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Let $(X,\mathfrak B(X))$ and $(Y,\mathfrak B(Y))$ be measurable spaces and further let $\mu$ be a measure on $\mathfrak B(X)$ and let $K$ be a kernel, i.e. for any $x\in X$ we have $K_x$ is a measure on $\mathfrak B(Y)$ and the map $x\mapsto K_x(B)$ is $\mathfrak B(X)$-measurable for any $B\in \mathfrak B(Y)$. Let us further assume that $\mu$ is finite and measures $K_x$ are uniformly bounded. Then there exists a unique measure $P$ on $\mathfrak B(X)\otimes \mathfrak B(Y)$ such that $$ P(A\times B) = \int_AK_x(B)\mu(\mathrm dx) $$ for any measurable rectangle $A\times B$. Thus, $P$ can be considered as a certain product $\mu\otimes_{?}K$ and in case $K_x\equiv\nu$ we have $P = \mu\otimes \nu$ is just a product measure. We know that product measures are not enough to describe all possible measure over the product space, whereas the construction above is thanks to the disintegration theorems. I wonder what are the nice sources for the properties of the construction $\mu\otimes_{?}K$ - e.g. does it have its own name.

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Disintegration theorem. –  Did Jan 20 '13 at 13:33
    
@Did as far as I understand (and as stated in OP) this theorem proves an existence of K given P and $\mu$ and hence is somehow an inverse statement to the existence of P given K and $\mu$. I'm interested in the latter, which properties does it have, and in particular whether there is a special notation or a name for such product. Perhaps, a joint measure? Maybe you can suggest anything? –  Ilya Jan 20 '13 at 15:02
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In this German script about stochastic processes: www3.math.tu-berlin.de/Vorlesungen/WS11/WT2/… it is called semidirect product measure. The author then goes on concatenating several kernels in order to have notation for a proof of the Ionescu-Tulcea extension theorem. –  Thomas Jan 21 '13 at 7:36
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@Ilya: The definition is on page 19, but the section starts on page 17 with the definition of product $\sigma$-algebras and Markov kernels. –  Thomas Jan 21 '13 at 8:52
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@Ilya One can do this with a bit of trickery. It is possible to take products of kernels. Let $\kappa$ from $X$ to $Y$ be a prob kernel, $I_X$ the identity kernel from $X$ to $X$. Then you can construct a product kernel $\kappa\otimes I_X$ from $X$ to $X\times Y$ that does this. I'm in a train right now, but I will post the details later. –  Michael Greinecker Jan 21 '13 at 10:58

2 Answers 2

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The following is in terms of probability kernels, which I'm most familiar with, but much of it generalizes. If $(X,\mathcal{X})$, $(Y,\mathcal{Y})$, and $(Z,\mathcal{Z})$ are measurable spaces and $\kappa_1:X\times\mathcal{Y}\to[0,1]$ and $\kappa_2:Y\times\mathcal{Z}\to[0,1]$ are probability kernels, we can define a composition $\kappa_2\circ\kappa_1:X\times\mathcal{Z}\to [0,1]$ by $$\kappa_2\circ\kappa_1(x,B)=\int\kappa_2(y,B)~d\kappa_1(x,d(y)).$$ One can show that this composition is associative. Measurable spaces and probability kernels form a category. This category was called the category of probabilistic mappings by Lawvere in unpublished notes from 1962. The same category was named the category of statistical decision functions in the book Statistical Decision Rules and Optimal Inference by Cencov (based on his dissertation). A nice overview of material on this category can be found in the introduction of a paper by Culbertson and Sturtz.

If $f:X\to Y$ is a measurable mapping, there is a corresponding kernel $\kappa_f:X\times\mathcal{Y}\to[0,1]$ given by $$\kappa_f(x,B) = \begin{cases} 1 & \text{if } f(x)\in B\\ 0 & \text{if } f(x)\notin B. \end{cases}$$

This way, we can view measurable mappings as special kernels. If $(X_i,\mathcal{X}_i)$ and $(Y_i,\mathcal{Y}_i)$ are families of measurable spaces and $\kappa_i:X_i\times\mathcal{Y}_i\to[0,1]$ is a kernel for all $i$, then there is a product-kernel $$\kappa_\times:\prod_i X_i\times\otimes_i\mathcal{Y}_i\to [0,1]$$ such that $\kappa_\times(x,\cdot)=\otimes_i\kappa_i(x_i,\cdot)$.

We can also treat a probability measure as a constant-valued kernel. If we combine these ideas, we can take a probability measure (seen as a kernel) $\mu$ on $(X,\mathcal{X})$ and a kernel $\kappa:X\times\mathcal{Y}\to [0,1]$ to get a natural distribution on $\mathcal{X}\otimes\mathcal{Y}$. We let $p:X\to X\times X$ be given by $x\mapsto (x,x)$. Let $\kappa_p:X\to\mathcal{X}\otimes \mathcal{X}$ be the corresponding kernel. Let $\kappa_1:X\times\mathcal{X}$ be the kernel corresponding to the identity on $X$. We let $\kappa_\times$ be the product kernel of $\kappa_1$ and $\kappa$. Then $\kappa_\times\circ\kappa_p\circ\mu$ is essentially the natural distribution on $\mathcal{X}\otimes\mathcal{Y}$ such that $\mu$ is the marginal on $X$ and $\kappa$ gives the conditional probabilities on $Y$ given $x$. So the generalized product measure is really a special form of composing kernels.

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Thanks a lot for this nice answer! –  Ilya Jan 23 '13 at 15:32
    
@Ilya Thanks for the bounty! –  Michael Greinecker Jan 23 '13 at 15:40
    
I've taken a brief look on your references (as some of them are pretty long), but haven't find a construction with $\kappa_\times\circ \kappa_p\circ \mu$ for the joint distribution given conditionals. Did you come up with it by yourself, or have you seen it somewhere else as well? –  Ilya Jan 24 '13 at 12:24
    
@Ilya I came up with it, but I doubt I'm the first one to do so. –  Michael Greinecker Jan 24 '13 at 12:35

There are lecture notes from Erwin Bolthausen for a course in stochastic processes in discrete time, albeit in German: www.math.uzh.ch/index.php?file&key1=11678

On page 17 he introduces Markov kernels and on page 19 the construction from the OP is defined and called »semidirektes Produktmass«, which could be translated as »semidirect product measure«. Shortly after that a sort of wedge product between kernels is introduced. All this notation is needed for his proof of the Ionescu-Tulcea Theorem.

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For the sake of historical accuracy: The Ionescu-Tulcea theorem is not due to Alexandra Ionescu-Tulcea, as claimed in the PDF. It is due to Cassius Ionescu-Tulcea. –  Michael Greinecker Jan 21 '13 at 12:04
    
Wow, I feel cheated now. How did the author come to think it was her? Is that a common misattribution? –  Thomas Jan 21 '13 at 13:19
    
I'm not sure. But I was once curious about it. I guess the story would be more fun with Alexandra Below in the lead. –  Michael Greinecker Jan 21 '13 at 16:23

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