TITLE: The density of the square of the exponentially distributed random variable
QUESTION [6 upvotes]: Let $\xi$ - a random variable with an exponential distribution $p_{\xi} (x) = \lambda e^{-\lambda x}$
$$
p_{\xi} (x) = 
\begin{cases} 
0, & x < 0, \\ \lambda e^{-\lambda x}, & x \ge 0.
\end{cases}
$$
I want to find the density of a square $\xi^2$. Can I say that $p_{\xi^2} (x) = p_{\xi}^2 (x)$
or should I look for density according to the definition $p_{\xi^2} (x)  = \big ( F_{\xi^2} (x) \big )' = (\mathbb{P} \big ( \xi^2 < x) \big )'$ ?

REPLY [5 votes]: Your second approach is right; your first is not.
\begin{align}
p_{\xi^2}(x) & = \left(F_{\xi^2}(x)\right)' = \overbrace{\left(\mathbb P(\xi^2\le x)\right)' = \left( \mathbb P(\xi\le\sqrt{x}) \right)'}^{\text{This step works because $\xi$ is always $\ge\,0$.}} \\[15pt]
& =\frac{d}{dx} F_{\xi}(\sqrt{x}) = f_{\xi}(\sqrt{x})\frac d{dx}\sqrt{x} = \text{etc.}
\end{align}