Square root of a Beta(1,1) random variable If $X^2 \sim \text{Beta}(1,1)$, is there a closed form for the distribution of $X$? If yes, what does it look like?
And if this is not too much to ask, is there a general way to find the distribution of $X$ for other related situations, such as $X^2 \sim \text{Beta}(0.5, 0.5)$ or $X^3 \sim \text{Beta}(1,1)$?
 A: If $X^{2}\sim\operatorname{Beta}(1,1)$ (which is a uniform distribution), then $X^p\sim\operatorname{Kumaraswamy}(1/p, 1)$ (see the Wikipedia page). The PDF of the resulting Kumaraswamy distribution is given by
$$
f(x)=\frac{x^{\frac{1}{p}-1}}{p}\qquad 0<x<1
$$
So for the example with $p=1/2$ we have $(X^2)^{1/2}=X\sim\operatorname{Kumaraswamy}(2, 1)$ with PDF $f(x) = 2x$ for $0<x<1$.
Generally, if $X\sim\operatorname{Beta}(\alpha,\beta)$, then $X^p$ with $p>0$ has PDF
$$
f(x)=\frac{x^{\alpha/p-1}\left(1 - x^{1/p}\right)^{\beta- 1}}{pB(\alpha, \beta)}\qquad 0<x<1
$$
where $B(\alpha, \beta)$ is the beta function.
For example, if $X^{2}\sim\operatorname{Beta}(0.5, 0.5)$, then $X$ has PDF (setting $p=1/2, \alpha = \beta=1/2$)
$$
f(x)=\frac{2}{\pi\sqrt{1 - x^{2}}}\qquad 0<x<1
$$
A: You ask for a general method.  Here is one.
When $X^p$ has a Beta$(\alpha,\beta)$ distribution for $p\gt 0,$ this means for all $0\lt y \lt 1$ that
$$F_X(y^{1/p}) = \Pr(X \le y^{1/p}) = \Pr(X^p \le y) = \frac{1}{B(\alpha,\beta)}\int_0^y t^{\alpha-1}(1-t)^{\beta-1}\,\mathrm{d}t.$$
Differentiating with respect to $y$ via the Chain Rule (at the left) and Fundamental Theorem of Calculus (at the right) reveals that
$$f_X(y^{1/p}) \frac{y^{1/p-1}}{p} = \frac{\mathrm{d}}{\mathrm{d}y} F_X(y^{1/p}) = \frac{1}{B(\alpha,\beta)}y^{\alpha-1}(1-y)^{\beta-1}$$
Solve this for $f_X,$ writing $x=y^{1/p}$ (which also lies in the interval $(0,1)$) to obtain

$$\begin{aligned}
f_X(x) &= \frac{1}{B(\alpha,\beta)}y^{\alpha-1}(1-y)^{\beta-1}\,p\, y^{1-1/p}\\
&= \frac{1}{B(\alpha,\beta)}\left(x^p\right)^{\alpha-1}(1-\left(x^p\right))^{\beta-1}\,p\, \left(x^p\right)^{1-1/p}\\
&=\frac{p}{B(\alpha,\beta)}\,x^{p\alpha-1}(1-x^p)^{\beta-1}.
\end{aligned}$$

That gives you closed forms for the density $f_X$ and, via integration, the distribution $F_X.$
Here are histograms created by drawing a sample of a million values from each of four Beta distributions and taking their $p^\text{th}$ roots for various $p.$  Over them are plotted in red the graphs of $f_X$ to demonstrate its correctness.

Here is the R code to use if you would like to view more examples.
#
# Describe a collection of distributions.
#
Theta <- rbind(c(alpha=3, beta=7, p=5),
               c(1/2, 3/2, 5),
               c(3/2, 1/2, 1/5),
               c(1, 1, 1/2))
n <- 1e6 # Sample size
#
# Sample and plot each distribution.
#
par(mfrow=c(2,2))
apply(Theta, 1, function(theta) {
  alpha <- theta[1]; beta <- theta[2]; p <- theta[3]
  Y <- rbeta(n, alpha, beta) 
  X <- Y^(1/p)
  
  hist(X, breaks=80, freq=FALSE, col="Gray")
  curve(p / beta(alpha,beta) * x^(p*alpha-1) * (1-x^p)^(beta-1), 
        n=501, lwd=2, col="Red", add=TRUE)
  mtext(bquote(paste(alpha==.(alpha), ", ", beta==.(beta),  ", and ", p==.(p))), 
        side=3, line=0)
})
par(mfrow=c(1,1))

