It is true that $\mathbf X \sim F_X \Rightarrow F_X(\mathbf X) \sim U_{[0;1]}$; does the converse hold for multivariate $\mathbf X$?

For a univariate real-valued random variable I am pretty sure that the converse holds.

Consider a multivariate $\mathbf X$ with values in $\mathbb R^n$ with measure $\mu(X)$ and its multivariate CDF is:

$$F_X(a_0,\dots, a_n) = \int_{-\infty}^{a_0} \dots \int_{-\infty}^{a_n} \mu(d\mathbf x)$$

, then the univariate random variable $\tilde X = F_X(\mathbf X)$ is still distributed uniformly in $U_{[0;1]}$.

Is it true that if for some other $\mathbf Y$ holds $F_X(\mathbf Y) \sim U_{[0; 1]}$, then $F_X = F_Y$?

Or, in other words, can we use uniformity of $F_X(\mathbf Y)$ to test if $F_X = F_Y$ for multivariate $\mathbf X$ and $\mathbf Y$?

UPD

@whuber answered in the comment below that the "$\tilde X \sim U_{[0;1]}$" part is actually wrong, so the question does not make much sense and one interested in multivariate variables and uniformity should read about Copulas.

• You implicitly assume all variables are non-discrete; for discrete variables, it is not the case that $F_X(X)$ is uniform. Notice that the distribution function implied by "$U_{[0;1]}$" is the identity map on the open interval $(0,1)$. – whuber Oct 18 '17 at 20:01
• By using copulas, you may restrict your study to the case where the marginals of $X$ are uniform on the interval $[0,1]$. Plotting a few copulas will make the answers immediately apparent: in particular, $\tilde X$ is not necessarily uniform. – whuber Oct 30 '17 at 19:54

I think I came up with a counterexample, though it is kind of intuitive then rigorous. Consider any reasonable $F(x_0, x_1)$ on $X^2$, then for each number $a$ on $[0;1]$ there must be some "large" subset $H_a \subset X^2$ such that $F(x) = a \ \ \forall x \in H_a$, because "whatever positive direction you go from any point with $F(x_s) < a$, you will eventually hit $F(.) = a$ because you should eventually hit $F=1$". Therefore, if we take that set $H_a$ and do any type of perturbation of points there, distribution of $F(X)$ will not change. For example, choose some single line $l\in X^2$ and put all points in $H_a$ so that they lie on this line for each value of $a\in[0;1]$ such that $H_a$ intersects with $l$, we whould get a different distribution, but $F(X)=U[0;1]$ would still hold.
I am not sure about "repeat this process for each $a$" part, because $[0;1]$ is continuum and each $H_a$ has zero measure.
• It isn't at all evident what you mean by a "perturbation of points there" or how it could affect $F$. – whuber Oct 30 '17 at 19:54