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Several threads (e.g. Integration of the product of pdf & cdf of normal distribution ) have shown that

$E[\Phi(x)]=\Phi(\mu/\sqrt{\sigma^2+1})$ when $x\sim N(\mu,\sigma^2)$.

I'd like to compute $Var(\Phi(x))$ for such an $x$, ideally without numerically integrating. Does anyone have any ideas?

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closed as unclear what you're asking by Chris Godsil, Alexey Ustinov, Wolfgang, Alex Degtyarev, Franz Lemmermeyer Apr 10 '16 at 17:25

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ could you define $\Phi$ in your question? $\endgroup$ – Amir Sagiv Apr 10 '16 at 12:51
  • $\begingroup$ Sorry, yes, $\Phi(\cdot)$ denotes the normal CDF $\endgroup$ – autoregress Dec 13 '16 at 1:27
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I think I figured it out, actually. Just as for the mean we can reinterpret $E[\Phi(x)]$ as $E[\mathbf 1[y<x]]$ for $y\sim N(0,1)$, we can reinterpret $E[\Phi(x)^2]$ as $E[\mathbf 1[y<x,y^\prime<x]]$ for independent standard normals $y,y^\prime$. Therefore,

$Var(\Phi(x)) = E[\Phi(x)^2]-E[\Phi(x)]^2=\Phi(\frac{\mu}{\sqrt{\sigma^2+1}},\frac{\mu}{\sqrt{\sigma^2+1}},\frac{\sigma^2}{\sigma^2+1})-\Phi(\frac{\mu}{\sqrt{\sigma^2+1}})^2$

Where $\Phi(h,k,r)$ is the bivariate normal CDF with correlation $r$.

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