I'm reading Steven E. Shreve's "Stochastic calculus for finance II", and find myself not really understand the concept of "filtration".
Yes, the definition of filtration is straight forward, it's set of $\sigma$-algebra. However, when it comes to the Martingale Representation and Girsanov Theorem below, I'm lost on the different of a filtration generated by the Brownian motion or not.
First it's Theorem 5.3.1 (Martingale representation, one dimension):
Let $W(t)$, $0 \leq t \leq T$, be a Brownian motion on a probability space $(\Omega,\mathscr F, \mathbb P)$, and let $\mathscr F(t)$, $0 \leq t \leq T$, be the filtration generated by this Brownian motion. Let $M(t)$, $0 \leq t \leq T$, be a martingale with respect to this filtration (i.e., for every $t$, $M(t)$ is $\mathscr F(t)$-measurable and for $0 \leq s \leq t \leq T$, $\mathbb E [M(t) | \mathscr F(s)] = M(s)$).
Then there is an adapted process $\Gamma(u)$, $0 \leq u \leq T$, such that
$$M(t) = M(0) + \int_0^t \Gamma(u) d W(u), 0 \leq t \leq T \tag{5.3.1} $$
Then Shreve says, "
The assumption that the filtration in Theorem 5.3.1 is the one generated
by the Brownian motion is more restrictive than the assumption of Girsanov's
Theorem, Theorem 5.2.3, in which the filtration can be larger than the one
generated by the Brownian motion.
If we include this extra restriction in Girsanov's Theorem, then we obtain the following corollary. The first paragraph
of this corollary is just a repeat of Girsanov's Theorem; the second part contains the new assertion" (the bold part "the filtration generated by this Brownian motion" I highlighted below, is the difference comparing to original Girsanov Theorem 5.2.3):
Corollary 5.3.2. Let $W(t)$, $0 \leq t \leq T$, be a Brownian motion on a probability
space $(\Omega,\mathscr F, \mathbb P)$, and let $\mathscr F(t)$, $0 \leq t \leq T$, be the filtration generated by this
Brownian motion. Let $\Theta(t)$, $0 \leq t \leq T$, be an adapted process, define
$Z(t) = \exp\left\{ – \int_0^t \Theta(u) d W(u) – \frac{1}{2} \int_0^t \Theta^2(u) d u \right\}$,
$\widetilde W(t) = W(t)+ \int_0^t \Theta(u) d u$, and assume that $\mathbb E \int_0^T \Theta^2(u) d u < \infty$. Set $Z = Z(T)$. Then $\mathbb E Z = 1$, and
under the probability measure $\widetilde P$ given by
$$\widetilde P(A) = \int_A Z(\omega) d P(\omega), \forall A \in \mathscr F \tag{5.2.1}$$
, the process $\widetilde W(t)$, $ 0 \leq t \leq T$, is a Brownian motion.
Now let $\widetilde M(t)$, $0 \leq t \leq T$, be a martingale under $\widetilde{\mathbb P}$. Then there is an
adapted process $\widetilde \Gamma(u)$, $0 \leq u \leq T$, such that
$$\widetilde M(t) = \widetilde M(0) + \int_0^t \widetilde \Gamma(u) d \widetilde W(u), 0 \leq t \leq T. \tag{5.3.2}$$
Shreve says: "*Corollary 5.3.2 is not a trivial consequence of the Martingale Representation Theorem, Theorem 5.3.1, with $\widetilde W(t)$ replacing $W(t)$ because the filtration
$\mathscr F(t)$ in this corollary is generated by the process $W(t)$, not the $\widetilde P$-Brownian
motion $\widetilde W(t)$".
My problem is I could not visualize why the difference matters? I could not understand, if $\widetilde{\mathbb P}$ is defined based on $\mathbb P$, how different could they be?
Is there any example that could explain why Shreve "makes a big fuzz" here?
Best Answer
You have $$\widetilde{W}_t=W_t+\int\Theta(u)du$$ which is in general not a Brownian motion, because it has a drift component.
But 5.3.1 states
$$M_t=M_0+\int \Gamma(u)dW_u\tag{5.3.1}$$
, which holds only for a Brownian motion $W$ (and $M_t$ martingale).
So one cannot trivially replace $W_t$ and $W_t+\int\Theta(u)du=\widetilde{W}_t$ in 5.3.2 aswell by setting
$$\widetilde M_t = \widetilde M_0 + \int_0^t \widetilde \Gamma(u) d \widetilde W_u\tag{5.3.2}$$
(because $\widetilde W_t$ is not in general a Brownian motion).
5.3.2 holds only under the special change of measure defined as $$Z_t = \exp\left\{ - \int_0^t \Theta(u) d W(u) - \frac{1}{2} \int_0^t \Theta^2(u) d u \right\}$$
Then $\widetilde M$ is a martingale, and $\widetilde W$ becomes a Brownian motion (proof is not trivial).
But still the filtration of $W_t$ and $\widetilde{W}_t=W_t+\int\Theta(u)du$ is obviously not the same.