Definition of Left-Closable Martingale

Question

I am currently studying martingales with Resnick's book A Probability Path. He defines a martingale as closed on the right if there is an $X \in L_1$ such that $X_n = \mathbb{E}[X \mid \mathcal{B}_n]$ for all $n$. One can also find that definition here, defining it as "right-closable."

What I am curious about is the "on the right." I did a little bit (perhaps not enough) searching around on the internet and I can't seem to find any definition of what it might mean to be closed "on the left." Does anyone happen to know of such a definition or where I might find one?

Answer 1

I haven't run into this terminology before, but here is my guess. The term closed probably refers to the set of indices $n$ where the martingale is defined. That is, consider the domain $$ \{ n : X_{n} \ \mathrm{is \ defined} \} = \mathbb{Z}_{\geq 0}, $$ which is not (topologically) closed in the extended reals because it has a hole on the right (at $n=+\infty$). Therefore, we define $X_{\infty}$ such that $$ X_{n} = \mathbb{E}[X_{\infty} | \mathcal{B}_{n} ], $$ which is just what you call $X$. The domain then is closed.

From this, it should be clear that "closed on the left" is not a particularly useful concept, since the domain is always closed on the left (at $n=0$). I assume that for this reason the term "closed" is used instead, excluding the "on the right".

Notice also that essentially $ X_{\infty} = \lim_{n \to \infty} X_{n}, $ although the actual nature of this convergence depends on what conditions we have on the $X_{n}$. See Doob's Martingale Convergence Theorem for this. This limit is useful in many applications such as filtering theory. Lastly, if we consider continuous time martigales $(X_{t} : t \in T)$, a more general notion of closedness may be useful, but in that case the left/right distinction is irrelevant since $T$ may well be a subset of $\mathbb{R}^{n}$ for $n > 1$.

Answer 2

To build on Jacob Maibach's answer on the positive integers being open on the right, later on in Resnick's book he discusses reversed/backwards martingales: given a DECREASING family of $\sigma$-fields $\{\mathcal{B}_n\}_{n\ge0}$, $\{X_n\}_{n\ge0}$ is a reversed martingale if $X_n$ is $\mathcal{B}_n$ measurable and $\mathbb{E}[X_n \mid \mathcal{B}_{n+1} ] = X_{n+1}$ for $n\ge0$.

But one could also write this as a "normal" martingale with a negative index set: let $\mathcal{B}'_n = \mathcal{B}_{-n}$ and $X_n' = X_{-n}$ for $n \ge 0$. Then, $\{\mathcal{B}'_n\}_{n\le0}$ is a filtration and $\{X_n'\}_{n\le0}$ is a martingale with respect to the previous filtration. Here, "time [is] flowing as usual from left to right" (Resnick's words) so that this martingale is closed on the right by $X_0'$ and in fact, $\mathbb{E}[X_0' \mid \mathcal{B}'_n] = X_n'$.

All that to say, defining a notion of closure on the left in light of Jacob's answer and in the context of reversible martingales (martingales with index set $\{n \in \mathbb{Z} : n \le 0\}$) is to find a limit $\lim_{n\to-\infty} X_n' =: X'_{-\infty}$ almost surely that is measurable with respect to $\mathcal{B}'_{-\infty} := \bigcap_{n\le0} \mathcal{B}'_n$, integrable, and satisfies $\mathbb{E}[X_n' \mid \mathcal{B}'_{-\infty}] = X'_{-\infty}$. Indeed, this can always be done according to the Reversed/Backwards Martingale Convergence Theorem (Theorem 10.15.1 in Resnick's book).

See also this question for more discussion on reversed martingales.