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How can one evaluate the expectation of the squared normal CDF in closed-form?

$$\mathbb{E}\left[\Phi\left(aZ+b\right)^{2}\right] = \int_{-\infty}^{\infty}\Phi\left(az+b\right)^{2}\phi(z)\,dz$$

Here, $a$, $b$ are real numbers, $Z\sim\mathcal{N}(0,1)$, and $\phi(\cdot)$ and $\Phi(\cdot)$ are the density and distribution functions of a standard normal random variable, respectively.

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  • $\begingroup$ Well where do you get stuck? Have you tried to evaluate it? Maybe use the fact that $\text{Var}(g(X))=E[g(X)^2]-(E[g(X)])^2$ $\endgroup$
    – stoched
    Commented Mar 23, 2015 at 0:33
  • $\begingroup$ I tried to evaluate the integral, using integration by parts and other (simple) techniques, but that did not lead me anywhere. Also, I actually started from the variance to get here. I found a similar question (stats.stackexchange.com/questions/61080/…), but extending to the squared CDF does not seem to be trivial. $\endgroup$
    – Andrei
    Commented Mar 23, 2015 at 0:35
  • $\begingroup$ Have you considered using polar coordinates? $\endgroup$
    – StatsStudent
    Commented Mar 23, 2015 at 0:57
  • $\begingroup$ No I haven't, can you detail a bit? $\endgroup$
    – Andrei
    Commented Mar 23, 2015 at 1:35
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    $\begingroup$ If $b=0$ and $a=1$, then $\Phi(Z)$ is uniformly distributed between 0 and 1. Its second moment is then $1/3$. I recall trying to calculate something like what you ask for general $a$ and $b$, but I found no closed-form solutions. $\endgroup$
    – StijnDeVuyst
    Commented Mar 23, 2015 at 2:32

1 Answer 1

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As noted in my comment above, check Wikipedia for a list of integrals of Gaussian functions. Using your notation, it gives $$\int_{-\infty}^{\infty} \Phi (az+b)^2 \phi(z) dz=\Phi \left( {{b} \over {\sqrt{1+a^2}} }\right) - 2T \left( {{b} \over {\sqrt{1+a^2}}} \ , {{1} \over {\sqrt{1+2a^2}}} \right) ,$$ where $T(h,q)$ is Owen's T function defined by $$T(h,q)=\phi(h) \int_0^q {{\phi(hx) } \over {1+x^2}} dx$$

If you plug in $a=1,b=0$ you will get $1 \over 3$ as the comments indicate you should.

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