In class, we talked about Bessel process as a process which solves the SDE: $$ dB=\frac{n-1}{2B}dt+\sum_{i=1}^n\frac{W_i}{B}dW_i $$ Where $W_1,W_2,...,W_n$ are independent, standard Brownian motions. We then showed that $B=||W||=(\sum_{i=1}^n(W_i)^2)^{1/2}$ solves this equation. As a side note, the professor said that $X_t=\intop_0^t\sum_{i=1}^n\frac{W_i}{B}dW_i$ is a BM and that we can show it using Levy’s theorem.
I tried to prove that myself but got stuck at showing that $X_t^2-t$ is a martingale. Wherever I looked (books, lecture notes, or other questions here) it was like "you can easily see that $X_t$ is BM by Levy’s theorem". Well, turned out it's not that easy for me.
Can someone here please help me see that "easy" proof? Thank you!
I'll show $X_t^2-t$ is a martingale. Note $X_t$ satisfies $dX_t=\sum_{i=1}^n\tfrac{W_i(t)}{B(t)}dW_i(t)$. Since the $W_i$ are independent $(dW_i)(dW_j)=\delta_{ij}dt$ where $\delta_{ij}$ is 1 if $i=j$ and 0 otherwise. It follows that \begin{align*} (dX_t)^2 = \sum_{i=1}^n \tfrac{W_i(t)^2}{B(t)^2}dt \end{align*} or, equivalently, \begin{align*} X_t^2 = \sum_{i=1}^n\int_0^t \tfrac{W_i(s)^2}{B(s)^2}ds = \int_0^t \sum_{i=1}^n\tfrac{W_i(s)^2}{B(s)^2}ds = t. \end{align*} The latter holds since $\sum_{i=1}^n\tfrac{W_i(s)^2}{B(s)^2}=1$ for all $s>0$ almost surely. Thus $X_t^2-t=0$ for all $t$ and so \begin{align*} \mathbb{E}(X_t^2-t\vert\mathcal{F}_s) = 0 = X_s^2-s \end{align*} for any $s<t$. Here $\{\mathcal{F}_t\}$ is the filtration generated by the Brownian motions. Since $X_t^2-t$ is also integrable for all $t$ (it is $0$) and adapted to $\{\mathcal{F}_t\}$, it is a martingale.