I'm really trying very hard to understand the intuition behind percentile point function. From wikipedia, It is the inverse of cdf.

From the CDF, a point on its curve indicates the percentage of points less than or equal to that point. I hope this is the right understanding ?

For ppf to be inverse of cdf, the definition of a point on its curve would indicate the value of the variable with probability less than or equal to the probability at that point. Is it the right understanding ?

How would the definition change in context of beta distribution, specifically beta.ppf() ?

  • $\begingroup$ I have updated my answer adding an example that illustrates the discrete case. By the way, if you found my answer useful, please consider accepting/upvoting it. $\endgroup$
    – utobi
    Oct 28, 2022 at 18:38

1 Answer 1


For a random variable $X$ with cumulative distribution function $F(x) = P(X\leq x)$, the usual definition of the quantile function is $$ Q(p) = \inf\{x: F(x)\geq p\},\quad p\in(0,1) .$$

Now if $X$ is absolutely continuous, then $F$ is continuous, monotone increasing and, in this case, $Q = F^{-1}$.

Now, in the continuous case, for any $p\in (0,1)$, the $p$th quantile, say $\xi_p = Q(p)$, can be seen as

the particular value of the r.v. $X$ which leaves probability mass $p$ on the left and probability mass $1-p$ on the right.

In other words, the $p$th quantile $\xi_p$ is s.t.

$$ p = P(X\leq \xi_{p}) = F(\xi_{p}). $$

This interpretation holds for any distribution, the beta included.

For the discrete case consider the following example. Let $X$ be a discrete random variable with probability density function

$$f(x) = \begin{cases} 1/3 &\text{if } x=-1\\ 1/6 &\text{if } x=0\\ 1/2 &\text{if } x=1\\ 0 &\text{otherwise} \end{cases}. $$

The distribution function is $$F(x) = \begin{cases} 0 &\text{if } x \in (-\infty, -1)\\ 1/3 &\text{if } x \in [-1, 0)\\ 3/6 &\text{if } x \in [0, 1)\\ 1 &\text{if } x \in [1, \infty) \end{cases}, $$

and the quantile function is

$$ Q(p)= \begin{cases} -1 & \text{if } p\in(0,1/3]\\ 0 & \text{if } p\in(1/3,3/6]\\ 1 & \text{if } p\in(3/6,1)\\ \end{cases}. $$

You can if you wish set $Q(0) = -\infty$ and $Q(1) = \infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.