Stochastic differential of Bessel process Let $ \underline{B}_{t}=(B_1(t), \dots, B_d(t))$ be a $d$-dimensional Brownian motion.
How to calculate the stochastic differential of $ \Vert{\underline{B}_t}\Vert$?
$\Vert . \Vert$ denotes the $d$-dimensional Euclidean norm, i.e. $\Vert \underline{X} \Vert=\sqrt{x_1^2+ \dots + x_d^2}$.
$ \Vert{\underline{B}_t}\Vert$ is also called as Bessel-process.
 A: Because each $B_t^{(i)}$ is independent of the others, the only quadratic variation terms will be $d\left[B^{(i)},B^{(i)}\right]_t$. To avoid confusion in notation, I will say $B_t$ is an $n$-dimensional BM. Applying Ito's lemma to $\Vert B_t\Vert$,
$$
\begin{align*}
\frac{\partial}{\partial B_t^{(i)}}\left\Vert B_t\right\Vert &=\frac{B_t^{(i)}}{\left\Vert B_t\right\Vert} \\
\frac{\partial^2}{\partial \left(B_t^{(i)}\right)^2}\left\Vert B_t\right\Vert &= \frac{\Vert B_t\Vert - \frac{\left(B_t^{(i)}\right)^2}{\Vert B_t\Vert}}{\Vert B_t\Vert^2} \\
&= \frac{\Vert B_t\Vert^2 - \left(B_t^{(i)}\right)^2}{\Vert B_t\Vert^3} 
\end{align*}
$$
$$
\begin{align*}
d\Vert B_t\Vert &= \frac{1}{\Vert B_t\Vert}\left(\sum_{i=1}^nB_t^{(i)}\,dB_t^{(i)}\right)+\frac{1}{2\Vert B_t\Vert^3}\left(n\Vert B_t\Vert^2\, dt- \sum_{i=1}^n \left(B_t^{(i)}\right)^2\, dt\right) \\
&= \frac{1}{\Vert B_t\Vert}\left(\sum_{i=1}^nB_t^{(i)}\,dB_t^{(i)}\right)-\frac{\Vert B_t\Vert^2}{2\Vert B_t\Vert^3}\, dt + \frac{n}{2\Vert B_t \Vert}\, dt \\
&= \frac{1}{\Vert B_t\Vert}\left(\sum_{i=1}^nB_t^{(i)}\,dB_t^{(i)}\right)+\frac{n-1}{2\Vert B_t\Vert}\, dt \\
&= \frac{1}{\Vert B_t\Vert}\left(\frac{1}{2}(n-1)\, dt + \sum_{i=1}^nB_t^{(i)}\,dB_t^{(i)}\right).
\end{align*}
$$
