I am trying to integrate the product of the standard normal CDF ($\Phi(x)$) and a shifted and scaled normal survival function (SF) given by $1-\Phi\left(\frac{x-\mu}{\sigma}\right)$, i.e. the following:
$$\int^{\infty}_{-\infty}\Phi\left(x\right)\left[1-\Phi\left(\frac{x-\mu}{\sigma}\right)\right]\mathrm{d}x$$
where $\mu, \sigma > 0$. I figured this could be made equivalent to the integration of $\mathrm{erf}(x)\cdot\mathrm{erfc}((x-\mu)/\sigma)$, however looking in the D.B. Owen Table of Normal Integrals or the M. Geller Table of Integrals of the Error Functions didn't turn up anything applicable. Any suggestions would be appreciated.
Best Answer
Letting $$ f \colon \mathbb{R} \times \mathbb{R}^+ \to \mathbb{R}, \, f(\mu,\sigma) = \int \limits_{-\infty}^\infty \Phi(x) \left[1 - \Phi \left(\frac{x - \mu}{\sigma}\right)\right] \, \mathrm{d} x \, , $$ denote your integral, we have $$ \partial_1 f(\mu,\sigma) = \frac{1}{\sigma} \int \limits_{-\infty}^\infty \Phi(x) \phi \left(\frac{x - \mu}{\sigma}\right) \, \mathrm{d} x = \Phi\left(\frac{\mu}{\sqrt{1+\sigma^2}}\right) .$$ Here $\phi$ is the standard normal PDF and the last step follows from this question. Since $x \mapsto x \Phi(x) + \phi(x)$ is an antiderivative of $\Phi$ (and $f(-\infty,\sigma) = 0$), we obtain $$ f(\mu,\sigma) = \int \limits_{-\infty}^\mu \Phi\left(\frac{\nu}{\sqrt{1+\sigma^2}}\right) \, \mathrm{d} \nu = \mu \Phi\left(\frac{\mu}{\sqrt{1+\sigma^2}}\right) + \sqrt{1 + \sigma^2} \phi\left(\frac{\mu}{\sqrt{1+\sigma^2}}\right) .$$