TITLE: Moment bounds on exponential martingale
QUESTION [6 upvotes]: 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 $X_s?$

REPLY [3 votes]: 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 $q= 1 + \sqrt{1-\frac{1}{p}}$.