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It is straightforward to show

$$\int_{-\infty}^{\infty}\Phi(a+bx) \phi( x ) dx = \Phi\left(\frac{a}{\sqrt{1+b^2}}\right)$$

but what value does

$$\int_{c}^{\infty}\Phi(a+bx) \phi( x ) dx$$

have for $c<\infty$, e.g., for $c=0$? Even a good approximation would really help me!

StijnDeVuyst
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Jenny Reininger
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  • Can you please explane what the different phis stand for? – cdalitz Dec 01 '20 at 13:13
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    @Stijn By comparing the lists of definite and indefinite integrals there, it looks likely this was just an inadvertent omission. – whuber Dec 01 '20 at 16:02
  • @whuber: yes. Maybe we should add it on Wikipedia although I have no time left today. I spent it all on breaking my head the last few hours on how to prove the result, even with the help of Stephane's answer. Maybe later. – StijnDeVuyst Dec 01 '20 at 19:48
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    @Stijn Re proving the result: letting $(X,Z)$ be independent standard Normal variates, write $\Pr(c\le X, Z\le a+bX)$ as an integral to show it equals the integral in question, then work it out in terms of the distribution of $(X,Y)=(X,Z-bX)$ for which the event is $c\le X$ and $Y\le a.$ (This strategy converts a simple integral over a "complicated" region to a more complicated integral over a very simple region.) Note that the correlation of $(X,Y)$ is $\rho(X,Y)=-b/\sqrt{(1)(1+b^2)}$ and compare with Stephane's answer (taking note that it solves the problem in terms of $w$ rather than $c$). – whuber Dec 01 '20 at 20:01
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    Thanks, @whuber, now I can finally go to sleep :-) – StijnDeVuyst Dec 01 '20 at 20:10

1 Answers1

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The integral $$ \int_{-\infty}^w \phi(x)\Phi(a+bx)\mathrm{d}x $$ can be expressed with the help of the bivariate normal function (I don't remember the usual name of this function but I define it below):

$$ \int_{-\infty}^w \phi(x)\Phi(a+bx)\mathrm{d}x = B_{\frac{-b}{\sqrt{1+b^2}}}\left(\frac{a}{\sqrt{1+b^2}}, w\right) $$ with $$ B_\rho(g_1,g_2) = \Pr(G_1 \leqslant g_1, G_2 \leqslant g_2) $$ where the random pair $(G_1,G_2) \sim \mathcal{N}_2\left(\mathbf{0}, \begin{pmatrix}1 & \rho \\ \rho & 1 \end{pmatrix}\right)$.

It can also be expressed with the help of the Owen $T$-function: $$ \frac{1}{2}\left(\Phi\biggl(\frac{a}{\sqrt{1+b^2}}\biggr) + \Phi(w) - 2T\biggl(w, \frac{a+bw}{w}\biggr) - 2T\biggl(\frac{-a}{\sqrt{1+b^2}}, \frac{ab+w(1+b^2)}{a}\biggr) - \mathbb{1}_{]-\infty,0]}(a) \right). $$

I found these equalities in my archives but I don't remember how I derived them or where I found them. Perhaps they are given in Owen's paper Table of normal integrals (1980).

R demonstration:

a = 2; b = 3; w = 1
integrate(function(x) dnorm(x)*pnorm(a+b*x), lower=-Inf, upper=w)$value
# 0.5778001
library(mvtnorm)
rho <- -b/sqrt(1+b^2)
pmvnorm(upper=c(a/sqrt(1+b^2), w), sigma=cbind(c(1,rho),c(rho,1)))
# 0.5778001
library(OwenQ)
1/2*(pnorm(a/sqrt(1+b^2))  + pnorm(w) - 2*OwenT(w, (b*w+a)/w) - 
       2*OwenT(-a/sqrt(1+b^2), (a*b+w*(1+b^2))/a) - (a <= 0))
# 0.5778001
Stéphane Laurent
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    +1 One easy derivation interprets the integral in terms of an ordinary regression in a bivariate Normal setting. – whuber Dec 01 '20 at 15:21
  • Thanks! The first is in Owen's Table as (10,010.1) and the definition of the bivariate CDF as (3.1). There is even the indefinite integral (10,010.1). But how can I derive this rule? – Jenny Reininger Dec 16 '20 at 11:40
  • I dont think this derivation is correct. `a = -8; b = 1.; w = -5` using `integrate` results in `5.001087e-46` but the OwenT version produces `-0.5`. – jvdillon Feb 04 '21 at 02:25
  • FYI, the condition in the OwenT derivation is wrong. It should be `-(a/w<0)` and not `-(a<=0)`. The conclusion can be derived from 10,010.3 of Owen's paper and using the property $T(u,v) + T(uv,1/v) = \frac{1}{2} (\Phi(u) + \Phi(uv)) - \Phi(u) \Phi(uv) - \frac{1}{2} [v<0]$. (Here $v=a/(w \sqrt{1+b^2}).$) – jvdillon Feb 04 '21 at 05:38