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Is the disintegration of two borelian probabilities measures is associative ? It means if $\mu = \mu_{y}^{1} \oplus h_{\#}^{1}\mu$ and $ h_{\#}^{1}\mu = \mu_{y}^{2} \oplus h_{\#}^{2} h_{\#}^{1}\mu$. Then do we have $$ \mu = \mu_{y}^{1} \oplus \mu_{y}^{2} \oplus (h^{2} \circ h^{1})_{\#} \mu $$ Where $\mu$ is a borelian probability over $\mathbb{R}^{d}$ and I note $\mu = \mu_{y} \oplus h_{\#}\mu$ the disintegration of $\mu$ according to $h_{\#} \mu$ that gives the kernel $\mu_{y}$ with $y \in \mathbb{R}^{d}$.

EDIT : To be clearer, we have $\mu = \mu^{1,2} \oplus(h^{2} \circ h^{1})_{\#} \mu$. The question is $$ \mu^{1,2}_{y}(A) =? \int \mu_{x}^{1}(A) d\mu_{y}^{2}(x) $$

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  • $\begingroup$ I have problems with your notation. What is $h_2 \circ h_1$? I think if you formulate (if possible) your question with stochastic kernels the answer is simply true. Have a look f.i. in the old book of Bertsekas/Shreve (1978), Stochastic Optimal Control: The discrete Time Case, in particular ch. 7.4.3 Stochastic Kernels. $\endgroup$
    – Dieter Kadelka
    Commented Feb 27, 2020 at 15:17
  • $\begingroup$ I edit thank you. Here the things is we have to check if $$ \mu_{y}^{1}\oplus \mu_{y}^{2} (\mathbb{R}^{n} - (h^{2}\circ h^{1})^{-1}(\{y\}))=0 $$ And $$ y \rightarrow \mu_{x}^{1} \oplus \mu_{y}^{2} $$ Is measurable. $\endgroup$
    – CechMS
    Commented Feb 27, 2020 at 15:22
  • $\begingroup$ Obious edit : $$ y \rightarrow \mu_{x}^{1} \oplus \mu_{y}^{2} (A) $$ Is measurable for any borel set $A$. $\endgroup$
    – CechMS
    Commented Feb 27, 2020 at 15:34

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Disintegration is associative, in the following sense: Suppose that we have a disintegration $$\mu(dx)=\int_Y(\mu h_1^{-1})(dy)\mu_{h_1}(y,dx)$$ of a probability measure $\mu$ with a kernel $\mu_{h_1}$ -- meaning that $\int_X\mu(dx)f(x)=\int_Y(\mu h_1^{-1})(dy)\int_X\mu_{h_1}(y,dx)f(x)$ for all nonnegative measurable functions $f$ on $X$, and that we further have a disintegration $$(\mu h_1^{-1})(dy)=\int_Z(\mu h_1^{-1}h_2^{-1})(dz)(\mu h_1^{-1})_{h_2}(z,dy) \\ =\int_Z(\mu(h_2\circ h_1)^{-1})(dz)(\mu h_1^{-1})_{h_2}(z,dy)$$ of the measure $\mu h_1^{-1}$. Then we have the disintegration $$\mu(dx) =\int_Z(\mu(h_2\circ h_1)^{-1})(dz)(\mu_{h_1}*(\mu h_1^{-1})_{h_2})(z,dx),$$ where $$(\mu_{h_1}*(\mu h_1^{-1})_{h_2})(z,dx):=\int_Y (\mu h_1^{-1})_{h_2}(z,dy)\; \mu_{h_1}(y,dx).$$

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  • $\begingroup$ We have $\mu = \mu^{1,2} \oplus(h^{2} \circ h^{1})_{\#} \mu$. The question is $$ \mu^{1,2}_{y}(A) =? \int \mu_{x}^{1}(A) d\mu_{y}^{2}(x) $$ $\endgroup$
    – CechMS
    Commented Feb 27, 2020 at 15:51
  • $\begingroup$ @CechMS : I think you never said what you mean by $\oplus$, $\mu_x^1$, $\mu^{1,2}$, ... . $\endgroup$
    – Iosif Pinelis
    Commented Feb 27, 2020 at 16:02
  • $\begingroup$ The identity $\mu = \mu_{y} \oplus h_{\#}\mu$ means that $\{ \mu_{y} ; y \in \mathbb{R}^{n} \}$ is a probability kernel such that : $$ \mu(A) = \int \mu_{y} (A) d h_{\#} \mu $$ And $$ \mu_{y}(\mathbb{R}^{n} - h^{-1}(\{ y\}) = 0 $$ We write $\mu = \mu_{y} \oplus \nu$ everytime you have $\nu = h_{\#} \mu$ otherwise even over $\mathbb{R}^{n}$ it could not exist. $\endgroup$
    – CechMS
    Commented Feb 28, 2020 at 7:06
  • $\begingroup$ It is still unknown what you mean by $\mu_x^1$, $\mu^{1,2}$, ... . Also, the first displayed formula in your latest comment, with just one entry of $y$, is apparently written not the way it should be. $\endgroup$
    – Iosif Pinelis
    Commented Mar 1, 2020 at 13:54

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