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Suppose that $S$ follows a geometric brownian motion: $$ dS(u) = r S(u)du + S(u)\sigma(u,S(u))dW(u) , $$ with $r$ a deterministic constant, and let the process $Z$ be defined by: $$ Z(t) = \int_0^t g(u, S(u))du, $$ where $g$ is a deterministic function. I would like to find the stochastic process followed by $Z$.

Following the idea shown here, I compute: $$ S(t) = S(0) + \int_0^t \alpha(u,S(u))S(u)du + \int_0^t \sigma(u,S(u))S(u)dW(u), $$ and for $Z$: $$ Z(t) = \int_0^t g(u, S(0)+ \int_0^u \alpha(v,S(v))S(v)dv + \int_0^u\sigma(v,S(v))S(v)dW(v))du, $$ where all the integrals are deterministic, except $\int_0^u \sigma(v,S(v))S(v)dW(v)$ which is an Ito integral. At this point, I am not sure on how to differentiate appropriately the equation above in order to get the stochastic process for $Z$.

Naively, I would just take advantage of Ito's lemma to write: $$ dZ(t) = \left(g(t, S(t)) + \int_0^t g_S(u,S(u))du\cdot \alpha(t,S(t))S(t) + \frac12 \int_0^t g_{SS}(u,S(u))du \cdot\sigma^2(t,S(t))S(t)^2 \right)dt + \int_0^t g_S(u,S(u))du \cdot\sigma(t,S(t))S(t)dW(t), $$ but this does not seem to lead to the correct answer.

What is the satisfactory way of formalising the stochastic process for $Z$?

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  • $\begingroup$ I do not think you applied Ito correctly. It should be $dZ_t = g(t, S(t))dt$. $\endgroup$
    – Gordon
    Commented Sep 13, 2019 at 15:31
  • $\begingroup$ @Gordon Is it because $Z(t)$ does not depend explicitly on $S(t)$, but only on $t$? $\endgroup$
    – J. D.
    Commented Sep 13, 2019 at 16:17
  • $\begingroup$ You can only take derivative with respect to $t$. Your integrand does not involve $t$, then you can not take derivative within the integral. $\endgroup$
    – Gordon
    Commented Sep 13, 2019 at 17:35

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