Let $B_{t} = (B_{1}(t), \ldots B_{d}(t))$ be a d-dimensional Brownian motion with $d>3$. Fix $0 < r < R < \infty$. Define $$A_{r}(R) = \{x \in \mathbb{R}^{d} : r < |x| < R\},$$ where, $|x|:= \sqrt{x_{1}^2 + \dots + x_{d}^2}$.
a) Write down the SDE for the process of the radial component $X(t) = |B(t)|$ and its generator $\mathcal{L}$.
b) Define the stopping times $\tau_{r} := inf\{t: |B(t)| = r\}, \tau_{R} := inf\{t: |B(t)| = rR\}, \tau := \tau_{r} \land \tau_{r}$. Show that $Y(t) = f(X(t \land \tau))$ is a martingale where, $f(x) := \frac{1}{x^{d-2}}$.
c) Explain why $Z(t) = f(X(t))$ is a local martingale.
$\textbf{My approach:}$
a) I do not have any idea how to start a). I am confused about the radial component part. Do, we just find an SDE that satisfies $X(t) = |B(t)|$?
b) Logically, we know that $B(t)$ will eventually move out of the ball with radius R. We know that $f(X(t \land \tau)$ is a stopped process and BM is a martingale. So, a stopped process is a martingale. Is it okay to argue like this?
c) How do I find a sequence of times to show that this is a local martingale?
Any help will be appreciated.