$\langle X\rangle_t = t?$

Question

Suppose $B_t$ is a standard Brownian motion in $\mathbb{R}^d$ and $X_t = |B_t|$. Do we have that$$\langle X\rangle_t = t?$$

Answer 1

Since $|X_t|^2=\sum_{i=1}^d (B_t^{(i)})^2$ where $B_t^{(i)}$ are a family of iid Brownian motions, and since $(B_t^{(i)})^2-t$ is a martingale, it follows also that $|X_t|^2-td$ is also a martingale. Thus $\langle X_t\rangle=td$.