Let $(M_t)_{t\geq 0}$ be a continuous martingale such that $M_0=0$ almost surely.
There exists an increasing process $(\langle M\rangle_t)_{t\geq 0}$ which is called the bracket of $M$ such that $M^2-\langle M\rangle$ is a martingale.
If we look at the special case where $M$ is a Brownian motion, then for every $t\geq 0$, $\langle M\rangle_t=t$. Then, for every $t>0$ the quantity $\frac{M_t^2}{\langle M\rangle_t}$ is distributed as the square of a standard Gaussian random variable which is also $2\Gamma(1/2,1)$.
My question is the following: is this always true that $\frac{M_t^2}{\langle M\rangle_t}$ is distributed as $2\Gamma(1/2,1)$? I am under the impression that this is true for the geometric Brownian motion. However I do not see any general proof. If it is false, does someone know a counter example?