# How to derive the probabilistic interpretation of the AUC?

Why is the area under the ROC curve the probability that a classifier will rank a randomly chosen "positive" instance (from the retrieved predictions) higher than a randomly chosen "positive" one (from the original positive class)? How does one prove this statement mathematically using integral, giving the CDFs and PDFs of the true positive and negative class distributions?

First thing, let's try to define the area under the ROC curve formally. Some assumptions and definitions:

• We have a probabilistic classifier that outputs a "score" s(x), where x are the features, and s is a generic increasing monotonic function of the estimated probability p(class = 1|x).

• $f_{k}(s)$, with $k = \{0, 1\}$ := pdf of the scores for class k, with CDF $F_{k}(s)$

• The classification of a new observation is obtained compraing the score s to a threshold t

Furthermore, for mathematical convenience, let's consider the positive class (event detected) k = 0, and negative k = 1. In this setting we can define:

• Recall (aka Sensitivity, aka TPR): $F_{0}(t)$ (proportion of positive cases classified as positive)
• Specificity (aka TNR): $1 - F_{1}(t)$ (proportion of negative cases classified as negative)
• FPR (aka Fall-out): 1 - TNR = $F_{1}(t)$

The ROC curve is then a plot of $F_{0}(t)$ against $F_{1}(t)$. Setting $v = F_1(s)$, we can formally define the area under the ROC curve as: $$AUC =\int_{0}^{1} F_{0}(F_{1}^{-1}(v)) dv$$ Changing variable ($dv = f_{1}(s)ds$): $$AUC =\int_{ - \infty}^{\infty} F_{0}(s) f_{1}(s)ds$$

This formula can easiliy be seen to be the probability that a randomly drawn member of class 0 will produce a score lower than the score of a randomly drawn member of class 1.

This proof is taken from: https://pdfs.semanticscholar.org/1fcb/f15898db36990f651c1e5cdc0b405855de2c.pdf

@alebu's answer is great. But its notation is nonstandard and uses 0 for the positive class and 1 for the negative class. Below are the results for the standard notation (0 for the negative class and 1 for the positive class):

Pdf and cdf of the score for negative class: $$f_0(s)$$ and $$F_0(s)$$

Pdf and cdf of the score for positive class: $$f_1(s)$$ and $$F_1(s)$$

FPR = $$x(s) = 1-F_0(s)$$

TPR = $$y(s) = 1-F_1(s)$$

\begin{align} \text{AUC} &= \int_0^1 y(x) dx\\ &= \int_0^1 y(x(\tau)) dx(\tau) \\ &= \int_{+\infty}^{-\infty} y(\tau) x'(\tau) d\tau \\ &= \int_{+\infty}^{-\infty} \big( 1-F_1(\tau) \big) \big( -f_0(\tau) \big) d\tau \\ &= \int_{-\infty}^{+\infty} \big( 1-F_1(\tau) \big) f_0(\tau) d\tau \end{align}

where $$\tau$$ stands for threshold. One can apply the interpretation in @alebu's answer to the last expression.

The way to calculate AUC-ROC is to plot out the TPR and FPR as the threshold, $\tau$ is changed and calculate the area under that curve. But, why is this area under the curve the same as this probability? Let's assume the following:

1. $A$ is the distribution of scores the model produces for data points that are actually in the positive class.
2. $B$ is the distribution of scores the model produces for data points that are actually in the negative class (we want this to be to the left of $A$).
3. $\tau$ is the cutoff threshold. If a data point get's a score greater than this, it's predicted as belonging to the positive class. Otherwise, it's predicted to be in the negative class.

Note that the TPR (recall) is given by: $P(A>\tau)$ and the FPR (fallout) is given be: $P(B>\tau)$.

Now, we plot the TPR on the y-axis and FPR on the x-axis, draw the curve for various $\tau$ and calculate the area under this curve ($AUC$).

We get:

$$AUC = \int_0^1 TPR(x)dx = \int_0^1 P(A>\tau(x))dx$$ where $x$ is the FPR. Now, one way to calculate this integral is to consider $x$ as belonging to a uniform distribution. In that case, it simply becomes the expectation of the $TPR$.

$$AUC = E_x[P(A>\tau(x))] \tag{1}$$ if we consider $x \sim U[0,1)$ .

Now, $x$ here was just the $FPR$

$$x=FPR = P(B>\tau(x))$$ Since we considered $x$ to be from a uniform distribution,

$$P(B>\tau(x)) \sim U$$ $$=> P(B<\tau(x)) \sim (1-U) \sim U$$ \begin{equation}=> F_B(\tau(x)) \sim U \tag{2}\end{equation}

But we know from the inverse transform law that for any random variable $X$, if $F_X(Y) \sim U$ then $Y \sim X$. This follows since taking any random variable and applying its own CDF to it leads to the uniform.

$$F_X(X) = P(F_X(x)<X) =P(X<F_X^{-1}(X))=F_XF_X^{-1}(X)=X$$ and this only holds for uniform.

Using this fact in equation (2) gives us: $$\tau(x) \sim B$$

Substituting this into equation (1) we get:

$$AUC=E_x(P(A>B))=P(A>B)$$

In other words, the area under the curve is the probability that a random positive sample will have a higher score than a random negative sample.