Can we simplify further the following function?
$\int_{-\infty}^z \phi(x)\,\Phi(\beta\, x)\,dx$,
Where $\phi(x)$ is the pdf of standard normal distribution, i.e., $\phi(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$.
Also $\Phi(x)$ is the CDF of standard normal distribution, i.e., $\Phi(z)=\int_{-\infty}^z\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\,dx$.
$\beta>0$.
Let's define $$ f_z(\beta):=\int_{-\infty}^z \phi(x)\,\Phi(\beta\, x)\,dx $$ For $\beta=1$ $$ f_z(1)=\int_{-\infty}^z \phi(x)\,\Phi( x)\,dx=\int_{-\infty}^z \Phi( x)\,\Phi(x)'\,\,dx=\frac12 \Phi(z)^2 $$ For any other $\beta$, I am almost sure that there is not an elementary expression. One can try some manipulation to put it in a different form, but still non explicit. For example one can find explicitly the derivative of $f_z(\beta)$ with respect to beta: $$ \frac{\partial}{\partial\beta}f_z(\beta)=\int_{-\infty}^z \phi(x)\,\phi(\beta\, x)\,x\,dx=\frac{1}{\pi}\int_{-\infty}^ze^{-\frac{x^2(1+\beta^2)}{2}}\frac{x}{2}\,dx=\frac{1}{\pi(1+\beta^2)}\int_{-\infty}^{z\sqrt{(1+\beta^2)}}e^{-\frac{u^2}{2}}\frac{u}{2}\,du $$ Put $$ I_z(\beta):=\int_{-\infty}^{z\sqrt{(1+\beta^2)}}e^{-\frac{u^2}{2}}\frac{u}{2}\,du $$ For example, for $z>0$ $$ I_z(\beta):=\left(\int_{-\infty}^{0}+\int_{0}^{z\sqrt{(1+\beta^2)}}\right)e^{-\frac{u^2}{2}}\frac{u}{2}\,du=\int_0^{\infty} e^{-t/2}dt+\int_0^{z^2(1+\beta^2)}e^{-t/2}dt=\\ =2+2(1-e^{-z^2(1+\beta^2)/2}) $$ then $$ \frac{\partial}{\partial\beta}f_z(\beta)=\frac{2}{\pi(1+\beta^2)}(2-e^{-z^2(1+\beta^2)/2}) $$ and $$ f_z(\beta)=\frac12 \Phi(z)^2+\int_1^\beta\frac{2}{\pi(1+b^2)}(2-e^{-z^2(1+b^2)/2})\, db=\\ =\frac12 \Phi(z)^2+\frac{4}{\pi}\left[\arctan(b)\right]_1^\beta-\frac{2}{\pi}e^{-z^2/2}\int_1^\beta\frac{e^{-(zb)^2/2}}{1+b^2}\, db $$ Of course the integral cannot be explicitly evaluated. Nonetheless, it can be useful to evaluate the function near $\beta=1$ or for some asymptotycs in $z$.