# Very basic doubt about Itô's lemma

While trying obtain the dynamics of $X_t = \exp( \int_t ^T \phi_s ds)$, where $\phi$ is an Ito process following $$d\phi_t = \mu dt+ \sigma dW_t \ ,$$ I had some doubt concerning the application of Itô's lemma.

If $g_T(t,x) := \int_t ^T x(s) ds$ which is a functional of $x$ depending of $t$ and of the parameter $T$, then $X_t = f_T(t,\phi) :=\exp(g_T(t,\phi))$, so Itô's lemma gives us that $$dX_t =df_T(t,\phi) = (\partial_t +\frac{\sigma}{2}\partial_{xx}^2) f_T(t,\phi) dt +\partial_x f_T(t,\phi) dW_t$$ My doubt is precisely about the computation of $\partial_x f_T(t,\phi)$. Is it zero or am I making a mistake?

Thus, the dynamic of $X$ must be

$$dX_t =\partial_t f_T(t,\phi) dt =-\phi_tX_tdt$$

Could someone enlight me please?

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Should it be $X_T=\exp{(\int_t^T\phi_s ds)}$ or does the process $X$ depends on the lower bound of the integral? –  user8 Mar 18 '13 at 7:49
What's the definition of the dynamic of a stochastic process? What have you tried? –  saz Mar 18 '13 at 7:55
Very basic question: what did you try? (And how come the formatting of this question is so far from being acceptable, being your question #27 on the site?) –  Did Mar 18 '13 at 10:44
You say, you have a basic doubt, but you didn't specify what do you doubt about. –  Ilya Mar 18 '13 at 12:32
If $\phi$ is a diffusion and $dX_t=\sigma(X_t,\phi_t)dW_t+b(X_t,\phi_t)dt$, then $X$ is a diffusion. But if your $X$ is not a diffusion, to try to describe it as one is doomed. Here, a problem is that $X_t$ is not adapted. –  Did Mar 19 '13 at 13:48

The point is that Ito Lemma applies to the situation where you have a function of an Ito process, rather than a functional. E.g. you can apply it in case $h(\phi_t)$, but you don't apply Ito Lemma to compute the differential of $\int_0^t\phi_s\mathrm ds$. In the latter case you rather have $$\mathrm d\left(\int_0^t\phi_s\mathrm ds\right) = \phi_t\mathrm dt.$$ Similarly, for any continuous function $\psi \in C([0,\infty))$ you have that $$g(t):=\int_t^T\psi(s)\mathrm ds \in C^1([0,\infty))$$ and thus $f(t):=\exp(g(t))$ belongs to this class as well. As a result, $$\mathrm df(t) = f'_t(t)\mathrm dt = f(t)g'_t(t)\mathrm dt = -f(t)\psi(t)\mathrm dt.$$