Are the integral and differential definitions of Ito process equivalent?

Question

I think an ito process $X_t$ can be defined as

$$X_t := X_0 + \int_0^t\sigma_s dB_s + \int_0^t\mu_s ds.$$

(Is this an Ito drift-difussion process?)
(Why use the subscript $s$? Eg. why is it $\sigma_s$ and not just $\sigma$?)

And I've also seen:

$$dX_t := \sigma_t dB_t + \mu_t dt$$

Are the 2 definitions equivalent, and why (not)?
(One seems to be the integral of the other. Which would mean one is the derivative of the other, which would mean there is an Ito derivative?)

Answer 1

Is this an Ito drift-difussion process?

yes

Why use the subscript $s$? Eg. why is it $\sigma_s$ and not just $\sigma$?

because it depends on time too eg. see Solution to General Linear SDE

And I've also seen: $$dX_t := \sigma_t dB_t + \mu_t dt$$ Are the 2 definitions equivalent, and why (not)?
(One seems to be the integral of the other. Which would mean one is the derivative of the other, which would mean there is an Ito derivative?)

Yes, they are equivalent. And in general for semimartingales, they always mean them in integral form (think of weak derivatives in Sobolev). The main issues are

• Brownian motion is not differentiable Proving Brownian motion is not monotone
• generally any martingale that is differentiable is actually constant Does there exist an almost surely differentiable martingale?

Indeed, if a martingale is a.s. everywhere differentiable, then its quadratic variation is a.s $0$. So, by the Burkholder--Davis--Gundy inequality, the martingale is a.s. constant.