Geometric Brownian motion and exact distributions

Definition Let $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t \ge 0}, \mathbb{P})$ be a filtered probability space carrying a standard Brownian motion $W$, and fix constants $\mu \in \mathbb{R}$ and $\sigma \ge 0$ and a starting price $S_0 > 0$. A geometric Brownian motion is a continuous adapted process $S$ with

which means that for every $t \ge 0$

Time is measured in years, $\mu$ is a rate per year and $\sigma$ a volatility per square root of a year. The coefficients $\mu x$ and $\sigma x$ are globally Lipschitz with linear growth, so the existence and uniqueness theorem of the lesson on stochastic differential equations gives a strong solution that is pathwise unique. The meaning of $\mu$ depends on the measure. Under $\mathbb{P}$ it is an assumed physical drift of the ex-dividend price. Under a pricing measure $\mathbb{Q}$ with the money-market numeraire $B_t = e^{rt}$, a stock with a constant continuous dividend yield $q$ follows the same equation with $\mu$ replaced by $r - q$ and $W$ by a $\mathbb{Q}$-Brownian motion $W^{\mathbb{Q}}$, as in the lesson on Girsanov’s theorem. The volatility $\sigma$ is the same under both measures, because an equivalent change of measure of this kind changes only the drift.

Definition For constants $\mu$ and $\sigma$ and a starting price $S_0 > 0$, the unique strong solution of the GBM equation is

To find it, apply Itô’s formula to $\ln x$, whose first derivative is $1/x$ and second derivative $-1/x^2$, while $S$ is positive. With $d[S]_t = \sigma^2 S_t^2\,dt$,

and integrating from $0$ to $t$ gives $\ln(S_t/S_0) = \nu t + \sigma W_t$. That step assumed positivity, so the candidate is then checked directly. For $f(t, w) = S_0 \exp(\nu t + \sigma w)$ the partial derivatives are $f_t = \nu f$, $f_w = \sigma f$ and $f_{ww} = \sigma^2 f$, and Itô’s formula for $S_t = f(t, W_t)$ gives

Pathwise uniqueness identifies this process as the solution. It is strictly positive at every time on every path, since the exponential of a finite number is positive, and its logarithm $\ln S_t = \ln S_0 + \nu t + \sigma W_t$ is a Brownian motion with drift $\nu$ and volatility $\sigma$. With deterministic coefficients $\mu(t)$ and $\sigma(t)$ satisfying $\int_0^T \lvert \mu(u) \rvert\,du < \infty$ and $\int_0^T \sigma(u)^2\,du < \infty$, the same steps give

which is still normal, so the price is still lognormal. With random adapted coefficients the same exponential still solves the equation, but the stochastic integral is in general no longer normal and the price is no longer lognormal. Jumps, default and stochastic volatility are outside this model.

Definition Let $\sigma > 0$ and $t > 0$. Since $W_t \sim N(0, t)$, the exact solution gives

so $S_t$ is lognormal, with CDF $\mathbb{P}(S_t \le s) = \Phi\bigl(z_t(s)\bigr)$ and density

for $s > 0$, and $p_t(s) = 0$ for $s \le 0$, where

The density is the derivative of the CDF, and the chain rule contributes $z_t'(s) = 1/(s\sigma\sqrt{t})$, which contains the Jacobian $1/s$ of the logarithm. The same solution started at a later date gives the conditional transition. For $0 \le t < u$,

where $W_u - W_t \sim N(0, u - t)$ is independent of $\mathcal{F}_t$. Conditionally on $\mathcal{F}_t$, therefore, $S_u$ has the density above with $S_0$ replaced by $S_t$ and $t$ by $u - t$. This conditional law depends on the past only through $S_t$ and on time only through $u - t$, so $S$ is a time-homogeneous Markov process. When $\sigma = 0$ or $t = 0$ the price $S_t = S_0 e^{\mu t}$ is a constant, a single atom with no density, and the density formula does not apply.

Definition For every real number $p$ and every $t \ge 0$, the power $S_t^p$ has the finite mean

To prove it, write $S_t^p = S_0^p \exp(p\nu t + p\sigma W_t)$ and use the Gaussian moment generating function $\mathbb{E}[e^{a W_t}] = e^{a^2 t/2}$, valid for every real $a$, with $a = p\sigma$. The exponent becomes $p\nu t + \tfrac12 p^2 \sigma^2 t = \kappa(p)\, t$. Negative powers are included, because $S_0 > 0$ and the Gaussian moment generating function is finite everywhere. In the same way $\mathbb{E}[S_u^p \mid \mathcal{F}_t] = S_t^p\, e^{\kappa(p)(u - t)}$ for $t \le u$. The cases $p = 1$ and $p = 2$ give

Since $\ln S_t$ is normal and the exponential is increasing, the median of $S_t$ is $S_0 e^{\nu t}$, strictly below the mean whenever $\sigma > 0$ and $t > 0$. Under a pricing measure with drift $r - q$, the case $p = 1$ gives $\mathbb{E}^{\mathbb{Q}}[S_T \mid \mathcal{F}_t] = S_t e^{(r - q)(T - t)}$, the forward price. All power moments are finite, yet $\mathbb{E}[e^{\theta S_t}] = \infty$ for every $\theta > 0$ when $\sigma > 0$ and $t > 0$, because the lognormal right tail decays more slowly than every exponential $e^{-\theta s}$.

Definition Let $0 = t_0 < t_1 < \dots < t_n = T$ be a time grid with steps $h_k = t_{k+1} - t_k$, and let $Z_1, \dots, Z_n$ be independent standard normal variables. The exact sampling scheme sets $\hat S_0 = S_0$ and

By the transition formula and the independence of Brownian increments, the vector $(\hat S_1, \dots, \hat S_n)$ has exactly the joint law of $(S_{t_1}, \dots, S_{t_n})$, whatever the step sizes. A Monte Carlo average of a function of these grid values therefore has sampling error only, with no time-discretization bias. The Euler scheme

approximates the equation over each step instead. Its law differs from the exact one, its one-step conditional mean is $\tilde S_k(1 + \mu h_k)$ in place of $\tilde S_k e^{\mu h_k}$, and it is negative whenever $Z_{k+1} < -(1 + \mu h_k)/(\sigma\sqrt{h_k})$, an event of positive probability when $\sigma > 0$. Exactness holds at the grid dates only. Functionals of the continuous path, such as its running minimum or a continuously monitored barrier, are not determined by the grid values, and the minimum over the grid is never below the minimum of the path.