Let X have a logistic distribution with p.d.f.
f(x) = e^(-x)/(1 + e^(-x))^2 -infinty <x<infinity
Show that
Y = 1/(1 + e^(-x))
has a U (0, 1) distribution. (Hint: Find P (Y ≤ y) = P(1/(1 + e^(-x)<=y)
First note that $\displaystyle 0 < y < 1$ since $\displaystyle 0 < \frac{1}{1 + e^{-x}} < 1$ for $\displaystyle -\infty < x < + \infty$.
Then the cdf of Y is given by
$\displaystyle F(y) = \Pr(Y \leq y) = \Pr \left( \frac{1}{1 + e^{-X}} \leq y \right) = \Pr\left(1 + e^{-X} \geq \frac{1}{y} \right)$
$\displaystyle = \Pr\left( X \leq \ln \left( \frac{y}{1 - y}\right)\right)$
$\displaystyle = \int_{-\infty}^{\ln \left( \frac{y}{1 - y}\right)} f(x) \, dx$.
Your job is to calculate this integral (for 0 < y < 1 the answer is y, by the way) and then calculate $\displaystyle f(y) = \frac{dF}{dy}$. Note also that details are important ....