Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
In which context such an integral appears? –  Davide Giraudo May 24 '12 at 15:30

2 Answers 2

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 $

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.