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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

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

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2 Answers

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

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

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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.

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