Start with a random variable $r \sim \mathcal{N}(0,1)$.

Now consider the random variable $\sigma(r)$ formed by passing it through a standard logistic function $\sigma(x) = \frac{1}{1 + e^{-x}}$. I would like to know the mean and variance of $\sigma(r)$.

The mean value is 0.5 by symmetry, but the variance is trickier. By definition $$\mathrm{Var}[\sigma(r)] = \mathbb{E}[\sigma(r)^2] - \mathbb{E}[\sigma(r)]^2$$ and $\mathbb{E}[\sigma(r)]^2 = 0.25$.

Is there a way to calculate $\mathbb{E}[\sigma(r)^2]$ analytically?

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    $\begingroup$ With is the computation of the mean so obvious? $\endgroup$ – Davide Giraudo Jan 30 '19 at 11:40
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    $\begingroup$ I'm pretty sure it follows from the symmetry of $r$ and $\sigma$. $\endgroup$ – prdnr Jan 30 '19 at 11:41
  • $\begingroup$ I see. The key point is that $\sigma(x)+\sigma(-x)=1$, but I do not know whether a similar relation holds for $\sigma^2$. $\endgroup$ – Davide Giraudo Jan 30 '19 at 14:58
  • $\begingroup$ Yes, $\sigma^2$ is not symmetric so there I doubt there's any similar identity $\endgroup$ – prdnr Jan 30 '19 at 15:16
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    $\begingroup$ According to the wiki page for the logit-normal (the distribution you're describing), there are no analytical formula for the moments en.wikipedia.org/wiki/Logit-normal_distribution. However, you could easily get a monte carlo approximation with a few lines of R. $\endgroup$ – aleshing Jan 30 '19 at 15:57

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