Consider the exponential martingale used in the Girsanov transformation of
measure:
$$Z(t) = \exp\Big(\int\_0^tXdW - \frac{1}{2}\int\_0^t|X|^2ds\Big)$$
so that $Z$ solves the sde $dZ = ZXdW$ where $W$ is a one dimensional 
Brownian motion. Under certain conditions (e.g. Novikov) on $X,$ $Z$ is a martingale. Many things are known as well, like 
$$W\_t - \int\_0^t X\_sds$$
is a Brownian motion under $dQ/dP = Z(t)$ where $P$ is the original 
measure attached to $W.$
I'm interested in moments of $Z$ given in terms of moments of $X.$
Using the sde above, we can see that
$$Z\_t^2 = 1 + 2\int\_0^tZ\_sdZ\_s + \int\_0^tZ\_s^2X\_s^2ds$$
which shows
$$E Z\_t^2 = 1 + \int\_0^t E(Z\_s^2X\_s^2)ds$$
where I was hoping to apply a Gronwall inequality to get a bound on
$E Z\_t^2.$ It seems we are unable to do this unless we know $X\_s$ is bounded and can apply an $L\_1, L\_\infty$ bound.


Does anyone have any reference or knowledge on moment bounds of this exponential martingale in terms of moment bounds of $X\_s?$



Moment bounds on exponential martingale
There are a number of ways to bound moments of $Z$ in terms of exponential moments of $X$. For some sharp results, see Theorem 1.5 of Kazamaki's book, "Continuous exponential martingales and BMO," as well as Remark 1.2 thereafter (page 8).


For a more pedestrian bound we need nothing more than Holder's inequality and the fact that a stochastic exponential always has expectation at most $1$. Let $p,q,q^\* > 1$ with $1/q + 1/q^\* =1$. Then
\begin{align\*}
\mathbb{E}\left[Z\_t^p\right] &= \mathbb{E}\left[\exp\left(p\int\_0^tX\_sdW\_s - \frac{qp^2}{2}\int\_0^t|X\_s|^2ds\right)\exp\left(\frac{p(qp-1)}{2}\int\_0^t|X\_s|^2ds\right)\right] \\
 &\le \mathbb{E}\left[\exp\left(\frac{pq^\*(qp-1)}{2}\int\_0^t|X\_s|^2ds\right)\right]^{1/q^\*}.
\end{align\*}
A simple exercise shows that $q^\*(qp-1) = \frac{q(qp-1)}{q-1}$ is minimized by $q= 1 + \sqrt{1-\frac{1}{p}}$.


