I am doing an exercise regarding the Girsanov theorem, $B_t$ is standard brownian motion. I already find that $$M_t=\exp(Y_t)\quad Y_t=\int_{0}^{t}-2mB_sdB_s-\frac{1}{2} \int_{0}^{t}4m^2B_s^2ds $$ $dM_t=(-2mB_t)M_tdB_t$, $M_t$ is already a local martingale. I need to further show that $M_t$ is actually a martingale, there are two ways(based on my knowledge) to show martingale. If we can show any of following:
- $\mathbb{E}\left[M_{t}\right]=1$
- $\mathbb{E}\left[\exp \left\{\frac{\langle Y\rangle_{t}}{2}\right\}\right]<\infty$
I want to try second way, we need to show $$\mathbb{E}\left[\exp \left\{\frac{1}{2}\int_{0}^{t}4m^2B_s^2ds\right\}\right]<\infty$$
But I don't know how to proceed, can anybody help?
I'd do it using Kazamaki's criterion.
We want to show that for any $t>0$, $\mathbb{E}[e^{\frac{1}{2}\int_0^t-2mB_sdB_s}]<+\infty$.
This is equivalent to $\mathbb{E}[e^{-m\int_0^tB_sdB_s}] < +\infty$.
It should be easy to conclude for $m>0$ using that $\int_0^t B_sdB_s = \frac{B_t^2-t}{2}$.