Stochastic differential equation for $|x_t|$ if $(x_t)$ solves $dx_t=b(x_t)dt+dw_t$ Let ${\mathbf{b}}:\mathbb{R}^n\to \mathbb{R}^n$ be a smooth vector field.  Let ${\mathbf{x}}(t)\in \mathbb{R}^n$ satisfy
$$
d \mathbf{x}(t) = \mathbf{b}(\mathbf{x}(t))dt + d\mathbf{W}_t
$$
where $\mathbf{W}_t\in \mathbb{R}^n$ is an $n$-dimensional Brownian motion.  What is the SDE for $x(t):= |\mathbf{x}(t)|$? Since $|\cdot|$ is not smooth at the origin, must this be modified with a local-time term?
 A: In dimension $n=1$, you certainly have to worry about hitting the origin, and so you need to use Tanaka's formula rather than Ito's, which indeed gives you a local time term: $$|x(t)|=|x(0)|+\int_0^t\text{sgn}(x(s))dW_s+\int_0^t\text{sgn}(x(s))b(x(s))ds+\frac{1}{2}L^0_t(x)$$
In dimensions $n \ge 2$, we will show that the process $\bf x$ almost surely never hits the origin after time $0$, and therefore you don't need any modification and you may apply Ito's formula as usual. To prove this, we will show that the law of $(\mathbf x(t))_{t \in [0,T]}$ is absolutely continuous with respect to the law of $(\mathbf W(t))_{t \in [0,T]}$, for any fixed time horizon $T \geq 0$. This would imply that $\bf x$ never hits the origin, because it is a well-known fact that $\bf W$ never hits the origin in dimensions $d \ge 2$.
Let $(\mathbf W(t))_{t \in [0,T]}$ be a Brownian motion started from $\mathbf x(0)$. We define a local martingale $$M_t:= \int_0^t \mathbf b(\mathbf W(s)) \cdot d \mathbf W(s),\;\;\;\;\;\; 0 \leq t \leq T$$ where the $\cdot$ denotes the usual dot product. We will assume that $\bf b$ is bounded and Lipchitz, so that $M$ and its exponential $\mathcal E(M)$ are actually martingales, and $\bf x$ actually exists for all time.
Let $\nu$ and $\mu$ denote the laws of $(\mathbf x(t))_{t \in [0,T]}$ and $(\mathbf W(t))_{t \in [0,T]}$, respectively. These are probability measures considered on the space $C([0,T], \Bbb R^n)$, equipped with its Borel $\sigma$-algebra. 
I claim that there is an explicit formula for the Radon-Nikodym derivative, namely \begin{equation}\frac{d\nu}{d\mu} (\mathbf W(t))_{t \in [0,T]} = e^{M_T-\frac{1}{2}\langle M \rangle_T} \tag{1}\end{equation} which makes sense because $M_t$ is an adapted process w.r.t the natural filtration of $\bf W$ (i.e., $M$ is a deterministic functional of $\bf W$). 
Proof of Formula (1): This is basically just a use of Girsanov's theorem. Namely, define $Q$ to be the probability measure on $\mathcal F_T$ given by $$Q(A)=E_P[1_A \cdot e^{M_T-\frac{1}{2}\langle M \rangle_T}].$$ Write $\mathbf W=(W^1,W^2,...,W^n)$ and similarly for $\bf b$. By Girsanov's theorem, for every $1 \leq i \leq n$ the process $$W^i_t-\langle W^i,M \rangle_t = W^i_t - \int_0^t b^i(\mathbf W(s))ds =: B^i_t $$ is a $Q$-local martingale. Moreover, it is easily checked that $$\langle B^i,B^j \rangle_t=\delta_{ij}t.$$ So by Levy's characterization theorem, we conclude that $\mathbf B=(B^1,...,B^n)$ is actually a Brownian motion (started from $\mathbf x(0)$ since $\mathbf W$ was also started from $\mathbf x(0)$), with respect to $Q$. Furthermore $W$ solves the SDE $$d\mathbf W_t = \mathbf b( \mathbf W(t))dt+ d \mathbf B_t,$$ which is the same form of the SDE which $\mathbf x$ solves. So by uniqueness in law we actually conclude that $\mathbf W$ is distributed the same under $Q$ as $\mathbf x$ is under $P$, and thus for any bounded measurable $f:C[0,1] \to \Bbb R$ we have $$E_P[f(\mathbf x)] = E_Q[f(\mathbf W)] = E_P [f(\mathbf W) e^{M_T-\frac{1}{2} \langle M \rangle_T}]$$ which proves equation $(1)$.
