Local martingale in change of measure

Question

$h$ is a bounded function, and $X=\{X_t;t\in [0,T]\}$ is defined by $dX_t=f(X_t)dt+\sigma(X_t) dV_t$ we defined $Z$ by

$$ Z_t = \exp\left(-\int_0^th(X_s)dV_s -\frac{1}{2}\int_0^th(X_s)^2 ds \right) $$

$V$ and $W$ Brownian motions. I can show that $Z_t=1-\int_0^th(X_s)Z_sdW_s$, and I have also showed that $\mathbb{E}[\int_0^t (h(X_s)Z_s)^2ds]<+\infty$. Seems to me that that shows that $\int_0^th(X_s)Z_sdW_s$ is a martingale, but it says in my notes that this shows that $Z_t$ is a local martingale. How has this been shown?

Answer 1

Note that if $M_t$ is a continuous local martingale, and $h_t$ an $\mathbf{F}^M$-predictable function, the stochastic exponential defined by $$ \mathcal{E}(hM)_t:=\exp\left(\int_0^t h_tdM_t - \frac{1}{2}\int_0^th_t^2d\langle M\rangle_t\right) $$ is a positive local martingale, (and it is a martingale iff $\mathbb{E}[\mathcal{E}(hM)_t]=1\;\forall t$).

However there are some different conditions that can ensure that the stochastic exponential is actually a true martingale. In a general setting where $h$ is a bounded function and $W_t$ is a Brownian Motion, you have the following:

Let $\zeta_t=\int_0^t h(s)dW_s$, where $h$ satisfies $$ \sup_{s\le t}\,|h(s)|\le C. $$ Then the process $Z_t=\exp(\zeta_t -\frac{1}{2}\langle\zeta\rangle_t)$ is a martingale.

Check eg. Revuz & Yor (1999).