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$$S_t=S_0\exp\bigg(\bigg(\mu-\frac{\sigma^2}{2}\bigg)t+\sigma W_t\bigg).$$

What I tried to do is applying Ito's formula to above equation to derive SDE.

If I define $g = S_t$ then, $dg = g_t*dt+g_{W_t}*dW_t+\frac{1}{2}g_{tt}(dt)^2+\frac{1}{2}g_{W_tW_t}(dW_t)^2+\frac{1}{2}g_{tW_t}(dtdW_t)$
And the solution of $g_t$ is $(μ-\frac{1}{2}σ^2)*g$

But I think $W_t$ is dependent on time, so the solution is wrong because it just deals $W_t$ independently with $t$

Above solution totally makes $W_t$ constant on time

Why $g_t = (μ-\frac{1}{2}σ^2)*g$ ??

Jessie
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user13232877
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1 Answers1

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You need to disentangle the notation. In SDE one often uses the old notation dating back to the times of Leibniz where the argument is in the subscript. So $S_t$ is $S$ at time $t$, or $S_t(\omega)=S(t,\omega)$.

Next you used subscripts for partial derivatives. In using both, it is customary to put a separation sign before the derivative subscripts. So if $g(x)=x$, the $g_{,x}(S_t)=1$. But that is probably not what you wanted. If you meant $$ g(t,x)=S_0\exp((μ−\tfrac12σ^2)t+σx), $$ then the Ito formula gives for $S_t=g(t,W_t)$ $$ dS_t=(g_{,t}(t,W_t)+\tfrac12g_{,xx}(t,W_t))\,dt+g_{,x}(t,W_t)\,dW_t \\ =[(μ−\tfrac12σ^2)+\tfrac12σ^2]S_t\,dt+σS_t\,dW_t. $$

Lutz Lehmann
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  • That's right! $g(t,x)=S_0exp((μ−\frac{1}{2}σ^2)t+σx)$ It is what I meant. Then What I want to know is $g_{,t}(t,W_t)$. $W_t$ is dependent on $t$. But the outcome seems to treat $W_t$ as constant when we calculate $g_{,t}(t,W_t)$ – user13232877 Mar 30 '21 at 14:49
  • That was answered in your previous question. The derivation subscript denotes the position of the argument for the partial derivative. What you are talking about is the total derivative of the composite function, but for SDE that is given by the Ito formula, by all terms of it. – Lutz Lehmann Mar 30 '21 at 15:04
  • Now I understand. You mean my notation is misleading. My thought : $g_t(t,W)$ means total derivative But actually (Your explanation) it is not the total derivative it is partial derivative about the variable '$t$'. And correct notation about partial derivative is $g_{,t}$ also if I want to do total derivative (That's what I applied to $W_t$ using the variable '$t$'), what I need to do is just "calculate with Ito formula"!. Wonderful. Thanks Sincerely, – user13232877 Mar 30 '21 at 15:12
  • You need a notation that is unambiguous, what exactly that is does not matter much. Keeping within the major convention of course helps to find and keep readers. You could also use $\partial_tg$, $\partial_x^2g$ etc. to separate the two uses of subscripts. – Lutz Lehmann Mar 30 '21 at 15:17
  • It's an honor to have opportunity to correct my mistake. Thank you Lutz Lehmann, I need to keep in mind it. Have a nice day :) – user13232877 Mar 30 '21 at 15:26