3
$\begingroup$

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?

Improve this question
$\endgroup$
2
  • $\begingroup$ could you define $\Phi$ in your question? $\endgroup$
    – Amir Sagiv
    Commented Apr 10, 2016 at 12:51
  • $\begingroup$ Sorry, yes, $\Phi(\cdot)$ denotes the normal CDF $\endgroup$
    – autoregress
    Commented Dec 13, 2016 at 1:27

1 Answer 1

Reset to default
3
$\begingroup$

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 $r$.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .