Brownian and Brackets

Question

A continuous martingale with deterministic bracket must be a Brownian motion.

Is this statement ture or not, please? If true, how to show it? If not, what is a counter example?

Answer 1

A continuous martingale is a Brownian motion if and only if its quadratic variation over each interval $[0, t]$ is equal to $t$.

Thus, if $M$ is a continuous martingale and $\langle M\rangle_t=u(t)$ for some deterministic function $u$, then $M$ is a time change of a Brownian motion in the sense that $M_t=B_{u(t)}$ for every $t$, where $B$ is a Brownian motion.

Note that each martingale as above such that $u$ is $C^1$ can be realized as $$M_t=M_0+\int_0^t\sqrt{u'(s)}\,\mathrm d\beta_s,$$ for some (other) Brownian motion $\beta$.

To sum up, if you ask that $M$ is exactly a Brownian motion, $M=2B$ disproves the claim, but if you ask only that $M$ is a Brownian motion with a change of time, there is no counterexample.

Answer 2

Hint: consider the martingales $$ M(t) = \int 1_{[0,t]}(s) a(s) dB(s) $$with $a\in L^2(R^+)$. Then $$ \langle M\rangle(t) = \int 1_{[0,t]}(s) a(s)^2 ds $$