Does martingale convergence hold for arbitrary time? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:54:40Z http://mathoverflow.net/feeds/question/114074 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114074/does-martingale-convergence-hold-for-arbitrary-time Does martingale convergence hold for arbitrary time? Joel Moreira 2012-11-21T16:18:58Z 2012-11-21T18:21:09Z <p>Let <code>$\{\mathcal B_i:i\in I\}$</code> 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 $I$.</p> <p>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. </p> <p>In what senses (if any) does the family <code>$\{\mathbb{E}[X\mid\mathcal B_i];i\in I\}$</code> approximate $X$? More precisely: </p> <p>Is it true that for every $\epsilon>0$ there is some $i\in I$ such that $\|X-\mathbb{E}[X\mid\mathcal B_i]\|_{L^2}&lt;\epsilon$? ($L^2$ convergence)</p> <p>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 <code>$|X(\omega)-\mathbb{E}[X\mid\mathcal B_i](\omega)|&lt;\epsilon$</code>? (almost sure convergence)</p> http://mathoverflow.net/questions/114074/does-martingale-convergence-hold-for-arbitrary-time/114086#114086 Answer by Gerald Edgar for Does martingale convergence hold for arbitrary time? Gerald Edgar 2012-11-21T18:21:09Z 2012-11-21T18:21:09Z <p>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. </p> <p>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.</p> <p>[plug: G. Edgar, L. Sucheston, <a href="http://www.math.osu.edu/~edgar.2/books/stdp.html" rel="nofollow"><em>Stopping Times and Directed Processes</em></a> (Cambridge University Press 1992)]</p>