Deriving Geometric Brownian Motion with Itô’s Lemma

# Close form solution for Geometric Brownian Motion


# 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.

Your answer and detailed explanation will be greatly appreciated.

Thanks in advance!

## Answer by Malick (score 0)

https://quant.stackexchange.com/a/32582

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.

## Answer by Jan Sila (score -1)

https://quant.stackexchange.com/a/28273

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.