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instead of using Wikipedia's definition: $$ {d}(f(X_t,t)) = \frac{\partial f}{\partial t}(X_t,t)\,\mathrm{d}t + \frac{\partial f}{\partial x}(X_t,t) \, \mathrm{d}X_t + \frac{1}{2} \frac{\partial^2 f}{\partial x^2}(X_t,t)\sigma_t^2 \, \mathrm{d}t.$$

Can I write it like that:

$$ d(f(X_t,t)) = \frac{\partial f}{\partial Xt} \, \mathrm{d}Xt + \frac{1}{2} \frac{\partial^2 f}{\partial Xt^2} \, \mathrm{d}Xt^2$$

Thanks

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    $\begingroup$ What do you think could be wrong with your expression $df = (\cdots) df + (\cdots) (df)^2$? $\endgroup$ – ilovevolatility Sep 1 at 10:12
  • $\begingroup$ I edited the equation, sorry ! $\endgroup$ – Gryz Sep 1 at 10:15
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    $\begingroup$ No, it's actually even more incorrect :) If a function depends on two variables then from standard calculus $df(x,y) = f_x dx + f_y dy$ where subscripts denote partial differentiation. Ito calculus is a generalization of this, and the correct expression is as given in for instance wiki. $\endgroup$ – ilovevolatility Sep 1 at 10:20
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    $\begingroup$ Almost, you forgot the dependence on $t$, and if you include that you get the correct formula. $\endgroup$ – ilovevolatility Sep 1 at 10:24
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    $\begingroup$ If f is just a function of the stochastic process $X_t$, then yes, you don't need to write time derivative. This is the case when solving GBM SDE - $\ln X_t$. But if f is a function of t as well then you will need to add time derivative. This is the case when dealing with say Europrean call/put options. $\endgroup$ – Magic is in the chain Sep 1 at 18:56
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Ito Lemma (as 'Taylor expansion'): For $X$ an Ito process and $f = f(t, x) ∈ C^{1,2}(\mathbb{R}^2)$ a deterministic function, the stochastic process $$Y_t = f(t,X_t)$$ is an Ito process and we have $$df (t,X_t) = \partial_tf(t,X_t)\,dt + \partial_xf(t,X_t)\,dX_t + \frac{1}{2} \partial_{xx}^2f(t,X_t)(dX_t)^2. $$

Note: Functions

$$g(t,x)= \partial_tf(t,x), $$

$$h(t,x) = \partial_xf(t,x), $$

$$k(t,x) = \partial_{xx}^2f(t,x) $$ are also deterministic. So:

$$df (t,X_t) = g(t,X_t)\,dt + h(t,X_t)\,dX_t + \frac{1}{2} k(t,X_t)(dX_t)^2. $$

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