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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!

Donkey_JOHN
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    Consider $d\ln S(t)$, and see what you can have. Should $r-\frac{1}{2}\sigma^2$ be $\mu-\frac{1}{2}\sigma^2$? – Gordon Jul 25 '16 at 19:44
  • Have you come across Ito lemma / Ito calculus?. As Gordon suggests: divide the top equation by St, so you get dSt/St which is same as differential of ln(St)...d ln(St)=dSt/St just chain rule. then use Ito lemma on ln(St) – Jan Sila Jul 25 '16 at 20:06
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    @Donkey_JOHN Let $f(t,x)\in C^{1,2}([0,\infty)\times\mathbb{R}) $ then $$df(t,W_t)=\frac{\partial f}{\partial t}(t,W_t)dt+\frac{\partial f}{\partial x}(t,W_t)dW_t+\frac{1}{2}\frac{\partial ^2f}{\partial x^2}(t,W_t)d[W_t,W_t]$$ –  Jul 25 '16 at 20:44
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    @Donkey_JOHN Also $d[W_t,W_t]=dt$ and $d[t,W_t]=0$ and $d[t,t]=0$ –  Jul 25 '16 at 20:45
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    Let $f(s)\in C^{2}(\mathbb{R}) $ then $$df(S_t)=f'(S_t)dS_t+\frac{1}{2}f''(S_t)d[S_t,S_t]$$ –  Jul 25 '16 at 21:03
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    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 $d[S_t,S_t]=\sigma^2S_t^2 dt$ –  Jul 25 '16 at 21:09
  • For more details read it http://faculty.math.tsinghua.edu.cn/~zliang/paper/Stochastic%20Differential%20Equations%20%20An%20Introduction%20with%20Applications.pdf –  Jul 25 '16 at 21:14
  • @BehrouzMaleki Thanks for your help, it solves my problem perfectly, and the paper you shared might be too deep for my level at the moment. Another question, what if you take the derivatives respect to both $t$ and $S(t)$? will it give you the same result? – Donkey_JOHN Jul 26 '16 at 09:18
  • $\ln S_t$ is not time-dependent in Ito lemma. Please check it http://quant.stackexchange.com/questions/28082/is-there-a-better-more-rigorous-explanation-for-why-this-partial-derivative-is/28083#28083 –  Jul 26 '16 at 09:40

2 Answers2

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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.

Community
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Malick
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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.

Jan Sila
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    Nice answer is actually here by @SRKX as well. http://quant.stackexchange.com/questions/1330/what-is-itos-lemma-used-for-in-quantitative-finance/1345#1345 – Jan Sila Jul 25 '16 at 20:19
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    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. – Gordon Jul 25 '16 at 20:23
  • you are right, there dln(St)/dt=0, my bad. Too hasty. – Jan Sila Jul 25 '16 at 20:27
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    Thanks for correction, but why do you still say that "$dln(St)=\frac{dSt}{St}$ by chain rule"? – Gordon Jul 25 '16 at 20:31
  • @Jan Sila your answer is not correct. –  Jul 25 '16 at 20:31
  • Donkey_JOHN is newbie, so you should explain correctly. What means chain rule? –  Jul 25 '16 at 20:33
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    Also $\frac{df^2}{dW_t^2}?$ –  Jul 25 '16 at 20:35
  • the $\frac{df^{2}}{dW_t^{2}}$ was typo. But what is meant to be wrong in my answer @BehrouzMaleki? I edited it to make more clear.. – Jan Sila Jul 25 '16 at 22:44
  • Change $\frac{df}{d(.)}\to \frac{\partial f}{\partial (.)}$. –  Jul 25 '16 at 22:47
  • look at comments of Donkey_JOHN's question –  Jul 25 '16 at 22:49
  • chain rule is so wrong –  Jul 25 '16 at 22:50
  • Fair enough, with $\partial$ but what is wrong with chain rule? They are composite functions, arent they? – Jan Sila Jul 25 '16 at 22:53
  • Ito calculus and Leibniz calculus are different –  Jul 25 '16 at 22:56
  • Heard about it, but probably have to read up on it. Our lecturer actually said exactly this, that dln(St)=dSt/St – Jan Sila Jul 26 '16 at 00:05
  • @JanSila Thanks for the link you shared, that solves my problem, you have to use the logorithm to make it easy to calculate. Thank you very much for your help! – Donkey_JOHN Jul 26 '16 at 09:12
  • @Donkey_JOHN ${d\ln y}=\frac{\frac{dy}{dx}}{y}$ but $d{\ln S_t}\ne\frac{dS_t}{S_t}$. Right? –  Jul 26 '16 at 09:22
  • @Donkey_JOHN Indeed $S_t$ is a stochastic process whereas $y$ is a deterministic function. –  Jul 26 '16 at 09:36
  • You are welcome @Donkey_JOHN, sorry for the confusion. Hope it is all right and acceptable answer now, I changed the notation and the bits commented by others now. :) – Jan Sila Jul 26 '16 at 15:28
  • it is good answer ( +1) –  Jul 26 '16 at 16:48
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    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! – Jan Sila Jul 26 '16 at 17:22
  • In $dln(S_T)=ln(S_T)-ln(S_t)=ln(\frac{S_T}{S_t})= \cdots$, the term $dln(S_T)$ is not needed. Moreover, if you put it there, the identity does not hole anymore. That is, $dln(S_T)\ne ln(S_T)-ln(S_t)$. – Gordon Jul 26 '16 at 17:55
  • @Gordon. could you further explain why the parameter tt in the process S(t)S(t) not considered, in my understanding, the process is time-dependent, so when you take the Ito's lemma, you should take both parameters S(t)S(t) and tt into account. I know my comments are wrong, but just wondering why, as it is not very intuitive for me. Thank you! – Donkey_JOHN Jul 27 '16 at 18:22
  • @Donkey_JOHN: It is based in Ito's lemma, no other particular reason. Ito's formula is in that form, that is, only the $t$ as an independent variable is considered for the derivative $\frac{\partial f}{\partial t}$. – Gordon Jul 27 '16 at 18:31