Is it possible to construct any random variable on the Euclidean Probability space?

Question

Let $(\Omega,\mathscr A,P)$ be an arbitrary probability space, and let $X:\Omega\to\mathbb R$ be a random variable. Then, one can generate a random variable $Y$ from the probability space $\big([0,1],\mathscr B,\lambda\big)$ (where $\mathscr B$ denotes the Borel algebra and $\lambda$ denotes the Lebesgue measure) to $\mathbb R$ whose probability distribution $\mu_Y$ is equal to that of $X$ by means of the quantile function by letting \begin{align*} Y(x):=\sup\big\{y\in\mathbb R:x\leq F_X(y)\big\},&&x\in(0,1) \end{align*} where $F_X$ is the cumulative distribution function of $X$.

This has many important theoretical and practical implications. For instance, it shows that any cumulative distribution function can be realized by a random variable (which allows one to make statements such as let $X$ be a random variable with distribution $F$); it can make some technical proofs easier, as $\big([0,1],\mathscr B,\lambda\big)$ has many nice properties (separability, regularity, $[0,1]$ is compact), and it gives rise to the so-called inverse transform sampling, which is very useful in simulations.

Question. Does there exist similar results for random variables $X$ taking values in spaces other than $\mathbb R$ (such as $\mathbb R^n$)? For example, if one has a random variable $X:\Omega\to\mathbb C$, is it always possible to construct a random variable $Y$ defined on the unit square $[0,1]\times[0,1]\subset\mathbb C$ such that $X$ and $Y$ have the same probability distribution ($\mu_X=\mu_Y$)?

Answer 1

Your question is a little bit unprecise, because of the fuzziness of the word "construct". Since all Borel $\sigma$-algebras of complete separable metric spaces are equivalent, your desired $Y$ exists (and in fact, you can even take $Y$ from $[0,1]$ to $\mathbb{R}^n$, for that matter). I guess you would like to have special properties like in the one-dimensional case.

Then I guess that you have to make at least some arbitrary choices, because you do not have a very good linear order on $\mathbb{R}^n$ anymore. You could use desintegration and the Knothe-Rosenblatt rearrangment: taking as you did $n=2$ to simplify, let $X_1$ be the first coordinate of $X$ and construct $Y_1$ from $[0,1]$ to $\mathbb{R}$ having the same law as $X_1$, in the monotonic way you describe. Then, let $(\mu_a)$ be the defined-almost-everywhere family of laws of the second coordinate of $X$, conditional to $X_1=a$. Construct, again monotonically, independent random variables $Y_a$ from $[0,1]$ having law $\mu_a$. Then take $Y$ to have first coordinate $Y_1$, and second coordinate $Y_a$ conditionally to $Y_1=a$. Then by construction the law of $Y$ is equal to the law of $X$, and you keep some kind of monotonicity.

However, this construction gives a prominent role to the coordinates and their order, so even an isometric modification of $X$ (i.e. compose $X$ with a displacement) would yield a completely different $Y$ (different from the original $Y$ composed with the same displacement).