TITLE: Integral with Normal Distributions
QUESTION [5 upvotes]: I know that the following equality is true for any $a$ and $\sigma$ (I have solved it numerically):
$$\int\nolimits_{-\infty}^{+\infty}\Phi\left(\frac{a-x}{\sigma}\right)\frac1{\sigma} \phi\left(\frac{x-a}{\sigma}\right)\mathrm dx=\frac12$$
Where $\Phi$ and $\phi$ respectively denote the CDF and PDF of the standard normal distribution, $a$ is any given real number, and $\sigma$ is a positive real number.
I have tried for a while but I am unable to prove it. Can anyone help me?

REPLY [4 votes]: This follows from a basic change of variables, for which one just needs to know that $\Phi'=\phi$ and what the limits of $\Phi$ at $\pm\infty$ are, but certainly not the precise form of $\Phi$. 
To wit, consider $z=\displaystyle\Phi\left(\frac{x-a}{\sigma}\right)$, then $\mathrm{d}z=\displaystyle\phi\left(\frac{x-a}{\sigma}\right)\frac{\mathrm{d}x}{\sigma}=\displaystyle\phi\left(\frac{a-x}{\sigma}\right)\frac{\mathrm{d}x}{\sigma}$ because $\Phi'=\phi$ (first equality) and because $\phi$ is even (second equality). 
The bounds of the integral in $z$ are $\Phi(-\infty)=0$ for $x=-\infty$ and $\Phi(+\infty)=1$ for $x=+\infty$, hence the integral is $\displaystyle\int_0^1z\mathrm{d}z=\left[\frac{z^2}2\right]_0^1=\frac12$.