Convolution of Gaussian and error function I am trying to evaluate the following integral:
$$
\int_{-\infty}^{\infty}e^{-\frac{x^2}{2}}\Phi(x-t)dx
$$
where
$$
\Phi(y) = \frac{1}{2} + \frac{1}{2}erf\left(\frac{y}{\sqrt{2}}\right)
$$
I have tried integration by parts, a substitution, but nothing useful came out of this
 A: $$ \begin{align} \int_{-\infty}^{\infty} e^{-\frac{x^2} {2}} \Phi(x-t) dx
& = \sqrt{2\pi}\int_{-\infty}^{\infty} \phi(x) \Phi(x-t) dx \\
& = \sqrt{2\pi}\Pr\{Z_1 \leq Z_2 - t\} \\
& = \sqrt{2\pi}\Pr\left\{\frac {Z_1 - Z_2} {\sqrt{2}} \leq 
- \frac {t} {\sqrt{2}}\right\} \\
& = \sqrt{2\pi}\Phi\left(- \frac {t} {\sqrt{2}}\right)
\end{align} 
$$
where $\phi$ and $\Phi$ is the standard normal pdf and CDF respectively, $Z_1, Z_2$ are independent standard normal random variables.
A: The key is to express as integral of a gaussian.
$$
I(t) = \int_{-\infty}^\infty \exp\left(-\frac{x^2}{2} \right)~\Phi(x-t)~dx\\
= \int_{-\infty}^\infty \exp\left(-\frac{x^2}{2} \right)~\left[
1-\Phi(t-x)
\right]~dx\\
=\sqrt{2\pi}-\int_{-\infty}^\infty{
 \exp\left(-\frac{x^2}{2} \right)~
\Phi(t-x)
~dx
}
$$
Plug in 
$$
\Phi(t-x)=\int_{\infty}^{t}{
\frac{1}{\sqrt{2\pi}}\exp\left( -\frac{(y-x)^2}{2} \right)
~dy
}
$$
to get
$$
I(t)=\sqrt{2\pi}-\int_{-\infty}^\infty{
 \exp\left(-\frac{x^2}{2} \right)~
\int_{\infty}^{t}{
\frac{1}{\sqrt{2\pi}}\exp\left( -\frac{(y-x)^2}{2} \right)
~dy
}
~dx
}\\
=\sqrt{2\pi}-
\int_{\infty}^{t}{
\frac{1}{\sqrt{2}}\exp \left(-\frac{y^2}{4} \right)
~dy
}\\
=\sqrt{2\pi}-
\sqrt{2\pi}\Phi\left(\frac{t}{\sqrt{2}} \right)\\
=\sqrt{2\pi}\Phi\left( -\frac{t}{\sqrt{2}} \right)
$$
