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Let $ (\mathbb{X} _i, \mathscr{X}_i) $ and $ (\mathbb {Y} _i, \mathscr {Y} _i) $ measurable spaces with $ i = 1, 2 $. Let $ \gamma_i: \mathscr {X}_i\times\mathbb {Y}_i\longrightarrow [0,1] $ a probability kernel from $ (\mathbb {Y}_i, \mathscr {Y}_i) $ to $ (\mathbb {X}_i , \mathscr {X}_i) $, i.e., $$ \gamma_i(X_i|\;\;\;\;):\mathbb {Y}_i\longrightarrow [0,1] $$ is a $\mathscr{Y}_i$-measurable function for all $X_i\in\mathscr{X}_i$ and $$ \gamma_i(\;\;\;\;|\omega_i): \mathscr {X}_i\longrightarrow [0,1] $$ is a probability for all $\omega_i\in\mathbb{Y}_i$.

Let $ \mathbb {X} = \mathbb {X} ^ 1 \times \mathbb{X}^ 2 $, $ \mathbb {Y} = \mathbb{Y}^1\times\mathbb{Y}^2$. Denote by $\mathscr{X}=\sigma(\mathscr{X}^1\times\mathscr{X}^2)$ a $\sigma$-algebra generated by the algebra $\mathscr{X}^1\times\mathscr{X}^2$ and $\mathscr{Y}=\sigma(\mathscr{Y}^1\times\mathscr{Y}^2)$ a $\sigma$-algebra generated by the algebra $\mathscr{Y^1}\times\mathscr{Y}^2$. Set $\gamma_1 \times \gamma_2: \mathscr {X}\times\mathbb {Y} \longrightarrow [0,1] $ for $ X \in \mathscr {X} $ and $ \omega = (\omega_1, \omega_2) \in \mathbb{Y}_1 \times \mathbb{Y}_2 = \mathbb{Y} $ making $$ \gamma_1\times\gamma_2(\;\cdot\;|\omega)=\gamma_1(\;\cdot\;|\omega_1)\times\gamma_2(\;\cdot\;|\omega_2) $$ and $$ \gamma_1\times\gamma_2(X|\omega)=\gamma_1(\;\cdot\;|\omega_1)\times\gamma_2(\;\cdot\;|\omega_2)(X) $$ It is interesting to raise the following question. Under what conditions can we say that $ \gamma = \gamma_1\times\gamma_2$ is a probability kernel from $ (\mathbb {Y}, \mathscr {Y}) $ to $ (\mathbb {X} , \mathscr {X}) $?

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2 Answers 2

up vote 3 down vote accepted

The result holds in full generality. I'm a bit lazy, so I will use my own notation.

Proposition: Let $(X_1,\mathcal{X}_1)$, $(X_2,\mathcal{X}_2)$, $(Y_1,\mathcal{Y}_1)$, and $(Y_2,\mathcal{Y}_2)$ all be measurable spaces and let $K_1:X_1\times\mathcal{Y}_1\to[0,1]$ and $K_2:X_2\times\mathcal{Y}_2\to[0,1]$ be probability kernels. Then the function $$K:X_1\times X_2\times\mathcal{Y_1}\otimes\mathcal{Y}_2\to[0,1]$$ given by $K\big((x_1,x_2),A\big)=K_1(x_1,\cdot)\otimes K(x_2,\cdot)(A)$ is a probability kernel.

Proof: We only have to verify the $\mathcal{X}_1\otimes\mathcal{X}_2$-measurability. We use the monotone class theorem. Our $\pi$-system will be the measurable rectangles and our $\delta$-system $D$ the family of all measurable sets $A$ in $\mathcal{Y}_1\otimes\mathcal{Y}_2$ such that $K(\cdot,A)$ is measurable. We first show that the measurable rectangles are in $D$. Let $A_1\times A_2$ be a measurable rectangle and $r\in [0,1]$. We let $$R=\{(q_1,q_2)\in\mathbb{Q}_+\times\mathbb{Q}_+:q_1 q_2<r\}.$$ Then $$K(\cdot,A_1\times A_2)^{-1}\big([0,r)\big)=\bigcup_{(q_1,q_2)\in R}\bigg(K_1(\cdot,A_1)^{-1}\big([0,q_1)\big)\times K_2(\cdot,A_2)^{-1}\big([0,q_2)\big)\bigg),$$ which proves our first claim. Now we show that $D$ is actually closed under proper differences. So let $r\in[0,1]$ $A,B\in D$ and $B\subseteq A$. Let $$Q=\{(q_1,q_2)\in\mathbb{Q}_+\times\mathbb{Q}_+:q_1-q_2<r\}$$

Then $$K(\cdot,A\backslash B)^{-1}\big([0,r)\big)=\bigcup_{(q_1,q_2)\in Q} \bigg(K(\cdot,A)^{-1}\big([0,q_1)\big)\cap K(\cdot,B)^{-1}\big((q_2,1]\big)\bigg),$$ which proves our second claim. Finally, we have to show that $D$ is closed under increasing limits. So let $(A_n)$ be an inreasing sequence in $D$, $r\in [0,1]$ and let $A=\bigcup_n A_n$. Then $$K(\cdot,A)^{-1}\big((r,1]\big)=\bigcup_n K(\cdot,A_n)^{-1}\big((r,1]\big),$$ by the continuity of probability measures and we have proven the last claim.

Edit: It is worth pointing out that the result generalizes to arbirtrary products of kernels. The finiteness of the product was only used in the first step with the measurable rectangles. Now the product $\sigma$-algebra is essentially generated by finite rectangles, so the proof generalizes.

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Thank you for answering this question Greinecker. –  Elias Dec 29 '11 at 16:12
    
Greinecker, I could not understand how you got this equality. $$K(\cdot,A\backslash B)^{-1}\big([0,1)\big)=\bigcup_{(q_1,q_2)\in Q} \bigg(K(\cdot,A)^{-1}\big([0,q_1)\big)\times K(\cdot,B)^{-1}\big((q_2,1]\big)\bigg),$$ Can you clarify? –  Elias Jan 12 '12 at 22:42
    
It's a typo, the interval appearing on the left should be $[0,r)$, so the event that $A\backslash B$ is smaller than $r$ is true if $A$ is not too big and $B$ not too small. This is true when the probability of $A$ minus the probability of $B$ is smaller than $r$. Since the inequality is strict, whenever the inequality is satisfied, it is satified for rational numbers serving as a "buffer". –  Michael Greinecker Jan 12 '12 at 23:25
    
Also $\times$ aon the right has to be replaced by $\cap$. –  Michael Greinecker Jan 12 '12 at 23:32
    
Greinecker, thank you. –  Elias Jan 13 '12 at 11:51

I think that this construction need more work. My point is the left hand side of
$$ \gamma_1\times\gamma_2(\;\cdot\;|\omega)=\gamma_1(\;\cdot\;|\omega_1)\times\gamma_2(\;\cdot\;|\omega_2) $$ is well defined for a non-rectangle set, but what about the rhs ?

Perhaps I miss something.

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