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I am currently reading John Hull's book and am a bit confused about the Ito's lemma when it is applied to the stock price. Given $dS=\mu Sdt+\sigma Sdz$, by applying Ito's lemma to $G=\ln S$, we have $$dG=(\mu-\frac{\sigma^2}{2})dt + \sigma dz$$

However, it seems that from $dS=\mu Sdt+\sigma Sdz$, if we rewrite it as $\frac{dS}{S}=\mu dt+\sigma dz$, then can we get $dG=d(\ln S)=\frac{dS}{S}=\mu dt+\sigma dz$?

I think the problem should be $d(\ln S)=\frac{dS}{S}$ does not hold, but I am wondering why this is the case?

Thank you!

username123
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    That's the entire point of Ito's Lemma. Let $f(x)=\ln(x)$. You identified the problem to be $\mathrm{d}f(X)=f'(X)\mathrm{d}X$. This is not true for stochastic processes, we instead have $\mathrm{d}f(X)=f'(X)\mathrm{d}X+\frac{1}{2}f''(x)(\mathrm{d}X)^2$. That's why that equation does not hold. Recall that $(\mathrm{d}X)^2=\sigma^2 X^2 \mathrm{d}t$. – Kevin May 23 '20 at 19:48
  • @KeSchn Thanks for your comment! Do you have any intuitive understanding why the noise term would affect the drift rate? I do not really get it since it seems to me that noise has neither positive nor negative impact on the return. – username123 May 23 '20 at 20:11
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    $f$ is not linear so random noise down does not cancel random noise up. – Bob Jansen May 23 '20 at 20:22
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    Here is an intuition I use. Instead of the positive S process in continuous time, consider the first step of a discrete time process that starts at one and can go to e or 1/e with equal probability. The expected value of S is cosh(1) >1 so the expected change in S is positive. Now let L = ln S start at 0 and go to 1 or -1 with equal probability. The expected value is zero so the expected change in L is also zero. We see that the expected change in L = ln S is lower than the expected change in S since 0 < cosh(1) - 1. The Ito term captures this negative concavity effect. – Peter Carr May 24 '20 at 21:32

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