# Differentiation of a stochastic process by Ito's formula

Let $$W_t$$ denote a Brownian Motion, use ito's formula to compute the differential of the following stochastic process: $$X_t=e^{t/2}\cos(W_t)$$

Ito's formula is supposed to be used when we know a diffusion differential equation like $$dx=\mu dt+\sigma dW_t$$, then we can apply it to solve a $$dG(x,t)$$ by: $$dG=\left(\frac{\partial G}{\partial t}+\frac{\partial G}{\partial x}\mu+\frac{1}{2}\frac{\partial^2G}{\partial x^2}\sigma^2\right)dt+\frac{1}{2}\frac{\partial G}{\partial x}\sigma dW_t.$$ But for this question, why we can solve the $$dX_t$$ directly?

Also, the answer to this question suggests that we should use the following Ito's formula to solve $$dX_t$$: $$dX(t,W_t)=\frac{\partial X(t,W_t)}{\partial t}dt+\frac{\partial X(t,W_t)}{\partial W}dW_t+\frac{1}{2}\frac{\partial ^2 X(t,W_t)}{\partial W^2}dt$$

What's the logic behind this Ito formula?

• You should try to use precise language and notation. Personally, I don't really see what your point is here. Mar 26, 2021 at 12:56
• The form $dX(t,W_t)=\dots$ that you wrote is just a special case of the more general form you wrote further up with $\mu \equiv 0$ and $\sigma \equiv 1$.
– Ian
Mar 26, 2021 at 13:00

I think you have a slight misunderstanding of Itô's lemma. What it says (in a simplified version) is that for any drift-diffusion process $$(Y_t)_t$$ which is solution of $$dY_t = \mu dt +\sigma dW_t$$, and any twice-differentiable function $$f(t,x)$$, it holds that the process $$(f(t,Y_t))_t$$ is a solution of the SDE : $$df(t,Y_t) = \frac{\partial f}{\partial t}(t,Y_t)dt+\frac{\partial f}{\partial x}(t,Y_t)dW_t+\frac{1}{2}\frac{\partial ^2 f}{\partial x^2}(t,Y_t)\cdot\sigma^2dt$$
In your case, the process $$Y_t := W_t$$ is indeed a drift-diffusion process and $$f:(t,x)\mapsto e^{t/2}\cos(x)$$ is twice-differentiable w.r.t. its arguments so Itô's lemma applies.
The usual form of It's formula you note is applied like chain rule, when you need to find $$dG(x,t)$$ but you only know the dynamics of $$dx$$. The "regular-Calculus" analog of this is when you know $$x(t)$$ and $$x'(t)$$ but need to find $$\frac{d}{dx} g(x(t))$$.