Finding the twice differentiable function in Itos Lemma (I am an undergrad in Econ and new to this forum, so I'm sorry if this will be easy for you guys)
Im currently struggling with Stochastic Calculus, resp. Itô's Lemma. I understand that once we have an Itô Process $X_t$ that satisfies the Stochastic Differential Equation, we have to find a twice differentiable function. I posted an example, and wanted to ask if anyone knows how to get the:
$$Z_t := f(t, W_t) := t \cdot W_t$$
Thanks In advance, any form of help is very appreciated
 A: Itô's formula states that
$$f(t,W_t)-f(0,W_0) = \int_0^t \partial_x f(s,W_s) \, dW_s + \int_0^t \left( \frac{1}{2} \partial_x^2 f(s,W_s) + \frac{\partial}{\partial s} f(s,W_s) \right) \, ds \tag{1}$$
for any (nice) function $f$. Equivalently,
$$\int_0^t \partial_x f(s,W_s) \, dW_s = f(t,W_t)- f(0,W_0) - \int_0^t \left( \frac{1}{2} \partial_x^2 f(s,W_s) + \frac{\partial}{\partial s} f(s,W_s) \right) \, ds. \tag{2}$$
Now let's consider the stochastic integral $\int_0^t s \, dW_s$. If we can find a function $f$ such that
$$\partial_x f(s,x) = s \tag{3}$$
then the left-hand side of $(2)$ is nothing but $\int_0^t s \, dW_s$, i.e. exactly the stochastic integral which we are looking for. Integrating the equation $(3)$ with respect to $x$ yields
$$f(s,x) = sx + c$$
for some constant $c$; let's choose $c:=0$. Then $$\partial_x^2 f(s,x) = 0 \qquad \partial_s f(s,x) = x,$$ and therefore (2) gives $$\int_0^t s \, dW_s = t W_t - \int_0^t W_s \, ds.$$
See e.g. this question and this question for further examples.
