Geometric Brownian Motion definition 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?
 A: 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
$$\ln(\frac{x_t}{x_0})=(\mu-\frac{1}{2}\sigma^2)(t-0)+\sigma (B_t-B_0)\\\
\ln(\frac{x_t}{x_0})=(\mu-\frac{1}{2}\sigma^2)t+\sigma B_t\\\frac{x_t}{x_0}=e^{(\mu-\frac{1}{2}\sigma^2)t+\sigma B_t}\\
x_t=x_0e^{(\mu-\frac{1}{2}\sigma^2)t+\sigma B_t}$$
