6
$\begingroup$

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 $X_s?$

$\endgroup$
  • $\begingroup$ Search for "L^p bound" instead of moment bound? $\endgroup$ – Henry.L Apr 19 '17 at 20:22
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, Dan, this is a nice reference. I checked it out of the library today. $\endgroup$ – user253775 Apr 27 '17 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.