Turns out I wrote a [medium article][1] just for that.

TL;DR : to go through the end of the demonstration, one needs to use the **convolution theorem**.

If you don't want to change sites, here is the full trick.
We want to show that, for a given binary classifier, :

$$ROC-AUC = P\left(X_1>X_0\right) = P\left(X_1-X_0>0\right)$$

where :
- **X₁** is a continuous random variable giving the “score” output by our binary classifier for a randomly chosen **positive** sample
- **X₀** is a continuous random variable giving the “score” output by our binary classifier for a randomly chosen **negative** sample

Definitions and preliminary results

First, some definitions :

- Let **X₁** and **X₀** be defined as above
- Let **f₁** and **f₀** be, respectively, the density function of X₁ and X₀
- Let **F₁** and **F₀** be, respectively, the repartition function of X₁ and X₀
- **True Positive Rate (TPR)** and **False Positive Rate (FPR)** have their usual meaning, i.e. :

$$TPR=\frac{TP}{P}\:\:\,FPR=\frac{FP}{N}$$

We can already observe that, for a classifier threshold T, a randomly chosen positive sample would be correctly classified (true positive) if X₁>T. So, for a randomly chosen positive sample, the probability of correctly classifying it is P(X₁>T). By definition of the TPR, it corresponds to the probability of correctly classifying a randomly chosen positive sample, so **TPR(T) = P(X₁>T) = 1- P(X₁⩽ T) = 1-F₁(T)**. (1)

This also means, by definition of the density function, that :

$$TPR(T) = \int\limits_{T}^{+\infty} f_1(x)\: \mathrm{d}x$$

Similarly, we can show that FPR(T) = 1- F₀(T) (2)
Demonstration

Now let’s dig into the calculus!

By definition of the ROC, we have that :

$$ROC-AUC = \int\limits_0^1 TPR(FPR)\: \mathrm{d}FPR$$
$$= \int\limits_0^1 TPR(FPR^{-1}(x))\: \mathrm{d}x$$

By using this change in variable :

$$T=FPR^{-1}(x)\iff\ x=FPR(T)$$

the integral becomes :

$$\int\limits_{+\infty}^{-\infty} TPR(T) \times FPR'(T)\: \mathrm{d}T$$

Now, thanks to (2) we know that we can express this integral as :

$$\int\limits_{+\infty}^{-\infty} TPR(T) \times (-f_0(T))\: \mathrm{d}T = \int\limits_{-\infty}^{+\infty} TPR(T) \times f_0(T)\: \mathrm{d}T$$

Thanks to (1) we know that this can be expressed as :

$$\int\limits_{-\infty}^{+\infty} \int\limits_{T}^{+\infty} f_1(x)\: \mathrm{d}x \times f_0(T)\: \mathrm{d}T$$

By using this change in variable for the inner integral :

$$v=x-T$$

the integral becomes :

$$\int\limits_{-\infty}^{+\infty} \int\limits_{0}^{+\infty} f_1(v+T)\: \mathrm{d}v \times f_0(T)\: \mathrm{d}T$$
$$= \int\limits_{0}^{+\infty} \int\limits_{-\infty}^{+\infty}    f_0(T)\: \mathrm{d}T \times \: f_1(v+T)\: \mathrm{d}v$$

and by using this change in variable for the inner integral :

$$u=v+T$$

it becomes :

$$\int\limits_{0}^{+\infty} \int\limits_{-\infty}^{+\infty} f_1(u)\:  \times f_0(u-v)\: \mathrm{d}u \: \mathrm{d}v$$

Do you get where we’re going? Yes, right to the convolution theorem!
First, let’s point out that since f₀(t) is a density function of X₀, **f₀(-t) is a density function of (-X₀)**.
Then, according to the convolution theorem and assuming the convergence, **a density of X₁- X₀=X₁+(- X₀) is** :

$$\int\limits_{-\infty}^{+\infty} f_1(u)\:  \times f_0(u-v)\: \mathrm{d}u$$

This means that :

$$P\left(X_1>X_0\right)=P\left(X_1-X_0>0\right)$$
$$=\int\limits_{0}^{+\infty} \int\limits_{-\infty}^{+\infty} f_1(u)\:  \times f_0(u-v)\: \mathrm{d}u \: \mathrm{d}v$$

And eventually we have that :

$$P\left(X_1>X_0\right) = ROC - AUC$$


  [1]: https://medium.com/@nathanaim/mathematics-behind-roc-auc-interpretation-e4e6f202a015