# If $X$ is standard normal, how to compute the PDF of $Y=X/(1+ e^{-X})$?

If $X \sim N(0,1)$, how can we compute the PDF of $\frac{X}{1+ e^{-X}}$?

I am wondering if there is perhaps an easy way without jacobians. Does anyone have any ideaS?

• It seems that $\frac{X}{1+ e^{-X}}$ is not a one to one function. In this case, how can we determine the Jacobian? Jul 23, 2017 at 23:36
• When $h$ is not one-to-one but regular enough, the PDF $g$ of $Y=h(X)$ is $$g(y)=\sum_{x:h(x)=y}\frac{f_X(x)}{|h'(x)|}$$
– Did
Jul 23, 2017 at 23:43
• Is the summation above over points of the region where $h$ is one-to-one? Jul 23, 2017 at 23:57
• Huh? The summation is over every $x$ such that $h(x)=y$ (as written in the formula), whether there is a unique such point or several of them.
– Did
Jul 24, 2017 at 0:02
• @Did is there a link to the above formula and/or more expalantion? Oct 16, 2017 at 8:14