Good evening,
Let $\mu$ and $\nu$ two measures on $X$ and $Y$.
Do you know when it exists a measurable function $h : X \rightarrow Y$ such as $ \nu = h\text{#}\mu$ with $ h\text{#}\mu(B) = \mu(h^{-1}(B))$.
Thank you, have a good day.
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1maybe you can apply the theorem of Radon-Nikodym (Wikipedia) in some way; if you had a function $f : X \to Y$ such that $f # \mu \ll \nu$, then this theorem tells you, there is such an $h$ you're looking for. – Targon Jun 06 '19 at 09:03
3 Answers
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Here's two ideas to begin.
1) 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.
2) If $X = Y = R^{n}$ and $\mu := f.\mathcal{L}$ et $\mu$ two probabilties then it exists $\phi$ convex such as $\nabla{\phi} \text{#} \mu = \nu$.
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I think I have solution for discrete measures. Let $\mu = \sum_{i=1}^{n} \delta_{x_{i}} a_{i}$ and $\nu = \sum_{i=1}^{n} \delta_{y_{i}} b_{i}$.
A such $h$ exists if, and only if, $b_{i} = \sum_{J(i)} a_{j}$ with $J(i)$ a disjoint union of $\{1,...,n\}$. In that case $h$ is easy to compute.
Do you agree ?
I was told it's, in general, a really hard problem. So I would feel glad if someone has references (books, articles, ...) or maybe can show me some speciale cases :).
Thank you for your help, have a nice day.
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On this thread : When does a measurable function exist with a given distribution?
@MichaelGreinecker gave us a proof of the following result : Let $(A,\mathcal{A},P)$ an atomless probability space, $(B,\mathcal{B},\mu)$ a polish space.
It exists a mesurable function $h:A \rightarrow B$ such as $h\text{#}P= \mu $