Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mathcal{A}$ and $\mathcal{B}$ be two sub-$\sigma$-algebras in a measure space. To each one, there is a conditional expectation associated, respectively $E^\mathcal{A}$ and $E^\mathcal{B}$. Given the two $\sigma$-algebras, we can form a third one, $\sigma(\mathcal{A},\mathcal{B})$, generated by both, and consequently, its conditional expectation $E^{\mathcal{A},\mathcal{B}}$.

My question is: knowing only the first two conditional expectations as projection operators in the measurable function space, can we obtain the third one as an algebraic expression of the first two, as the limit of a polynomial, for example?

A first try was to think about them as geometrical projections and try to find a complementar conditional expectation and calculate it in a similar fashion to $A\cup B=(A^C\cap B^C)^C$, and define the intersection as the limit of $(E^\mathcal{A}E^\mathcal{B})^n$. But the expected complementar $1-E^\mathcal{A}+E$ fails to be a conditional expectation.

Thank you!

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

Clearly not. Let the measure space be the uniform measure on {$1,2,3,4$}. $A$ allows you to discern whether the number is greater or less than $2.5$ or not. $B$ allows you to discern whether the number is 0 or 1 mod $2$. Let $f$ be $x^2-5x+6$, then $E^Af=1$, $E^B f=1$, $E^{A,B}f=f$. One can't write $f$ as any polynomial or olgebraic expression in $1$ and $1$.

Edit: If you want $A \cap B$ instead of $A \cup B$ you can use $\lim \dots E^A E^B E^A E^B E^A E^B$, if the limit exists, since that is the projection onto the subspace fixed by both $E_A$ and $E_B$, which is the subspace of functions defined over both $A$ and $B$, which is the subspace you want to project onto.

share|improve this answer
add comment

By restricting to algebraic expressions you treat conditional expectations as merely projections, perhaps self-adjoint ones. That is, you miss too much structure.

share|improve this answer
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.