# Close form solution for Geometric Brownian Motion

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.

• 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
• @Donkey_JOHN Let $f(t,x)\in C^{1,2}([0,\infty)\times\mathbb{R})$ then $$df(t,W_t)=\frac{\partial f}{\partial t}(t,W_t)dt+\frac{\partial f}{\partial x}(t,W_t)dW_t+\frac{1}{2}\frac{\partial ^2f}{\partial x^2}(t,W_t)d[W_t,W_t]$$ – user16651 Jul 25 '16 at 20:44
• @Donkey_JOHN Also $d[W_t,W_t]=dt$ and $d[t,W_t]=0$ and $d[t,t]=0$ – user16651 Jul 25 '16 at 20:45
• Let $f(s)\in C^{2}(\mathbb{R})$ then $$df(S_t)=f'(S_t)dS_t+\frac{1}{2}f''(S_t)d[S_t,S_t]$$ – user16651 Jul 25 '16 at 21:03
• 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 $d[S_t,S_t]=\sigma^2S_t^2 dt$ – user16651 Jul 25 '16 at 21:09

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.

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.

• Nice answer is actually here by @SRKX as well. quant.stackexchange.com/questions/1330/… – Jan Sila Jul 25 '16 at 20:19
• 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
• 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
• Also $\frac{df^2}{dW_t^2}?$ – user16651 Jul 25 '16 at 20:35
• 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