# Empirical multivariate probability integral transform

Is there a 'simple' way to obtain a non-parametric empirical multivariate probability integral transform?

Univariate case

The probability integral transform relates to the transform of any random variables with continuous cumulative distribution function into an uniformly distributed random variables : Let $X$ be a random variable and $F_X$ its continuous cumulative distribution functions, then $F_X(X) \sim U(0,1)$.

Indeed $F_{F_X(X)}(x) = P\left(F_X(X) \leq x\right) = P\left(X \leq F_X^{-1}(x)\right) = F_X\left(F_X^{-1}(x)\right) = x.$

It can be conveniently estimated from a sample of real values $\{x_1,\ldots,x_n\}$ with rank statistics: $\frac{1}{n} \sum_{i=1}^n \mathbf{1}(x_i \leq x_j) = \text{rank}(x_j)/n$ yields values uniformely distributed in $U(0,1)$.

Multivariate case

Now consider a random vector $X = (X_1,\ldots,X_d)$ with marginal cumulative distribution functions $F_1,\ldots,F_d$. Using the probability integral transform we obtain $U = (U_1,\ldots,U_d) = (F_1(X_1),\ldots,F_d(X_d))$ whose joint distribution is the copula $C$. The analogous of the probability integral transform for the multivariate case is the Kendall distribution function $K(t) := P(C(U) \leq t)$. We have $K(C(U)) \sim U(0,1)$.

From a sample $\{x_1,\ldots,x_n\}$ where $\forall j \in \{1,\ldots,n\}, x_j = (x_j^1,\ldots,x_j^d)$ are $d$-dim. vectors, I want to estimate $K(C(U))$ in a 'simple' non-parametric way like the one with rank statistics of the $x_j$ in the univariate case.

• This may be relevant. – GeoMatt22 Apr 21 '17 at 22:57