Find the pdf of a new random variable that is a power of a uniform distribution

Question

I'm not sure how to find the pdf for this example

The random variable $U$ follows the distribution $U(0, 1)$ find the pdf of a new random variable $X = U^\alpha$ for $\alpha > 0$.

Answer 1

Using the cumulative distribution function (cdf) of $V \stackrel{\rm def}{=}U^\alpha$, denoted $F_V$: for every $t\geq 0$, $$ F_V(t) = \mathbb{P}\{ V \leq t \} = \mathbb{P}\{ V^{1/\alpha} \leq t^{1/\alpha} \} = \mathbb{P}\{ U \leq t^{1/\alpha} \} = \begin{cases} t^{1/\alpha} &\text{ if } 0\leq t\leq 1\\ 1 &\text{ if } t\geq 1\end{cases} $$ and $F_V(t) = 0$ if $t<0$. Then, the probability density function of $F_V$ is, by differentiation, $$ f_V(t) = \begin{cases} \frac{1}{\alpha}t^{1/\alpha-1} &\text{ if } 0\leq t\leq 1\\ 0 &\text{ otherwise.}\end{cases} $$

Sanity check: for $\alpha=1$, we get $f_V(t) = \mathbf{1}_{(0,1)}(t)$ ($V=U$ is uniform on $[0,1)$)