Probability integral transform

One use for the probability integral transform in statistical data analysis is to provide the basis for testing whether a set of observations can reasonably be modelled as arising from a specified distribution. Specifically, the probability integral transform is applied to construct an equivalent set of values, and a test is then made of whether a uniform distribution is appropriate for the constructed dataset. Examples of this are P-P plots and Kolmogorov-Smirnov tests.

A second use for the transformation is in the theory related to copulas which are a means of both defining and working with distributions for statistically dependent multivariate data. Here the problem of defining or manipulating a joint probability distribution for a set of random variables is simplified or reduced in apparent complexity by applying the probability integral transform to each of the components and then working with a joint distribution for which the marginal variables have uniform distributions.

A third use is based on applying the inverse of the probability integral transform to convert random variables from a uniform distribution to have a selected distribution: this is known as inverse transform sampling.

Suppose that a random variable X has a continuous distribution for which the cumulative distribution function (CDF) is F_X. Then the random variable Y defined as

${\displaystyle Y=F_{X}(X)\,,}$

has a standard uniform distribution.[1]

Given any random continuous variable ${\displaystyle X}$, define ${\displaystyle Y=F_{X}(X)}$. Then:

${\displaystyle {\begin{aligned}F_{Y}(y)&=\operatorname {P} (Y\leq y)\\&=\operatorname {P} (F_{X}(X)\leq y)\\&=\operatorname {P} (X\leq F_{X}^{-1}(y))\\&=F_{X}(F_{X}^{-1}(y))\\&=y\end{aligned}}}$

${\displaystyle F_{Y}}$ is just the CDF of a ${\displaystyle \mathrm {Uniform} (0,1)}$ random variable. Thus, ${\displaystyle Y}$ has a uniform distribution on the interval ${\displaystyle [0,1]}$.

For an illustrative example, let X be a random variable with a standard normal distribution ${\displaystyle {\mathcal {N}}(0,1)}$. Then its CDF is

${\displaystyle \Phi (x)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{x}e^{-t^{2}/2}\,dt={\frac {1}{2}}{\Big [}\,1+\operatorname {erf} {\Big (}{\frac {x}{\sqrt {2}}}{\Big )}\,{\Big ]},\quad x\in \mathbb {R} ,\,}$

where ${\displaystyle \operatorname {erf} (),}$ is the error function. Then the new random variable Y, defined by Y=Φ(X), is uniformly distributed.

If X has an exponential distribution with unit mean, then its CDF is

${\displaystyle F(x)=1-\exp(-x),}$

and the immediate result of the probability integral transform is that

${\displaystyle Y=1-\exp(-X)}$

has a uniform distribution. The symmetry of the uniform distribution can then be used to show that

${\displaystyle Y'=\exp(-X)}$

also has a uniform distribution.

• Inverse transform sampling