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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?

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  • $\begingroup$ It seems that $\frac{X}{1+ e^{-X}}$ is not a one to one function. In this case, how can we determine the Jacobian? $\endgroup$
    – user321627
    Jul 23, 2017 at 23:36
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    $\begingroup$ When $h$ is not one-to-one but regular enough, the PDF $g$ of $Y=h(X)$ is $$g(y)=\sum_{x:h(x)=y}\frac{f_X(x)}{|h'(x)|}$$ $\endgroup$
    – Did
    Jul 23, 2017 at 23:43
  • $\begingroup$ Is the summation above over points of the region where $h$ is one-to-one? $\endgroup$
    – user321627
    Jul 23, 2017 at 23:57
  • $\begingroup$ 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. $\endgroup$
    – Did
    Jul 24, 2017 at 0:02
  • $\begingroup$ @Did is there a link to the above formula and/or more expalantion? $\endgroup$
    – user321627
    Oct 16, 2017 at 8:14

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