# Deriving Geometric Brownian Motion's solution?

The Black Scholes model assumes the following underlying dynamics, known as 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.

By applying Ito's Lemma on $f(t,x) = \ln x$ we get $$d(\ln S_t) = \frac{1}{S_t}dS_t - \frac{1}{2}\frac{1}{S_t^2}dS_tdS_t$$ Substituting the expression for $dS_t$ we obtain $$d(\ln S_t) = (\mu-\frac{1}{2}\sigma^2)dt + \sigma dW_t$$

This is an expression that you can integrate in a straightforward way if you know $\mu$ and $\sigma$. That leads to the expression for $S_t$ you had in your question.

Although very few SDEs have exact solutions, those that do are often found using some ansatz -- as is common for deterministic ODEs.

Without making the a priori choice of $\ln(S_t)$ you would use an ansatz like

$$S_t = X_te^{\sigma W_t},$$

where $X_t$ is a deterministic function. Apply Ito's lemma to $f(t,w) = A_te^{\sigma w}.$ Then

$$d(X_te^{\sigma W_t}) = \frac{dX_t}{dt}e^{\sigma W_t}dt + \sigma X_te^{\sigma W_t}dW_t+\frac1{2}\sigma^2X_te^{\sigma W_t}dt.$$

Comparing with the SDE for $S_t$, we find

$$\frac1{X_t}\frac{dX_t}{dt}+\frac1{2}\sigma^2= \mu,$$

with solution

$$X_t = X_0e^{(\mu - \frac1{2}\sigma^2)t}.$$

Then $S_0 = X_0e^{\sigma W_0}=X_0$ and the solution is

$$S_t = S_0e^{(\mu - \frac1{2}\sigma^2)t}e^{\sigma W_t}.$$

• I solve the differential equation from the beginning. T is the maturation time of the credit. In the Vasicek model the solution is the same except that there is an exponent term that is $\sqrt{T}$ obtained $e^{\sigma W_t \sqrt{T}}$. Do you know how to get that missing term in this case? Apr 21, 2019 at 12:23