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For a basic stochastic differential equation $$dS_t=S_t(μdt+σdW_t),$$ a classic method to solve it is by letting $f = \ln S$, and when using Ito formula to this $$df=d\left(\ln(Y(t)\right)=\left(\frac{∂f}{∂t}+\frac{∂f}{∂Y}+\frac 12\frac{∂^2f}{∂Y^2}\right)dt+\frac{∂f}{∂Y}dW(t)$$ as to this answer "Stochastic Differential Equation solution for Geometric Brownian Motion".

why $\partial f/\partial t = 0$? Although $f = \ln S$, seems not appear $t$, but $S$ is a function of $t$?

why don't need to the rule to composite function? like $\partial f/\partial t = \partial f / \partial s \cdot \partial s / \partial t$, and I think this is not $0$.

for example in classic calculus, if $f(x)=2x^2+3x$, $g(f(x))=\sin(f(x))$, then, $dg(x)/dx=dg(x)/df(x) \cdot df(x)/dx$, and $d(g(x))/d(f(x)) = \cos(f(x))$, $d(f(x))/dx = 4x+3$, so the final result is $d(g(x))/dx = \cos(2x^2+3x) \cdot (4x+3)$. So why here no need to $\partial f/\partial s \cdot \partial s/ \partial t$?

FD_bfa
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Wang Jing
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 06 '23 at 13:16
  • what about now? Is it clear enough? – Wang Jing Apr 06 '23 at 13:43
  • The function $f$ you are looking at is $S_t=f(t,W_t)=S_0\exp(\sigma W_t-\sigma^2 t/2+\mu t),.$ Now apply Ito. – Kurt G. Apr 06 '23 at 14:35

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You consider the function $$ f(t,x)=\ln(x). $$ This function is independent of $t$, so the partial derivative for the first argument is zero. $$ Y_t=f(t,S_t) $$ is now depending on $t$ via composition, but the partial derivative of $f$ for its first argument is still zero.

Lutz Lehmann
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  • still a little bit confused. for example in classic calculus, if $f(x)=2x^2+3x$, $g(f(x))=sin(f(x))$, then, $dg(x)/dx=dg(x)/df(x) \cdot df(x)/dx$, and $d(g(x))/d(f(x)) = cos(f(x))$, $d(f(x))/dx = 4x+3$, so the final result is $d(g(x))/dx = cos(2x^2+3x) \cdot (4x+3)$. So why here no need to $\partial f/\partial s \cdot \partial s/ \partial t$? – Wang Jing Apr 06 '23 at 13:51
  • Because here you first consider a Taylor expansion of $f$ and then insert $S_t$ and $dS_t$. The Ito formula is the equivalent to the total differential in smooth calculus. Only that here you get a contribution from the quadratic terms of the Taylor expansion as $(dW_t)^2$ is on average $dt$ (while $dt^2$ or $dt,dW_t$ are "infinitesimally" smaller than $dt$). – Lutz Lehmann Apr 06 '23 at 15:46