Random variable $\xi : (\Omega, \mathcal{F}, \mathbb{P}) \rightarrow \mathbb{R}$ has Laplace distribution with density function $f_{\xi} (x)= \frac{1}{2}\exp \{-|x| \}$ where $x \in \mathbb{R}$. I need to find a distribution function of random variable $\xi^2$.
I have no idea what to do. But I think I could find a distribution function of $F_{\xi}(x)$ because I have a density function and I know that $$F_{\xi}(x)=\int_{-\infty}^x\frac{1}{2}\exp \{-|u| \} du$$
So I found that
when $x<0$ we have $$F_{\xi}(x)=\frac{e^x}{2}$$
when $x \geq 0$ we have $$F_{\xi}(x)=1-\frac{e^{-x}}{2}$$ but what I have to do next?
You can find an expression for $F_{\xi^2}(x)$ now.
Evidently $F_{\xi^2}(x)=0$ for $x\leq 0$, and for $x>0$ we have:$$F_{\xi^2}(x)=P(\xi^2\leq x)=P(-\sqrt x\leq\xi\leq\sqrt x)=F_{\xi}(\sqrt x)-F_{\xi}(-\sqrt x)$$
If next to the CDF you also want a PDF then find it as derivative of the CDF.