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I am a bit confused about how the geometric brownian motion process is commonly defined.

On this reference it seems to imply that the $\mu$ and $\sigma$ are the mean and the standard deviation of the normal distribution where the logarithm of the ratios of consecutive points are drawn from:

$GBM(t) = e^{X(t)}$, where $X(t) \sim BM(\mu, \sigma)$ and BM is a brownian motion random process.

I also found other references which seem to define it as follows:

$GBM(t) = e^{X(t)}$, where $X(t) \sim BM(\mu - \sigma^2/2, \sigma)$

In case I am not missing something important, and there are indeed different ways to model this process, what is the most common?

mvc
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  • Do you mean $GBM(t) = e^{X(t)}$ ? – Khosrotash Jul 26 '20 at 16:32
  • It depends on which interpretation --- Ito or Stratonovich, you interpret the SDE $dS_t=\mu S_t dt + \sigma S_t dW_t$. – user10354138 Jul 26 '20 at 16:59
  • @Khosrotash, yes, sorry. Edited. – mvc Jul 26 '20 at 17:05
  • @user10354138 Thanks for the pointers, I will investigate. – mvc Jul 26 '20 at 17:06
  • I now understand that the $-\sigma^2/2$ term in the second definition is some kind of correction to make the mean and the median of GBM(t) coincide. Thank you both for the directions. I have now a follow up question https://math.stackexchange.com/questions/3770340/limit-of-geometric-brownian-motion. – mvc Jul 26 '20 at 22:45

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AS a hint; if you apply $It\hat{ o}$ derivation to $g(t,x)=ln(x) $ when $ dx_t=\mu x_t dt+\sigma x_tB_t$ you will have no doubt.( hope it will) $$dy=\frac{\partial g}{\partial g}dt+\frac{\partial g}{\partial x}dB_t+\frac12 \frac{\partial^2 g}{\partial^2 x}(dx_t)^2\\$$ $$dy=\\0dt+\frac 1x dx+\frac12(-\frac1{x^2})(dx)^2=\\0dt+\frac 1x \underbrace{dx}_{dx_t=\mu x_t dt+\sigma x_tB_t}+\frac12(-\frac1{x^2})\underbrace{(dx)^2}_{dx_t=\mu x_t dt+\sigma x_tB_t}=\\ 0+\frac1x(\mu x_t dt+\sigma x_tB_t)+\frac{-1}{2x^2}(\mu x_t dt+\sigma x_tB_t)^2=\\ \mu dt+\sigma dB_t+\frac{-1x^2}{2x^2}(\mu dt+\sigma B_t)^2=\\ \mu dt+\sigma dB_t+\frac{-1}{2}(\mu^2 (dt)^2+(\sigma B_t)^2+2\mu\sigma dtdB_t)=\\ $$note that $$(dt)^2\to 0\\dt.dB_t\to 0$$so $$dy=\mu dt+\sigma dB_t+\frac{-1}{2}(\mu^2 (dt)^2\downarrow_0+\sigma^2(B_t)^2\downarrow_{dt}+2\mu\sigma dtdB_t\downarrow_0)\\= \mu dt+\sigma dB_t+\frac{-1}{2}\sigma^2dt\\ =(\mu-\frac{1}{2}\sigma^2)dt+\sigma dB_t $$ remember $y=ln(x_t) $ so $$d(ln(x_t))=(\mu-\frac{1}{2}\sigma^2)dt+\sigma dB_t$$ apply integration to both sides $$\int^{t}_{0}d(ln(x_s))=\int^{t}_{0}(\mu-\frac{1}{2}\sigma^2)ds+\int^{t}_{0}\sigma dB_s\\ \ln(x_t)-ln(x_0)=\int^{t}_{0}(\mu-\frac{1}{2}\sigma^2)ds+\int^{t}_{0}\sigma dB_s$$in the end $$\ln(\frac{x_t}{x_0})=(\mu-\frac{1}{2}\sigma^2)(t-0)+\sigma (B_t-B_0)\\\ \ln(\frac{x_t}{x_0})=(\mu-\frac{1}{2}\sigma^2)t+\sigma B_t\\\frac{x_t}{x_0}=e^{(\mu-\frac{1}{2}\sigma^2)t+\sigma B_t}\\ x_t=x_0e^{(\mu-\frac{1}{2}\sigma^2)t+\sigma B_t}$$

Khosrotash
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    Thanks @Khosrotash! I think I will need a couple of days to digest your answer :) Is there any reason why the formula you got in the end is not exactly the one in the question (note the sign of the $\sigma^2$ term in the exponential) ? – mvc Jul 26 '20 at 18:01
  • I show you the progress of finding gbm formula from begining. – Khosrotash Jul 26 '20 at 19:05