As an introduction to the "Transformation of Random Variables" in my lecture notes, we have the following example:
$\textit{Suppose $Y$ has density function $f(y) = 2y$, for $0<y<1$. Let $U = Y^2$. What is the pdf of $U$?}$
The solution is the following:
$\textit{Let $G$ and $g$ be the cdf and pdf of $U$. Then}$
$$\textit{ $G(u) = P(U\leq u) = P(Y^2 \leq u) = P(Y \leq\sqrt{u}) = F(\sqrt{u})$}$$
$$\textit{$g(u) = \frac{dG(u)}{du} = \frac{dF(\sqrt{u})}{du} = f(\sqrt{u})\frac{1}{2\sqrt{u}} = 2\sqrt{u}\frac{1}{2\sqrt{u}}= 1\,\,\,\,\,\,\,\,$ for $\,\,\,\,\,\,\,\,0<u<1$}$$
$\textit{So $U$ is a uniform random variable on $(0,1)$}$
I understand most of it I think. But I have two doubts:
- Do we have $0<u<1$ because for $y\in(0,1)$ we have that $y^2 = u \in (0,1)$? Or is there another reason? Maybe we just decided to assign that interval to $u$?
- How do we conclude that $U$ is a uniform random variable? Because the pdf is constant?
Thank you
Best Answer
Regarding your first question, yes: $0 < y < 1$ clearly implies $0 < y^2 < 1$, hence $0 < u < 1$.
Regarding the second, $U$ is uniform on $(0,1)$ because we saw that $f_U(u) = 1$ for $0 < u < 1$, which is precisely the definition of a (continuous) uniform random variable on this interval.