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If the stock price S follows the geometric brownian motion: $$dS=\mu Sdt+\sigma Sdz$$ $$\frac{dS}S=\mu dt+\sigma dz$$

Where $dz=\epsilon\sqrt{dt}$ is a wiener process.

Integrating this to get $S_T$ as a function of $S_0$

$$\int^T_0\frac{1}Sds=\int^T_0\mu dt +\int^T_0\sigma dz$$ $$=ln(S_T/S_0)=\mu(T-0)+\sigma (z_T-z_0)$$

$$S_T=S_0e^{\mu T+\sigma (z_T-z_0)}$$ Why do others allow $(z_T-z_0)$ to be a standard Wiener process?

And why is this not the same as: $$S_T =S_0 e^{(\mu -0.5\sigma^2)T+\sigma \epsilon \sqrt{T}}$$

And what is the difference between the above version, and this version: $$S_T =S_0+dS=S_0+\mu S_0 T+\sigma S_0 \epsilon \sqrt{T}$$?

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closed as off-topic by skoestlmeier, LocalVolatility, Gordon, Bob Jansen May 16 at 19:58

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Basic financial questions are off-topic as they are assumed to be common knowledge for those studying or working in the field of quantitative finance." – skoestlmeier, LocalVolatility, Gordon, Bob Jansen
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Your integration of 1/S dS is incorrect for a stochastic process. You must use stochastic calculus. That would give you the adjustment term, somewhat like a convexity adjustment.

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