Disintegration of a push foward measure.

Question

Let $\mu$ and $\nu$ be two probability measures on $\mathbb{R}^{d}$.

Let $T : \mathbb{R}^{d} \rightarrow \mathbb{R}^{d}$ be a measurable map such that $T_{\#} \mu = \nu$.

I can disintegrate $\gamma := (id,T)_{\#} \mu$ according to $(h, h_{\#} \gamma$), I get a familly of measure $\gamma_{y}$ concentred on $h^{-1}(\{y\})$ as usually. But this time $\gamma$ has a special form I'm expecting the disintegration too.

General questions :

1) Do we have $\gamma_{y} := (id,S)_{\#}\omega$ with $\omega$ a measure. I think $\omega$ could be a disintegration of $\mu$.

2) Can I say something about the marginals of $\gamma_{y}$ ?

Any help would be appreciated, thanks and regards.

EDIT : I think I will never have any answer so I will start a bounty and ask an other question. The first question was general and now I'm going to talk about my specific problem, of course a general answer will be appreciated.

Let $\mu$ and $\nu$ be two probabilites on $\mathbb{R}^{d}$. We suppose that $\mu$ is absolutely continuous with respect to the Lebesgue measure.

Let $\gamma := (id, T)_{\#} \mu$ with $T : \mathbb{R}^{d} \rightarrow \mathbb{R}^{d}$ such that $T_{\#} \mu = \nu$. And $T$ has the form $$ T(x_{1},...,x_{d}) = (T^{1}(x_{1},...,x_{d}),....,T^{d-1}(x_{d-1},x_{d}),T^{d}(x_{d})) $$ We will see how $T$ is built later.

We have $ \gamma \in \Pi(\mu,\nu)$ is the subset of the probability measure $\pi$ such as $$ \pi (A\times Y) =\mu(a) \text{ and } \pi(X\times B)=\nu(B) $$

Let $\mu^{k} := \pi_{k\#} \mu$ with $\pi_{k}(x_{1},...,x_{d}) = (x_{k},...,x_{d})$

Let $\pi_{d-1,d}(x_{1},...,x_{d},y_{1},...,y_{d}) = (x_{d-1},x_{d},y_{d-1},y_{d})$

So $\gamma^{d-1,d} = \pi_{d-1,1 \#} \gamma = (id,\hat{T}^{d-1}) \in \Pi(\mu^{d-1}, \nu^{d-1})$ with $$ \hat{T}^{d-1}(x_{d-1},x_{d}) = (T^{d-1}(x_{d-1},x_{d}),T^{d}(x_{d})) $$

Now how $T$ is built. We need a optimal transport's result because it is the context even if my question is, in my opinion, only about disintegration.

Let $c(x,y) = |x-y|^{2}$, it exists measure in $\gamma \in \Pi(\mu,\nu)$ such that $$ \int c d\gamma = \inf \{ \int{c d\pi} ; \pi \in \Pi(\mu,\nu) \} $$ And we have $\gamma = (id,T)_{\#} \mu$ with $T_{\#} \mu = \nu$. We say that $T$ is the optimal transport between $\mu$ ans $\nu$ for $c$.

Now let $T^{d}$ the optimal transport between $\mu^{d}$ and $\nu^{d}$ for $ |x_{d} - y_{d}|^{2}$

Now let disintegration $\mu^{d-1}$ according $\mu^{d}$ we obtain a family of measures $\mu^{d}_{x_{d}}$ on $\mathbb{R}$.

We can defined $T^{d-1}(x_{d-1},x_{d}) := T_{x_{d}}^{d-1}(x_{d-1})$ with $T_{x_{d}}^{d-1}$ the optimal transport between $\mu^{d}_{x_{d}}$ and $\nu_{T^{d}(x_{d})}^{d}$ for $|x_{d-1} - y_{d-1}|^{2}$ (we also disintegrated $\nu$ the same way).

It is enough to go to the question but I add details if you need to.

We disintegrate $\gamma^{d-1,d}$ according $\gamma^{d}$ and we obtain a family of meausre $\gamma^{d-1,d}_{(x_{d},y_{d})}$ on $\mathbb{R}^{2}$

I found $\gamma^{d-1,d}_{(x_{d},y_{d})} = (id, T^{d-1}_{x_{d}})_{\#} \mu_{x_{d}} \in \Pi(\mu_{x_{d}}^{d-1}, \nu_{T^{d}(x_{d})}^{d-1})$ for a.e $(x_{d},y_{d})$.

The question is 3) Do you think it's just a coincidence? I think there's a general fact behind it about the disintegration of optimal plan. Do you know what it could be ?

I think the measure with the form $(id,T)$ could admit a specific disintegration form which will lead me to the result I'm looking for. So if you are accustomed with disintegration, I would love to read your advice.

Answer 1

I found something, not exactly what I need, maybe it could be a good start.

If $\gamma = \gamma_{y} \oplus \alpha_{\#} \gamma$ the disintegration of $\gamma$ according to $(\alpha, \alpha_{\#})$. Then $\beta_{\#}\gamma = \beta_{\#} \gamma_{y} \oplus \alpha_{\#} \gamma$

Now let $ \beta_{\#}\gamma = \mu_{y} \oplus \delta_{\#} \beta_{\#}\gamma$ the disintegration of $\beta_{\#} \gamma$ according to $(\delta, \delta_{\#} \beta_{\#} \gamma)$ then

$$ \mu_{y} = \beta_{\#} \gamma_{y} $$

I hope someone will answer...

Answer 2

I assume $h:\mathbb{R}^d\times\mathbb{R}^d\rightarrow\mathbb{R}^d$ is some given measurable map. Answer to 1):

$\gamma_y=(id,T)_\#\omega_y$ for almost every $y\in\mathbb{R}^d$, where $\omega_y$ is the disintegration of $\mu$ wrt the measure $h_\#\gamma=h\circ(id,T)_\#\mu$.

Proof: Denote $\beta_y=(id,T)_\#\omega_y$. For any measurable $f:\mathbb{R}^d\times\mathbb{R}^d\rightarrow\mathbb{R}$: $$\iint f(x_1,x_2)\text{d}\beta_y(x_1,x_2)\text{d}h_\#\gamma(y)=\iint f(x,Tx)\text{d}\omega_y(x)\text{d}h_\#\gamma(y)=\int f(x,Tx)\text{d}\mu(x)=\int f(x_1,x_2)\text{d}\gamma(x_1,x_2)$$ Therefore $\beta_y$ is the disintegration of $\gamma$ wrt $h_\#\gamma$, but so is $\gamma_y$. By the disintegration theorem, the disintegration is unique thus $\beta_y=\gamma_y$ almost everywhere.