My Math Forum relationship between dS/S and d(ln(S))

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 July 1st, 2008, 12:15 AM #1 Newbie   Joined: Jul 2008 Posts: 1 Thanks: 0 relationship between dS/S and d(ln(S)) Hi. I have a question about the equation for monte carlo simulations of a stock price. The basic equation is: dS / S = mu dt + sigma dz Using this equation above, Ito's lemma gives the equation for d(ln(S)) as below d(ln(S)) = (mu - (sigma ^ 2) / 2) dt + sigma dz I understand how Ito's lemma works and that this is the correct answer, but what I don't understand is why dS / S no longer is equal to d(ln(S)), as it is in usual calculus. Why does this break down in stochastic calculus?
 July 1st, 2008, 04:00 AM #2 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: relationship between dS/S and d(ln(S)) I'm going to assume it's because $d\log{S}\neq \frac{d\log{S}}{S}$, but I really need to relearn calculus...
 July 1st, 2008, 04:13 AM #3 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 Re: relationship between dS/S and d(ln(S)) I think you don't understand what $d(\ln S)$ stands for. It is not a derivation of the natural logarithm! It is a differential of size $\ln S$. When you do integrals, you have the $dx$ part that most people don't care about or they don't know what it's there for, but it represents the law by which you sum up your function values. For example, you can find: $\int e^{x^2} d\left(e^{x^2}\right)= \frac{x^2}{2}$ Because you sum a function by the same law the function does by, and therefor you get a linear progression. Basically, I'm trying to say that it's not the same!

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# d ln(s)

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