From Wikipedia
the probability integral transform or transformation relates to the result that data values that are modelled as being random variables from any given continuous distribution can be converted to random variables having a uniform distribution. This holds exactly provided that the distribution being used is the true distribution of the random variables; if the distribution is one fitted to the data the result will hold approximately in large samples.
- I was wondering if the probability integral transform can be generalized to a random variable with a discrete distribution? Or any distribution?
Is the probability integral transform an integral transform or some generalization of it?
an integral transform is any transform T of the following form: $$ (Tf)(u) = \int \limits_{t_1}^{t_2} K(t, u)\, f(t)\, dt $$
Thanks and regards!
I will answer the first question. When a random variable $X$ is discrete, the range of its cumulative distribution function $F_X$ is countable, and thus $F_X (X)$ could not be distributed as uniform $(0, 1)$. However, the converse of your result still holds when $X$ is discrete. The generalized inverse of $F_X$ (or quantile function) defined as \begin{eqnarray*} F_X^{- 1} (u) & = & \inf_{} \{ x : F_X (x) \geqslant u \} \end{eqnarray*} satisfies the property that if $U$ is distributed uniform $(0, 1)$ then $F_X^{- 1} (U)$ has the same distribution as $X$. For more details, see chapter 7 in Shorack's 2000 book "Probability for statisticians".