Timeline for Disintegration associative
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 1, 2020 at 13:54 | comment | added | Iosif Pinelis | 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. | |
| Feb 28, 2020 at 7:06 | comment | added | CechMS | 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. | |
| Feb 27, 2020 at 16:02 | comment | added | Iosif Pinelis | @CechMS : I think you never said what you mean by $\oplus$, $\mu_x^1$, $\mu^{1,2}$, ... . | |
| Feb 27, 2020 at 15:51 | comment | added | CechMS | We have $\mu = \mu^{1,2} \oplus(h^{2} \circ h^{1})_{\#} \mu$. The question is $$ \mu^{1,2}_{y}(A) =? \int \mu_{x}^{1}(A) d\mu_{y}^{2}(x) $$ | |
| Feb 27, 2020 at 15:38 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |