# Why is ROC AUC equivalent to the probability that two randomly-selected samples are correctly ranked? [duplicate]

I found there are two ways to understand what AUC stands for but I couldn't get why these two interpretations are equivalent mathematically.

In the first interpretation, AUC is the area under the ROC curve. Picking points from 0 to 1 as threshold and calculate sensitivity and specificity accordingly. When we plot them against each other, we get ROC curve.

The second one is that the AUC of a classifier is equal to the probability that the classifier will rank a randomly chosen positive example higher than a randomly chosen negative example, i.e. P(score(x+)>score(x−)). (from What does AUC stand for and what is it?)

It's easy to see once you obtained a closed-form formula for AUC.

Since we have finite number of samples $$\{(x_i, y_i)\}_{i=1}^N$$, we'll have finite number of points on the ROC curve. We do linear interpolation in between.

First, some definitions. Suppose we'd like to evaluate an algorithm $$A(x)$$ that outputs a probability of $$x$$ lying in the positive class $$+1$$. Let's define $$N_+$$ as the number of samples in the positive class $$+1$$ and $$N_-$$ as the number of samples in the negative class $$-1$$. Now, for a threshold $$\tau$$ let's define False-Positive-Rate (FPR, aka 1-specificity) and True-Positive-Rate (TPR, aka sensitivity):

$$\text{TPR}(\tau) = \frac{\sum_{i=1}^N [y_i = +1] [A(x_i) \ge \tau]}{N_+} \quad \text{and} \quad \text{FPR}(\tau) = \frac{\sum_{i=1}^N [y_i = -1] [A(x_i) \ge \tau]}{N_-}$$

(where $$[\text{boolean expression}]$$ is 1 if expression is true, and 0 otherwise). Then, ROC curve is built from points of the form $$(\text{FPR}(\tau), \text{TPR}(\tau))$$ for different values of $$\tau$$. Moreover, it's easy to see that if we order our samples $$x_{(i)}$$ (note the parentheses) according to the algorithm's output $$A(x_i)$$, then neither $$\text{TPR}$$ nor $$\text{FPR}$$ changes for $$\tau$$ between consecutive samples $$A(x_{(i)}) < \tau < A(x_{(i+1)})$$. So it's enough to evaluate FPR and TPR only for $$\tau \in \{A(x_{(1)}), \dots, A(x_{(N)})\}$$. For $$k^{\text{th}}$$ point we have

$$\text{TPR}_k = \frac{\sum_{i=k}^N [y_{(i)} = +1]}{N_+} \quad \text{and} \quad \text{FPR}_k = \frac{\sum_{i=k}^N [y_{(i)} = -1]}{N_-}$$

(Note both sequences are non-increasing in $$k$$). These sequences define x and y coordinates of points on the ROC curve. Next, we linearly interpolate these points to get the curve itself and calculate area under the curve (Using a formula for area of a trapezoid):

\begin{align*} \text{AUC} &= \sum_{k=1}^{N-1} \frac{\text{TPR}_{k+1} + \text{TPR}_{k}}{2} (\text{FPR}_{k} - \text{FPR}_{k+1}) \\ &= \sum_{k=1}^{N-1} \frac{\sum_{i=k+1}^N [y_{(i)} = +1] + \tfrac{1}{2} [y_{(k)} = +1]}{N_+} \frac{[y_{(k)} = -1]}{N_-} \\ &= \frac{1}{N_+ N_-} \sum_{k=1}^{N-1} \sum_{i=k+1}^N [y_{(i)} = +1] [y_{(k)} = -1] = \frac{1}{N_+ N_-} \sum_{k < i} [y_{(k)} < y_{(i)}] \end{align*}

Here I used the fact that $$[y = -1] [y = +1] = 0$$ for any $$y$$.

So there you have it: AUC is proportional to the number of correctly ordered pairs, which is proportional to the probability of random pair of samples being ranked according to their labels.

EDIT (6 years later): Since for $$a, b \in \{-1, +1\}$$ we have $$[a < b] = 1$$ only when $$a = -1$$ and $$b = +1$$, it's easy to see that $$\frac{1}{N_+ N_-} \sum_{k < i} [y_{(k)} < y_{(i)}] = \frac{1}{N_+ N_-} \sum_{\substack{k < i \\ y_{(i)} = 1 \\ y_{(k)} = -1}} [y_{(k)} < y_{(i)}]$$

In essence, we form all possible negative-positive pairs and see what fraction of them is correctly ordered according to our algorithm $$A$$, that is, $$A($$positive sample$$)\; > A($$negative sample$$)$$.

• Thanks @Barmaley.exe. This is really helpful. Finally connect the dots. A following up question though, is how AUC is equivalent to Mann Whitney U test. Jan 15, 2016 at 19:48
• I don't have an answer to that, but either way it'd not be a good idea to answer another (mostly unrelated) question here. I suggest you create a new thread. Jan 15, 2016 at 20:09
• This is all covered in Hanley and McNeil 1982: pubs.rsna.org/doi/10.1148/radiology.143.1.7063747 (Wilcoxon-Mann-Whitney equivalence etc.). Jan 13, 2019 at 13:17
• Can you please clarify how did the numerator before the last step is rearranged? where did $1/2[y_(k) = +1]$ go? May 11, 2020 at 20:51
• @christopher, $[y_{(k)} = +1] [y_{(k)} = -1]$ is always 0. May 11, 2020 at 21:19