# Transforming a martingale into Brownian motion by integration

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 ?

• Isn't this question answered by the answer in the other question you just asked? math.stackexchange.com/questions/4415737/… Mar 29, 2022 at 22:35
• 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. Mar 29, 2022 at 23:38
• @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. Mar 30, 2022 at 6:50