Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Stock price has a classic model based on GBM:

$$dS = \mu S dt + \sigma S dW$$

based on this call options values could be solve -- Black-Scholes formula.

But, what is the solution for the Stock price itself? is it

$$S(t) = S(0) e^{\mu t + \sigma W}$$ ?

share|cite|improve this question
up vote 3 down vote accepted

Your guess would have been true if $dW$ were just an ordinary differential. But this is not the case. $dW$ behaves somewhat different from the ordinary differential, in that intuitively we have $dW \approx \sqrt{dt}$ unlike to ordinary case. That's why the Ito,s formula is so important.

Now let us make a heuristic calculation. Let $Z_t = f(S_t) = \log S_t$, where $f(x) = \log x$. Then by the Ito's formula, we have

$$ Z_t - Z_0 = f(S_t) - f(S_0) = \int_{0}^{t} f'(S_s) \; dS_s + \int_{0}^{t} \frac{1}{2} f''(S_s) \; dS_s^2,$$

where $dS^2$ is a formal symbol given by the product rule

$$dW_s^2 = ds \quad \text{and} \quad dW_s ds = ds dW_s = 0 = ds^2.$$

Thus $dS_s^2 = \sigma^2 S_s^2 dW_s^2$, and hence

$$ \begin{align*} Z_t - Z_0 & = \int_{0}^{t} \frac{1}{S_s} \; (\mu S_s ds + \sigma S_s dW_s) + \int_{0}^{t} \left( -\frac{1}{2S_s^2}\right) \; (\sigma^2 S_s^2 ds) \\ & = \left( \mu - \frac{\sigma^2}{2}\right) t + \sigma W_t. \end{align*}$$

Plugging $Z_t = \log S_t$ back and exponentiating, we obtain

$$ S_t = S_0 \exp \left[ \left( \mu - \tfrac{\sigma^2}{2}\right) t + \sigma W_t \right]$$

share|cite|improve this answer

The SDE can solved analytically and the solution is the Geometric Brownian Motion which has the form: $$ S(t)=S(0)\exp\left(\left[\mu-\frac{\sigma^2}{2}\right]t+\sigma W(t)\right), $$ where $(W(t))_{t\geq 0}$ is a Brownian motion.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.