# 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?

• 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 ... May 23, 2022 at 9:51

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 $$\lambda>0$$, $$\mathbb{E}\left[ e^{-\lambda T}\right]=\frac{1}{\cosh\left(\sqrt{2\lambda}\right)}.$$
• 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. May 23, 2022 at 13:33