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 $B=(B_t)_{t\geq 0}$ $\mathcal F_t$ - brownian motion in $\mathbb R ^n, \ (n\geq 2)$ starting at zero, in a probability space $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb P)$. Then, consider $X_t ^x = x + B_t $ and $S_t = \left|X^x_t\right|$. I was trying to show that $(T_t)_{t\geq 0}$ defined by $$T_t := S_t -\left| x \right| -\frac {n-1}{2} \int _0 ^t \frac {1}{S_u}~du$$

is a real $\mathcal F_t$ - brownian motion.

It's not difficult to see that $W_t = S_t -\left| x \right|$ is a real $\mathcal F_t$ - brownian motion.

Inded, Ito's lemma implies that

$$ W_t = \int _0 ^t \frac{X^x_t}{\left| X^x_t \right| }~ dB_s $$ wich is well defined since $\mathbb P (\exists t>0 : S_t =\left| X^x_t \right| =0) =0$. Then by Pauls-Lévy Theorem we have that $(W_t)$ is a real $\mathcal F_t$ - brownian motion.

So, $$T_t = W_t -\frac {n-1}{2} \int _0 ^t \frac {1}{S_u}~du.$$

Have anybody a smarter idea that compute its mean and covariance in order to apply again Paul-Lévy theorem ? If not, any smart advices to simplify the calculation?

Thank's in advance.

Edit: Maybe this other result could helpful to crack this proof.

I could show that $\mathbb E \left\{ \int _0 ^t \frac {1}{S_u}~du\right\} =0$ since $M_t := \int _0 ^t \frac {1}{S_u}~du$ is a bounded local martingale. Then for the covariation we have $$ \mathbb E \left \{T_s T_t \right\}=t \wedge s -c (\mathbb E \left \{W_s M_t \right\} +\mathbb E \left \{M_s W_t \right\}) + \mathbb E \left \{M_s M_t \right\}$$

New Edit: Could someone please check the question at the end of my solution try at the answer ?

share|cite|improve this question
up vote 0 down vote accepted

I'd like to show that $M^\lambda = (M_t^\lambda)_{t\geq0}$ $$ M_t^\lambda := \exp\left(i\lambda T_t-\frac{\lambda^2}{2}t \right) $$ is a complex martingale so $T$ is a brownian motion.

Indeed, by Ito's lemma we have that

$$dM_t ^\lambda = \frac{\lambda^2}{2}M_t ^\lambda dt + i\lambda M_t ^\lambda dT_t - \frac{\lambda^2}{2} d\langle T \rangle_t$$

but also we have that

$$ \langle T \rangle_t = \langle W \rangle_t + \langle \int _0 ^ . \frac{1}{S_u}~du \rangle_t = t$$

so $$dM_t ^\lambda = i\lambda M_t ^\lambda dT_t$$ and $$\mathbb E \left\{ \int _0 ^t \lambda^2 (M_t^ \lambda)^2 d\langle T \rangle_t\right\}\leq \exp(\lambda^2 t)< +\infty$$ However, we must remember $T$ is a local martingale. Then even if $\phi_t := i\lambda M_t ^\lambda \in \mathbb H ^2(T)$ can we conclude that $M^\lambda$ is a martingale ?

share|cite|improve this answer
See here:… 1 is a dominating function for $M_t^\lambda$ – Chris Janjigian May 30 '13 at 1:09

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.