Law of the square of a martingale divided by its bracket 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?
 A: I think I found a counter example. Let us consider a Brownian motion $(B_t)_{t\geq 0}$ starting from $0$. Let $T=\inf\{t\geq 0, |B_t|=1\}$ which is a stopping time for the natural sigma-field associated with $B$.
By the stopping theorem, $(B_{t\wedge T})_{t\geq 0}$ is a continuous martingale starting from $o$. Moreover, the bracket of this martingale is $(t\wedge T)_{t\geq 0}$.
Let us assume, by contradiction, that for every $t\geq 0$, $\frac{B_{t\wedge T}^2}{t\wedge T}$ has distribution $2\Gamma(1/2,1)$. Then, making $t$ go to infinity, we obtain that $T^{-1}$ is distributed as $2\Gamma(1/2,1)$. A simple computation shows that this would imply that for every $\lambda>0$,
$$\mathbb{E}\left[ e^{-\lambda T}\right]=e^{-\sqrt{2\lambda}}.$$
However, this is well-known (See For example proposition 3.7 in the book "Continuous martingales and Brownian motion" of Revuz and Yor.) that the Laplace Transform of $T$ is such that for every $\lambda>0$,
$$ \mathbb{E}\left[ e^{-\lambda T}\right]=\frac{1}{\cosh\left(\sqrt{2\lambda}\right)}.$$
Therefore, there is a contradiction.
