# Showing that $P(X_1>X_2) = \int_{0}^1 P(X_1>X_2 | X_2=x) f_{X_2}(x) dx$

I am going through this post in trying to prove the probabilistic interpretation of the AUC for a ROC Curve (for a classifier):

The AUC for a ROC curve is the the probability of the classifier scoring a randomly chosen positive point higher than a randomly chosen negative one.

I was able to follow till the end, but am a little stuck on getting the final result. So far I have the following

\begin{align*} {AUC} &= \int_0^1 tpr(fpr) d(fpr)\\ &= \int_0^1 tpr(fpr(s))d(fpr(s))\\ &= \int_{+\infty}^{-\infty} tpr(s)fpr'(s)ds\\ &= \int_{+\infty}^{-\infty} \left(1-F_1(s)\right) \left(-f_{-1}(s)\right)ds\\ &= \int_{-\infty}^{+\infty} (1-F_1(s)) (f_{-1}(s))ds \\ &= \int_{-\infty}^{+\infty} Pr(ss) (f_{-1}(s))ds \\ \end{align*}

I am struggling in my confidence that

$$\int_{-\infty}^{+\infty} Pr(X_{-1}

is correct. I found the result in this post, but cannot find it elsewhere for continuous random variables. Is this a standard definition without proof? I have not seen it before and I can't find anything about conditional probabilities that makes me get it, could I get some help on that part?

Thanks

• should the integral of title from $-\infty$ to $\infty$, but not 0 to 1?
– Tan
Commented Jan 6, 2021 at 4:07
• Related references: De Schuymer, Bart, Hans De Meyer, and Bernard De Baets. "A fuzzy approach to stochastic dominance of random variables", in International Fuzzy Systems Association World Congress 2003; Montes Gutiérrez, I., "Comparison of alternatives under uncertainty and imprecision", doctoral thesis, Universidad de Oviedo, 2014. Commented Jan 6, 2021 at 4:49

$$\sum_{s=-\infty}^\infty P[X_{-1} and on the right $$\sum_{s=-\infty}^\infty P[X_{-1} so for each $$s$$ it's just the equality $$P(A|B)P(B)=P(A\cap B)$$
In the continuous case, probabilities with equalities turn into densities times $$ds$$, and sums into integrals.
It's probably easiest to use the law of iterated expectations. To do so we can write the probability as the expectation of an indicator: $$\mathbb{P}(X_1 > X_2) = \mathbb{E}[ \mathbb{I}\{X_1 > X_2 \} ]$$ The law of iterated expectations states that \begin{align*} \mathbb{E}[ \mathbb{I}\{X_1 > X_2 \} ] &= \mathbb{E}[ \mathbb{E}[\mathbb{I}\{X_1 > X_2 \} \mid X_2]] \\ &= \mathbb{E}[ \mathbb{P}(X_1 > X_2 \mid X_2)] \end{align*} The second line substitutes the definition of conditional probability as the expectation of an indicator function (similar to what we did before).
Finally, if $$X_2$$ has a continuous distribution, we can put the results together to show that $$\mathbb{P}(X_1 > X_2) = \mathbb{E}[ \mathbb{P}(X_1 > X_2 \mid X_2)] = \int \mathbb{P}(X_1 > X_2 \mid X_2 = x_2) f(x_2) dx_2$$ Continuity of $$X_2$$ is only used for the second equality, which substitutes the integral definition of the expectation. I like this proof because you don't have to take a stance on whether $$X_1$$ or $$X_2$$ are discrete/continuous until the last step.