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Let $X\colon \Omega \to \mathbb R$ a real valued random variable with continuous p.d.f. Then if $F_X$ is its c.d.f., the random variable $F_X(X)$ is uniformly distributed on $(0,1)$.

Now suppose $X\colon \Omega \to \mathbb R^d$ is multivariate, still with continuous p.d.f. My question is can we construct a continuous bijection $F\colon \mathbb R^d\to (0,1)^d$ such that $F(X)$ is uniformly distributed on $(0,1)^d$?

We could try $F= (F_{X_1},\dots,F_{X_d})$, where $F_{X_i}$ are the marginal c.d.fs. But this will only be uniformly distributed on $(0,1)^d$ if $X_1,\dots,X_d$ are independent. Otherwise there is additional information in the interdependence between components, captured by the copula $C\colon (0,1)^d\to (0,1)$.

Note the difference with these StackExchange questions: When is a measure the pushforward of another measure? When does a measurable function exist with a given distribution? which work under more general settings and do not require a (continuous) bijection.

Any help or references are appreciated. It feels like this problem must have been studied, but I haven't been able to find much online so far.

Nathaël
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  • Are you sure you don't mean you're looking for $F\colon {\mathbb R}^d\to [0,1]^d$ instead of $F\colon \mathbb R\to [0,1]^d$? – John Barber Dec 10 '24 at 15:42
  • Sorry, yes of course. It's corrected now – Nathaël Dec 10 '24 at 15:45
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    How about the conditional cdfs? So, $u_1(x_1) = F_{X_1}(x_1), u_2(x_1, x_2) = F_{X_2|X_1 = x_1}(x_2), \cdots, u_i(x_1^i) = F_{X_i|X_1^{i-1} = x_1^{i-1}}(x_i)$. Continuity should follow from continuity of the joint pdf, invertibility comes from a peeling argument (work out $x_1$ from $u_1$, then $x_2$ from $u_2$ and $x_1,$ and so on), and each $u_i$ should be indpendent of $x_1^{i-1}$ (and hence $u_1^{i-1}$), and uniform on $[0,1]$ given the same via the calculation $P(U_i \le u|U_1^{i-1} = u_1^{i-1}) = F_{X_i|X_{1}^{i-1} = x_1^{i-1}}(F_{X_i|X_{1}^{i-1} = x_1^{i-1}}^{-1}(u))$. – stochasticboy321 Dec 12 '24 at 00:02
  • Thank you, this is a beautiful solution! I find it interesting that this implies that any two measures on $\mathbb R^d$ with non-vanishing continuous p.d.f.s can be pushforwarded to one another through a $C^1$-diffeomorphism. – Nathaël Dec 13 '24 at 10:37

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