Be advised that I cross-posted this question on MathOverflow. You can find it in this link:


Assume we have the following stochastic process:

$$X_t=\int_0^t e^{B(s)^2}dB(s)\, ,0\leq t \leq 1$$

where $(B)_{t\geq 0}$ is a Brownian Motion.

I have to show that $X_t$ is not a martingale.

I know that if $t< \frac 1 4$ then $\int_0^t \mathbb E(e^{2B(s)^2})ds < \infty $ and then the process is a martingale, this makes me think that $X_t$ is actually a local martingale, but I don't see how to prove that it's not a proper martingale.

Thanks in advace.


After some calculation I have reduced the problem to show that

$$\mathbb E^{\mathcal F_u}\bigg(\int_{B(u)}^{B(t)}e^{s^2} ds-\int_u^t B(s)e^{B(s)^2}ds\bigg)\neq0$$

Still, I haven't been able to go on from here, maybe someone could find this illuminating.



You must log in to answer this question.

Browse other questions tagged .