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

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In which context such an integral appears? – Davide Giraudo May 24 '12 at 15:30

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 $

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Since $I(a) = \frac{1}{2}F(a)$ where $F(a)$ is the CDF of a standard Cauchy random variable, there ought to be some simple probabilistic description of how the integral came about. – Dilip Sarwate May 24 '12 at 16:36
@DilipSarwate Certainly, it came from computing $\mathbb{P}\left( \frac{Z_1}{Z_2} \leqslant a | Z_1 Z_2 > 0 \right)$, where $Z_1$ and $Z_2$ are standard normal random variables. – Sasha May 24 '12 at 17:04

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 . $$\begin{align*} \int_0^{\infty}\Phi(ax)\;\mathrm d\Phi(x) &= \int_0^{\infty}\Phi(ax)\phi(x)\mathrm\; dx\\ &= \int_0^{\infty}\left[ \int_{-\infty}^{ax}\phi(y)\;\mathrm dy\right] \phi(x)\mathrm\; dx\\ &= \int_0^{\infty} \int_{-\infty}^{ax}\phi(y) \phi(x)\;\mathrm dy\;\mathrm dx\\ &= \frac{1}{2\pi}\int_{r=0}^{\infty}\int_{\theta=-\pi/2}^{\arctan(a)} \exp(-r^2/2) \cdot r\;\mathrm d\theta\;\mathrm dr\\ &= \frac{\arctan(a)+\pi/2}{2\pi}\\ &= \frac{1}{4} + \frac{1}{2\pi}\arctan(a). \end{align*}$$

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Thank you very much, but just one more follow-up, do you know how to do if ax -> ax+b? i.e. $\int_0^\infty\Phi(ax+b)d\Phi(x)dx$ – pidig May 30 '12 at 11:54
@pidig It is $P\{Y \leq aX+b \mid X > 0\}$ for independent standard normal random variables $X$ and $Y$, so see if you can do something with that. – Dilip Sarwate May 30 '12 at 12:07
Thank you for your transformation, Dilip! in fact, that's the question when I received, may i ask you how did you proceed from this Probability? – pidig May 30 '12 at 12:18
Does anyone have an idea about ax+b version? – pidig Jun 2 '12 at 7:47
Use the method in Sasha's answer, maybe? – Dilip Sarwate Jun 2 '12 at 19:40

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