Showing log(Wt) is a martingale using Ito's Formula

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

• 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.
– saz
Apr 8, 2019 at 6:48

I do not completely understand the stochastic process nor its martingale condition. It seems very strange to me that you expect: $$\mathbb{E}[f(W_t)] = f(W_0) = \log(0)$$ However I think your initial approach was correct, although the derivation was not. Note that $$|W_t| = \sqrt{w^2_1(t)+w^2_2(t)}.$$ Assume that the Brownian vector has uncorrelated components (i.e. $$dw_1dw_2 = 0$$), hence using Ito's lemma: \begin{align} df = & \frac{\partial f}{\partial w_1} dw_1 + \frac{\partial f}{\partial w_1} dw_1 + \frac{1}{2}\frac{\partial^2 f}{\partial w^2_1} dw_1 + \frac{1}{2}\frac{\partial^2 f}{\partial w^2_2} dw_2 \,, \\ = & \frac{1}{|W_t|}W_t^\intercal dW_t + \frac{1}{2}\left(\frac{1}{|W_t|^2} + w_1\left( \frac{-2w_1}{|W_t|^4} \right) \right)dt + \frac{1}{2}\left(\frac{1}{|W_t|^2} + w_2\left( \frac{-2w_2}{|W_t|^4} \right) \right)dt \,, \\ = & \frac{1}{|W_t|}W_t^\intercal dW_t + \frac{1}{|W_t|^2} dt - \frac{w^2_1+w^2_2}{|W_t|^4} dt\,, \\ = & \frac{1}{|W_t|}W_t^\intercal dW_t \,, \end{align} with $$W_t^\intercal$$ the vector's transpose. This is an SDE with no drift term and hence a local martingale.
You might be able to complete the proof yourself for the remaining $$d$$'s.