I'm reading the book Bessel Processes, Schramm-Loewner Evolution, and the Dyson model of Makoto Katori. In chapter 1, the $D$-dimensional Bessel process $R_t^x$ is introduced and is defined as follows $$R_t=\|B_t^x\|_2$$ where $B_t^x$ is a $D$-dimensional Brownian motion starting at $x\in\mathbb R^D\setminus\{ 0\}$. One can prove using Ito's formula that $$R_t^x=\|x\|_2+\int^t_0\frac{1}{R_s^x}\sum_{i=1}^D B^{x_i}_s\,dB_s^{x_i}+\frac{D-1}{2}\int^t_0\frac{1}{R_s^x}\,ds$$ where $B_t^{x_i}$ is the $i$-th component of $B^x_t$. Also note that Ito's formula only holds until the first hitting time of zero, but for $D\geq 2$ it will never hit zero with probability $1$. Now one can easily argue using Levy's characterization of Brownian motion that the second term of the expression above is a $1$-dimensiona standard Brownian motion, that is $$\left(\int^t_0\frac{1}{R_s^x}\sum_{i=1}^D B^{x_i}_s\,dB_s^{x_i}\right)_{t\geq 0}$$ is a standard Brownian motion.
This is all basic, but now the confusion kicks in. Based on the observations above, the book just says that $R_t^x$ solves the SDE $$\tag{1}dR_t^x=dW_t+\frac{D-1}{2R_t^x}\,dt$$ where $W_t$ is a standard BM different than the $B_t^{x_i}$ for $i=1,...,D$. I be like "What? Wait a minute? Didn't we lose properties about the very definition of $R_t^x$ by saying that?". Let me formulate this. Of course we have this filtered probability space $(\Omega,\mathcal F,\mathcal F_t,\mathbb P)$ in the background. The problem is
Problem. Does the SDE in (1) with initial value $R_0^x=\|x\|_2$ has a strong unique solution in the sense that the solution of the SDE, say $R_t^x$ can be written as $ R_t^x=\|B_t^x\|_2$ for some $D$-dimensional BM $B_t^x$?
So you guys hopefully see that I'm only asking whether the definition of Bessel process and the SDE that it satisfies is consistent. I know that the solution of (1) is unique for $D\geq 2$ through uniqueness results from the book of Revuz-Yor for instance. But how do I make sure that the BM $W_t$ can be decomposed in terms of $B_t^{x_i}$...?
Many thanks! Please let me know if my question is unclear.
Suppose $B$ is a $d$-dimensional Brownain motion ($d\ge 2$). Define the radial part $R$ as you have, then $$ A_t:=\int_0^t R^{-2}_s\,ds,\qquad t\ge 0, $$ and finally the (time-changed) angular part $$ \Theta_u:={B_{\tau(u)}\over R_{\tau(u)}},\qquad u\ge 0, $$ where $$ \tau(u):=\inf\{t>0: A_t>u\}. $$ Then $R$ is the unique strong soution of (1), with $W$ defined as in the display just above (1), and $\Theta$ is a spherical Brownian motion independent of $R$. (The nature of $\Theta$ can be intuited from the polar-coordinates decomposition of the $d$-dimensioanl Laplacian.) This independence means that the full $d$-dimensional Brownian motion $B$ cannot be recovered from the Bessel process $R$ (though of course the 1-dimensional Brownian motion $W$ can).
Going the other way, starting with a 1-dimensional Brownian motion $W$ and an independent spherical Brownian motion $\Theta$, one can solve (1) to get $R$, and then invert the above construction using $R$ and $\Theta$ to build a $d$-dimensional Brownian motion $$ R_t\cdot\Theta_{A_t},\qquad t\ge 0. $$ For a nice discussion of these things see the first chapter of the first volume of the ** of **Rogers and Williams.