Convolution of Gaussian and error function

Question

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

Answer 1

$$ \begin{align} \int_{-\infty}^{\infty} e^{-\frac{x^2} {2}} \Phi(x-t) dx & = \sqrt{2\pi}\int_{-\infty}^{\infty} \phi(x) \Phi(x-t) dx \\ & = \sqrt{2\pi}\Pr\{Z_1 \leq Z_2 - t\} \\ & = \sqrt{2\pi}\Pr\left\{\frac {Z_1 - Z_2} {\sqrt{2}} \leq - \frac {t} {\sqrt{2}}\right\} \\ & = \sqrt{2\pi}\Phi\left(- \frac {t} {\sqrt{2}}\right) \end{align} $$ where $\phi$ and $\Phi$ is the standard normal pdf and CDF respectively, $Z_1, Z_2$ are independent standard normal random variables.

Answer 2

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 $$ I(t)=\sqrt{2\pi}-\int_{-\infty}^\infty{ \exp\left(-\frac{x^2}{2} \right)~ \int_{\infty}^{t}{ \frac{1}{\sqrt{2\pi}}\exp\left( -\frac{(y-x)^2}{2} \right) ~dy } ~dx }\\ =\sqrt{2\pi}- \int_{\infty}^{t}{ \frac{1}{\sqrt{2}}\exp \left(-\frac{y^2}{4} \right) ~dy }\\ =\sqrt{2\pi}- \sqrt{2\pi}\Phi\left(\frac{t}{\sqrt{2}} \right)\\ =\sqrt{2\pi}\Phi\left( -\frac{t}{\sqrt{2}} \right) $$