I know how to calculate this: $ \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \Phi(a x+b) \exp \left(-\frac{x^2}{2}\right) d x=\Phi\left(\frac{b}{\sqrt{1+a^2}}\right)$
but i am struggling with this:
$ \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^c \Phi(a x+b) \exp \left(-\frac{x^2}{2}\right) d x=\Phi_2\left(\frac{b}{\sqrt{1+a^2}}, c ;-\frac{a}{\sqrt{1+a^2}}\right)$.
where $\Phi_2\left(\cdot, \cdot ;\rho \right)$ is the bivariate cummulative Gaussian distribution with correlation $\rho$.
Can anyone help me?
I take it that $\Phi_2$ is the bivariate standard normal cdf with correlation $\rho$. Then, letting $X$ and $Y$ denote iid standard normal random variables, $$ \begin{align} &\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^c \Phi(a x+b) \exp \left(-\frac{x^2}{2}\right) d x\\&=\frac{1}{2 \pi} \int_{-\infty}^c \int_{-\infty}^{ax+b} \exp \left(-\frac{x^2+y^2}{2}\right) dy\,d x \\&=P(X\le c \cap Y\le aX+b) \\&=P(X\le c \cap Y-aX\le b) \\&=P(Z_1\le c\cap Z_2\le\frac{b}{\sqrt{1+a^2}}) \\&=\Phi_2(c,\frac{b}{\sqrt{1+a^2}}; \rho) \end{align} $$ where $Z_1=X$ and $Z_2=\frac{Y-aX}{\sqrt{1+a^2}}$ and $\rho=\operatorname{corr}(Z_1,Z_2)=-\frac a{\sqrt{1+a^2}}$.