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From my understanding, the Ito Integral is a random variable itself. Suppose we have $X_t=\int_0^t Z_udZ_u$. To find $dX_t$, I would think we can apply Ito's Lemma. However, how would the partial derivatives with respect to $t$ and $Z_t$ work?

Any comments are appreciated. Thanks

xyz1010
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    What's $(Z_t)_{t \geq 0}$...? An Itô process, a Brownian motion, ....? Note that $dX_t$ is defined such that $$X_t = X_0 + \int_0^t , dX_s,$$ so, here, we have $dX_t = Z_t , dZ_t$; no need for Itô's lemma. – saz Apr 24 '15 at 05:02
  • Thanks for your note. $Z_t$ here is a Brownian motion. I understand that $dX_t$ is a sympolic way of writing the Ito Integral. However, I was wondering if we can still apply Ito's Lemma in this case (although there is no need to). In other words, what if we have $F_t=exp(\int_0^t Z_udZ_u)$. Do we use Ito's Lemma in this case? If so, how would the partial derivatives work? – xyz1010 Apr 24 '15 at 20:37
  • Well, $(X_t){t \geq 0}$ is an Itô process, i.e. a process of the form $$dX_t = \sigma_t , dZ_t + b_t , dt.$$ (Here: $\sigma_t = Z_t$, $b_t = 0$). Therefore, we can apply Itô's lemma to $f(x) := \exp(x)$ and the Itô process $(X_t){t \geq 0}$: $$f(X_t)-f(X_0) = \int_0^t f'(X_s) , dX_s + \frac{1}{2} \int_0^t f''(X_s) , d\langle X \rangle_s.$$ – saz Apr 24 '15 at 20:41
  • Thanks very much! This has actually lifted the fog for me on so many things! Sorry to go off topic, I was wondering if you could help me with one other question. How do we go about getting $dX_t$ when $X_t=X_0 + \int_0^t f(s,t)ds + \int_0^t f(s,t)dZ_s$ where $Z_t$ is still a Brownian Motion and $f(s,t)$ is a deterministic function. Thanks for your help! – xyz1010 Apr 24 '15 at 21:39

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