Edit years later: No idea why I'm upvoted. I actually am not sure how I'm correct. But maybe I haven't forgotten conditional expectation as much as I thought I have.
We are given a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F}_t\}_{t \in [0,T]}, \mathbb{P})$, where $\{\mathscr{F}_t\}_{t \in [0,T]}$ is the filtration generated by standard $\mathbb P$-Brownian motion.
Let $dX_t = \theta_tdt +dW_t$ be an Ito process where $(\theta_t)_{t \in [0,T]}$ is $\mathscr{F}_t$-adapated and $E[\int_0^T \theta_s^2 ds] < \infty$ and
$$Y_t := X_tL_t, \ \ L_t = \exp Z_t, \ \ Z_t = -\int_0^t \theta_s dW_s - \frac{1}{2}\int_0^t\theta_s^2ds$$
It can be shown that $\{Y_t\}$ is a $(\mathscr{F}_t, \mathbb{P})$-martingale.
If $\frac{d \mathbb Q}{d \mathbb P} = L_T$, prove $E_{\mathbb Q}[X_t | \mathscr F_u] = X_u$, i.e. $\{X_t\}$ is a $(\mathscr{F}_t, \mathbb{Q})$-martingale.
What I tried:
Novikov's condition holds. Does this part use $\frac{d \mathbb Q}{d \mathbb P} = L_T$? If not, then where is the assumption used?
By Novikov's $L_t$ is a $(\mathscr{F}_t, \mathbb{P})$-martingale. Then we have that
$$E[X_tL_t | \mathscr F_u] = X_uE[L_t | \mathscr F_u]$$
$$ \to E[(X_t - X_u) L_t | \mathscr F_u] = 0$$
$$ \to E_{\mathbb Q}[(X_t - X_u) \frac{L_t}{L_T} | \mathscr F_u] = 0$$
$$ \to E_{\mathbb Q}[(X_t - X_u) \frac{L_t}{L_T} | \mathscr F_u] = 0$$
$$ \to E_{\mathbb Q}[(X_t - X_u) \exp(-Z_T + Z_t) | \mathscr F_u] = 0$$
Now what? I don't suppose $\exp(-Z_T + Z_t) = 1$...or is it?
Another thing:
$$E[Y_t | \mathscr F_u] = Y_u$$
$$\to E[X_t L_t | \mathscr F_u] = X_u L_u$$
$$\to E_{\mathbb P}[X_t L_t | \mathscr F_u] = X_u L_u$$
$$\to E_{\mathbb Q}[X_t \frac {L_t}{L_T} | \mathscr F_u] = X_u L_u$$
$$\to ? E_{\mathbb Q}[X_t | \mathscr F_u] E[\frac {L_t}{L_T} | \mathscr F_u] = X_u L_u$$
If so, I think we have $E[\frac {L_t}{L_T}| \mathscr F_u] = L_u \times$ some integral that will turn out to be 1 probably by mgf, but I don't think mgf applies as $\theta_t$ is not necessarily deterministic.
What to do?
Something else I tried:
$$E_{\mathbb Q}[X_t | \mathscr F_u] = E_{\mathbb Q}[\frac{Y_t}{L_t} | \mathscr F_u] $$
$$= E_{\mathbb P}[\frac{Y_t}{L_t L_T} | \mathscr F_u]$$
$$= E[\frac{Y_t}{L_t L_T} | \mathscr F_u]$$
$$= E[\frac{Y_t}{\exp Z_t \exp Z_T} | \mathscr F_u]$$
$$= \frac{1}{L_u^2} E[Y_t\exp (-Z_T+Z_t) | \mathscr F_u]$$
It looks like $\exp (-Z_T+Z_t)$ is independent of $\mathscr F_u$, but I don't think
$$E[Y_t\exp (-Z_T+Z_t) | \mathscr F_u] = E[Y_t| \mathscr F_u] E[\exp (-Z_T+Z_t) | \mathscr F_u]$$
Or is it? If so, why? If not, what to do?
