# In which sense does the quadratic variation depend on the considered filtration?

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge 0}$ be a complete right-continuous filtration on $(\Omega,\mathcal A,\operatorname P)$
• $X$ be an almost surely continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$

It's a classical result that there is an $\mathcal F$-adapted stochastic process $\langle X\rangle$ on $(\Omega,\mathcal A,\operatorname P)$ with

1. $\operatorname P\left[\langle X\rangle_0=0\right]=1$
2. $\langle X\rangle$ is almost surely continuous
3. $\langle X\rangle$ is almost surely increasing
4. $|X|^2-\langle X\rangle$ is an almost surely continuous local $\mathcal F$-martingale

$\langle X\rangle$ is unique up to indistinguishability and is called quadratic variation of $X$.

The question is: Do we need to call it the quadratic variation with respect to $\mathcal F$? In other words: If $(\mathcal G_t)_{t\ge 0}$ is another complete right-continuous filtration on $(\Omega,\mathcal A,\operatorname P)$ such that $X$ is a local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ and $\langle X\rangle'$ is the quadratic variation of $X$ with respect to $\mathcal G$, can we conclude that $\langle X\rangle$ and $\langle X\rangle'$ are indistinguishable?

• look at revuz & yor, under 'quadratic variation'. they show that under your assumptions the thing you are looking for is the limit of sums of squared etc Or protter who simple states that is is $X^2 - \int 2XdX$, which you can then show works. Neither depends on the filtration – user83457 Oct 25 '16 at 6:07