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?


2 Answers 2


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.

  • $\begingroup$ In Corollary 5.3.2, $\widetilde W-W$ is a drift term that is an integral of a process adapted to $\mathscr F$, unlike in your interesting example. Isn't the extra assumption that the drift term is an adapted process enough to say that the filtrations are the same? $\endgroup$
    – Kirill
    Commented Sep 13, 2014 at 5:04
  • $\begingroup$ I don't think so, take a look at my example. $d ( \beta - W ) = \theta dt$, where $\theta$ is $\mathscr{H}$-adapted, and still the $\sigma$-algebra generated by these two processes are different (there is a contention, but they are different ) $\endgroup$ Commented Sep 13, 2014 at 18:55
  • 2
    $\begingroup$ I don't think I quite understand: if $\theta$ is $\mathscr F$-adapted, then aren't the filtrations generated by $\beta$ and $W$ the same? $\endgroup$
    – Kirill
    Commented Sep 13, 2014 at 20:25
  • $\begingroup$ good point Kirill ! After I digest the posts and comments, I'm also lost... $\endgroup$
    – athos
    Commented Sep 17, 2014 at 16:40
  • 3
    $\begingroup$ thats the book ethos. krill, just interchange the role of the filtrations $\mathscr{H}$ and $\mathscr{F}$ and you get the counter example. $\endgroup$ Commented Sep 18, 2014 at 11:11

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.

  • $\begingroup$ I don't get it. The $Z_t$ induces the measure $\widetilde P$, under which naturally $\widetilde W$ is a Brownian motion (though the proof might not be trivial). Then $\widetilde M$ is delcared as Martingale under $\widetilde P$, then naturally exists $\widetilde \Gamma$ such that $d \widetilde M = \widetilde \Gamma d \widetilde W$. Why Shreve emphasizes here the filtration $\mathscr F$? It seems that $\mathscr F$ does not appear anywhere, how could it matter which Brownian motion generates $\mathscr F$? $\endgroup$
    – athos
    Commented Sep 10, 2014 at 9:23
  • $\begingroup$ @athos I think its simply because of the connection $\widetilde{W}_t=W_t+\int\Theta(u)du$, meaning that you cannot in general substitute them. 5.3.1 holds for $W$, but not necessarily $W+\int\Theta(u)du=\widetilde{W}$ aswell. The filtrations generated by $W$ and $W+\int\Theta(u)du$ are obviously different. $\endgroup$
    – emcor
    Commented Sep 10, 2014 at 9:44
  • $\begingroup$ Is it possible to give some examples that in general, without "the filtration generated by this Brownian motion", 5.3.2 part 2 does not hold, while with the "the filtration generated by this Brownian motion" it holds? $\endgroup$
    – athos
    Commented Sep 11, 2014 at 9:41
  • $\begingroup$ @athos If you choose a different change of measure, $\widetilde W$ is no longer Brownian Motion, so 5.3.2 would fail to hold under its filtration. $\endgroup$
    – emcor
    Commented Sep 12, 2014 at 19:36
  • 1
    $\begingroup$ @athos You can see the proof for 5.3.2 here in p.33(Exercise 5.5): ms.mcmaster.ca/~zhaog22/Volume2.pdf $\endgroup$
    – emcor
    Commented Sep 12, 2014 at 19:40

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