recently, i need to compute this kind of integral: $$ \int ^\infty _c \Phi(ax+b) \phi(x) dx$$ where a, b and c are all constants and $\Phi(x)$ denotes the CDF of standard normal distribution and $\phi(x)$ denotes the PDF of standard normal

I have looked up similar questions both in "math.stackexchange.com" and here, but i don't find any satisfactory answer. if "c" is negative infinity here, it would be relatively easy. but here it's not.

later i find if i can find a solution to: $$ \int ^\infty _c x^2\Phi(ax+b) \phi(x) dx$$ then the former integral could be calculated.

can someone help me? if no closed solution exists, is there any practical approximation to that integral?

  • $\begingroup$ If $a=1, b=0,$ then Maple produces $$1/2\,{\frac { \left( -1/4\,\sqrt {\pi }\sqrt {2} {{\rm erf}\left(1/2\,c\sqrt {2}\right)}-1/8\,\sqrt {\pi }\sqrt {2} \left( {{\rm erf}\left(1/2\,c\sqrt {2}\right)} \right) ^{2}+3/8\, \sqrt {\pi }\sqrt {2} \right) \sqrt {2}}{\sqrt {\pi }}}. $$ $\endgroup$ – Mark May 17 '13 at 17:03
  • $\begingroup$ The simplified result is $$-1/8\, \left( {{\rm erf}\left(1/2\,c\sqrt {2}\right)} \right) ^{2}-1/4 \,{{\rm erf}\left(1/2\,c\sqrt {2}\right)}+3/8. $$ $\endgroup$ – Mark May 17 '13 at 17:09
  • $\begingroup$ @Mark: That case is easy to compute by hand. It would be the probability that $X \gt c, X \gt Y$ where $X,Y \sim N(0,1)$. If I calculate correctly, that's $P(X \gt c) - P(Y \gt X \gt c) = P(X \gt c) - 1/2 P(X \gt c)^2$. The general case doesn't simplify like that. $\endgroup$ – Douglas Zare May 17 '13 at 18:56
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    $\begingroup$ This questions asks for the probability that a standard normal distribution is in a wedge-shaped region, or equivalently for the CDF of a two-dimensional Gaussian distribution with a general covariance matrix. Wikipedia says that there isn't an analytic expression for this, but that approximations are known. One related paper I remember is by Marsaglia, jstatsoft.org/v16/i04/paper, on the distribution of the ratio between two normal distributions, but that would be a double wedge, between two lines instead of two rays. $\endgroup$ – Douglas Zare May 17 '13 at 19:16
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    $\begingroup$ @Mark: I don't understand what you are confused about. I translated the integral into a statement about random variables. Inside the integral is a density for $X=x, Y \lt x$. Integrating from $x=c$ to $x = \infty$ means $X \gt c, Y \lt X$. $\endgroup$ – Douglas Zare May 17 '13 at 20:44

Here's the answer to the first integral:


where $\Phi_2\left(x,y,\rho\right)$ is the bivariate normal cdf with means zero, variances one, and correlation $\rho$.

I found it by differentiating with respect to $b$, then reintegrating. You can get the second integral the same way.

  • $\begingroup$ Someone posted an answer for $c=-\infty$ at mathoverflow.net/questions/101469/…. I wonder whether the beiruti's answer is consistent with that. IMO, something's wrong... try this chunk of code in R: library(mvtnorm) a<-1 b<-1 c<-1 dnorm(b/sqrt(1+a^2))-pmvnorm(lower=c(-Inf,-Inf),upper=c(c,b/sqrt(1+a^2)),mean=c(0,0),corr=matrix(c(1,-a/sqrt(1+a^2),-a/sqrt(1+a^2),1),2,2)) it returns a value < 0... Maybe the answer is: $\Phi(\frac{b}{\sqrt{1+a^2}})-\Phi_2[c,\frac{b}{\sqrt{1+a^2}},-\frac{a}{\sqrt{1+a^2}}]$ (swap $\phi$ with $\Phi$). $\endgroup$ – al cliver Oct 27 '14 at 14:31

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