Simplifying $\int_{-\infty}^z \phi(x)\,\Phi(\beta\, x)\,dx$, $\phi(x)$ pdf of normal, $\Phi(x)$ CDF of normal 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., $\Phi(z)=\int_{-\infty}^z\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\,dx$.
$\beta>0$.
 A: 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 $z$.
A: We have:
\begin{eqnarray}
I_\beta(z):=\int\limits_{-\infty}^z \phi(x) \Phi(\beta x) dx= \underbrace{\int\limits_{\mathbb R} \phi(x) \Phi(\beta x) dx}_{f_1(\beta)} - \underbrace{\int\limits_z^\infty \phi(x) \Phi(\beta x) dx}_{f_2(\beta)}
\end{eqnarray}
Now we have
\begin{eqnarray}
f_1^{'}(\beta)=0 \quad \mbox{subject to $f_1(1)=1/2$} \\
\Rightarrow
f_1(\beta)=1/2
\end{eqnarray}
The second integral is expressed via the Owen-T function https://en.wikipedia.org/wiki/Owen%27s_T_function. We have:
\begin{eqnarray}
f_2(\beta)= \int\limits_{{\mathbb R}^2} 1_{x> z} \underbrace{1_{\beta x > x_1 > -\infty }}_{1_{\beta x > x_1 > 0} + 1_{0 > x_1}}
\phi(x) \phi(x_1) dx dx_1 = T(z,\beta) + \frac{1}{2}(1-\Phi(z))
\end{eqnarray}
Therefore we have:
\begin{equation}
I_\beta(z) = -T(z,\beta) + \frac{1}{2} \Phi(z)
\end{equation}
