Let $(\Omega, \mathcal{F}, \mathcal{P})$ be a probability space for $X\in L^1(\mathcal{P})$ and $(\mathcal{F}_n)$ a filtration. Then $$Y_n = \mathbb{E}(X|\mathcal{F}_n)$$ is called Doob Martingale. The class of Doob Martingales are uniformly integrable and together convergence in probability, by Vitali's convergence theorem, they converge in $L^1$. Many things imply convergence in probablity, e.g. $L^1$ boundedness implies convergence almost surely by Doob's convergence theorem, which in turn implies convergence in probability.
The the limit is $Y_\infty = \mathbb{E}(X|\mathcal{F}_\infty)$, where $\mathcal{F}_\infty = \sigma( \bigcup_{n\in\mathbb{N}} \mathcal{F}_n)$.
My question is, does it hold that $Y_\infty = X$ almost surely for any filtration, or do we require $ \mathcal{F}_n\nearrow \mathcal{F}$?
Thank you for your time.
Best Answer
As you pointed out, $Y_{\infty} = \mathbb{E}[X|\mathcal F_\infty]$. In order to have $X = Y_\infty = \mathbb{E}[X|\mathcal F_\infty]$, we would of course require that $X$ is $\mathcal F_\infty$ measurable. That doesn't necessarily mean that we need $\mathcal F_\infty = \mathcal F$, just that $\sigma(X) \subseteq \mathcal F_\infty$.
As a example to show $Y_\infty$ may not be $X$, one could take the trivial filtration $\mathcal F_n = \{\emptyset, \Omega\}$ for all $n$. Then $Y_\infty = Y_n = \mathbb{E}[X]$, which is not equal to $X$ for any non-constant random variable.
As an example to show we don't necessarily need $\mathcal F_\infty = \mathcal F$, one can take $X$ and $Z$ to be two independent non-trivial random variables, $\mathcal F = \sigma(X,Z)$, and $\mathcal F_n = \sigma(X)$ for all $n$. Then $\mathcal F_\infty = \sigma(X) \ne \mathcal F$, but $Y_\infty = X$.