Inverse CDF of a function I am trying to find the Inverse CDF  (quantile function) of this function to create an random number generator: 
$f(p_a) = (\beta + 1) p_a^{\beta} \text{ where } \beta \geq 1 \text{ and  } 0 \leq p_a \leq 1$
Any clue about how to do it? The idea is to use randomly generated uniform numbers (from 0 to 1) to plug on the inverse CDF.
My background is not in mathematics, I have tried to get some solutions using Mathematica, but without success. 
 A: The function $$f(p_a) = (\beta + 1)p_a^\beta$$
is what we call a PDF (probability density function) in probability, reason being that when you integrate it with respect to $p_a$ from $0$ to $1$, it gives $1$. That is, $$\int_{0}^{1}f(p_a)\text{ d}p_a = 1\text{.}$$
The CDF (which we denote $F(p_a)$) is defined by integrating $f(p_a)$ from $0$ to $p_a$, where $0 \leq p_a \leq 1$.
Hence,
$$F(p_a) = \int_{0}^{p_a}f(t)\text{ d}t = \int_{0}^{p_a}(\beta+1)t^{\beta}\text{ d}t = (\beta+1)\dfrac{p_a^{\beta+1}}{\beta+1} = p_a^{\beta+1}\text{.}$$ 
Note that $F(p_a)$ is going to be your randomly-generated uniform numbers between $0$ and $1$ (reason being that it can actually be shown that $F(p_a)$ follows a uniform distribution in $[0, 1]$). So, to find the inverse, you will need to solve for $p_a$, which means
$$F(p_a) = p_a^{\beta+1} \implies [F(p_a)]^{1/(\beta+1)} = p_a\text{.}$$
Thus, $[F(p_a)]^{1/(\beta+1)}$ is your inverse CDF.
Note that in probability, this method is also known as inverse transform sampling.
