# Closed forms for various expectations involving the standard normal CDF

Suppose that $X\sim\mathcal{N}\left(0,1\right)$ (i.e., $X$ is a standard normal random variable) and $a,b,$ and $c$ are some real constants. Does any of the following expectations have a closed-form?

1. $\mathbb{E}\left[\log\Phi\left(aX\right)\right]$
2. $\mathbb{E}\left[\Phi\left(aX\right)\log\Phi\left(aX\right)\right]$
3. $\mathbb{E}\left[\Phi\left(aX\right)\Phi\left(bX+c\right)\right]$

I’ve tried the usual derivative trick followed by Stein’s lemma, but the expressions didn’t get much simpler. For the third one if $a=b$ and $c=0$ the closed-form solution is $$\mathbb{E}\left[\left(\Phi\left(aX\right)\right)^2\right]=\frac{\tan^{-1}\left(\sqrt{1+2a^2}\right)}{\pi}$$ but I couldn’t reach at any generalization.

#### Solutions Collecting From Web of "Closed forms for various expectations involving the standard normal CDF"

Cases 1. and 2. when $a=\pm1$ follow from the following remark: for every suitable function $u$,
$$\mathrm E(u'(\Phi(X)))=\int_{-\infty}^{+\infty} u'(\Phi(x))\varphi(x)\,\mathrm dx=\left[u(\Phi(x))\right]_{x=-\infty}^{x=+\infty}=u(1)-u(0).$$
For example, $u(t)=t\log(t)-t$ yields $u'(t)=\log t$, $u(1)=-1$ and $u(0)=0$ hence
$$\mathrm E(\log\Phi(aX))=-1,\quad a=\pm1.$$
Likewise, $u(t)=\frac12t^2\log t-\frac14t^2$ yields $u'(t)=t\log t$, $u(1)=-\frac14$ and $u(0)=0$ hence
$$\mathrm E(\Phi(aX)\log\Phi(aX))=-\tfrac14,\quad a=\pm1.$$
About case 3., $u(t)=\frac13t^3$ yields $u'(t)=t^2$, $u(1)=\frac13$ and $u(0)=0$ hence
$$\mathrm E(\Phi(aX)^2)=\tfrac13,\quad a=\pm1.$$

Another approach is to differentiate with respect to the parameter $a$ and to use Stein’s lemma. Consider $v(a)=\mathrm E(u(\Phi(aX)))$ for some suitable function $u$, then $v(0)=u(\frac12)$ and
$$v'(a)=\mathrm E(X\varphi(aX)u'(\Phi(aX)))=\mathrm E(Xg(X))$$ where $$g(x)=\varphi(ax)u'(\Phi(ax)).$$
Stein’s lemma and the identities $\varphi'(s)=-s\varphi(s)$ and $\Phi'(s)=\varphi(s)$ yield
$$v'(a)=\mathrm E(g'(X))=\mathrm E(-a^2X\varphi(aX)u'(\Phi(aX))+a\varphi(aX)^2u”(\Phi(aX))),$$
hence
$$v'(a)=-a^2v'(a)+a\mathrm E(\varphi(aX)^2u”(\Phi(aX))),$$
and
$$v'(a)=\frac1{1+a^2}\mathrm E(\varphi(aX)^2u”(\Phi(aX))).$$
If $u(t)=t^2$, $u”(t)=2$ hence one gets $v(0)=u(\frac12)=\frac14$ and
$$\mathrm E(\varphi(aX)^2)=\int_{-\infty}^{+\infty}\varphi(ax)^2\varphi(x)\,\mathrm dx=\frac1{2\pi}\int_{-\infty}^{+\infty}\varphi(\sqrt{1+2a^2}x)\,\mathrm dx=\frac1{2\pi\sqrt{1+2a^2}}$$
hence
$$v'(a)=\frac{a}{\pi (1+a^2)\sqrt{1+2a^2}}.$$
Integrating this, one gets
$$v(a)=\frac14+\frac1{\pi}\int_0^a\frac{x\mathrm dx}{(1+x^2)\sqrt{1+2x^2}}=\frac14+\frac1{\pi}\left[\arctan\sqrt{1+2x^2}\right]_{x=0}^{x=a},$$
and finally,
$$\mathrm E(\Phi(aX)^2)=\frac1\pi\arctan\sqrt{1+2a^2}.$$
Likewise, $u(a)=\mathrm E(\Phi(aX)\Phi(bX))$ yields
$$u'(a)=\mathrm E(X\varphi(aX)\Phi(bX))=\mathrm E(Xg(X))$$ where $$g(x)=\varphi(ax)\Phi(bx).$$
Stein’s lemma and the formula $g'(x)=-a^2x\varphi(ax)\Phi(bx)+b\varphi(ax)\varphi(bx)$ yield
$$(1+a^2)v'(a)=b\mathrm E(\varphi(aX)\varphi(bX)),$$
hence
$$v'(a)=\frac{b}{2\pi (1+a^2)\sqrt{1+a^2+b^2}}.$$
Finally,
$$\mathrm E(\Phi(aX)\Phi(bX))=\frac1\pi\arctan\sqrt{1+2b^2}+\frac{b}{2\pi}\int_b^a\frac{\mathrm dx}{(1+x^2)\sqrt{1+x^2+b^2}}.$$
Amongst several equivalent formulations, this means that

$$\mathrm E(\Phi(aX)\Phi(bX))=\frac14+\frac1{2\pi}\arctan\left(\frac{ab}{\sqrt{1+a^2+b^2}}\right).$$

The case $\mathrm E(\Phi(aX)\Phi(bX+c))$ might also be solvable with this method.