Take the tour ×  Consider the following $\mathcal F_t$-
(continouous) local martingale $$M_t = \int_0 ^t
\exp{((B_2(s)^2)} dB_1(s)$$ where $\left(B_t\right)_{t\geq0}
=\left(B_1(t),B_2(t)\right)_{t\geq0}$ is $\mathcal F_t$-Brownian
motion in $\mathbb R^2$, null at $t=0$.  How to show that
$\left(M_t \right)_{t\geq 0}\notin \mathcal{M }^2_c=\{\mathcal
{F}_t - $real continuous martingale square integrable, with $
M_0 =0 \}$ ?  share|improve this question Do you intend to show
that it is not continuous or that it is not square-integrable?
Or that it is not a martingale? –  GEdgar Jan 3 at 22:08 By
definition $M$ is a (continuous) local martingale (i.e, $M \in
\mathcal M_{c,loc}$) as you can see, so the problem is
primordially the square-integrability because we know it's
continous. Nonetheless, there is also an acessorie problem
concerning localization. –  Paul Jan 3 at 22:20 Thaks for your
remark GEdgar. I edited the text to make it clear. I expect I
was successefull. –  Paul Jan 3 at 22:24 add comment  1 Answer