C Some proofs on ROC curves

C.1 The ROC curve

Assume we have a continuous (in mathematical terminology “absolutely continuous”) result $Y$ from a diagnostic test and denote the true and false positive fraction for a given threshold $c$

$$\begin{eqnarray*} \mbox{TPF}(c) & = & \Pr(Y \geq c \,\vert\,D=1) \\ \mbox{FPF}(c) & = & \Pr(Y \geq c \,\vert\,D=0). \end{eqnarray*}$$

The test result $Y$ will have different distributions in the diseased ($D=1$) and non-diseased ($D=0$) population and we will denote the corresponding random variable by $Y_D$ and $Y_{\bar D}$, respectively.

Further define the ROC curve via

$$\begin{eqnarray*} \mbox{ROC}(.) & = & \{(\mbox{FPF}(c), \mbox{TPF}(c)), c \in (-\infty, \infty)\}, \end{eqnarray*}$$

the ROC curve are the points $(\mbox{FPF}(c), \mbox{TPF}(c))$ for all possible thresholds $c$. Now define the survivor functions in the diseased and non-diseased populations:

$$\begin{eqnarray*} S_D(y) & = & \Pr(Y \geq y \,\vert\,D=1) = \mbox{TPF}(y) \\ S_{\bar D}(y) & = & \Pr(Y \geq y \,\vert\,D=0) = \mbox{FPF}(y). \end{eqnarray*}$$

Note that the survivor functions are strictly monotone and hence invertible.

For a given threshold $c$, suppose $t=\mbox{FPF}(c)= S_{\bar D}(c)$, so $c=S_{\bar D}^{-1}(t)$ and we obtain $$\mbox{ROC}(t) = \mbox{TPF}(c) = S_D(c) = S_D(S_{\bar D}^{-1}(t)).$$

C.2 AUC

The area under the curve (AUC) is defined as $$\mbox{AUC} = \int_0^1 \mbox{ROC}(t)dt$$ and we have

$$\begin{equation} \tag{C.1} \mbox{AUC} = \Pr(Y_D > Y_{\bar D}), \end{equation}$$

if $Y_D$ and $Y_{\bar D}$ are independent test results from the diseased and non-diseased populations, respectively.

To proof this result, we first note that the derivative of a survivor function $S(y)$ of a continous test result $Y$ is closely related to the cumulative distribution function $F(y)$ of $Y$: $$S(y) = \Pr(Y \geq y) = 1 - \Pr(Y < y) = 1 - \Pr(Y \leq y) = 1 - F(y)$$ Now $$\begin{eqnarray*} {\frac{d\,F(y)}{d\,y}} = f(y) \end{eqnarray*}$$ where $f(y)$ denotes the density function of $Y$. Therefore $${\frac{d\,S(y)}{d\,y}} = -f(y).$$

Now consider $$\begin{eqnarray} \tag{C.2} \mbox{AUC} &=& \int_0^1 \mbox{ROC}(t)dt \nonumber \\ &=& \int_0^1 S_D(S_{\bar D}^{-1}(t))dt \nonumber \\ &=& \int_{\infty}^{-\infty} S_D(y) \left\{ -f_{\bar D}(y) dy \right\}\nonumber \\ &=& \int_{-\infty}^{\infty} S_D(y) f_{\bar D}(y) dy, \end{eqnarray}$$ where we have applied the substitution $t=S_{\bar D}(y)$, so $$dt = d S_{\bar D}(y) = -f_{\bar D}(y)dy.$$ Let $\mathsf{I}\{A\}$ denote the indicator function of an event $A$. Now $Y_D$ and $Y_{\bar D}$ are assumed to be independent, so $$\begin{eqnarray*} \Pr(Y_D \geq Y_{\bar D}) &=& \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \mathsf{I}\{y_D \geq y_{\bar D}\} f(y_D, y_{\bar D}) dy_D dy_{\bar D} \\ &=& \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \mathsf{I}\{y_D \geq y_{\bar D}\} f_D(y_D) f_{\bar D}(y_{\bar D}) dy_D dy_{\bar D} \\ &=& \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \left[ \mathsf{I}\{y_D \geq y_{\bar D}\} f_D(y_D)dy_D \right] f_{\bar D}(y_{\bar D}) dy_{\bar D} \\ &=& \int_{-\infty}^{\infty} \Pr(Y_D \geq y_{\bar D}) f_{\bar D}(y_{\bar D}) dy_{\bar D} \\ &=& \int_{-\infty}^{\infty} S_D(y_{\bar D}) f_{\bar D}(y_{\bar D}) dy_{\bar D} \end{eqnarray*}$$

where the last equation is equal to (C.2) with $y_{\bar D}=y$. This proofs equation (C.1).