We assume a three-class setting, where one or more classifier yields measurements \(X = x\) on a continuous scale for the groups of healthy (\(D^-\)), intermediate (\(D^0\)) and diseased (\(D^+\)) individuals. By convention, larger values of \(x\) represent a more severe status of the disease. The package trinROC provides statistical tests to assess the discriminatory power of such classifiers. These tests are:
In this document, we refer to the tests by the corresponding R function name. All tests can assess a single classifier, as well as compare two paired or unpaired classifiers.
As their names suggest, the underlying testing approach is different and either based on VUS or on ROC. Hence, their null hypotheses differ as well, as illustrated now.
VUS based statistical tests
Given two classifiers, boot.test and trinVUS.test are based on the null hypothesis \(VUS_1 = VUS_2\) with the \(Z\)-statistic
\[\begin{align*}
Z = \frac{\widehat{VUS}_1 - \widehat{VUS}_2}{\sqrt{\widehat{Var}(\widehat{VUS}_1) + \widehat{Var}(\widehat{VUS}_2) - 2 \widehat{Cov}(\widehat{VUS}_1,\widehat{VUS}_2)}}.
\end{align*}\]
If the data of the two classifiers is unpaired, the term \(Cov(VUS_1,VUS_2)\) is zero. If a single classifier is investigated, the null hypothesis is \(VUS_1 = 1/6\) with the \(Z\)-statistic
\[\begin{align*}
Z = \frac{\widehat{VUS}_1 - 1/6}{\sqrt{\widehat{Var}(\widehat{VUS}_1)}},
\end{align*}\]
which is equivalent to compared the VUS of the classifier to the volume of an uninformative classifier. Details about the estimators are given in the aforementioned papers.
The trinormal based ROC test
In the trinormal model, the ROC surface is given by
\[\begin{align*}
ROC(t_-,t_+) = \Phi \left(\frac{\Phi^{-1} (1-t_+) + d}{ c} \right) - \Phi \left(\frac{\Phi^{-1} (t_-)+b}{a} \right),
\end{align*}\]
where \(\Phi\) is the cdf of the standard normal distribution and \(a\), \(b\), \(c\) and \(d\) are functions of the means and standard deviations of the three groups:
\[\begin{align*}
a = \frac{{\sigma}_0}{{\sigma}_-}, \qquad b = \frac{ {\mu}_- - {\mu}_0}{{\sigma}_-}, \qquad
c = \frac{{\sigma}_0}{{\sigma}_+}, \qquad d = \frac{ {\mu}_+ - {\mu}_0}{{\sigma}_+}.
\end{align*}\]
Given two classifiers, trinROC.test investigates the shape of the two ROC surfaces. Hence, the resulting null hypothesis is \(a_1=a_2\), \(b_1=b_2\), \(c_1=c_2\) and \(d_1=d_2\), i.e., the the surfaces have the same shape. Under the null hypothesis, the test statistic
\[\begin{align*}
\chi^2 =
\begin{pmatrix} \widehat{a}_1 - \widehat{a}_2 &\widehat{b}_1 -\widehat{b}_2 & \widehat{c}_1-\widehat{c}_2 & \widehat{d}_1-\widehat{d}_2 \end{pmatrix}
{ \widehat{\boldsymbol{W}}}^{-1}
\begin{pmatrix} \widehat{a}_1 - \widehat{a}_2 \\
\widehat{b}_1 - \widehat{b}_2 \\ \widehat{c}_1-\widehat{c}_2 \\ \widehat{d}_1- \widehat{d}_2 \end{pmatrix},
\end{align*}\]
is distributed approximately as a chi-squared random variables with four degrees of freedom, where \(\widehat{\boldsymbol{W}}\) contains the corresponding estimated variances and covariances. The test rejects if \(\chi^2 > \chi_{\alpha}^2\), i.e., if the test statistics exceeds the chi-squared quantile with four degrees of freedom of a pre-defined confidence level \(\alpha\).
When the data is unpaired, \(\widehat{a}_1\), \(\widehat{b}_1\), \(\widehat{c}_1\), \(\widehat{d}_1\) are independent from \(\widehat{a}_2\), \(\widehat{b}_2\), \(\widehat{c}_2\), \(\widehat{d}_2\), respectively, and hence every combination of covariances between them is zero.
It is also possible to investigate a single classifier with the above method. Instead of an existing second classifier, we compare the estimates \(\widehat{a}_1\) , \(\widehat{b}_1\), \(\widehat{c}_1\) and \(\widehat{d}_1\) of a single marker with those from an artificial marker. If we want to detect whether a single marker is significantly better in allocating individuals to the three classes than a random allocation function we would set the parameters of our null hypothesis \(a_{Ho} = 1= c_{Ho}\), as we assume equal spread in the classes and \(b_{Ho} = 0 = d_{Ho}\), as we impose equal means. This yields the null hypothesis \(a_1=1\), \(b_1=0\), \(c_1=1\), \(d_1=0\), which leads to the chi-squared test
\[\begin{align*}
{\chi}^2 = &
\begin{pmatrix} \widehat{a}_1 -1 & \widehat{b}_1 & \widehat{c}_1 -1 & \widehat{d}_1 \end{pmatrix}
\widehat{\boldsymbol{W}}^{-1}
\begin{pmatrix} \widehat{a}_1 -1 \\ \widehat{b}_1 \\ \widehat{c}_1 - 1 \\ \widehat{d}_1 \end{pmatrix}.
\end{align*}\]
