Does martingale convergence hold for arbitrary time?

2012-11-21T16:18:58Z

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)

Answer by Gerald Edgar for Does martingale convergence hold for arbitrary time?

2012-11-21T18:21:09Z

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)]

Does martingale convergence hold for arbitrary time? - MathOverflow

Does martingale convergence hold for arbitrary time?

Answer by Gerald Edgar for Does martingale convergence hold for arbitrary time?