Integration of the product of pdf & cdf of normal distribution  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!
 A: We have $\phi(x)=\frac 1{\sqrt{2\pi}}\exp\left(-\frac{†^2}2\right)$ and $\Phi(x)=\int_{-\infty}^x\phi(t)dt$. We try to compute
$$ I(a,b):=\int\phi(x)\Phi\left(\frac{x-b}a\right)dx.$$
Using the dominated convergence theorem, we are allowed to take the derivative with respect to $b$ inside the integral. We have 
$$\partial_bI(a,b)=\int\phi(x)\left(-\frac 1a\right)\phi\left(\frac{x-b}a\right)dx$$
and 
\begin{align}
2\pi\phi(x)\phi\left(\frac{x-b}a\right)&=\exp\left(-\frac 12\left(x^2+\frac{x^2}{a^2}-2\frac{bx}{a^2}+\frac{b^2}{a^2}\right)\right)\\\
&=\exp\left(-\frac 12\frac{a^2+1}{a^2}\left(x^2-2\frac b{a^2+1}x+\frac{b^2}{a^2+1}\right)\right)\\\
&=\exp\left(-\frac 12\frac{a^2+1}{a^2}\left(x-\frac b{a^2+1}\right)^2-\frac 12\frac{a^2+1}{a^2}\left(\frac{b^2}{a^2+1}-\frac{b^2}{(a^2+1)^2}\right)\right)\\\
&=\exp\left(-\frac 12\frac{a^2+1}{a^2}\left(x-\frac b{a^2+1}\right)^2\right)\exp\left(-\frac{b^2}{2a^2}\frac{a^2+1-1}{a^2+1}\right)\\\
&=\exp\left(-\frac 12\frac{a^2+1}{a^2}\left(x-\frac b{a^2+1}\right)^2\right)\exp\left(-\frac{b^2}{2(a^2+1)}\right).
\end{align}
Integrating with respect to $x$, we get that 
$$\partial_b I(a,b)=-\frac 1{\sqrt{a^2+1}}\phi\left(\frac b{\sqrt{a^2+1}}\right).$$
Since $\lim_{b\to +\infty}I(a,b)=0$, we have 
\begin{align}I(a,b)&=\int_b^{+\infty}\frac 1{\sqrt{a^2+1}}\phi\left(\frac s{\sqrt{a^2+1}}\right)ds\\\
&=\int_{b/\sqrt{a^2+1}}^{+\infty}\phi(t)dt = 1 - \Phi(b/\sqrt{a^2+1}).
\end{align}
This can be expressed with the traditional erf function. 
A: 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 $a\ne0$, $(\ast)=\frac12$.
A: *

*The last equation should be integral from $b/\sqrt{a^2+1}$

*In I(a, b), a is supposed to be positive. When $a<0$, the answer will be $\int_{-\infty}^{b/\sqrt{a^2+1}} \phi(t) dt$

