$\int_0^\infty\Phi(\frac{-x}{\sqrt{2}})d\Phi(x)=?$ where $\Phi(x)$ is the cumulative distribution function of a standard normal random variable.
How to compute this integral involving a cdf?
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1In which context such an integral appears? – 2012-05-24
3 Answers
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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0@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. – 2012-05-24
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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0Use the method in Sasha's answer, maybe? – 2012-06-02
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.