Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathcal{A}_1,\mathcal{A}_2$ be $\sigma$-algebras on $\Omega$. Let $P$ be a probability on $\mathcal{A}_1$ and let $Q$ be a Markov kernel from $\mathcal{A}_1$ to $\mathcal{A}_2$. Set $K:=P\otimes Q$, so that $K$ is a probability measure on $\mathcal{A}_1\otimes\mathcal{A}_2$. Does there exist a probability measure $K'$ on $\sigma\left(\mathcal{A}_1,\mathcal{A}_2\right)$, that satisfies $$ K'\left(D\cap E\right)=K\left(D\times E\right) $$ for all $D\in\mathcal{A}_1$, $E\in\mathcal{A}_2$?

share|improve this question
1  
I've modified a text a bit (hopefully clarifying) if you don't mind. –  Ilya May 23 '13 at 7:32
1  
What are your thoughts? First, forget about the structure of $K$ and think of the following. You define $K':\mathscr A\to [0,1]$ for any $D\cap E$ where $D\in \mathscr A_1$, $E\in \mathscr A_2$ and $\mathscr A := \sigma(\mathscr A_1, \mathscr A_2)$. What are the properties of the collection $\{D\cap E\}$, e.g. is it a $\pi$-system (closed under intersections?). Furthermore, $$ K'(D\cap E) = \int_D Q(x,E)P(\mathrm dx). $$ Did you check whether it satisfies necessary conditions of being a measure? –  Ilya May 23 '13 at 7:35
    
@Ilya: Firstly i'd have to check consistency, namely that if $D_1,D_2\in\mathcal{A}_1$ and $E_1,E_2\in\mathcal{A}_2$ with $D_1\cap E_2=D_2\cap E_2$, $K\left(D_1\times E_1\right)=K\left(D_2\times E_2\right)$. I don't know how to do that. –  Evan Aad May 23 '13 at 7:39
1  
try to do this for a couple of cases when $\Omega$ is finite, say it has $3$ elements. –  Ilya May 23 '13 at 7:43
add comment

1 Answer 1

up vote 3 down vote accepted

The answer is: no. $K'$ is not even necessarily consistent.

Set $\Omega:=\left\{0,1\right\}$, and $\mathcal{A}_1:=\mathcal{A}_2:=\mathcal{P}\Omega$.

Define

$$ P(0):=P(1):=\frac{1}{2} $$

$$ Q\left(\omega,B\right):=P(B) $$

Then $K=P^2$.

Set $D_1:=D_2:=E_1:=\left\{1\right\}$, $E_2:=\Omega$.

Then $D_1\cap E_1=D_2\cap E_2$, but

$$ K\left(D_1\times E_1\right)=\frac{1}{4}\neq\frac{1}{2}=K\left(D_2\times E_2\right) $$

share|improve this answer
    
Well-done :) ${}$ –  Ilya May 23 '13 at 8:19
    
@Ilya: Thank you very much! –  Evan Aad May 23 '13 at 8:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.