How to derive the differential notation of Ito process

Question

Starting from this general formulation of Ito process \begin{equation} X_s = X_0 + \int_0^s g(X_r, r) dr + \int_0^s f(X_r, r) dW_r \quad \text{ for } s \in [0,t]. \end{equation} how can I get to this one? \begin{equation} dX_s = g(X_s, s)ds + f(X_s,s) dW_s \quad \text{ for } s \in [0,t], \end{equation}

Answer 1

Well, I didn't want to write an answer, but...
Writing $dy=f(x)\,dx$ is just another way of writing $\dfrac{dy}{dx}=f(x)$, or $\displaystyle y(x)=y_0+\int^x_0 f(t)\,dt$. Formally, everything is clear: $y\to dy$, $y_0\to0$ (constant), $\displaystyle \int^x_0 f(t)\,dt\to f(x)\,dx$, so the $d$ is just a linear operator, and that's all rather trivial. The non-trivial part is the chain rule: if you have $$\frac{df(x,y)}{dx}=\frac{\partial f}{\partial x}+y'\,\frac{\partial f}{\partial y},$$ that becomes $$df(x,y)=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy.$$ With Ito integrals, you add the trivial rule $\displaystyle \int^x_0 f(t)\,dW_t\to f(x)\,dW_x$, and the non-trivial rule $$d f(t,X_t)=\left(\frac{\partial f}{\partial t}+\mu_t\,\frac{\partial f}{\partial x}+\frac12\,\sigma^2_t\,\frac{\partial^2 f}{\partial x^2}\right)\,dt+\sigma_t\,\frac{\partial f}{\partial x}\,dW_t$$ whenever $$dX_t=\mu_t\,dt+\sigma_t\,dW_t,$$ that's just the equivalent of the chain rule in this context. Formally, you could say that you just take one more term in the Taylor expansion, and use $(dW_t)^2=dt$, but that can't really replace the proof of Ito's formula in a rigorous treatment.