1
$\begingroup$

Let's disintegrate $\mu$ and $\nu$, two probabilities on $\mathbb{R}^{d}$ , according to $$ \pi_{k} (x_{1},...,x_{d}) = (x_{k},...,x_{d}) $$ We get a family of measures and each measure $\mu_{k,d}^{+} =: \mu^{+}$ (resp $\nu^{+}_{k,d} =: \nu^{+}$) is concentrated on $\mathbb{R}^{k-1} \times \{x_{k} \} \times ... \times \{ x_{d} \}$ (resp $\mathbb{R}^{k-1} \times \{y_{k} \} \times ... \times \{ y_{d} \}$).

For each $(x_{+},y_{+}) := (x_{k},...,x_{d},y_{k},...,y_{d})$ we consider $\gamma^{+,+}$ the Knothe transport measure between $\mu^{+}$ and $\nu^{+}$. We assume it is realized by a map $T$, i.e. $$ \gamma^{+,+} = (\text{id},T)_{\#} \mu^{+} $$ with $T_{\#} \mu^{+} = \nu^{+}$

Finally let's consider $\gamma = \gamma^{+,+} \otimes \eta \in \mathcal{P}(\mathbb{R}^{2d})$ with $\eta \in \mathcal{P}(\mathbb{R}^{d-k+1} \times \mathbb{R}^{d-k+1})$ and

$$ \gamma(A) = \int{\gamma^{++}(A) d\eta} $$

In this definition we have something to prove,that's $\gamma$ is indeed a measure. The only thing to prove is that

$$ (x_{+},y_{+}) \rightarrow \gamma^{++}(A) $$ is measurable.

But I don't know how to prove it, do someone know how to do that ? I would be relieved to read it.

Thanks and regards.

Improve this question
$\endgroup$
6
  • $\begingroup$ Your question is ill-posed: there may be no $T$ such that $T_\#\mu^+=\nu^+$ (think of the case when $\mu^+$ is a Dirac mass, while $\nu^+$ is not); and there may be several such $T$ (in which case by making bad choices, $\gamma^{+,+}$ could be non-measurable). $\endgroup$
    – Benoît Kloeckner
    Commented Dec 14, 2019 at 13:19
  • $\begingroup$ Hi, thanks for your help (I checked your work on this forum, it's amazing by the way). 1) You're right but I assume it exists. For exemple let $\mu^{+}$ and $\nu^{+}$ be atomless. 2) You can skip the 2) if you want, it's just to say that this problem is related to the buiding of a "competitor measure", the first 14 sentences of this article : spire.sciencespo.fr/hdl:/2441/5rkqqmvrn4tl22s9mc4c7eg1p/… $\endgroup$
    – CechMS
    Commented Dec 15, 2019 at 10:49
  • $\begingroup$ I do not know what your 2) refers to. In my first comment, the multiplicity is the most important part; I guess you want to mean that you construct $\gamma^{+,+}$ in some specific way, but I cannot be sure and the answer to your question probably depends on this. $\endgroup$
    – Benoît Kloeckner
    Commented Dec 18, 2019 at 16:16
  • 1
    $\begingroup$ You are right. I'm constructing the Knothe transport between $\mu_{+}$ and $\nu_{+}$ which don't always exists. But I guess as in the article it must be a assumption. For exemple we can take two measure such that the disintegrations are density measures. $\endgroup$
    – CechMS
    Commented Dec 19, 2019 at 16:41
  • $\begingroup$ Ok, I took the liberty t slightly edit you question accordingly to this. $\endgroup$
    – Benoît Kloeckner
    Commented Dec 21, 2019 at 8:47

1 Answer 1

Reset to default
0
$\begingroup$

EDIT : As I think I wont have any answer I changed my question forgetting about Knothe Transport to speak to a larger audiance. In y first question I considerated $(x_{+},y_{+}) := (x_{k},...,x_{d},y_{k},...,y_{d})$ we consider $\gamma^{+,+}$ the Knothe Rosenblatt transport plan between $\mu^{+}$ and $\nu^{+}$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Done thank you. $\endgroup$
    – CechMS
    Commented Nov 12, 2019 at 12:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .