The relationship between Gini terminology and the ROC curve

Abstract

The objectives of this note are to correct a common error and to clarify the connection between the Gini terminology as used in the economic literature and the one used in the diagnostic and classification literature. More specifically, the connection between the area under the receiver operating characteristic (ROC) curve, which is frequently used in the diagnosis and classification literature, and the Gini terminology, which is mainly used in the economic literature, is clarified. It is shown that the area under the ROC curve is related to the covariance between the two vectors $Y=\{y_i\}_{i=1}^{n_0}$ and $\{i/{n_0}\}_{i=1}^{n_0}$. Here $y_i$ is the number of items classified to group 1 lying between the $(i-1)\mathrm{th}$ and the $i\mathrm{th}$ items classified to group 0, and $n_0$ is the number of items in group 0.

Appendix: Proofs of Theorems 1 and 2

Proof of Theorem 1:

Recall the definition of ${\tilde{t}}_i$ given just before the statement of Theorem 1 and denote $x_i=F_T(\tilde{t}_i)=i/n_0$, $i=1,\dots ,n_0$

$$\begin{aligned} \text {Cov}(Y,F_T(T))=\frac{1}{n_0}\sum _{i=1}^{n_0}(x_i-\bar{x})(y_i-\bar{y})=\frac{1}{n_0}\sum _{i=1}^{n_0}x_iy_i-\frac{n_0+1}{2n_0}\bar{y} \end{aligned}$$

(5)

We start with the first term of the right-hand side of Eq. (5).

$$\begin{aligned} \frac{1}{n_0}\sum _{i=1}^{n_0}x_iy_i= & {} \frac{1}{n_0}\sum _{i=1}^{n_0}\left( \sum _{j=1}^{i}x_j-\sum _{j=0}^{i-1}x_j\right) y_i =\frac{1}{n_0}\sum _{i=1}^{n_0}\left( \sum _{j=1}^{i}x_j-\sum _{j=1}^{i}x_{j-1}\right) y_i\nonumber \\= & {} \frac{1}{n_0}\sum _{i=1}^{n_0}\sum _{j=1}^{i}(x_j-x_{j-1})y_i =\frac{1}{n_0}\sum _{j=1}^{n_0}\sum _{i=j}^{n_0}(x_j-x_{j-1})y_i\nonumber \\= & {} \frac{1}{n_0}\sum _{j=1}^{n_0}(x_j-x_{j-1})\sum _{i=j}^{n_0}y_i =\frac{1}{n^2_0}\sum _{j=1}^{n_0}\left( \sum _{i=j}^{n_0}y_i\right) \nonumber \\= & {} \frac{1}{n^2_0}\sum _{j=1}^{n_0}\left( \sum _{i=1}^{n_0}y_i-\sum _{i=1}^{j-1}y_i\right) =\bar{y}-\frac{1}{n^2_0}\sum _{j=1}^{n_0}\sum _{i=1}^{j-1}y_i. \end{aligned}$$

(6)

So

$$\begin{aligned} cov(Y,F_T(T))= & {} \frac{1}{n_0}\sum _{i=1}^{n_0}x_iy_i-\frac{n_0+1}{2n_0}\bar{y}=\bar{y}-\frac{1}{n^2_0}\sum _{j=1}^{n_0}\sum _{i=1}^{j-1}y_i-\frac{n_0+1}{2n_0}\bar{y}\nonumber \\= & {} \frac{n_0-1}{2n_0}\bar{y}-\frac{1}{n^2_0}\sum _{j=1}^{n_0}\sum _{i=1}^{j-1}y_i= \frac{n_0+1}{2n^2_0}\sum _{i=1}^{n_0}y_i- \frac{1}{n^2_0}\sum _{j=1}^{n_0+1}\sum _{i=1}^{j-1}y_i . \end{aligned}$$

(7)

We still need to show that

$$\begin{aligned} \sum _{j=1}^{n_0+1}\sum _{i=1}^{j-1}y_i=\sum _{i=1}^{n_0}t_i. \end{aligned}$$

Indeed,

$$\begin{aligned} \sum _{j=1}^{n_0+1}\sum _{i=1}^{j-1}y_i= & {} \sum _{i=1}^{n_0}(n_0+1-i)y_i=\sum _{i=1}^{n_0}(n_0+1-i)(t_i-t_{i-1})\\= & {} \sum _{i=0}^{n_0}(n_0+1-i)t_i-\sum _{i=0}^{n_0-1}(n_0-i)t_i=\sum _{i=0}^{n_0-1}t_i+t_{n{_0}}=\sum _{i=1}^{n_0}t_i. \end{aligned}$$

Proof of Theorem 2:

Note that $\sum _{i=1}^{n_0}y_i=n_1$, so $\bar{y}=n_1/n_0$ and by Theorem 1,

$$\begin{aligned} -\text {Cov}(Y,F_T(T))+0.5\bar{y}=\frac{1}{n^2_0}\sum _{i=1}^{n_0}t_i-\frac{(n_0+1)n_1}{2n^2_0}+0.5\bar{y}=\frac{1}{n^2_0}\sum _{i=1}^{n_0}t_i-\frac{n_1}{2n^2_0}. \end{aligned}$$

Dividing both sides by $\bar{y}=\frac{n_1}{n_0}$ we get

$$\begin{aligned} -\frac{n_0}{n_1}\text {Cov}(Y,F_T(T))+0.5=\frac{\sum _{i=1}^{n_0}t_i}{n_1n_0}-\frac{1}{2n_0}=\hat{A}-\frac{1}{2n_0}. \end{aligned}$$

where the last equation follows (3). This completes the proof.