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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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    $\begingroup$ In which context such an integral appears? $\endgroup$ May 24 '12 at 15:30
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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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  • $\begingroup$ 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. $\endgroup$ May 24 '12 at 16:36
  • $\begingroup$ @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. $\endgroup$
    – Sasha
    May 24 '12 at 17:04
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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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  • $\begingroup$ Thank you very much, but just one more follow-up, do you know how to do if ax -> ax+b? i.e. $\int_0^\infty\Phi(ax+b)d\Phi(x)dx$ $\endgroup$
    – pidig
    May 30 '12 at 11:54
  • $\begingroup$ @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. $\endgroup$ May 30 '12 at 12:07
  • $\begingroup$ 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? $\endgroup$
    – pidig
    May 30 '12 at 12:18
  • $\begingroup$ Does anyone have an idea about ax+b version? $\endgroup$
    – pidig
    Jun 2 '12 at 7:47
  • $\begingroup$ Use the method in Sasha's answer, maybe? $\endgroup$ Jun 2 '12 at 19:40
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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.

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