# Continuous Martingale Characterization through Conditioning

Consider a filtration $$\mathcal{F}_t$$ and let $$Y$$ be integrable on the same probability space. Then $$M_t=\mathbb{E}[Y|\mathcal{F_t}]$$ is a martingale.

In Chapter 3, Exercise 8, Øksendal asks us to show a converse statement (3.8b) leveraging a Corollary (C.7) to show that if an $$\mathcal{F}_t$$-martingale is uniformly $$L^p(\mathbb{P})$$ for some $$p>1$$ then there exists an integrable $$Y$$ for which $$M_t=\mathbb{E}[Y|\mathcal{F_t}]$$.

I'm finding that I require the additional condition that $$M_t$$ is continuous; am I missing something here, or is the book simply missing the condition?

By (C.7), with continuity, we have that an integrable $$Y$$ exists such that $$M_t\mathbin{\overset{ L^1 }{\longrightarrow}}Y$$. Then, notice that $$\mathbb{E}[Y-M_t|\mathcal{F}_t]=0$$ if $$\mathbb{E}[1_H(Y-M_t)]=0$$ is for all events $$H\in \mathcal{F}_t$$. But $$1_H(M_s-M_t)\mathbin{\overset{ L^1 }{\longrightarrow}}1_H({Y-M_t})$$ as $$s\rightarrow\infty$$. So $$\mathbb{E}[{1_H(Y-M_t)}]=\lim_{s\rightarrow\infty}\mathbb{E}[{1_H(M_s-M_t)}]$$ where for $$s\ge t$$ $$\mathbb{E}[1_H({M_s-M_t})]=\mathbb{E}[1_H\mathbb{E}[{M_s-M_t}|{\mathcal{F}_t}]]=0\,\,.$$

The continuity assumption seems load-bearing. Without continuity, let $$\epsilon$$ be a Rademacher random variable, $$M_t=\epsilon 1\{t\ge 1\}$$, and an unnaturally large constant filtration $$\mathcal{H}_t=\sigma(\epsilon)$$, then $$M_t=0\neq \epsilon=\mathbb{E}[{\epsilon}|{\mathcal{H}_{t}}]$$ for any $$t<1$$. Is it possible to construct a counterexample with a natural filtration?

(3.8b) Problem

(C.7) Corollary

• I'm not sure that your counterexample is a martingale with respect to the filtration $\mathcal H_t$. Since $M_t = \epsilon$ for $t \ge 1$, you showed that $M_s = 0 \ne \epsilon = \mathbb{E}[M_t | \mathcal H_s]$ for any $s < 1 \le t$. Sep 21, 2021 at 18:31
• Oh yes, I don't know what I was thinking there.
– VF1
Sep 21, 2021 at 20:47

There is an important property of stochastic processes that seems to be missing from Okesndal's book, or at least if it is there I couldn't find it. Specifically, a sub-martingale $$M$$ with respect to a filtration $$(\mathcal F_t)$$ satisfying the usual conditions has a right-continuous modification if and only if $$t \mapsto \mathbb{E}[M_t]$$ is right-continuous. In particular, if $$M$$ is a martingale, then $$M$$ has a right-continuous modification.

By considering this modification and using Theorem C.6 instead of Corollary C.7, we can prove the statement in the problem without using any additional assumptions.

• Hrm, I suppose this is the "right" way to do things for this problem; but $\mathcal{F}_t$ is never specified as augmented or complete. Perhaps it is just assumed, which is reasonable.
– VF1
Sep 21, 2021 at 20:48

If we make a couple of very generous interpretations of the conditions, then it seems like we can prove continuity, but that doesn't seem like an honest reading of the problem:

• assuming $$p\ge 2$$ or at least $$M_t\in L^2(\mathbb{P})$$ pointwise for all $$t$$, and
• assuming $$\mathcal{F}_t$$ is the natural filtration of Brownian motion,

we can then apply the Martingale Representation Theorem (Thm 4.3.4 in the book), which says that $$M_t-\mathbb{E}[M_0]=\int_0^t v_t dB_t$$ for some $$v_t$$. Then, by Thm 3.2.5, there exists a modification of $$M_t$$ with continuous sample paths. This suffices to show the desired converse, but seems roundabout and depends on theorems from later in the book.