If $X \sim N(\mu_X, \sigma^2_X)$ and $Y= \frac{\exp(X)}{1+\exp(X)} $, what is the distribution of $Y$?
I thought logit-normal would fit the bill, however the logit of $Y$ is
$$\log\left({\frac{\exp(X)}{(\exp(X)+1)(1-\frac{\exp(X)}{\exp(X)+1})}}\right)$$
which is not $X$. Is there a known formula or do we need to derive explicitly by differentiating the CDF $P(g(X)\le x)$
The function $g(x) = e^x/(1+e^x)$ is monotone, so if $Y = g(X)$ where $X \operatorname{Normal}(\mu,\sigma^2)$, then $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dg^{-1}}{dy} \right|.$$ Note the support of $Y$ is $(0,1)$; $$g^{-1}(y) = \log \frac{y}{1-y},$$ and $$\frac{dg^{-1}}{dy} = \frac{1}{y(1-y)}.$$ Therefore, $$f_Y(y) = \frac{1}{\sqrt{2\pi} \sigma y(1-y)} \exp\left( - \frac{\left(\log \frac{y}{1-y} - \mu\right)^2}{2\sigma^2} \right), \quad 0 < y < 1.$$