The probability integral transform is a fundamental concept in statistics that connects the cumulative distribution function, the quantile function, and the uniform distribution. We motivate the need for a generalized inverse of the CDF and prove the result in this context.

Definitions

Suppose $U$ is the uniform distribution on $[0,1]$ , and $F$ is the cumulative distribution of the random variable, $Y$ , i.e.

and define the inverse cdf — the quantile function — as

Figure: an example cdf

Recall that distributions — cdfs — are right continuous and monotonically increasing. However, as the example shows, they may also be flat and discontinuous. Indeed, the flat sections motivate the definition of the inverse cdf as a minimum. Thus, in the example, while $\forall x \in [0,1]$ $F(x) = 0.5$ , $F^{-1}(0.5) = 0$.

The Main Result:

The key result is:

That is, $F^{-1}(U)$ and $Y$ have the same distribution.

Proof

The proof isn’t particularly complicated, but it relies on two identities that follow from the definition of the inverse cdf.

First,

This follows from the definition as $F^{-1}(u)$ is the smallest value of $y$ for which $F(y) \geq u$.

Second,

Using the definition, write the left hand side, with a change of index, as

The inequality follows from the following argument. From the example, we can see that $z$ can certainly be less than $y$, but because $F$ is right continuous, the minimum cannot exceed $y$.

The main result is a statement about probabilities. As such, we can proceed by showing the following:

In what follows, let $u = U(\omega)$.

The forward implication

We will show

Suppose $F^{-1}(u) \leq y$.

Then, since $F$ is monotonic, we can apply it to both sides of the inequality without issue:

Combining this with the first of the above identities,

Hence,

The reverse implication

The argument is similar. Note that $F^{-1}$ is also monotonic. Thus, $u \leq F(y)$ implies

and the result follows after applying the second of the above identities: