How can I calculate $\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$ Suppose $\phi(\cdot)$ and $\Phi(\cdot)$ are density function and distribution function of the standard normal distribution.
How can one calculate the integral:
$$\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$$
 A: Let $X$ and $Y$ be independent normal random variables with $X \sim N(a,b^2)$ and $Y$ a standard normal random variable.  Then, $$P\{X \leq Y \mid Y = w\} = P\{X \leq w\} = \Phi\left(\frac{w-a}{b}\right).$$ So, using the law of total probability, we get that
$$P\{X \leq Y\} 
= \int_{-\infty}^\infty P\{X \leq Y \mid Y = w\}\phi(w)\,\mathrm dw
= \int_{-\infty}^\infty \Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw.$$
Now, $P\{X \leq Y\} = P\{X-Y \leq 0\}$ can be expressed in terms of $\Phi(\cdot)$ by noting that $X-Y \sim N(a,b^2+1)$, and thus we get 
$$\int_{-\infty}^\infty \Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw = \Phi\left(\frac{-a}{\sqrt{b^2+1}}\right)$$
which is the same as the result in whuber's answer.
A: A more conventional notation is
$$y(\mu, \sigma) = \int\Phi\left(\frac{x-\mu}{\sigma}\right)\phi(x) dx = \Phi\left(\frac{-\mu}{\sqrt{1+\sigma^2}}\right).$$
This can be found by differentiating the integral with respect to $\mu$ and $\sigma$, producing elementary integrals which can be expressed in closed form:
$$\frac{\partial y}{\partial \mu}(\mu, \sigma) = -\frac{1}{\sqrt{2 \pi } \sqrt{\sigma ^2+1}}e^{-\frac{1}{2}\frac{\mu ^2}{\sigma ^2+1}},$$
$$\frac{\partial y}{\partial \sigma}(\mu, \sigma) = \frac{\mu\sigma  }{\sqrt{2 \pi } \left(\sigma ^2+1\right)^{3/2}}e^{-\frac{1}{2}\frac{\mu ^2}{\sigma ^2+1}}.$$
This system can be integrated, beginning with the initial condition $y(0,1)$ = $\int\Phi(x)\phi(x)dx$ = $1/2$, to obtain the given solution (which is easily checked by differentiation).
A: Here is another solution: We define
\begin{align*}
I(\gamma) & =\int_{-\infty}^{\infty}\Phi(\xi x+\gamma)\mathcal{N}(x|0,\sigma^{2})dx,
\end{align*}
which we can evaluate $\gamma=-\xi\mu$ to obtain our desired expression.
We know at least one function value of $I(\gamma)$, e.g., $I(0)=0$
due to symmetry. We take the derivative wrt to $\gamma$
\begin{align*}
\frac{dI}{d\gamma} & =\int_{-\infty}^{\infty}\mathcal{N}((\xi x+\gamma)|0,1)\mathcal{N}(x|0,\sigma^{2})dx\\
 & =\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\xi x+\gamma\right)^{2}\right)\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left(-\frac{x^{2}}{2\sigma^{2}}\right)dx.
\end{align*}
and complete the square
\begin{align*}
\left(\xi x+\gamma\right)^{2}+\frac{x^{2}}{\sigma^{2}} & =\underbrace{\left(\xi^{2}+\sigma^{-2}\right)}_{=a}x^{2}+\underbrace{-2\gamma\xi}_{=b}x+\underbrace{\gamma^{2}}_{=c} \\
&=a\left(x-\frac{b}{2a}\right)^{2}+\left(c-\frac{b^{2}}{4a}\right)
\left(c-\frac{b^{2}}{4a}\right)\\ 
& =\gamma^{2}-\frac{4\gamma^{2}\xi^{2}}{4\left(\xi^{2}+\sigma^{-2}\right)}\\
&=\gamma^{2}\left(1-\frac{\xi^{2}}{\xi^{2}+\sigma^{-2}}\right)\\
&=\gamma^{2}\left(\frac{1}{1+\xi^{2}\sigma^{2}}\right)
\end{align*}
Thus, 
\begin{align*}
\frac{dI}{d\gamma} & =\frac{1}{2\pi\sigma}\exp\left(-\frac{1}{2}\left(c-\frac{b^{2}}{4a}\right)\right)\sqrt{\frac{2\pi}{a}}\int_{-\infty}^{\infty}\sqrt{\frac{a}{2\pi}}\exp\left(-\frac{1}{2}a\left(x-\frac{b}{2a}\right)^{2}\right)dx\\
 & =\frac{1}{2\pi\sigma}\exp\left(-\frac{1}{2}\left(c-\frac{b^{2}}{4a}\right)\right)\sqrt{\frac{2\pi}{a}}\\ 
&=\frac{1}{\sqrt{2\pi\sigma^{2}a}}\exp\left(-\frac{1}{2}\left(c-\frac{b^{2}}{4a}\right)\right)\\
 & =\frac{1}{\sqrt{2\pi\left(1+\sigma^{2}\xi^{2}\right)}}\exp\left(-\frac{1}{2}\frac{\gamma^{2}}{1+\xi^{2}\sigma^{2}}\right)
\end{align*}
and integration yields 
$$
\begin{align*}
I(\gamma)
&=\int_{-\infty}^{\gamma}\frac{1}{\sqrt{2\pi\left(1+\sigma^{2}\xi^{2}\right)}}\exp\left(-\frac{1}{2}\frac{z^{2}}{1+\xi^{2}\sigma^{2}}\right)dz\\
&=\Phi\left(\frac{\gamma}{\sqrt{1+\xi^{2}\sigma^{2}}}\right)
\end{align*}
$$
which implies 
$$
\begin{align*}
\int_{-\infty}^{\infty}\Phi(\xi x)\mathcal{N}(x|\mu,\sigma^{2})dx
&=I(\xi\mu)\\
&=\Phi\left(\frac{\xi\mu}{\sqrt{1+\xi^{2}\sigma^{2}}}\right).
\end{align*}
$$
