# n-factor martingale representation theorem

Baxter & Rennie at pag. 162 state the following theorem.

Let $$W$$ be an $$n$$-dimensional $$\mathbb Q$$-Brownian motion and let $$M_t=(M_1(t),...,M_n(t))$$ be an $$n$$-dimensional $$\mathbb Q$$-martingale process, which has volatility matrix $$(\sigma_{ij}(t))$$, in that $$dM_j(t) = \Sigma_i \sigma_{ij}(t) dW_i(t)$$ and the matrix is non singular (for all $$t$$ with probability 1). Let $$N_t$$ be any one-dimensional $$\mathbb Q$$-martingale.

Then there exists an $$n$$-dimensional $$F$$-previsible process $$\phi_t = (\phi_1(t), ... \phi_n(t))$$ such that $$\int^T_0 (\Sigma_j \sigma_{ij}(t) d\phi_j(t))^2 dt <\infty$$, and the martingale $$N$$ can be written as $$N_t = N_0 + \Sigma_j \int_0^T \phi_j(s) dM_j(s)$$. Further $$\phi$$ is (essentially) unique.

My question is: because each of the $$M_j(t)$$ is a $$\mathbb Q$$-martingale, why can't I chose the $$\phi_j$$ to write $$N_t=N_0 + \int_0^T \phi_j(s) dM_j(s)$$ $$\forall j$$? In other words, what is the idea behind uniqueness in the $$n$$-dimensional case ($$n>1$$)?

• I think there's a missing hypothesis. $N_t$ can't be any old $\mathbb{Q}$-martingale; it has to be a martingale with respect to the filtration generated by the Brownian motion $W$. And if so, then I think that answers your question: given an $N_t$ which is a martingale wrt the $W$ filtration, you can't just apply the theorem to the one-dimensional Brownian motion $W^j$, because $N_t$ might not be a martingale with respect to the $W^j$ filtration. – Nate Eldredge Oct 16 '18 at 14:25
• I noticed in a Google Books preview that B&R define "martingale" with respect to a particular measure, without emphasizing that the filtration really plays a role. That seems unfortunate. Maybe there is a standing assumption that the probability space comes equipped with a Brownian motion $W$ and the filtration is the one generated by $W$, which would at least make the theorem correct but seems very awkward. I don't have the full book to check. – Nate Eldredge Oct 16 '18 at 23:42
• I think you are right. The filtration is the one generated by the Brownian motion $W$ mentioned in the statement. So now I am confused about what it means for a one-dimensional process $N_t$ to be martingale wrt (the filtration generated by) a $n$-dimensional Brownian motion. But I am sure you put me on the right path. Thanks! Please formulate it as an answer, so that I can close. – Luigi Scorzato Oct 17 '18 at 6:02

The statement and associated definitions in the book seem to gloss over an important assumption: the process $$N_t$$ needs to be a martingale with respect to the measure $$\mathbb{Q}$$ and the filtration $$\{\mathcal{F}_t^W\}$$ generated by the $$n$$-dimensional Brownian motion $$W(t)$$, i.e. $$\mathcal{F}_t^W = \sigma(W(s) : s \le t)$$. That is, for each $$t$$, $$N_t$$ needs to be $$\mathcal{F}_t^W$$-measurable (i.e. the process is $$\mathcal{F}_t^W$$-adapted), and for each $$s < t$$ we have $$E[N_t \mid \mathcal{F}_s^W] = N_s$$.
This explains why your objection doesn't apply. I think your thought was that if you have a representation of a process $$N_t$$ with respect to the Brownian motion $$W$$, you could apply the theorem again using the first coordinate $$W_1$$, which is indeed a one-dimensional Brownian motion, in place of $$W$$, and get a different representation. But an $$\mathcal{F}_t^W$$-martingale will typically not be an $$\mathcal{F}_t^{W_1}$$-martingale.
As a simple example, consider $$N(t) = W_2(t)$$. Then $$W_2(t)$$ is not measurable with respect to the $$\sigma$$-field $$\mathcal{F}^{W_1}_t$$: indeed, it is independent of this $$\sigma$$-field. So although $$W_2(t)$$ is a martingale with respect to $$\mathcal{F}_t^W$$, it is not an $$\mathcal{F}_t^{W_1}$$-martingale.