# Why is $dX_t=X_t(\mu_t dt+ \sigma_t dW_t)$ an Ito process?

In solving the SDE $dX_t=X_t(\mu_t dt+ \sigma_t dW_t)$ we pick $Y_t=ln X_t$ and then apply Ito's lemma on the twice differential function $f(x)=ln (x)$ .But then why is $X_t$ anIto's prcess given that both $\mu_t$ and $\sigma_t$are progressive bounded . In order for a process to be an Ito process it needs to be of the form $$dX_t=\alpha_t dt +\rho_t dW_t$$ where $\rho \in \Lambda^2_{loc }$ and and $\alpha_t$ is progressive such that the integral $\int_0^t | \alpha_s| ds < \infty$ Now in my SDE I do not know anything about $X$ yet, i.e whether it is an Ito process or not , so how can I be sure that $\alpha_t X_t$ and $\rho_t X_t$ are of the type above so that I can apply Ito's lemma?

Sorry I know this is probably a very stupid question but I am confused. Can some one help me? Thanks

• The usual trick is: First, apply Itô's formula (without caring whether you are allowed to do so) to get a candidate for the solution. Afterwards you show, using Itô's formula, that this candidate is indeed a solution to the SDE (and when applying Itô's formula this time, you really check that it is applicable; that's usually not difficult if you have a nice representation for your solution).
– saz
Sep 30, 2015 at 19:12

Following up on @saz's comment: In expanding $Y:=\log X$ using Ito's formula you express $Y$ in terms of $\mu$, $\sigma$, and $W$. This then leads to an explicit formula for (a putative solution) $X=e^Y$ in terms of those same three ingredients. One of your jobs is then to check that this formula does indeed provide a solution. The first thing to notice is that your proposed solution is pathwise continuous, hence locally bounded. This is enough to show that $X_t\mu_t$ and $X_t\sigma_t$ have the integrability properties required to apply Ito's formula.