Assume that there are
n participants in the sample and three test results for every subject. We represent the observed data as
\(Y=(Y_{t_{1},t_{2},t_{3}})\), where
\(Y_{t_{1},t_{2},t_{3}}\) is the number of subjects with
\(T_{1}=t_{1}\),
\(T_{2}=t_{2}\), and
\(T_{3}=t_{3}\); here
\(t_{1},t_{2},t_{3}=0,1\). For example,
\(Y_{111}\) denotes the number of subjects whose diagnostic results for all three tests indicate that the syndrome factor is present. Correspondingly,
\(p_{t_{1},t_{2},t_{3}}\) represents the joint probability of the outcome
\((T_{1}=t_{1},T_{2}=t_{2},T_{3}=t_{3})\), which is defined as follows:
$$\begin{aligned} p_{t_{1},t_{2},t_{3}} & = {} P(T_{1}=t_{1},T_{2}=t_{2},T_{3}=t_{3})\nonumber \\ &= {} P(T_{1}=t_{1},T_{2}=t_{2},T_{3}=t_{3}|D=1)\times P(D=1)\nonumber \\ & \quad + {} P(T_{1}=t_{1},T_{2}=t_{2},T_{3}=t_{3}|D=0)\times P(D=0)\nonumber \\& = {} P(T_{1}=t_{1},T_{2}=t_{2},T_{3}=t_{3}|D=1)\times \pi \nonumber \\ & \quad + {} P(T_{1}=t_{1},T_{2}=t_{2},T_{3}=t_{3}|D=0)\times (1-\pi ). \end{aligned}$$
(1)
Among the three tests in this study, the first two tests represent the diagnostic results of the SSDC and ISDS, respectively, and the last test represents the diagnostic result of the expert opinion, called the ESD. Since the two diagnostic scales consist of standardized questionnaires, while the ESD is based on the individual opinion of expert CM practitioners, it is reasonable to assume that the CM expert and the diagnostic scales err independently (i.e., they are conditionally independent, given the true CM syndrome status). Nevertheless, the two diagnostic scales do not err independently (i.e., they are conditionally dependent, given the true CM syndrome status). Such dependence is measured by the conditional dependence correlations, given the true CM syndrome status. Hence, we assume that
\(T_{3}\) is independent of
\(T_{1}\) and
\(T_{2}\) conditional on
D, while we allow
\(T_{1}\) and
\(T_{2}\) to be conditionally dependent, given
D. Let
\(C_{+}\) and
\(C_{-}\) denote the covariance between
\(T_{1}\) and
\(T_{2}\) among the CM syndrome positive and negative individuals, respectively. In other words,
\(C_{+}=cov(T_{1},T_{2}|D=1)\) and
\(C_{-}=cov(T_{1},T_{2}|D=0)\). Such a model has also been studied by Dendukuri and Joseph [
17]. To present the Bayesian method, we need to compute the likelihood function of the observed data. Note that we can respectively write
\(P(T_{1}=t_{1},T_{2}=t_{2}|D=1)\) and
\(P(T_{1}=t_{1},T_{2}=t_{2}|D=0)\) as follows:
$$\begin{aligned} P(T_{1}=t_{1},T_{2}=t_{2}|D=1)=\prod \limits _{i=1}^{2}Se_{i}^{t_{i}}(1-Se_{i})^{(1-t_{i})}+(-1)^{t_{1}+t_{2}}C_{+}, \\ P(T_{1}=t_{1},T_{2}=t_{2}|D=0)=\prod \limits _{i=1}^{2}Sp_{i}^{(1-t_{i})}(1-Sp_{i})^{t_{i}}+(-1)^{t_{1}+t_{2}}C_{-}. \end{aligned}$$
Consequently, we can rewrite the joint probability of the outcome (
T
1 =
t
1,
T
2 =
t
2,
T
3 =
t
3) as follows:
$$\begin{aligned}
p_{t_{1},t_{2},t_{3}} &= P(T_{1}=t_{1},T_{2}=t_{2}|D=1)\times
P(T_{3}=t_{3}|D=1)\times \pi \nonumber \\
& \quad + P(T_{1}=t_{1},T_{2}=t_{2}|D=0)\times
P(T_{3}=t_{3}|D=0)\times (1-\pi )\nonumber \\
& = \pi \left[\prod \limits_{i=1}^{2}
Se_{i}^{t_{i}}(1-Se_{i})^{(1-t_{i})} + (-1)^{t_{1}+t_{2}}
C_{+}\right]\\&\quad\times
[Se_{3}^{t_{3}}(1-Se_{3})^{(1-t_{3})}]\nonumber \\
&\quad +(1-\pi )\left[\prod \limits
_{i=1}^{2}Sp_{i}^{(1-t_{i})}(1-Sp_{i})^{t_{i}}+(-1)^{t_{1}+t_{2}}C_{-}\right]\nonumber
\\& \quad \times
[(1-Sp_{3})^{t_{3}}Sp_{3}^{(1-t_{3})} ].
\end{aligned}$$
(2)
Let
\(Y=(Y_{111},Y_{110},Y_{101},Y_{100},Y_{011},Y_{010},Y_{001},Y_{000})\), the observed data, and
\(\theta =(Se_{1},Sp_{1},Se_{2},Sp_{2},Se_{3},Sp_{3},\pi ,C_{+},C_{-})\), which represents the set of parameters in the model. According to (
2), the likelihood function based on the observed data is:
$$\begin{aligned} L(\theta |Y)= & {} \prod \limits _{t_{1},t_{2},t_{3}}p_{t_{1},t_{2},t_{3}}^{Y_{t_{1},t_{2},t_{3}}}\nonumber \\= & {} \prod \limits _{t_{1},t_{2},t_{3}}\Big \{ \pi \Big [\prod \limits _{i=1}^{2}Se_{i}^{t_{i}}(1-Se_{i})^{(1-t_{i})}+(-1)^{t_{1}+t_{2}}C_{+}\Big ]\nonumber \\\times & \,{} \Big [Se_{3}^{t_{3}}(1-Se_{3})^{(1-t_{3})}\Big ]+(1-\pi )\Big [\prod \limits _{i=1}^{2}Sp_{i}^{(1-t_{i})}(1-Sp_{i})^{t_{i}}\nonumber \\+ &\, {} (-1)^{t_{1}+t_{2}}C_{-}\Big ] \times \Big [(1-Sp_{3})^{t_{3}}Sp_{3}^{(1-t_{3})}\Big ]\Big \}^{Y_{t_{1},t_{2},t_{3}}} \end{aligned}$$
(3)
To use the Bayesian method to estimate the vector of the parameters,
\(\theta \), we need to specify a prior distribution for
\(\theta \). Let
\(f(\theta )\) denote the prior distribution of
\(\theta \). The Bayesian method combines the prior information about
\(\theta \) with the data we have collected, and then uses the Bayes theorem to obtain an interpretable posterior distribution for
\(\theta \). We can use the median of the posterior distribution to estimate
\(\theta \). According to the Bayes theorem, the joint posterior distribution
\(f(\theta |Y)\) of the parameter
\(\theta \) given the observed data
Y can be written as follows:
$$\begin{aligned} f(\theta |Y)=\frac{L(\theta |Y)f(\theta )}{\int L(\theta |Y)f(\theta )d\theta }=\frac{\mathcal {A}}{\mathcal {B}}, \end{aligned}$$
(4)
where
$$\begin{aligned} \mathcal {A}&= f(\theta )\prod \limits _{t_{1},t_{2},t_{3}}\Big \{ \pi \Big [\prod \limits _{i=1}^{2}Se_{i}^{t_{i}}(1-Se_{i})^{(1-t_{i})}+(-1)^{t_{1}+t_{2}}C_{+}\Big ]\\& \quad \times \Big [Se_{3}^{t_{3}}(1-Se_{3})^{(1-t_{3})}\Big ]+(1-\pi )\quad\times\Big [\prod \limits _{i=1}^{2}Sp_{i}^{(1-t_{i})}(1-Sp_{i})^{t_{i}}+(-1)^{t_{1}+t_{2}}C_{-}\Big ]\\& \quad \times \Big [(1-Sp_{3})^{t_{3}}Sp_{3}^{(1-t_{3})}\Big ]\Big \}^{Y_{t_{1},t_{2},t_{3}}}, \end{aligned}$$
and
$$\begin{aligned} \mathcal {B} &= {} \underbrace{\int \int \cdots \int }_{9}f(\theta )\prod \limits _{t_{1},t_{2},t_{3}}\\&\quad\times\Big \{ \pi \Big [\prod \limits _{i=1}^{2}Se_{i}^{t_{i}}(1-Se_{i})^{(1-t_{i})}+(-1)^{t_{1}+t_{2}}C_{+}\Big ] \\ & \quad \times {} \Big [Se_{3}^{t_{3}}(1-Se_{3})^{(1-t_{3})}\Big ]+(1-\pi )\Big [\prod \limits _{i=1}^{2}Sp_{i}^{(1-t_{i})}(1-Sp_{i})^{t_{i}}+(-1)^{t_{1}+t_{2}}C_{-}\Big ]\\ \\ & \quad \times {} \Big [(1-Sp_{3})^{t_{3}}Sp_{3}^{(1-t_{3})}\Big ]\Big \}^{Y_{t_{1},t_{2},t_{3}}} \underbrace{dSe_{1}dSe_{2}\cdots dC_{+}dC_{-}}_{9} \end{aligned}$$
Consequently, the marginal posterior density function for any component in
\(\theta \), such as
\(Sp_{2}\), given the data, can be expressed as:
$$\begin{aligned} f(Sp_{2}|Y)&=\underbrace{\int \int \cdots \int \int }_{8}f(\theta |Y)dSe_{1} \\ & \quad \times dSe_{2}dSe_{3}dSp_{1}dSp_{3}d\pi dC_{+}dC_{-} \end{aligned}$$
By estimating the median of the margin distribution about
\(Sp_{2}\), denoted by
\(\widehat{{Sp}_{2}}\), we obtain a Bayesian estimate,
\(\widehat{{Sp}_{2}}\), for
\(Sp_{2}\).