# Is every probability measure in the line induced by a random variable?

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 $$X$$?

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.

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 $$\mu$$.

• 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! Oct 3, 2019 at 19:18
• @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. Oct 3, 2019 at 19:35
• @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. Nov 3, 2023 at 13:43
• @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. Nov 3, 2023 at 13:45
• @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. 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.)

Reference:

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

• 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.) Oct 3, 2019 at 14:43
• @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. Oct 3, 2019 at 14:44
• 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. Oct 3, 2019 at 14:46
• @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". Oct 3, 2019 at 14:49