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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?

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    $\begingroup$ You should try to use precise language and notation. Personally, I don't really see what your point is here. $\endgroup$
    – Tobsn
    Mar 26, 2021 at 12:56
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    $\begingroup$ 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 $\sigma \equiv 1$. $\endgroup$
    – Ian
    Mar 26, 2021 at 13:00

2 Answers 2

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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.

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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 $\frac{d}{dx} g(x(t))$.

In the specific case you bring, you definitely could differentiate directly, as your second formula suggests.

Not sure what else you are asking, please post comments or update the question.

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