_l)), n = 1, …, N; (Ω, β_Ω,j(a_l))} (j ≠ i), the aggregated assessment B(y(a_l)), by using the ER rule presented in Definition 1, satisfies Property 1.

Proof

To verify the conclusion in this theorem, the combination of the assessment B(e_i(a_l)) = {(P(Ω), 1)} with the aggregated result of the first (i − 1) assessments is presented as follows.

Suppose that {(H_n, β_n,b(i-1)(a_l)), n = 1, …, N; (Ω, β_Ω,b(i-1)(a_l))} is the aggregated result of the first (i − 1) assessments generated by using Definition 1. The combination result of this and B(e_i(a_l)) = {(P(Ω), 1)} is defined as

$$ \left\{ {\left( {H_{n} ,\beta_{n,b(i)} \left( {a_{l} } \right)} \right),\quad n = \, 1, \ldots ,N;\;\left( {\varOmega ,\beta_{\varOmega ,b(i)} \left( {a_{l} } \right)} \right)} \right\}, $$

(A.1)

where

$$ \beta_{n,b\left( i \right)} \left( {a_{l} } \right) = \frac{{\hat{\beta }_{n,b(i)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i)} (a_{l} )}}, $$

(A.2)

$$ \beta_{\varOmega ,b\left( i \right)} \left( {a_{l} } \right) = \frac{{\hat{\beta }_{\varOmega ,b(i)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i)} (a_{l} )}}, $$

(A.3)

$$ \vec{\beta }_{n,b(i)} (a_{l} ) = \frac{{\hat{\beta }_{n,b(i)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i)} (a_{l} ) + \hat{\beta }_{P(\varOmega ),b(i)} (a_{l} )}}, $$

(A.4)

$$ \vec{\beta }_{\varOmega ,b(i)} (a_{l} ) = \frac{{\hat{\beta }_{\varOmega ,b(i)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i)} (a_{l} ) + \hat{\beta }_{P(\varOmega ),b(i)} (a_{l} )}} $$

(A.5)

$$ \vec{\beta }_{P(\varOmega ),b(i)} (a_{l} ) = \frac{{\hat{\beta }_{P(\varOmega ),b(i)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i)} (a_{l} ) + \hat{\beta }_{P(\varOmega ),b(i)} (a_{l} )}}, $$

(A.6)

$$ \hat{\beta }_{n,b(i)} (a_{l} ) = \left( {1 - w_{i} } \right) \cdot \vec{\beta }_{n,b(i - 1)} (a_{l} ) + \vec{\beta }_{n,b(i - 1)} (a_{l} ) \cdot w_{i} , $$

(A.7)

$$ \hat{\beta }_{\varOmega ,b(i)} (a_{l} ) = \left( {1 - w_{i} } \right) \cdot \vec{\beta }_{\varOmega ,b(i - 1)} (a_{l} ) + \vec{\beta }_{\varOmega ,b(i - 1)} (a_{l} ) \cdot w_{i} $$

(A.8)

and

$$ \hat{\beta }_{P(\varOmega ),b(i)} (a_{l} ) = \left( {1 - w_{i} } \right) \cdot \vec{\beta }_{P(\varOmega ),b(i - 1)} (a_{l} ) + \vec{\beta }_{P(\varOmega ),b(i - 1)} (a_{l} ) \cdot w_{i} . $$

(A.9)

From Eqs. (A.7)–(A.9), we have

$$ \hat{\beta }_{n,b(i)} (a_{l} ) = \vec{\beta }_{n,b(i - 1)} (a_{l} ), $$

$$ \hat{\beta }_{\varOmega ,b(i)} (a_{l} ) = \vec{\beta }_{\varOmega ,b(i - 1)} (a_{l} ), $$

and

$$ \hat{\beta }_{P(\varOmega ),b(i)} (a_{l} ) = \vec{\beta }_{P(\varOmega ),b(i - 1)} (a_{l} ). $$

Through $\hat{\beta }_{n,b(i)} (a_{l} )$, $\hat{\beta }_{\varOmega ,b(i)} (a_{l} )$, and $\hat{\beta }_{P(\varOmega ),b(i)} (a_{l} )$, it can be deduced from Eqs. (A.4)–(A.6) that

$$ \vec{\beta }_{n,b(i)} (a_{l} ) = \frac{{\hat{\beta }_{n,b(i - 1)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i - 1)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i - 1)} (a_{l} ) + \hat{\beta }_{P(\varOmega ),b(i - 1)} (a_{l} )}} = \vec{\beta }_{n,b(i - 1)} (a_{l} ), $$

$$ \vec{\beta }_{\varOmega ,b(i)} (a_{l} ) = \frac{{\hat{\beta }_{\varOmega ,b(i - 1)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i - 1)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i - 1)} (a_{l} ) + \hat{\beta }_{P(\varOmega ),b(i - 1)} (a_{l} )}} = \vec{\beta }_{\varOmega ,b(i - 1)} (a_{l} ),\;{\text{and}} $$

$$ \vec{\beta }_{P(\varOmega ),b(i)} (a_{l} ) = \frac{{\vec{\beta }_{P(\varOmega ),b(i - 1)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i - 1)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i - 1)} (a_{l} ) + \hat{\beta }_{P(\varOmega ),b(i - 1)} (a_{l} )}} = \vec{\beta }_{P(\varOmega ),b(i - 1)} (a_{l} ). $$

Finally, because $\vec{\beta }_{n,b(i)} (a_{l} )$ = $\vec{\beta }_{n,b(i - 1)} (a_{l} )$, $\vec{\beta }_{\varOmega ,b(i)} (a_{l} )$ = $\vec{\beta }_{\varOmega ,b(i - 1)} (a_{l} )$, and $\vec{\beta }_{P(\varOmega ),b(i)} (a_{l} )$ = $\vec{\beta }_{P(\varOmega ),b(i - 1)} (a_{l} )$, one has from Eqs. (A.2)–(A.3) that

$$\begin{aligned} \beta_{n,b\left( i \right)} \left( {a_{l} } \right) &= \frac{{\hat{\beta }_{n,b(i)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i)} (a_{l} )}} = \frac{{\hat{\beta }_{n,b(i - 1)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i - 1)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i - 1)} (a_{l} )}}\\ &= \beta_{n,b(i - 1)} \left( {a_{l} } \right)\;{\text{and}}\end{aligned} $$

$$\begin{aligned} \beta_{\varOmega ,b\left( i \right)} \left( {a_{l} } \right) &= \frac{{\hat{\beta }_{\varOmega ,b(i)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i)} (a_{l} )}} = \frac{{\hat{\beta }_{\varOmega ,b(i - 1)} (a_{l} )}}{{\sum\nolimits_{n = 1}^{N} {\hat{\beta }_{n,b(i - 1)} (a_{l} )} + \hat{\beta }_{\varOmega ,b(i - 1)} (a_{l} )}}\\ & = \beta_{{\varOmega ,b\left( {i - 1} \right)}} \left( {a_{l} } \right).\end{aligned} $$

The above analysis indicates that the aggregated assessment B(y(a_l)), by using Definition 1, satisfies Property 1. □