Martingale property of Brownian motion with resprect to a different filtration Let $W$ be a Brownian motion on $(\Omega,\mathcal F,\mathbb P)$ and let $N$ be a Poisson process on the same probability space. Denote by $\mathbb F$ the filtration that is generated by $(W,N)$. Now there is an exercise that says: 

Show that the stochastic exponential $\mathcal E(W)$ is a $\mathbb P$-martingale. 

It does not say with respect to which filtration, so I assume that it is $\mathbb F$. My question is: Do I have to assume that $W$ and $N$ are independent for the statement from the exercise to be true? In fact, the actual question is: Is $W$ itself an $\mathbb F$-martingale without the assumption that $W$ and $N$ are independent?
 A: I have a feeling that there's a much simpler reasoning for this, but nevertheless, here's a counterexample.
Let $F(x)=1-e^{-x}$ and $G=F^{-1}$. Let $W_t$ be a Brownian motion starting at zero on some probability space $(\Omega_1,\mathcal{F}_1,P_1)$. For  $T=G(1/2)*2$ denote $A=\{W_T-W_{T/2}>0\}\subset\mathcal{F}_1$. Define the random variable $\eta=I_A$, then it has the 1/2-Bernoulli distribution.
Next, take a Poisson process $N$ with rate $1$ on a probability space $(\Omega_2,\mathcal{F}_2,P_2)$. Since it is determined by the independent identically distributed differences $\{\tau_k\}$ between its jump times, we can construct it on the product space $\Omega_2=\prod\Omega_{2j}$, such that $\tau_k$ are extensions of the random variables on $\Omega_{2j}$ with the corresponding probability space structures. 
Take a uniformly distributed random variable $\xi$ on $[0,1]$, then $G(\xi)$ is exponentially distributed as $\tau_1$. Consider the binary expansion $\xi=\eta_1\eta_2\ldots$. Then $\eta_k$ are i.i.d and $\xi\buildrel d \over =\eta\eta_2\eta_3\ldots=\tilde{\xi}$, if we consider the product space $\tilde{\Omega}_1=[0,1]\times\Omega_1$ with the natural probability space structure. 
Now, substituting $G(\tilde{\xi})$ instead of $\tau_1$, we obtain a Poisson process $\tilde{N}$ on the probability space $\Omega=\Omega_1\times[0,1]\times\prod\limits_{j\ge 2}\Omega_{2j}$. 
Note that 
$$A=(\eta=1)=(\tilde{\xi}\ge 1/2)=(G(\tilde{\xi})\ge G(1/2))\in\sigma(\tilde{N}_{s},s\le T/2).$$
(Random variables and sets encountered before are naturally extended to the space $\Omega$ and denoted by the same letters).
Then we get the following contradiction with the martingale property, if we try try to consider $W$ as a $(W,\tilde{N})$-martingale:
$$
E(W_T-W_{T/2})I_A>0.
$$
