# Does there exist a mutivariate inverse?

Let $F:\mathbb{R}\rightarrow \mathbb{R}$ be a distribution function (CDF)

In this case, we can define the inverse $X$ of $F$, and it is a random variable on $(0,1)$ such that $F_X=F$.

Hence, every distribution (CDFs) can be viewed as the cdf of a random variable on $(0,1)$.

Is there an analogous result for joint distribution functions (CDFs)?

That is, for a fixed $n$, does there exists a probability space $(\Omega,\mathscr{F},P)$ such that every joint distribution function $F:\mathbb{R}^n\rightarrow \mathbb{R}$ is $F_X$ for some $n$-dimensional random vector $X$ on $(\Omega,\mathscr{F},P)$?

• It is not true that every CDF can be viewed as the CDF of some random variable on $(0,1)$. Perhaps you are referring to the fact that if $U$ is a Uniform random variable on $(0,1)$, then $F^{-1}(U)$ is a random variable whose CDF is $F$ (where $F^{-1}$ is the generalized inverse function, link.) Your final question's answer is true, every joint CDF is the CDF of some $n$-dimensional random vector. – Milind Apr 19 '18 at 15:55
• @Milind It is true. For example, see the proof for theorem 1.2.2 in durrett - probability and example – Rubertos Apr 19 '18 at 16:08
• @Milind and for the second question, I am not asking that. I know that. I am asking wheter there exists a “universal” probability space $\Omega$ that every joint distribution of a random vecot $X$ becomes a joint distribution of a random vector $X’$ on $\Omega$. – Rubertos Apr 19 '18 at 16:09
• Ah ok, Durrett 1.2.2 says that there exists a random variable defined on the probability space $(0,1)$ with Lebesgue measure and Borel measurable sets. So it is a difference of terminology; the phrase "$X$ is a random variable on the space $S$" usually is taken to mean $X$ takes values in $S$, not that $S$ is the probability space. – Milind Apr 19 '18 at 20:11

## 1 Answer

The answer to your question is yes; As in the 1-D case, you can take the probability space $([0,1], \mathcal F, \mathcal L)$, where $\mathcal F$ is the Borel sets and $\mathcal L$ is the Lebesgue measure, and for any CDF $F$ there exists $X: [0,1]\to\mathbb R^n$ such that $X$ has CDF $F$.

This is an instance of the more general fact that any Borel probability measure on a complete separable metric space is a pushforward of Lebesgue measure on $[0,1]$. See chapter 13 of Dudley's Real Analysis and Probability.

• Thank you, but I cannot find this content in Dudley’s book.. Could you tell me which page it is? – Rubertos Apr 19 '18 at 21:42