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
 A: 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.
