Geometric Brownian motion

This lesson builds the standard continuous-time model of a stock price, geometric Brownian motion (GBM). It combines the Brownian motion $W_t$ from "Random walks and Brownian motion" with the log returns and volatility of "Returns and volatility", and it is the continuous limit of the trees in "Multi-step trees and backward induction". The Black–Scholes formula in the next lesson is computed under this model.

Modelling returns instead of prices

The simplest idea would be to let the price itself follow a Brownian motion, $S_t = S_0 + \mu t + \sigma W_t$. This has two defects. First, $W_t$ is normally distributed, so the price can become negative. Second, a move of 1 means very different things for a stock at 10 and a stock at 1000. What should be random and comparable across stocks is the return, not the price change.

So we model the return over a short interval $dt$. The assumptions are:

• The expected return per unit time is a constant $\mu$ (the drift).
• The volatility of returns is a constant $\sigma$.
• The random part of the return comes from a Brownian motion: returns over non-overlapping intervals are independent and normally distributed.
• The stock pays no dividends and its price moves continuously, without jumps.

The model equation

These assumptions are written as

Read it as a recipe for one small step: over $dt$ the return has a deterministic part $\mu\,dt$ and a random part $\sigma\,dW_t$. Since $dW_t$ is normal with mean 0 and variance $dt$, the return over the step has mean $\mu\,dt$ and standard deviation $\sigma\sqrt{dt}$, which is the square-root-of-time rule from "Returns and volatility". Because both terms are proportional to $S_t$, price moves scale with the price level.

The solution

The equation has an explicit solution:

Taking logarithms, the log return over $[0, T]$ is

So the log price is a Brownian motion with a constant drift. A random variable whose logarithm is normal is called log-normal. Since $S_T$ is $S_0$ times an exponential, it is always positive: GBM fixes the first defect of the naive model.

The volatility correction

The drift of the log price is $\mu - \tfrac{1}{2}\sigma^2$, not $\mu$. The correction comes from a property of the normal distribution. If $Z \sim N(m, s^2)$, then

The exponential is convex, so large positive values of $Z$ are amplified more than negative ones are damped, and the mean of $e^Z$ exceeds $e^m$. Apply this with $m = (\mu - \tfrac{1}{2}\sigma^2)T$ and $s^2 = \sigma^2 T$:

The $-\tfrac{1}{2}\sigma^2$ is exactly what is needed so that the expected price grows at rate $\mu$, as assumed. The formal derivation uses Itô's lemma, the chain rule for functions of Brownian motion, which adds the term $-\tfrac{1}{2}\sigma^2\,dt$ when we pass from $dS/S$ to $d\ln S$.

An everyday version of the same effect is volatility drag. A price that rises 10% and then falls 10% ends at $1.1 \cdot 0.9 = 0.99$ of where it started: the average return is zero, but the compound growth is negative. The more volatile the returns, the larger the gap between the average return and the growth rate actually achieved.

Mean and median

Because the exponential is increasing, the median of $S_T$ is the exponential of the median of the log price, which equals its mean:

The mean lies above the median: the log-normal distribution is skewed to the right, with a long tail of large outcomes pulling the average up. A typical path ends near the median, below the mean.

Example: $S_0 = 100$, $\mu = 8\%$, $\sigma = 20\%$, $T = 1$. The mean is $100\,e^{0.08} \approx 108.33$, the median $100\,e^{0.06} \approx 106.18$. The log return is $N(0.06,\, 0.2^2)$, so the probability that the stock ends below 100 is $N(-0.06/0.2) = N(-0.3) \approx 38\%$, even though the expected return is positive. A 95% range for $S_T$ is $100\,e^{0.06 \pm 1.96 \cdot 0.2}$, about 71.7 to 157.1, which is not symmetric around the median.

The figure simulates 20 paths over one year from $S_0 = 100$, each built from many small log-normal steps, and draws the mean curve $S_0e^{\mu t}$ and the median curve $S_0e^{(\mu - \sigma^2/2)t}$. Raise $\sigma$ and note that the paths fan out, the median falls further below the mean, and more paths end below their start. Press "New paths" to draw a fresh sample.

Link to the binomial tree

In a CRR tree with $u = e^{\sigma\sqrt{\Delta t}}$ and $d = 1/u$, each step adds $\pm\sigma\sqrt{\Delta t}$ to the log price. That is the scaled coin-flip walk of "Random walks and Brownian motion". As the number of steps grows, the log price converges to a Brownian motion with volatility $\sigma$ plus a drift, and the terminal stock price becomes log-normal. GBM is the model that the trees approximate, which is why tree prices converge as $n$ grows.

The risk-neutral version

"Risk-neutral probabilities" showed that for pricing we use the measure $\mathbb{Q}$ under which the stock earns the risk-free rate. In GBM this means replacing $\mu$ by $r$ and $W$ by a Brownian motion $W^{\mathbb{Q}}$ under $\mathbb{Q}$:

The volatility $\sigma$ stays the same; only the drift changes. Then $\mathbb{E}^{\mathbb{Q}}[S_T] = S_0 e^{rT}$, and the discounted price $e^{-rt}S_t$ is a martingale under $\mathbb{Q}$, exactly as in the one-step tree. The real drift $\mu$ disappears from option prices, just as the real probability $p$ did.

Summary

• GBM models returns: $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$, with constant drift $\mu$ and volatility $\sigma$.
• Its solution is $S_T = S_0\exp((\mu - \tfrac{1}{2}\sigma^2)T + \sigma W_T)$: log-normal and always positive.
• The $-\tfrac{1}{2}\sigma^2$ term makes the mean $S_0e^{\mu T}$; the median $S_0e^{(\mu - \sigma^2/2)T}$ is lower.
• For pricing, replace $\mu$ by $r$; the volatility is unchanged.

Under $\mathbb{Q}$, $\ln S_T \sim N\big(\ln S_0 + (r - \tfrac{1}{2}\sigma^2)T,\; \sigma^2 T\big)$. The next lesson integrates the call payoff against this distribution to obtain the Black–Scholes formula.