3
$\begingroup$

Let $\{\mathcal B_i:i\in I\}$ be a family of $\sigma$-algebras (over the same set $\Omega$) which are totally ordered by inclusion, in the sense that for any $i,j\in I$ either $\mathcal B_i\subset\mathcal B_j$ or $\mathcal B_j\subset\mathcal B_i$. I think this can be called a filtration with time $I$.

Let $\mathcal B=\sigma\left(\bigcup_{i\in I}\mathcal B_i\right)$ be the $\sigma$-algebra generated by all the $\mathcal B_i$, let $\mu$ be a probability on $\mathcal B$ and let $X\in L^2(\mathcal B,\mu)$ be a bounded random variable.

In what senses (if any) does the family $\{\mathbb{E}[X\mid\mathcal B_i];i\in I\}$ approximate $X$? More precisely:

Is it true that for every $\epsilon>0$ there is some $i\in I$ such that $\|X-\mathbb{E}[X\mid\mathcal B_i]\|_{L^2}<\epsilon$? ($L^2$ convergence)

Is it true that there exists a set $A\in\mathcal B$ with $\mu(A)=1$ and for every $\epsilon>0$ there is some $i\in I$ such that for each $\omega\in A$ we have $|X(\omega)-\mathbb{E}[X\mid\mathcal B_i](\omega)|<\epsilon$? (almost sure convergence)

$\endgroup$
2
  • $\begingroup$ I think you'll want to look at Williams' Probability with Martingales: books.google.com/books?id=e9saZ0YSi-AC $\endgroup$ Nov 21, 2012 at 17:49
  • $\begingroup$ That book, although quite nice for learning, doesn't seem to deal with uncountable time, which is what I was looking for. $\endgroup$ Nov 29, 2012 at 2:32

1 Answer 1

4
$\begingroup$

The sigma algebra generated by $X$ is countably generated. Thus $X$ is measurable for $$ \sigma\left(\bigcup_{k=1}^\infty \mathcal{B}_{i_k}\right) $$ for some increasing sequence $i_1 \le i_2 \le \dots$ in $I$. So some of your desired results follow from the usual martingale convergence theory.

The almost sure convergence will fail, in general, for uncountable $I$. A reasonable replacement is called "essential convergence". That is what Sucheston and I used for such things in our book.

[plug: G. Edgar, L. Sucheston, Stopping Times and Directed Processes (Cambridge University Press 1992)]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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