# Can $<.>$ of a martingale determine it only?

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!

• 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? – CodeGolf Jan 27 '14 at 19:54
• But here $B$ is just a martingale(not necessary Brownian martingale). As is well known that not all martingales admit a representation. – CodeGolf Jan 27 '14 at 23:02
• 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 $B$? – CodeGolf Jan 27 '14 at 23:06
• Does someone have an idea? – CodeGolf Jan 28 '14 at 13:55

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.