Integral with linear function, Normal PDF, Normal CDF

Question

I am trying to calculate the following integral: $$\int_a^\infty x \Phi(cx+d) \phi\left(\frac{x-\mu}{\sigma}\right) dx,$$ where $\Phi$, $\phi$ denote the CDF and PDF of the standard Normal $N(0,1)$.

I found this table of integrals but it includes only the indefinite integral: https://www.tandfonline.com/doi/pdf/10.1080/03610918008812164?casa_token=0E6SYYGFkkUAAAAA:cd6Lp-PTXsVRR20JaMQcNV8twies8FQAV0sO0DgOcAMD-L8aKBjxwpbqjClYlcFdIIxNUTCrnySg

Can you please help me? Thanks!

Answer 1

The indefinite integral $\int x \Phi(\alpha x + \beta) \phi\left(x\right) \mathrm{d}x$ is given in Wikipedia. Let's denote it by $I(\alpha,\beta,x)$.

The change of variables $y=\frac{x-\mu}{\sigma}$ in $\int_a^\infty x \Phi(cx+d) \phi\left(\frac{x-\mu}{\sigma}\right) \mathrm{d}x$ gives $$\sigma\int_{\frac{a-\mu}{\sigma}}^\infty (\mu + \sigma y) \Phi(c\sigma y +c\mu + d) \phi(y) \mathrm{d}y \\ = \sigma\mu\int_{\frac{a-\mu}{\sigma}}^\infty\Phi(c\sigma y +c\mu + d)\phi(y)\mathrm{d}y + \sigma^2\int_{\frac{a-\mu}{\sigma}}^\infty y\Phi(c\sigma y +c\mu + d)\phi(y)\mathrm{d}y.$$

The second integral is ${\bigl[I(c\sigma,c\mu+d,x)\bigr]}_{\frac{a-\mu}{\sigma}}^\infty$. The first integral can be found here.

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Edit

I checked with R and it is correct.

I <- function(a, b, x) {
  t <- sqrt(1+b^2)
  b/t * dnorm(a/t) * pnorm(x*t + a*b/t) - dnorm(x) * pnorm(a + b*x)
}
J <- function(a, b, w) {
  t <- sqrt(1+b^2)
  rho <- -b/t
  library(mvtnorm)
  pmvnorm(upper = c(a/t, w), sigma = cbind(c(1, rho), c(rho, 1)))
}
K <- function(a, b, w) {
  J(a, b, Inf) - J(a, b, w)
}

a <- 1
mu <- 1; sigma <- 1; c <- 2; d <- -1

mu*sigma * K(c*mu+d, c*sigma, (a-mu)/sigma) + 
  sigma^2*(I(c*mu+d, c*sigma, Inf) - I(c*mu+d, c*sigma, (a-mu)/sigma))
# 0.8796307

f <- function(x) { 
 x * pnorm(c*x+d) * dnorm((x-mu)/sigma)  
}
integrate(f, lower = a, upper = Inf)
# 0.8796307 with absolute error < 1.7e-08

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Edit

So if I didn't do a mistake, the final result is $$ \sigma \left(\mu\Phi\bigl(\frac{\alpha}{t}\bigr) - B_{\frac{-\beta}{t}}\Bigl(\frac{\alpha}{t}, v\Bigr) + \sigma \Bigl(\frac{\beta}{t} \phi\bigl(\frac{\alpha}{t} \bigr) \Bigl(1 - \Phi\bigl(tv+\frac{\alpha\beta}{t}\bigr)\Bigr) - \phi(v)\Phi(\alpha + \beta v)\Bigr)\right) $$ where $t = \sqrt{1+c^2\sigma^2}$, $\alpha = c\mu+d$, $\beta = c\sigma$, $v = \frac{a-\mu}{\sigma}$, and $B_\rho$ is the bivariate normal function.

To be checked...

Answer 2

the definite integral has a closed-form expression for $\mu=0$,

\begin{align} & \int_a^\infty x \Phi(cx+d) \phi\left(\frac{x}{\sigma}\right) dx \\[8pt] = {} & \frac{c \sigma^3 e^{-\frac{d^2}{2 c^2 \sigma^2+2}}}{2 \sqrt{2 \pi } \sqrt{c^2 \sigma^2+1}}\left[1-\operatorname{erf}\left(\frac{c \sigma^2 (c a+d)+a}{\sigma \sqrt{2 c^2 \sigma^2+2}}\right)\right] \\[8pt] & -\sigma^2(2 \sqrt{2 \pi })^{-1}e^{-\frac{a^2}{2 \sigma^2}} \left[\operatorname{erfc}\left(\frac{c a+d}{\sqrt{2}}\right)-2\right] \end{align}