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I encountered the following slide in a lecture on Ito's Lemma.

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The lecturer explained that $$\frac{\partial V}{\partial t} = 0$$ because the first two derivatives on the slide already took into account time into the change of the value of V.

I'm not convinced. If $V = \log S(t)$ is a function of time, why wouldn't we have to use the chain rule for the third derivative on the slide?

$$\frac{\partial V}{\partial t} = \frac{\partial V}{\partial S(t)} \cdot \frac{\partial S(t)}{\partial t} = S^{-1} \cdot \frac{\partial S(t)}{\partial t} = ...$$

I'm not sure where to go from here to show that it is in fact 0.

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  • $\begingroup$ No , it is not function of time. $\endgroup$ – user16651 Jul 14 '16 at 17:37
  • $\begingroup$ Indeed, $V(x)=\ln x$ and $V \in C^{2}[(0,+\infty)]$ $\endgroup$ – user16651 Jul 14 '16 at 17:43
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A process indeed depends on time $t$. However, in Ito's lemma, only derivatives with respect to independent time variable $t$ is considered. That is, for a process of the form $f(S_t, t)$, $\frac{\partial f}{\partial t}$ is the derivative with respect to the second, that is, the independent, $t$ variable, however, the parameter $t$ in the process $S_t$ is not considered. Ito's lemma takes a particular form, which can not be understood in the normal calculus sense.

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  • $\begingroup$ could you further explain why the parameter $t$ in the process $S(t)$ not considered, in my understanding, the process is time-dependent, so when you take the Ito's lemma, you should take both parameters $S(t)$ and $t$ into account. I know my comments are wrong, but just wondering why, as it is not very intuitive for me. Thank you! $\endgroup$ – Donkey_JOHN Jul 26 '16 at 9:48
  • $\begingroup$ I think I have already explained. In Ito's lemma, for a process of the form $f(S_t, t)$, $\frac{\partial f}{\partial t}$ is the derivative with respect to the second, that is, the independent, $t$ variable.Please read the Ito's lemma carefully. $\endgroup$ – Gordon Jul 26 '16 at 13:32

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