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I have a very fundamental problem, please help me out. I am little confused with the derivation for the close form solution for the Geometric Brownian Motion, from the very fundamental stock model: $$\begin{equation} dS(t)=\mu S(t)dt+\sigma S(t)dW(t) \end{equation} $$ The close form of the above model is following: $$ \begin{equation} S(T)=S(t)\exp((\mu-\frac1 2\sigma^2)(T-t)+\sigma(W(T)-W(t))) \end{equation} $$

I believe this is quite straightforward for most of you guys, but I really dont know how did you get the $(\mu-\frac 1 2 \sigma^2)$ term. It is clear for me the other way round (from bottom to top), but I fail to derive directly from the top to bottom. I checked some material online, it was saying something with the drift term, which some terms are artificially added during the derivation.

Your answer and detailed explanation will be greatly appreciated.

Thanks in advance!

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    $\begingroup$ Consider $d\ln S(t)$, and see what you can have. Should $r-\frac{1}{2}\sigma^2$ be $\mu-\frac{1}{2}\sigma^2$? $\endgroup$ – Gordon Jul 25 '16 at 19:44
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    $\begingroup$ @Donkey_JOHN Let $f(t,x)\in C^{1,2}([0,\infty)\times\mathbb{R}) $ then $$df(t,W_t)=\frac{\partial f}{\partial t}(t,W_t)dt+\frac{\partial f}{\partial x}(t,W_t)dW_t+\frac{1}{2}\frac{\partial ^2f}{\partial x^2}(t,W_t)d[W_t,W_t]$$ $\endgroup$ – user16651 Jul 25 '16 at 20:44
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    $\begingroup$ @Donkey_JOHN Also $d[W_t,W_t]=dt$ and $d[t,W_t]=0$ and $d[t,t]=0$ $\endgroup$ – user16651 Jul 25 '16 at 20:45
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    $\begingroup$ Let $f(s)\in C^{2}(\mathbb{R}) $ then $$df(S_t)=f'(S_t)dS_t+\frac{1}{2}f''(S_t)d[S_t,S_t]$$ $\endgroup$ – user16651 Jul 25 '16 at 21:03
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    $\begingroup$ Set $f(s)=\ln s\in C(\mathbb{R} ^+) $ we have $f'(S_t)=\frac{1}{S_t}$ and $f''(S_t)=-\frac{1}{S_t^2}$ and $d[S_t,S_t]=\sigma^2S_t^2 dt$ $\endgroup$ – user16651 Jul 25 '16 at 21:09
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To get this term you need to take the log of S and to use Ito’s lemma, you can find a detailed explanation in this answer.

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Have you come across Ito lemma / Ito calculus?. As Gordon suggests: divide the top equation by St, so you get $\frac{dS_t}{S_t}$ and we look at 'by intuition' what does $dln(S_t)$ look like in the Ito world.

Ito states: $df(W,t)=\frac{\partial{f}}{\partial{W}}dW+\frac{\partial{f}}{\partial{t}}dt+\frac{1}{2}\frac{\partial^{2}f}{\partial{W^{2}}}dW^{2}$

Now from Ito: $d ln(S_t)=\frac{1}{S_t}dS_t - \frac{1}{2}\cdot \frac{1}{S_t^{2}}\cdot dS_t^{2} = \mu dt+\sigma dW_t-\frac{\sigma^2}{2}dt=(\mu-\frac{\sigma^2}{2})dt+\sigma dW_t$

We use in the second equality that $dS_t^{2}=\mu^{2}S_t^{2}d_t^{2}+2\mu\sigma dt\cdot dW_t+\sigma^{2}dW^{2}_t=\sigma^{2}dW^{2}_t=\sigma^{2}dt$ and substitute the original equation for $dS_t$.

From that it follows that

$ln(S_T)-ln(S_t)=ln(\frac{S_T}{S_t})=(\mu-\frac{\sigma^2}{2})(T-t)+\sigma(W_T-W_t)$ Then by taking $exp(x)$ of both sides of the last equality and multiply by $S_t$ you get the final formula.

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    $\begingroup$ Nice answer is actually here by @SRKX as well. quant.stackexchange.com/questions/1330/… $\endgroup$ – Jan Sila Jul 25 '16 at 20:19
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    $\begingroup$ The claim $d\ln S_t =\frac{dS_t}{S_t}$ and $dln(S_t,t)=\frac{1}{S_t}\times dS_t+\frac{1}{S_t}\times dt - \frac{1}{2}\times\frac{1}{S_t^{2}}\times dS_t^{2} $ do not appear correct to me. $\endgroup$ – Gordon Jul 25 '16 at 20:23
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    $\begingroup$ Thanks for correction, but why do you still say that "$dln(St)=\frac{dSt}{St}$ by chain rule"? $\endgroup$ – Gordon Jul 25 '16 at 20:31
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    $\begingroup$ Also $\frac{df^2}{dW_t^2}?$ $\endgroup$ – user16651 Jul 25 '16 at 20:35
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    $\begingroup$ Thanks @BehrouzMaleki ! Funny how sometimes you think you understand stuff, but you find out you dont only if you try to explain to someone else. Thanks a lot for the comments! $\endgroup$ – Jan Sila Jul 26 '16 at 17:22

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