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 $(\mathcal{F}_t^{W^*})$
Best Answer
As mentioned in the question, $W^Q_t = W^P_t - \int_0^t{\gamma(t)} ds$ is $Q$-Brownian motion where $Q$ is equivalent to P. Let $\Gamma(t) = \int_0^t{\gamma(t)} ds$, then $\cal F^{W^Q}_t$ will be identical to $\cal F^{W^P}_t$ only if $\Gamma(t)$ is a $\cal F^{W^P}_t$- measurable process or in other words $\Gamma(t)$ is an adapted process with respect to $\cal F^{W^P}_t$.
Under this condition, clearly $W^P_t-\Gamma(t)$ will be adapted with respect to $\cal F^{W^P}_t$ and this filtration is the smallest $\sigma$-algebra with this property. Thus it will be identical to $\cal F^{W^Q}_t$.
As an interpretation linked with mathematical finance, $\cal F^{W^P}_t$ corresponds to public information present in the capital markets till time $t$ and if we use a stochastic process($\gamma(t)$) which relies on private information, we cannot perform the change of measure using Girsanov theorem and calculated Arbitrage Free Prices will not be correct/consistent.