Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|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 '13 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 '13 at 22:20
Thaks for your remark GEdgar. I edited the text to make it clear. I expect I was successefull. – Paul Jan 3 '13 at 22:24
up vote 1 down vote accepted

Let $\mathcal G_t=\sigma(B_2(s),s\leqslant t)$. Since $(B_1(s))_{s\leqslant t}$ is independent of $\mathcal G_t$ and $(B_2(s))_{s\leqslant t}$ is measurable with respect to $\mathcal G_t$, Itô's isometry indicates that $$ \mathbb E(M_t^2\mid\mathcal G_t)=\int_0^t\mathrm e^{2B_2(s)^2}\,\mathrm ds. $$ Everything is nonnegative, hence $$ \mathbb E(M_t^2)=\int_0^t\mathbb E(\mathrm e^{2B_2(s)^2})\,\mathrm ds. $$ Since $B(s)$ is centered normal with variance $s$, $\mathbb E(\mathrm e^{2B_2(s)^2})$ is infinite for every $s\geqslant\frac14$. Thus, $\mathbb E(M_t^2)$ is infinite for every $t\gt\frac14$ and in particular, $(M_t)_{t\geqslant0}$ is not a square integrable martingale.

(Note that $\mathbb E(M_t^2)=\frac12(1-\sqrt{1-4t})$ for every $t\leqslant\frac14$, hence $(M_t)_{t\leqslant1/4}$ is a square integrable martingale.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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