# How do I understand the intuition behind percentile point function?

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() ?

• 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. Oct 28, 2022 at 18:38

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 $$Q(1) = \infty$$.