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Let $(\mathcal F_t)$ be an arbitrary filtration on $[0,T]$ and $(X_t)$ a sample continuous stochastic process on $[0,T]$ that is a martingale with respect to $(\mathcal F_t)$. The quadratic variation of $(X_t)$ is given by a deterministic function $[X]_t=g(t) \neq 0$. We suppose that $g$ is smooth almost everywhere on $(0,T)$.

I would like to use $(X_t)$ to construct a Brownian motion adapted to $(\mathcal F_t)$ using the Levy characterisation theorem. The latter says that if a process $(Y_t)$ is an $(\mathcal F_t)$-martingale with $[Y]_t=t$, then $(Y_t)$ is a Brownian motion adapted to $(\mathcal F_t)$.

So if I take $$Y_t= \int_0^t \frac{1}{\sqrt{g'(s)}} dX_s,$$

Then $[Y]_t= \int_0^t \frac{1}{g'(s)} d[X]_s = \int_0^t \frac{g'(s)}{g'(s)} ds =t$. But is $(Y_t)$ an $(\mathcal F_t)$-martingale ?

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    $\begingroup$ Isn't this question answered by the answer in the other question you just asked? math.stackexchange.com/questions/4415737/… $\endgroup$
    – Jose Avilez
    Mar 29, 2022 at 22:35
  • $\begingroup$ I do not think so. The author of the only answer claims we need $(X_t)$ to be $L^2$ for the general result (also, in the mentioned textbooks, the filtration is not arbitrary). Here, I don't assume $L^2$, and I am asking if it works in the very particular case. But maybe I am wrong, please feel free to add an answer if you think you've got it. $\endgroup$
    – W. Volante
    Mar 29, 2022 at 23:38
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    $\begingroup$ @W.Volante . Here is another question you asked recently. Before asking the next question, it would be nice to accept the answers if acceptable, or comment why they are not. $\endgroup$
    – Kurt G.
    Mar 30, 2022 at 6:50

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