0
$\begingroup$

Let d > 1 and let $W_t$ denote a standard d-dimensional Brownian motion starting at $ x \neq 0$. Let $M_t = log|W_t|$ if d = 2 and $ M_t =| W_t|^{2-d}$ if d > 2. Show that $M_t$ is a martingale.

So I have been looking at this question for the a good hour or two because of my inexperience with stochastic integrals and here is my attempted solution:

We have that $W_t$ is a standard Brownian Motion excluding the fact that $W_0 \neq 0$

$$ M_t = \left\{ \begin{array}{ll} log|W_t|& \quad d =2 \\ |W_t|^{2-d} & \quad d \geq 2 \end{array} \right. $$

So I first consider the case d=2 :

$ f(W_t) = log(W_t) $

$f'(W_t) = |W_t|^{-1} $

$f''(W_t) = -|W_t|^{-2}$

So I use Ito's formula

$ f(W_t) = f(W_0) + \int_0^t\frac{1}{|W_s|}dW_s + \int_0^t\frac{1}{2|W_s|^2}ds$

$ f(W_t) = log|W_t| + \int_0^t\frac{1}{2|W_s|^2}ds$

Now I gather that I must show the expectation of the rightmost term is 0 but dont know how to handle the $ W_t$ with the $dt$

Similarly for $ d \geq 2$ :

$ f(W_t) = |W_t|^{2-d} $

$f'(W_t) = (2-d)|W_t|^{1-d} $

$f''(W_t) = -(2-d)|(1-d)|W_t|^{-d}$

Likewise, using Ito's formula

$ f(W_t) = f(W_0) + \int_0^t (2-d)|W_s|^{1-d}dW_s + \frac{1}{2}(2-d)(1-d)\int_0^t |W_s|^{-d}ds$

$ f(W_t) = |W_t|^{2-d} + \frac{1}{2}(2-d)(1-d)\int_0^t |W_s|^{-d}ds$

Similarly I want to show the expectation of the rightmost term is 0 but I don't know how to arrive to such a conclusion

$\endgroup$
  • $\begingroup$ As far as I remember, the processes are only local martingales and not true martingales, take e.g. a look at the book by Revuz & Yor. $\endgroup$ – saz Apr 8 at 6:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.