When does the predictable $\sigma$-algebra $\mathcal{P}$ coincide with the optional $\sigma$-algebra $\mathcal{O}$?

Question

The setup of my question is the following: Suppose that we have a measurable space $(\Omega,\mathcal{F})$ and a filtration $\mathbf{F} = (\mathcal{F}_t)_{t \geq 0}$ on it. Let $\mathcal{P}(\mathbf{F})$ be the predictable $\sigma$-algebra, that is, the $\sigma$-algebra generated by all real-valued left-continuous processes adapted to the filtration $\mathbf{F}$. Let $\mathcal{O}(\mathbf{F})$ be the optional $\sigma$-algebra, that is, the $\sigma$-algebra generated by all real-valued right-continuous process with left-limits adapted to the filtration $\mathbf{F}$.

I was trying to find conditions on the underlying measurable space $(\Omega,\mathcal{F})$ or the filtration $\mathbf{F}$, under which the predicatble $\sigma$-algebra $\mathcal{P}(\mathbf{F})$ coincides with the optional $\sigma$-algebra $\mathcal{O}(\mathbf{F})$. For example, if the filtered space $(\Omega,\mathcal{F})$ supports a process $X=(X_t)_{t \geq 0}$ with values in a metric space $(E,d)$ for example, such that the paths $\mathbb{R}_+ \ni t \mapsto X_t \in E$ are continuous, and the filtration $\mathbf{F} = (\mathcal{F}_t)_{t \geq 0}$ is generated by $X$, that is, $$\mathcal{F}_t = \sigma(X_r : 0 \leq r \leq t),$$ does the equality $$\mathcal{P}(\mathbf{F}) = \mathcal{O}(\mathbf{F})$$ hold?

If this condition is too weak, does there exist some other general condition under which the predictable $\sigma$-algebra coincides with the optional $\sigma$-algebra?

Answer 1

(Answering your comment)

Off the top of my head, I'd look in vol.2 of Probabilités et Potentiel (Dellacherie & Meyer) or in Limit Theorems for Stochastic Processes (Jacod & Shiryaev). Another convenient resource is the blog https://almostsure.wordpress.com of Geo. Lowther.

The key is that for a bounded rc martingale $M$, the predictable projection ${}^p\!M$ coincides with the left limit process $(M_{t-})$. Thus, if $\mathcal P =\mathcal O$ then each such $M$ coincides with its predictable projection and so is left continuous, hence continuous.

Conversely, if each bounded rc martingale is continuous, then each such martingale is predictable. In particular, for a bounded r.v. $Z$ the rc version of the martingale $t\mapsto E[Z|\mathcal F_t]$ must be predictable. Using this, Theorem IV-62 of Dellacheie & Meyer cited above yields that every stopping time is predictable, and this in turn implies that $\mathcal O =\mathcal P$.

Answer 2

A sufficient (and necessary) condition is that each bounded right-continuous martingale is continuous. This is true, for example, if the filtration is that of a Brownian motion.

The condition you suggest is too weak. Example: Let $U$ be uniformly distributed on $(0,1)$ and let $\xi$ be an independent random variable taking the two values $1$ and $-1$ with equal likelihood, both defined on some probability space $(\Omega,\mathcal F,\Bbb P)$. Take your filtration to be $\mathcal F_t:=\cap_{\epsilon>0}\mathcal F_{t+\epsilon}^o$, where $\mathcal F_t^o$ is defined by $$ \mathcal F^o_t:=\sigma\{1_{\{U\le s\}}, 0\le s\le t; \xi1_{\{U\le t\}}),\qquad t\ge 0. $$ Notice that $U$ is an $(\mathcal F_t)$-stopping time. Now define a continuous-path process $X$ by
$$ X_t=\cases{0,&$0\le t<U$,\cr \xi(t-U),&$U\le t$.\cr} $$ This process generates $(\mathcal F_t)$ (modulo null sets) but the (right-continuous) martingale $$ M_t:=\Bbb E[X_1|\mathcal F_t] = \cases{0,&$0\le t<U$,\cr X_1,&$U\le t$\cr} $$ is not continuous.