1
$\begingroup$

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 ($n>1$)?

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Oct 16 '18 at 14:25
  • 1
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Oct 16 '18 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. $\endgroup$ – Luigi Scorzato Oct 17 '18 at 6:02
1
$\begingroup$

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.

$\endgroup$

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.