3
$\begingroup$

Let $\Omega$ be the space of continuous functions defined on $[0,1]$. Define the canonical process $B$ by

$$B_t(\omega)=\omega_t,~ \forall\omega\in\Omega$$

Let us equip $\Omega$ with the usual filtration $(\mathcal{F}_t)$ given by

$$\mathcal{F}_t:=\sigma(B_s, s\leq t)$$

Given a probability measure $\mathbb{P}$ under which $B$ is a martingale. Thus we get a probabilistic space $(\Omega,\mathcal{F}_1,\mathbb{P})$.

Now let $M$ be some continuous martingale defined on $(\Omega,\mathcal{F}_1,\mathbb{P})$ with $M_0=0$. Suppose that

$$<M, B>_t=0,~ \forall 0\leq t\leq 1$$

Can we conclude that $M_t=0$ for every $t\in [0,1]$? Many thanks for the answer!

$\endgroup$
4
  • $\begingroup$ Since $M$ is a martingale with respect to the filtration generated by $B$, thus there exists some measurable function $f: [0,1]\times\Omega\to\mathbb{R}$ such that $M_t=f(t,B^t)$ where $B^t$ is the process stopped at $t$. In order to show $M\equiv 0$ we start with the $M$ of the form $M_t=f(t, B_t-B_{t_n}, B_{t_n}-B_{t_{n-1}}, ..., B_{t_1}-B_{t_0})$. If we can prove $M\equiv 0$ for such $f$'s, then by an argument of monotone class theorem. Firstly we suppose $f$ is $\mathcal{C}^2$, then Ito's formula gives the result, but I do not know how to prove for any $f$. Does some have an idea? $\endgroup$ – CodeGolf Jan 27 '14 at 19:54
  • $\begingroup$ But here $B$ is just a martingale(not necessary Brownian martingale). As is well known that not all martingales admit a representation. $\endgroup$ – CodeGolf Jan 27 '14 at 23:02
  • $\begingroup$ In other words, the question is whether any martingale $M$ wrt the filtration generated by another martingale $B$(arbitrary) can be written as the integral of $B$? $\endgroup$ – CodeGolf Jan 27 '14 at 23:06
  • $\begingroup$ Does someone have an idea? $\endgroup$ – CodeGolf Jan 28 '14 at 13:55
2
$\begingroup$

1) If $M_t$ is a martingale adapted to the Brownian filtration $\mathcal{F}_B$ with $M_0=0$, then $\langle B,M \rangle_t=0$ implies $M=0$. As you point out, this is a consequence of Ito's representation theorem.

2) If we do not require $M$ to be adapted to $\mathcal{F}_B$, then this result does not hold anymore because any martingale independent from $B$ will satisfy $\langle M,B \rangle=0$. This is because the product of two independent martingales is still a martingale.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.