It is a basic fact in probability theory that, for every random variable $$ X: (\Omega, \mathcal{F}, \mathbb{P}) \to (\mathbb{R}, \mathcal{B}), $$ we have an associated measure $\mathbb{P}_X$ on the borelians of $\mathbb{R}$ given by

$$ \mathbb{P}_X(B) := \mathbb{P}(X \in B) = \mathbb{P}(X^{-1}(B)), \forall B \in \mathbb{B}. $$ $\mathbb{P}_X$ is called the measure induced by $X$.

My question: is the opposite also true? That is, given a probability measure on the borelians of $\mathbb{R}$, is it induced by a random variable? More precisely:

Is there a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ such that, given a probability measure $\mu$ in $(\mathbb{R}, \mathcal{B})$, one can find a random variable $X: (\Omega, \mathcal{F}, \mathbb{P}) \to (\mathbb{R}, \mathcal{B})$ such that $\mu$ is induced by $X$?

I aprreciate your concern!


What I really want is a probability space that can be used for global representation of probability measures as measures induced by random variables. I have a guess that the space $\Omega = \{f: \mathbb{R} \to \mathbb{R} | f \:\text{is Borel measurable}\}$ with the $\sigma$-algebra of cilinders and some appropriate measure is the space desired, but I'm not pretty sure on how to prove it.


2 Answers 2


Yes, trivially: consider the probability space $\Omega = \mathbb{R}$, $\mathcal{F} = \mathcal{B}$, $\mathbb{P} = \mu$, and let $X : \Omega = \mathbb{R} \to \mathbb{R}$ be the identity map $X(\omega) = \omega$.

Alternatively, use inverse transform sampling. Let $F(x) = \mu((-\infty, x])$ be the cumulative distribution function of $\mu$, and let $G(t) = \inf\{x : F(x) \ge t\}$ be its "inverse". (If $F$ is actually 1-1 then $G$ is truly the inverse of $F$.) Now let $U$ be a Uniform(0,1) random variable on any probability space $(\Omega, \mathcal{F}, \mathbb{P})$ (e.g. the identity map on $[0,1]$ with Lebesgue measure) and set $X = G(U)$. It is then easy to see that the measure induced by $X$ is again $\mu$. Proof: we have $\mathbb{P}(X \le x) = \mathbb{P}(G(U) \le x) = \mathbb{P}(U \le F(x)) = F(x)$. So the measure induced by $X$ has $F$ as its cdf, and this uniquely determines it as being $\mu$.

  • $\begingroup$ Thank youj very much. Notice that the first par of your answer is not quite what i was looking for, for i want a fixed probability space $(\Omega, \mathcal{F}, \mathbb{P})$ where i can define all the random variables $X$ which will eventually represent the different probability measures of $\mathbb{R}$. But the second part gives it, and a i aprreciate! $\endgroup$
    – Brisão
    Oct 3, 2019 at 19:18
  • $\begingroup$ @Brisão: You're welcome. More generally, if $(\Omega, \mathcal{F})$ is any uncountable standard Borel space and $\mathbb{P}$ is atomless, then every probability measure on $\mathbb{R}$ (or indeed on any standard Borel space $\Omega'$) is induced by some random variable $X : \Omega \to \mathbb{R}$ (respectively, $X : \Omega \to \Omega'$). So any such probability space is "universal" in the sense that you seem to be looking for. $\endgroup$ Oct 3, 2019 at 19:35
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    $\begingroup$ @FlorianEnte: If you look carefully at the argument, I expect you will find there is no need for $\xi$ to be defined on the same probability space as the $\xi_n$, since there doesn't seem to be any use made of a joint distribution between the $\xi_n$ and $\xi$. For instance, weak convergence of random variables makes perfect sense even if they are defined on different probability spaces. So you could define $\xi$ trivially on the probability space $(M, \mathcal{B}, P^*)$ instead. $\endgroup$ Nov 3, 2023 at 13:43
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    $\begingroup$ @FlorianEnte: Alternatively, "enhance" $(\Omega, \mathcal{F}, P)$ by replacing it with the product space $(\Omega \times M, \mathcal{F} \otimes \mathcal{B}, P \otimes P^*)$. Then the $\xi_n$ make sense as functions of the first coordinate, and you can define $\xi$ trivially as the identity function of the second coordinate. This makes $\xi$ independent of the $\xi_n$, which is fine since the joint distribution should be irrelevant. $\endgroup$ Nov 3, 2023 at 13:45
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    $\begingroup$ @FlorianEnte: Or third, if $(\Omega, \mathcal{F}, P)$ is standard Borel (probably a safe assumption to add), then it can support a random variable in any standard Borel space with any distribution. The weak topology on $M$ is Polish, I believe, so $(M, \mathcal{B})$ is standard Borel, and then there does exist a $\xi : \Omega \to M$ with distribution $P^*$. Here the existence of $\xi$ is kind of abstract, and you can't say anything about its joint distribution with the $\xi_n$, but again we should not need to. $\endgroup$ Nov 3, 2023 at 13:48

Yes. Let $(\Omega ,\mathcal F,\mathbb P) = ([0,1],\mathcal B([0,1]),\mathcal L)$ where $\mathcal L$ is the Lebesgue measure on $[0,1]$. Fix your measure $\mu$, and write $F(x) = \mu((-\infty,x])$.

Let $X(\omega)= \inf\{z : F(z) \ge \omega \}$. This is known as the Skorokhod Representation of the random variable $X$.

I'll leave you to verify that the distribution of $X$ is exactly given by $F$. (However, if this gives you trouble, see section 3.12 of the reference below.)


Williams, D. (1991). Probability with martingales. Cambridge university press.

  • $\begingroup$ I guess we had the same idea. But I always thought "Skorokhod representation" usually referred to representing $\mu$ by a random variable of the form $B_\tau$, where $B_t$ is Brownian motion and $\tau$ is a stopping time. (There is also the so-called "Skorokhod representation theorem" where you convert a sequence converging in distribution to a sequence converging almost surely.) $\endgroup$ Oct 3, 2019 at 14:43
  • $\begingroup$ @Nate Yes, it doesn't appear to be the commonly used term for this. (For example, a Google query for "Skorkohod representation" doesn't seem to be consistent with my post.) However, it is what Williams calls it, so I defer to them. $\endgroup$ Oct 3, 2019 at 14:44
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    $\begingroup$ Oh, I think I get it. This is the trick that you use to prove the "Skorokhod representation theorem" I mentioned above. You just do it for a whole sequence of random variables instead of just one. $\endgroup$ Oct 3, 2019 at 14:46
  • $\begingroup$ @Nate Huh, good point. That's nice to know. Always nice to unify concepts in my head with the same name but ostensibly refer to different things. Though, I am also used to calling the first idea in your first comment the "Skorokhod embedding problem". $\endgroup$ Oct 3, 2019 at 14:49

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