TITLE: Integration of the product of pdf & cdf of normal distribution 
QUESTION [7 upvotes]: Denote the pdf of normal distribution as $\phi(x)$ and cdf as $\Phi(x)$. Does anyone know how to calculate $\int \phi(x) \Phi(\frac{x -b}{a}) dx$? Notice that when $a = 1$ and $b = 0$ the answer is $1/2\Phi(x)^2$. Thank you!

REPLY [13 votes]: This might be a setting where relying on the probabilistic meaning of the functions $\phi$ and $\Phi$ saves ink and tedious computations. 
Let $X$ and $Y$ denote standard normal random variables. Then $\int\limits_{-\infty}^\infty u(x)\phi(x)\mathrm dx=E(u(X))$ for every suitable function $u$  and $\Phi(x)=P(Y\leqslant x)$ for every real number $x$. 
Using this for the function $u:x\mapsto\Phi((x-b)/a)$ and assuming furthermore that $X$ and $Y$ are independent, one sees that the integral to be computed is
$$
(\ast)=E(\Phi((X-a)/b))=P(Y\leqslant(X-b)/a).
$$
Thus,
$$
(\ast)=P(Z\geqslant b),
$$
where $Z=X-aY$ (this step uses the fact that $a\gt0$). Now, the random variable $Z$ is normal as a linear combination of independent gaussian random variables, with mean $0$ and variance $1+a^2$, hence $Z=\sqrt{a^2+1}\cdot T$, where $T$ is standard normal. Thus,
$$
(\ast)=P(T\geqslant b/\sqrt{a^2+1})=1-\Phi\left(b/\sqrt{a^2+1}\right).
$$
Likewise, if $a\lt0$, then $(\ast)=\Phi\left(b/\sqrt{a^2+1}\right).$
In particular, if $b=0$ then, for every $a\ne0$, $(\ast)=\frac12$.