Let $X_t$ and $\xi_t$ be two stochastic processes and let $\mathcal{M}_t$ be a sigma-algebra generated by $X_t$ and $\mathcal{N}_t$ a sigma-algebra generated by $\xi_t$. I'm trying to show that $\mathcal{M}_t = \mathcal{N}_t$.
I have shown that $$X_t = x e^{\xi_t - \frac{\sigma^2 t}{2}}$$ and so $$\xi_t =\frac{\sigma^2 t}{2} + \log {\frac{X_t}{x}}.$$
Thus it should (Why is this exactly? I have hard time seeing this from definition.) follow that $\mathcal{M}_t \subset \mathcal{N}_t$ and that $\mathcal{N}_t \subset \mathcal{M}_t$ i.e. $\mathcal{M}_t = \mathcal{N}_t$.
So if this is the case, does it mean in general that if I can come up with a measurable functions $f,g$ such that $f(X_t)=g(\xi_t)$ it follows that the filtrations generated by the processes are equal?
A more broad question is: If I have two processes given by SDEs what tools do I have to show that the filtrations of the processes are equal?
Consider that for each measurable function $f$ it holds that $$\sigma(f(X)) \subseteq \sigma(X)$$ what follows directly from the definition of the measurability of $f$ But if $f$ is measurable then you only know that the pre-image of a mesurable set is again a measurable set… but you don't know if also each measurable set can be written as a preimage of a measurable set under $f$.
That's why in general you just have $$\sigma(f(X)) \subseteq \sigma(X)$$
In your case you have $X_t = f(\xi_t)$ for a measurable function $f$ hence you have $$\sigma(X_t) \subseteq \sigma(\xi_t)$$ and you have $\xi_t = g(X_t)$ for a measurable $g$ so you get $$\sigma(\xi_t) \subseteq \sigma(X_t)$$ hence it follows equality.
It's not enough to have $f(X_t) = g(\xi_t)$ for measurable $f,g$ to get $$\sigma(X_t) = \sigma(\xi_t)$$
To see this take $X_t,\xi_t$ s.t. $\sigma(X_t) \not= \sigma(\xi_t)$ and take $f = g \equiv 0$
But it's enough to have $$X_t = f(\xi_t), \xi_t = g(X_t)$$ for measurable $f,g$ as written above…