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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?)

Xander Henderson
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étale-cohomology
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    The second is just a notation whose definition is the first – Rhys Steele Aug 30 '23 at 21:07
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    To add to the succint first comment and answer your last question: There is *no* Ito derivative. – Kurt G. Aug 31 '23 at 00:51
  • It looks like you have edited this post a couple of times to include the tag [tag:ito-calculus]. This tag doesn't already exist on the site, and it is generally considered bad form to create a new tag without consulting the community, first. New tags should be proposed in the tag management thread. For further discussion, https://chat.stackexchange.com/rooms/3740/tagging . – Xander Henderson Sep 05 '23 at 23:35

1 Answers1

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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

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.

Thomas Kojar
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