stochastic calculus - How to get Geometric Brownian Motion's closed-form solution in Black-Scholes model?

The Black Scholes model assumes the following dynamics for the underlying, well known as the Geometric Brownian Motion: $$dS_t=S_t(\mu dt+\sigma dW_t)$$

Then the solution is given: $$S_t=S_0\,e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t}$$

It can be shown by Ito Lemma on function $f(t,W_t)=\ln S_t$ that this solution is correct as it leads to above dynamics.

But how do we solve the above SDE originally to find this solution?

Guessing the above solution to apply Ito seems unlikely to me.

Answer

If by 'solve' you mean how do we know that $\ln S_t$ is the right change of variable, then you can go by the following (not rigorous) line of thought:

• Ito's fomula suggests that given an SDE $$dX_t = \mu(X_t,t)dt+\sigma(X_t,t)dW_t$$ and a function $f(x,t)$: the SDE for the process $Y_t=f(X_t,t)$ will satisfy $$dY_t = [f_t(X_t,t) + f_x(X_t,t)\mu(X_t,t) + \frac{1}{2}f_{xx}(X_t,t)\sigma^2(X_t,t)]dt+f_x(X_t,t)\sigma(X_t,t)dW_t$$


• Now the SDE for the spot price is, as you wrote, given by $$dS_t = \mu S_tdt+\sigma S_t dW_t$$: hence a transformation $f$ applied to this SDE will have dynamics given by $$dY_t = [...]dt+f_x(S_t,t)\sigma S_t dW_t$$


• We want the transformation to kill the dependence of volatility on the spot, therefore we want to 'solve' something like '$f_x(S_t,t)\sigma S_t = const.$' which essentially means '$f_x(x,t) = \frac{1}{x}$'. This points towards the guess $$f(x,t)=\ln x$$.