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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 ($n>1$)?

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    $\begingroup$ 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. $\endgroup$ Oct 16, 2018 at 14:25
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    $\begingroup$ 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. $\endgroup$ Oct 16, 2018 at 23:42
  • $\begingroup$ 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. $\endgroup$ Oct 17, 2018 at 6:02

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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.

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