[Math] How to compute this integral involving a cdf

Question

$\int_0^\infty\Phi(\frac{-x}{\sqrt{2}})d\Phi(x)=?$ where $\Phi(x)$ is the cumulative distribution function of a standard normal random variable.

Answer 1

Consider $I(a) = \int_0^\infty \Phi(a x) \mathrm{d} \Phi(x)$. Differentiate with respect to $a$, and denote $\phi(x) = \Phi^\prime(x)$: $$ I^\prime(a) = \int_0^\infty x \phi(a x) \phi(x) \mathrm{d} x = \frac{1}{2 \pi} \int_0^\infty x \mathrm{e}^{-\frac{(1+a^2) x^2}{2}} \mathrm{d} x = \frac{1}{2 \pi} \frac{1}{1+ a^2} $$ Now, noting that $I(0) = \int_0^\infty \frac{1}{2} \mathrm{d} \Phi(x) = \frac{1}{4}$: $$ I\left(a\right) = \frac{1}{4} + \frac{1}{2 \pi} \int_{0}^{a} \frac{\mathrm{d} a}{1+a^2} = \frac{1}{4} + \frac{1}{2 \pi} \arctan(a) $$ Now $I\left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{4} - \frac{1}{2 \pi} \arctan\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2 \pi} \arctan\left(\sqrt{2}\right) \approx 0.152043 $