2
$\begingroup$

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 problem

(C.7) Corollary corollary

$\endgroup$
2
  • $\begingroup$ 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 $s < 1 \le t$. $\endgroup$ Sep 21, 2021 at 18:31
  • $\begingroup$ Oh yes, I don't know what I was thinking there. $\endgroup$
    – VF1
    Sep 21, 2021 at 20:47

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – VF1
    Sep 21, 2021 at 20:48
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .