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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?

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  • $\begingroup$ It works for Wiener integrals, but I doubt that it can be extended beyond this case. For example for a geometric Brownian motion I doubt that this ratio could follow a ch-2 law, as its numerator is log normal and its denominator is the time integral of a log normal process (so in spirit an infinite sum of log normal variables), I don't see how this could result into a Chi-2 law, even though both numerator and denominator have the same expected values by Ito's isometry. I can't find an analytically tractable counterexample though ... $\endgroup$
    – TheBridge
    Commented May 23, 2022 at 9:51

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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.

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  • $\begingroup$ Yes it's seems ok to me, on another hand looking at Dambis-Dubins-Schwarz theorem, you have that $X_t= B_{<X>_t}$ so $X_t^2= B_{<X>_t}^2$, and then $X_t^2$ has the same law as ${<X>_t}^2.B_1^2$. So if you have independence between ${<X>_t}^2$ and $B_1^2$ you have the result you want (for example this proves my claim for Wiener integrals as ${<X>_t}^2$ is then deterministic) as you can divide both terms by ${<X>_t}^2$ and get what you want i.e. $X^2_t/{<X>_t}^2$ = $B_1^2$ in law. So there might be some other cases (apart form Wiener integrals) where it's true but they must be quite ad hoc IMO. $\endgroup$
    – TheBridge
    Commented May 23, 2022 at 13:33

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