# Generated sigma algebra from Brownian Motion

Suppose that we have a Brownian motion and we define the P-augmented filtration by

$$\mathcal{F}^W_t:=\sigma(\mathcal{F}^0_t \cup \mathcal{N})$$ where $\mathcal{F}_t^0:=\sigma(W_s;s\le t)$ and $\mathcal{N}:=\{B\subset \Omega|\exists A\in\mathcal{F}^0_\infty \mbox{ with }B\subset A \mbox{ and } P(A)=0\}$

Now suppose we have an equivalent measure of Girsanov type, i.e. the density has the form $$\frac{dQ}{dP}=\mathcal{E}(\gamma \cdot W)$$

where $\mathcal{E}$ is the stochastic exponential. Moreover to be an equivalent measure we assume that $\mathcal{E}(\gamma\cdot W)_\infty >0$ and it is an uniformly integrable martingale. Hence we know, $W^*=W-\int\gamma ds$ is a $Q$-Brownian Motion. If we now define the Q-augmented filtration of $W^*$, is this the same as the P-augmented filtration for $W$, i.e.

$$\mathcal{F}^W_t=\mathcal{F}^{W^*}_t$$

If $\gamma$ would be deterministic, I would intuitively say yes, but $\gamma$ is usually assumed to be an predictable process, so I'm not sure about it.

As motivation:

This question shows up in a script about mathematical finance: Let $(\Omega,\mathcal{F},\{\mathcal{F}_t^W\},P)$ be a filtered prob. space. Suppose we have equivalent martingale measure for all the risky asset $S^1,\dots,S^n$, i.e. $Q$ is equivalent to $P$ and under $Q$, every $S^i$ is a martingale. We had the following martingale representation theorem:

let $(\Omega,\mathcal{F},(\mathcal{F}_t^W),P)$ as above, then every local $P$ martingale $M$ has a continuous version and there exists a pred. process $b$(nice enough) such that $$M(t)=M(0)+\int_0^t b(s)dW(s)$$

Am I right about the following: Saying that $Q$ is an EMM, means $S^i$ is a martingale with respect to $Q$ and the filtration given in the setup, here $(\mathcal{F}_t^W)$?

In the script they say, we can write every $S^i$ as:

$$dS^i_t=b(t)dW^*(t)$$

which comes from the martingale representation. So the thing I do not see is, we have a $Q$ martingale, with respect to the filtration $(\mathcal{F}_t^W)$ but we want to use the martingale rep. with $W^*$. Hence we must either know that $S^i$ is a $(\mathcal{F}_t^{W^*})$ martingale (w.r.t to $Q$), or the two filtrations should be equal. If $\mathcal{F}_t^W=\mathcal{F}_t^{W^*}$ then trivially $S^i$ is also a martingale w.r.t to $Q$ and $(\mathcal{F}_t^{W^*})$

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Note that even when $\gamma$ is constant, $Q$ is not equivalent to $P$. –  Ilya Jan 4 '13 at 10:38
@Ilya You are right. I also assumed that $\mathcal{E}(\gamma \cdot W)_\infty$>0 and it is a uniformly integrable martingale. This are sufficient conditions that we have an equivalent martingale measure. I will add this, in the question. –  user20869 Jan 4 '13 at 10:41
One needs to check whether $\mathscr F^{0,W}_t = \mathscr F^{0,W^*}_t$ for all $t$. Then by equivalence of measures you will have that $\mathscr N$ is the same in both cases, which yields the desired result. It clearly holds when $\gamma$ does no depend on $\omega$ (i.e. deterministic) as in such case it acts like a constant shift on the space of trajectories. Let me take a look on the general case. –  Ilya Jan 4 '13 at 12:35
@Ilya That was also my argument for a deterministic $\omega$. I will add the motivation, why this question shows up. maybe it is not necessary that $\mathcal{F}_t^{0,W}=\mathcal{F}_t^{0,W^{*}}$ –  user20869 Jan 4 '13 at 12:49
@hulik One approach: Try to show that the set of all predictable processes $\gamma$ for which this property holds is a monotone class and from there use the monotone class theorem. –  Ben Derrett Jan 4 '13 at 23:58