Original Question:
Derive an SDE for $B^2(t)$, where $B(t)$ is standard Brownian Motion.
Attempt at an answer:
Apply Ito's calculus over $f(t,b):= B^2(t)$.
$df(t,b) = \frac{\partial f(t,b)}{\partial t}dt + \frac{\partial f(t,b)}{\partial b}dB^2(t) + \frac12 \frac{\partial^2 f(t,b)}{\partial b^2}d\langle\, B^2\rangle_t$
$ = 2B(t)dB^2(t) + d\langle\, B^2\rangle_t.$
MY Question:
What does $d\langle\, B^2\rangle_t$ equal, why, and how do I arrive at it?