How to compute this integral involving a cdf? $\int_0^\infty\Phi(\frac{-x}{\sqrt{2}})d\Phi(x)=?$ where $\Phi(x)$ is the cumulative distribution function of a standard normal random variable.
 A: 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 $
A: Sasha's comment following his answer suggests a different calculation that does not require knowledge of the antiderivative of $(1+a^2)^{-1}$, only pie-cutting
or using the circular symmetry of the joint density of two independent standard
normal random variables .
$$\begin{align*}
\int_0^{\infty}\Phi(ax)\;\mathrm d\Phi(x)
&= \int_0^{\infty}\Phi(ax)\phi(x)\mathrm\; dx\\
&= \int_0^{\infty}\left[ \int_{-\infty}^{ax}\phi(y)\;\mathrm dy\right]
\phi(x)\mathrm\; dx\\
&= \int_0^{\infty} \int_{-\infty}^{ax}\phi(y)
\phi(x)\;\mathrm dy\;\mathrm dx\\
&= \frac{1}{2\pi}\int_{r=0}^{\infty}\int_{\theta=-\pi/2}^{\arctan(a)}
\exp(-r^2/2) \cdot r\;\mathrm d\theta\;\mathrm dr\\
&= \frac{\arctan(a)+\pi/2}{2\pi}\\
&= \frac{1}{4} + \frac{1}{2\pi}\arctan(a).
\end{align*}$$
A: Consider a slightly more general integral:
\begin{eqnarray}
\int\limits_b^\infty \phi(\xi) \Phi(a \xi) d\xi &=& \int\limits_{{\mathbb R}^2} 1_{\xi > b} \underbrace{1_{a \xi> \xi_1 > -\infty}}_{1_{b \xi > \xi_1 \ge 0} + 1_{0> \xi_1}} \phi(\xi)\phi(\xi_1) d\xi d\xi_1\\
&=& T(b,a)+\frac{1}{2} \left( 1 - \Phi(b)\right)
\end{eqnarray}
where $T(.,.)$ is the Owen'sT-function https://en.wikipedia.org/wiki/Owen%27s_T_function .
Now setting $b=0$ we get:
\begin{eqnarray}
rhs = T(0,a)+\frac{1}{4} = \frac{1}{2\pi} \arctan(a)+\frac{1}{4}
\end{eqnarray}
as expected.
