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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 ?

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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 ?

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See here: math.stackexchange.com/questions/406054/… 1 is a dominating function for $M_t^\lambda$ –  Chris Janjigian May 30 '13 at 1:09

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