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I have the following definition for an


Ito process:

For $a(\omega, t), b(\omega, t)$ real valued, adapted stochastic processes that respectively satisfy the conditions $$ P(\int_0^t \vert a(\omega, s)\vert ds < \infty ) = 1 \quad \text{and} \quad P(\int_0^t b(\omega, s)^2 ds < \infty ) = 1 $$

and Ito-process has the form $$ X_t = x_0 + \int_0^t a(\omega, s) ds + \int_0^t b(\omega, s) dB_s. $$



Now for another stochastic process $f(\omega, t)$ that satisfies $$ P(\int_0^t \vert f(\omega, s) a(\omega, s)\vert ds < \infty ) = 1 \quad \text{and} \quad P(\int_0^t (f(\omega, s)b(\omega, s))^2 ds < \infty ) = 1 $$

the stochastic integral of $f$ with respect to $X$ is defined as $$ \int_0^t f(\omega, s) dX_s := \int_0^t f(\omega, s)a(\omega, s) ds + \int_0^t f(\omega, s)b(\omega, s) dB_s $$



For the Ito integral I have seen the interpretation of say the cumulative gains or losses of a gaming/investing strategy $f(\omega,t)$ w.r.t. an underlying asset price (the Brownian motion). It makes sense to me to extend the original definition to more complicated asset price models (here an Ito-process), but I don't see how the above defined stochastic integral is

a) consistent with the formal definition of the Ito integral - i.e. whether this integral can also be derived as a limit of sums of this form $$ \lim_{\vert \mathcal{P} \vert \to 0} \sum_{t_i \in \mathcal{P}} f(t_i) (X_{t_i} - X_{t_{i-1}}) $$ b) consistent with the interpretation given above

So essentially I'm asking: why does this definition make sense?

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  • $\begingroup$ The equivalence of these two formulations is a (nontrivial) theorem. Usually the latter is taken to be primary. You can find a proof of the equivalence in, say, Baldi’s Stochastic Calculus, Chapter 8. $\endgroup$
    – Nate River
    Commented Jul 15, 2022 at 9:23
  • $\begingroup$ @NateRiver Thank you so much, the Chapter you mentioned was really helping my understanding and the book in general is fantastic. $\endgroup$
    – lpnorm
    Commented Jul 26, 2022 at 7:11

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