# Filtration and measure change

I asked this question in math stackexchange but to no avail. So i'm trying the luck here.

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?

• I think your question is good for this site, but you should delete it in Mathematics; cross-posting is not tolerated by SE sites. – SRKX Sep 10 '14 at 6:45
• Well, then leave it there, and I'll delete it here as cross-posting is not tolerated, see this post. – SRKX Sep 10 '14 at 7:19
• @SRKX I think there can be a Mathematical and a Financial answer to the same question, and I think this question fits on both Math and Quant.SE. – emcor Sep 10 '14 at 7:37
• @emcor I'm not here to debate this. Cross-posting is not allowed (except in very special case). It's my role to apply the community's rules and they were discussed in SE Meta extensively. It's not even my decision. – SRKX Sep 10 '14 at 7:43
• ok, how about like this: i'll delete the post here. and see what will come out on math SE (not to waste the 400 bonus points). afterwards, if there's no good answer, i'll post here again. – athos Sep 10 '14 at 8:34

First of all, a filtration $( \mathscr{F}_t )_{t \geq 0 }$ is a "set" of sigma algebras indexed usually by time t that are increasing. That is, for every $t>0$, $\mathscr{F}_t$ is a sigma algebra and $\mathscr{F}_t \subseteq \mathscr{F}_T$ for all $0\leq t \leq T$. The canonical example, is the filtration generated by a process, say Brownian Motion $W$: The filtration $( \mathscr{F}_t )_{t \geq 0 }$ is such that $\mathscr{F}_0$ is the minimum $\sigma$-algebra such that $W_0$ is measurable with respect to it (that is if $W_0$ is $\mathscr{G}$-measurable, then $\mathscr{F}_0 \subset \mathscr{G}$ ) and $\mathscr{F}_t$ is the minimum $\sigma$-algebra such that all increments of $W$ up to time $t$ are measurable with respect to it. Having said that, the interpretation of filtration is that of the flow of information: as time progresses you know at least as much information as before. In particular they are useful to define the concept of a martingale.

We say $(M_t)_{t\geq0}$ is a martingale if the conditional expectation at time $t$ given the information up to time $s$ is the process at time $s$; that is, the best we can say about the process at time $t$ with the information up tp time $s < t$ is the process itself at time $s$, but which information and how to define it formally, this is where the concept of filtration comes in play. On a filtered probability space (a probability space with a filtration ) $( \Omega, \mathscr{F}, \mathbb{P}, (\mathscr{F} )_{t \geq 0 } )$, the process $( M_t )_{ t \geq 0 }$ adapted to the filtration $( \mathscr{F}_t )_{t\geq 0}$ is a martingale if $M_0$ is integrable ( $M_0 \in L^1(\mathbb{P})$ ), and $$\mathbb{E} \left [ M_t \right \vert \left. \mathscr{F}_s \right] = M_s, \quad \text{ for every }s \leq t$$ In particular, a process might be martingale with respect to one filtration and not with respect to another. Also, it is not necessarily the case in which, say the filtration generated by Brownian Motion $W$, and the filtration generated by $$X_t = W_t + \int_0^t \theta_s ds$$ are the same. I will give a famous example below.

I think the comment that Shreve wants to make is that the process $\widetilde M$ is a martingale in the filtered probability space $( \Omega, \mathscr{F},\widetilde{ \mathbb{P} }, (\mathscr{F} )_{t \geq 0 } )$ (hence with respect to the filtration generated by $W$ ) still, it admits a stochastic representation representation like in Theorem 5.3.1 with $\widetilde W$. As you mentioned if the filtrations generated by $W$ and $\widetilde W$ were the same, then the result would be trivial, but they are in general not the same as I will give an example next.

This is a famous example due to Ito, and it can be found on the stochastic integration book by Protter, on the second section of the last chapter. Consider brownian motion $W$ with its respective filtration $(\mathscr{F}_t)_{t\geq 0}$. Now for every $t>0$, consider the filtration $\mathcal{G}_t$ that is the minimum filtration such that $\mathcal{F}_t \subseteq \mathcal{H}_t$ and $W_1$ is $\mathscr{H}_t$-measurable. It is easy to see that $(\mathscr{H}_t)_{t \geq 0 }$ is a filtration different from $( \mathscr{F}_t )_{ t \geq 0 }$ (since $W$ is not a martingale with respect to $\mathscr{H}$ ). What Ito showed was that the process $$\beta_t = W_t - \int_0^t \frac{W_1 - W_s}{1-s}ds, t \in [0,1]$$ is a martingale with respect to the $\mathscr{H}$ filtration, and in fact (due to Levy's theorem) a Brownian Motion. So there you have two completely different filtrations for processes related by a simple drift addition.

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 different.

• you beat me to it and Exercise 5.5 shows exactly that. – Matt Sep 12 '14 at 6:07