About logistic distribution and showing $Y=\frac{1}{1+e^{-X}}$ has an uniform distribution over (0,1)

Question

Let X have a logistic distribution with PDF $f_X(x)=\frac{e^{-x}}{(1+e^{-x})^2}$, $-\infty<x<\infty$. Show $Y=\frac{1}{1+e^{-X}}$ has an uniform distribution over (0,1).

Let $Y=u(x)=\frac{1}{1+e^{-X}} $Since this is monotone decreasing, Can use $f_Y(y)=\frac{1}{|u'(x)|}f_X(x)$. What's the general technique of approaching?

Answer 1

This is a special case of the fact that if a continuous random variable $X$ of support $\Bbb R$ has CDF $F$ then $F(X)\sim U(0,\,1)$, which is trivial because$$P(F(X)\le f)=P(X\le F^{-1}(f))=F(F^{-1}(f))=f$$for all $f\in[0,\,1]$.