# Moment bounds on exponential martingale

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 $X_s?$

• Search for "L^p bound" instead of moment bound? – Henry.L Apr 19 '17 at 20:22

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 $q= 1 + \sqrt{1-\frac{1}{p}}$.