probability density function of $X^2$ and $e^X$ based on $X$ Let $X$ be a random variable with probability density function $f_{X} \left( t \right)$, which for every $t<0$ satisfies: $f_{X} \left( t \right)=0$, and for every $t \geq 0$ satisfies: $f_{X} \left( t \right)=25te^{-5t}$.
I was asked to calculate the probability density function of $X^2$ and $e^X$.
This is my first time I encountered this kind of questions, and I don't really know how to get through it.
I only found that for every $t<0$ it holds that: $F_{X}\left(t \right) = 0$, and for every $t \geq 0$ it holds that: $$F_{X}\left(t\right)=\mathbb{P}\left(X \leq t\right)=\int_{0}^{t}25xe^{-5x}dx=1-e^{-5t}\left(5t+1\right)$$ using integration by parts. I know that I somehow need to use $F_{X^2}\left(t\right)=\mathbb{P}\left(X^{2} \leq t\right)=\mathbb{P}\left(|X| \leq \sqrt{t}\right)$ but I don't know how to proceed from here and how to to the same for $e^X$. Any help would be appreciated.
 A: There is a very useful theorem which applies to monotone transformations of random variables.  If $X$ has density $f_X(x)$, then for a function $g$ that is monotone on the support of $X$ and whose inverse $g^{-1}$ is differentiable, the random variable $Y = g(X)$ has density
$$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right|. \tag{1}$$
Let us apply this theorem to a simple example.  Suppose $$X \sim \operatorname{Exponential}(\lambda = 2), \\ f_X(x) = 2e^{-2x}, \quad x \ge 0.$$  Then if we wanted to find the density of $Y = X^2$, we let $g(x) = x^2$, hence $Y = g(X) = X^2$, and note that because the support of $X$ is $[0, \infty)$, $g$ is monotone increasing on this interval, and we have $$g^{-1}(y) = \sqrt{y},$$ which is differentiable:  $$\left|\frac{dg^{-1}}{dy}\right| = \left|\frac{1}{2\sqrt{y}}\right| = \frac{1}{2\sqrt{y}}.$$  Consequently, the density of $Y$ is
$$f_Y(y) = 2e^{-2\sqrt{y}} \cdot \frac{1}{2\sqrt{y}} = \frac{e^{-2\sqrt{y}}}{\sqrt{y}}, \quad y \ge 0.$$  The support of $Y$ is the same as the support of $X$, because if $X \ge 0$, then $Y = X^2 \ge 0$.
Now consider how we would use this theorem to answer your question: in your case, $$f_X(t) = 25 t e^{-5t}, \quad t \ge 0.$$  For $g(x) = x^2$, you would get
$$f_Y(t) = f_X(g^{-1}(t)) \left|\frac{dg^{-1}}{dt}\right| = f_X(\sqrt{t}) \left|\frac{d}{dt}\left[\sqrt{t}\right]\right|. \tag{2}$$  Can you finish the rest of the calculation?  Don't forget to specify the support of $Y$.
Last of all, try the same thing for $g(x) = e^x$.  In this case, what is $g^{-1}(t)$?  What is $dg^{-1}/dt$?  What is the support of the transformed variable?
