Distribution of Y=F(X) where F is the cdf of continuous X and X is a Multivariate r.v.

Question

Here is a well known fact (from this question):

Let $X$ be a random variable (r.v.) with a continuous and strictly increasing c.d.f. function $F$. Define a new random variable $Y$ by $Y=F(X)$. Then $Y$ has a uniform distribution on the interval $[0,1]$.

My question is:

Does this fact only hold for uni-variate case (i.e. $X \in \mathbb{R}$)? What if $X \in \mathbb{R}^d$?

p.s. the proof in this note seems assuming X is univariate...

Answer 1

Did a simple Matlab simulation, it turns out this conjecture is wrong.

Here is the code:

clear all;clc;close all;

n_pwr = 5;

N = 10^n_pwr;

n_bin = 2^n_pwr;

%% Univariate Case

mu = 0;

Sigma = 10;

%% 2-Dim Case

% mu = [1 -1];

% Sigma = [.9 .4; .4 .3];

%% 3-Dim Case

% mu = [1 -1 100];

% Sigma = [.9 0 0; % 0 .3 0; % 0 0 1];

X = mvnrnd(mu, Sigma, N);

Y = mvncdf(X, mu, Sigma);

f1 = figure;

hist(Y,n_bin);

print(f1, '-dpng', 'distr_CDF.png')

For 1D case:clearly uniformly distributed

2D case: looks like a sigmoid function

3D case: quite different from the 3D case

Now we have a better question: what kind of distribution is it?

Answer 2

Suppose that $X=(X_1,\ldots,X_d)$ and that each $X_k$ has a continuous distribution function $F_k$. Then $U_k:=F_k(X_k)$ is uniformly distributed on $[0,1]$ for $k=1,\ldots,d$.

Notice that $Y\le F(X_1,\infty,\ldots,\infty) =F_1(X_1) = U_1$, so the only way that $Y$ can be uniformly distributed is when $Y=U_1$ a.s. Considering the other coordinates as well, you see that if $Y$ is uniform on $[0,1]$ then $Y=U_1=U_2=\cdots=U_d$ a.s. This degenerate case aside, $Y$ will not be uniform.

Example: If the $X_k$ are independent, then $Y=U_1U_2\cdots U_d$ is the product of independent uniforms.