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Let's suppose (A,X,P) and (B,Y,Q) are two probability spaces (A,B underlying spaces, X,Y sigma-algebras, P,Q probability measures, respectively).

Under what (topological and/or measure theoretic) conditions on these two spaces does there exist a measurable map M: A -> B such that

P(M in b) = Q(b) for an arbitrary element b of Y.

Thanks a lot for your help.

Paulo Henrique
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Frank
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1 Answers1

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A sufficient condition is that $(B,\mathcal{Y})$ is a Polish (separable and completely metrizable) space with its Borel $\sigma$-algebra and $(A,\mathcal{X},P)$ is atomless.

This result is widely known, but it hard to find an exact reference for it. It follows from Proposition 9.1.2 and Theorem 13.1.1 in Dudley 2004, Real Analysis and Probability for the case that $A=[0,1]$, $\mathcal{X}$ the Borel sets and $P$ the uniform distribution (Lebesgue measure). Now on every atomless probability space, one can find a uniformly distributed random variable with values in $[0,1]$. I will sketch a proof of this below.

It is possible to generalize the result above slightly and allow for $(B,\mathcal{Y})$ to be a Souslin space with its Borel $\sigma$-algebra. To get this as a corollary to the previous result, use Theorem 9.1.5 in Bogachev 2006, Measure Theory.

Lemma: Let $(\Omega,\Sigma,\mu)$ be an atomless probability space. Then there is a random variable $f:\Omega\to [0,1]$ with $\mu\circ f^{-1}$ being the uniform distribution.

Proof: Using that finite atomless measures have convex range, we can construct a sequence of independen random variables $(g_n)$ with $g_n:\Omega\to\{0,1\}$ that put equal probability on both numbers. This induces a random variable $g:\Omega\to\{0,1\}^\mathbb{N}$ with fair coin-flipping measure as its distribution. The function $\{0,1\}^\mathbb{N}\ni(x_n)\mapsto\sum_n x_n/2^n$ is measurable and composing it with $g$ provides us with $f$.

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Michael Greinecker
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  • Excellent, thanks a lot, this is exactly what I needed. – Frank Feb 04 '14 at 15:13
  • It's absolutely clear to me that on any non-atomic probability space, there is a $\mathcal U_{[0,:1]}$-distributed random variable. But how can we conclude that there is a $B$-valued random variable $M$ on $(A,\mathcal X,\operatorname P)$ with $M_\ast\operatorname P=Q$? Asked for that here. That's only clear to me when $(B,\mathcal Y)=(\mathbb R,\mathcal B(\mathbb R))$. – 0xbadf00d Jan 03 '19 at 22:40
  • @NateEldredge Don't get it. Isn't it the target space we're assuming to be Borel? Why should $(\Omega,\Sigma)$ be Borel? – 0xbadf00d Jan 03 '19 at 22:49
  • @0xbadf00d The term is "standard Borel", a measurable space from which one can find a measurable bijection with measurable inverse onto $[0,1]$ with the Borel $\sigma$-algebra. – Michael Greinecker Jan 04 '19 at 00:24
  • @MichaelGreinecker And why is every non-atomic probability space standard Borel? And how do you conclude the existence of $M$ in the question? – 0xbadf00d Jan 04 '19 at 10:14
  • @0xbadf00d They are not. But every Polish space is standard Borel. – Michael Greinecker Jan 04 '19 at 10:17
  • @MichaelGreinecker Could you point out how you conclude the existence of $M$? – 0xbadf00d Jan 04 '19 at 10:34
  • @0xbadf00d Note the reference to Dudley's book. – Michael Greinecker Jan 04 '19 at 11:13
  • @MichaelGreinecker Sorry, maybe I'm too stupid, but I still don't get how you conclude. It's clear to me that if $(\Omega,\mathcal A,\operatorname P)$ is a probability space on which there is a $\mathcal U_{[0,:1]}$-distributed random variable $X$ and $F:\mathbb R\to[0,1]$ is any distribution function, then $F$ is the distribution function of $Y:=F^{-1}(X)$. However, how does this help here, where the target measure is defined on an arbitrary measurable space (and hence has no distribution function)? – 0xbadf00d Jan 04 '19 at 15:09
  • @0xbadf00d: No, you're right. I had mixed up the domain and codomain. I removed the comment. – Nate Eldredge Jan 04 '19 at 15:19
  • @NateEldredge Don't worry, it's easy to mix them up here. Do you have any idea how we can conclude (see my question)? What I'm actually after is the existence of a coupling in terms of random variables: https://math.stackexchange.com/questions/2973927/successful-couplings-and-total-variation-convergence-to-equilibrium-for-time-hom – 0xbadf00d Jan 04 '19 at 15:22
  • @MichaelGreinecker : I don't understand your proof either, are you sur about it ? –  Jun 24 '19 at 13:52
  • @CechMS Yes. Is there a specific point that is unclear? – Michael Greinecker Jun 24 '19 at 14:13
  • First of all. [9.1.2] For any distribution function $F$, $X_{F}$ is a random variable with the distribution function $F$. –  Jun 25 '19 at 07:33
  • [13.1.1] If $X$ and $Y$ are two separable metric spaces which are Borel subsets of their completions then $X ~ Y$ if and only if $X$ and $Y$ have the same cardinality, which moreover is either finite, countable, or c (the cardinal of the continum). –  Jun 25 '19 at 07:36
  • [9.1.5] Let $X$ and $Y$ two Souslin spaces and let $f : X \rightarrow Y$ be a borel mapping such that $f(X) = Y$. Then, for every Borel measure $\nu$ on $Y$, there exists a Borel measure $\mu$ on $X$ such that $\nu = \mu \circ f^{-1}$ and $| \mu | = | \nu |$. If $f$ is a one to one mapping, then $\mu$ is unique. –  Jun 25 '19 at 07:39
  • Now the problem is : Let $(A,\mathcal{X},P)$ an atomless probability space, $(B,\mathcal{Y}, \mu)$ a polish space. We would like to show that it exists a mesurable function $h : A \rightarrow B$ such as $h \text{#} P = \mu$. –  Jun 25 '19 at 07:50
  • What is your first step please ? –  Jun 25 '19 at 07:51
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    @CechMS We reduce the problem to the case in which $B$ is the real line using [13.1.1]. Then we construct a uniformly distributed random variable $f:A\to [0,1]$, which is what I have sketched. Then we construct a random variable $g:[0,1]\to \mathbb{R}=B$ with distribution $\mu$ when $[0,1]$ is endowed with the uniform distribution using [9.1.2]. Then $g\circ f$ does the trick. – Michael Greinecker Jun 25 '19 at 07:58
  • Thank you so much for your help, I have some questions. –  Jun 25 '19 at 12:53
  • How to be sur that the cardinality of $B$ is $c$ ? Maybe we could just add this hypothesis because a polish space can be discret.
    • –  Jun 25 '19 at 12:55
  • How do you construct $g_{n}$ in your $\mathbf{Lemma}$ ?
    • –  Jun 25 '19 at 12:55
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    @CechMS If the Polish space is countable, it is already isomorphic to a Borel subset of $[0,1]$. To construct $g_n$. Split up the probability space into two pieces of equal measure, let $g_1$ be $1$ on one of the pieces and $0$ on the other. Then split up each of these pieces again into two pieces of equal measure and let $g_2$ be 1 on one of the subpieces and 0 on the other. Continuing this way gives you the desired random variables. – Michael Greinecker Jun 25 '19 at 13:36
  • Let us continue this discussion in chat. –  Jun 26 '19 at 08:10
  • @MichaelGreinecker : I worked on the Linderstrauss proof of Lyapounov theorem. I proved your lemma, existence of $g_{n}$ and of the uniform random variable on $[0,1]$. I just have to work on Borel isomorphis now to fully understand the proof. –  Jun 28 '19 at 12:52
  • @MichaelGreinecker : I have some questions. –  Jun 28 '19 at 12:59
  • How to relax the assumption atomless ? The result is obviously false in the case. For exemple if $X = Y = {0,1}$, $\mu = 1/3 \delta_{0} + 2/3 \delta_{1}$, $\nu = 1/2(\delta_{0} + \delta_{1})$ then there is no such application. But it's easy to solve the transport problem between discret measure so maybe we can say something general ?
    • –  Jun 28 '19 at 13:02