Given the process $X_t=\int_0^t g_s\,dW_s$, compute $dX_t$ using the Ito formula.
The Ito formula says that, given:
$$f(t,X_t)=f(0,X_0)+\int_0^t\frac{\partial f}{\partial s}(s,X_s)\,ds+\int_0^t\frac{\partial f}{\partial x}(s,X_s)\,dX_s+\frac{1}{2}\int_0^t\frac{\partial^2 f}{\partial x^2}(s,X_s)\,d\langle X \rangle_s$$
it holds:
$$df(t,X_t)=\frac{\partial f}{\partial t}(t,X_t)\,dt+\frac{\partial f}{\partial x}(t,X_t)\,dX_t+\frac{1}{2}\frac{\partial^2 f}{\partial x^2}(t,X_t)\,d\langle X \rangle_t$$
I already know the result, that is $dX_t=g_t\,dW_t$, but I would like to understand how Ito formula works. I guess that first and third terms in the formula for $df$ are zero, but I don't get why. Why is $\frac{\partial f}{\partial t}(t,X_t)=0$ ? Since $X_t$ is a definite integral, I thought that it should be equal to $G(t)-G(0)$, where $G$ is the primitive of $g$, and so, since $G(t)$ is a function of time, its derivative is not zero.
So the question is how to compute these terms:
$$\frac{\partial f}{\partial t}(t,X_t),\quad\frac{\partial f}{\partial x}(t,X_t),\quad \frac{\partial^2 f}{\partial x^2}(t,X_t)\quad ?$$
Best Answer
You should put $f(t,x) = x$. It follows immediately that $$ \frac{\partial f}{\partial t}(t,x) = 0, \;\frac{\partial f}{\partial x}(t,x) = 1, \; \frac{\partial^2 f}{\partial x^2}(t,x) = 0. $$ By Ito's formula, we have $$ X_t = \int_0^t dX_s = \int_0^t g_sdW_s, $$ which is an obvious result.