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!
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
a = -8; b = 1.; w = -5 using integrate results in 5.001087e-46 but the OwenT version produces -0.5.
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-(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 $v=a/(w \sqrt{1+b^2}).$)
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