Indefinite integral of product of CDF and PDF of standard normal distribution

Question

Is there a solution to:

$\int ^\infty _x \Phi(z) \phi(z) dz$

where $z$ ~ $N(0,1)$ and $\Phi$ and $\phi$ refer to the CDF and PDF?

Many thanks.

Answer 1

Firstly, note that $$\frac{\mathrm{d}\Phi(z)}{\mathrm{d}z}=\phi(z)$$ then, integrating by parts, using $v(z)=\Phi(z)$ and $\mathrm{d}u(z)=\phi(z)$: $$\int_x^\infty \Phi(z)\phi(z)\ \mathrm{d}z=\left[\Phi(z)^2\right]^\infty_x-\int_x^\infty \phi(z)\Phi(z)\ \mathrm{d}z$$ as $\mathrm{d}v(z)=\phi(z)$, $u(z)=\Phi(z)$, so $$2\int_x^\infty \Phi(z)\phi(z)\ \mathrm{d}z=\left(1-\Phi(x)^2)\right)$$ as $\lim_{z\rightarrow\infty}\Phi(z)=1$, so $$\int_x^\infty \Phi(z)\phi(z)\ \mathrm{d}z=\frac{1}{2}\left(1-\Phi(x)^2\right).$$

Remark. You could also solve this integral by making a substitution $u(z)=\Phi(z)$, then $\mathrm{d}u=\phi(z)\ \mathrm{d}z$, giving the same answer (check it to see).

Just checked this remark and the limits on the integral don't work, so disregard this.

Answer 2

Your Remark was correct. Set $y=\Phi(z)$ then

$$\int_x^\infty \Phi (z)\phi(z) dz =\int_{\Phi (x)}^1 y dy = \frac{1}{2}(1-\Phi(x)^2)$$