Distribution of $Z= X^2$ when $X$ has pdf $ f_X(x)=\frac {e^{-x}}{(1+e^{-x})^2}$

Question

Suppose you have $$ f_X(x)=\frac {e^{-x}}{(1+e^{-x})^2} \quad \forall \,x \in \mathbb R. $$

You are given that $$Z= X^2.$$ In order to find the pdf and distribution function of $Z$, this is what I did:

\begin{align} F_Z(z)=P(Z \leq z) &= P(X^2 \leq z) \\&= P( |X| \leq \sqrt{z} ) \\&= P( -\sqrt{z} \leq X \leq \sqrt{z}) \\&= F_X(\sqrt{z})-F_X(-\sqrt{z}) \end{align}

For any $$z<0,$$ we have that $$P(Z<0)=0 $$ since $$ Z \geq 0. $$ Now suppose that $$ z \geq 0$$ Hence, we need to integrate the pdf of x to obtain the distribution function of x.

This will be

$$ F_X(x)= \int_{-\infty}^x f_X(x) dx = \int_{-\infty}^x \frac {e^{-x}}{(1+e^{-x})^2} dx. $$

However, I'm not too sure where to take it from here. Any tips for how to proceed? I know that $$Z=X^2$$ isn't a bijection so you cant use the formula for transformation directly either.

Answer 1

Hint: check that $\frac{1}{1+e^{-t}}$ is the antiderivative of your integrand.

(Sorry, I only know this from past experience dealing with the logistic distribution. Perhaps someone else knows of a method from scratch.)