What Distribution Has This CDF?

Question

Does the probability distribution with the CDF:

$$F(x)=1-(1-x^b)^\frac{1}{b}$$ where $x\in[0,1]$ and $b\in[0,\infty)$ have a name? Is this a known distribution?

Answer 1

The first thing to notice is that the support of such a distribution is on $X \in [0,1]$, thus ruling out most options. The density is $$f_X(x) = x^{b-1} (1-x^b)^{1/b - 1},$$ which is vaguely reminiscent of a beta distribution. Indeed, if we take the usual beta distribution $Y \sim \operatorname{Beta}(r,s)$ with density $$f_Y(y) = \frac{\Gamma(r+s)}{\Gamma(r)\Gamma(s)} y^{r-1} (1-y)^{s-1}, \quad y \in [0,1]$$ and transform it with $X = Y^{1/b}$, we obtain $$\begin{align} f_X(x) &= f_Y(x^b) bx^{b-1} \\ &= \frac{\Gamma(r+s)}{\Gamma(r)\Gamma(s)} x^{b(r-1)} (1 - x^b)^{s-1} bx^{b-1} \\ &= b \frac{\Gamma(r+s)}{\Gamma(r)\Gamma(s)} x^{br - 1}(1 - x^b)^{s-1}. \end{align}$$

Then the specific choice of parameters $r = 1$, $s = 1/b$, yields the target density:

$$f_X(x) = b\frac{\Gamma(1/b+1)}{\Gamma(1)\Gamma(1/b)} x^{b-1} (1-x^b)^{1/b-1} = x^{b-1}(1-x^b)^{1/b-1},$$

since $\Gamma(1/b+1) = \frac{1}{b} \Gamma(1/b)$. So we can recognize $X$ as following a transformed $\operatorname{Beta}(1,1/b)$ distribution where the transformation is $X = Y^{1/b}$. This also happens to be a special case of the Kumaraswamy distribution which is parameterized as $$f_X(x \mid a, b) = ab x^{a-1} (1-x^a)^{b-1}.$$ In your special case, we have $$X \sim \operatorname{Kumaraswamy}(b, 1/b).$$