Chapter 08

Medium

Geometric Brownian Motion

00 · Symbol Glossary

$S_t$asset price at time $t$

The process this chapter models. $S_t>0$ for all $t$ — geometric Brownian motion is built specifically to never hit zero, unlike a plain Brownian motion.

$dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$the GBM stochastic differential equation

Both drift and diffusion scale with the current level $S_t$ — a $1\%$ move is equally likely at any price level, which is the defining feature of multiplicative noise (Chapter 07).

$\mu$drift rate (expected return)

The instantaneous expected percentage growth rate. Note carefully: $\mu$ is not the growth rate of $\mathbb{E}[\log S_t]$ — Section 03 shows those two differ by $\sigma^2/2$ .

$\sigma$volatility

The instantaneous standard deviation of percentage changes. In this chapter $\sigma$ is a constant; letting it depend on $S_t$ or $t$ produces local- and stochastic-volatility models, outside this chapter's scope.

01 · The Model

Definition — Geometric Brownian Motion

$S_t$ follows geometric Brownian motion if it solves

dS_t = \mu S_t\,dt + \sigma S_t\,dW_t, \qquad S_0>0 \text{ given}

with constants $\mu\in\mathbb{R}$ and $\sigma>0$ .

Plain language: over an instant $dt$ , $S_t$ changes by a random percentage — expected value $\mu\,dt$ , standard deviation $\sigma\sqrt{dt}$ — rather than a random absolute amount. That is exactly the property a stock price should have: a $1 move means something different for a $5 stock than for a $500 stock, but a $1\%$ move means the same thing for both. Both $\mu$ and $\sigma$ satisfy the Lipschitz and linear-growth conditions of Chapter 07 trivially (they are linear in $x$ ), so a unique strong solution exists — the question is only what it looks like in closed form.

02 · Solving via Itô's Lemma on $\log S_t$

The SDE for $S_t$ has multiplicative noise, which Chapter 07 flagged as the case to attack with a log transform.

Step-by-step — Solving the GBM equation

Apply Itô's Lemma to $f(S_t)=\log S_t$ : with

f_x=1/x

f_{xx}=-1/x^2

, and

\mu_t=\mu S_t

\sigma_t=\sigma S_t

from the SDE,

d(\log S_t) = \left(\mu - \tfrac12\sigma^2\right)dt + \sigma\,dW_t

— the identity already derived in Chapter 06, Exercise 6.2.

Integrate: the right side is a constant drift plus

\sigma

times Brownian motion, both of which integrate directly (no further Itô correction, since the coefficients are now constants, not functions of

S_t

\log S_t - \log S_0 = \left(\mu-\tfrac12\sigma^2\right)t + \sigma W_t

Exponentiate:

S_t = S_0\exp\!\left(\left(\mu-\tfrac12\sigma^2\right)t + \sigma W_t\right)

Sanity-check positivity: since

\exp(\cdot)>0

always and

S_0>0

by assumption,

S_t>0

for every

t

with probability one — the promised property.

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

❌ Solving GBM as if it were a plain exponential ODE

Ignoring the noise term, the deterministic analogue $dS_t=\mu S_t\,dt$ solves to $S_t=S_0 e^{\mu t}$ — and it is tempting to graft $\sigma W_t$ onto this by writing $S_t = S_0 e^{\mu t + \sigma W_t}$ , dropping the $-\sigma^2/2$ correction.

Why it breaks: this is exactly the naive-chain-rule error flagged in Chapter 06 — it discards the $\tfrac12 f_{xx}\,dt$ term from applying Itô's Lemma to $\log S_t$ , treating the SDE as if ordinary calculus applied to it.

Consequence: the resulting process has drift $\mu+\sigma^2/2$ under expectation, not $\mu$ (Section 03 makes this precise) — an asset priced with this formula would be systematically overvalued relative to the model's own stated drift, more so as $\sigma$ grows.

03 · The Lognormal Distribution

Because $\log S_t$ is $\log S_0$ plus a deterministic drift plus $\sigma W_t$ , and $W_t\sim N(0,t)$ , $\log S_t$ is exactly Gaussian.

Definition — Distribution of $S_t$

\log S_t \;\sim\; N\!\left(\log S_0 + \left(\mu-\tfrac{\sigma^2}{2}\right)t,\; \sigma^2 t\right)

so $S_t$ is lognormally distributed. Its moments are

\mathbb{E}[S_t] = S_0 e^{\mu t}, \qquad \mathrm{Var}(S_t) = S_0^2 e^{2\mu t}\left(e^{\sigma^2 t}-1\right)

Plain language: $\mathbb{E}[S_t]=S_0e^{\mu t}$ recovers the deterministic growth rate $\mu$ exactly — so $\mu$ is the expected-value growth rate after all, but the median (and every individual path's typical behavior) grows at the smaller rate $\mu-\sigma^2/2$ . The gap between the two is entirely a consequence of Jensen's inequality applied to the convex function $\exp$ : averaging after exponentiating gives a larger number than exponentiating the average. Higher $\sigma$ pulls the median further below the mean without changing the mean itself, which is exactly the FailBlock's warning in numerical form.

Example — Median vs. mean for a volatile stock

$S_0=100$ , $\mu=0.08$ , $\sigma=0.40$ , $t=1$ . Mean: $100e^{0.08}\approx 108.33$ . Median (i.e. $S_0e^{(\mu-\sigma^2/2)t}$ ): $100e^{0.08-0.08}=100$ . At $40\%$ annual volatility, the median outcome after one year is unchanged from today, even though the mean is up $8\%$ — the average is pulled upward by a thin right tail of large outcomes, not by typical paths.

04 · Why GBM Is the Default Stock Price Model

GBM is not the only reasonable model, but three properties make it the default starting point. First, $S_t>0$ always — prices with limited liability cannot go negative, and GBM enforces that structurally rather than by truncation. Second, returns are stationary and scale-free: the distribution of $S_{t+h}/S_t$ depends only on $h$ , not on the current level $S_t$ or on $t$ itself, matching the rough empirical fact that a stock's percentage volatility does not depend on whether it trades at $10 or $1,000. Third, it is the unique diffusion consistent with these two properties under constant $\mu,\sigma$ — any model with proportional drift and proportional volatility, driven by Brownian motion, is GBM by construction.

These are also exactly the properties that make $\log S_t$ , not $S_t$ , the natural modeling object — the same conclusion Time Series Chapter 01 reaches from a purely empirical, discrete-time argument about stabilizing variance.

Where GBM is known to be wrong

Real returns have heavier tails than the lognormal predicts (large single-day moves are more frequent than GBM implies), volatility clusters and is not constant (the GARCH phenomenon of the Time Series subject, Chapter 06 there), and $\sigma$ implied from option prices varies by strike (the "volatility smile," addressed by local- and stochastic-volatility extensions, outside this chapter). GBM remains the default because it is the simplest model consistent with no-arbitrage and lognormal option pricing (Chapter 12), not because it is empirically exact.

05 · Exercises

EXERCISE 8.1

Apply the closed-form solution of Section 02 directly with the given numbers.

$S_1 = 50\exp[(0.10-0.5(0.30)^2)(1)+0.30 W_1] = 50\exp[0.055+0.30W_1]$ . This is a specific number only once $W_1$ (a $N(0,1)$ draw) is specified; as a random variable, $\log S_1\sim N(\log 50+0.055,\,0.09)$ .

For $S_0=50$ , $\mu=0.10$ , $\sigma=0.30$ , write $S_1$ explicitly in terms of $W_1$ using the closed-form solution of Section 02.

EXERCISE 8.2

Use the moment formulas of Section 03; the drift $\mu$ that keeps $\mathbb{E}[S_t]$ constant is the one that makes $S_t$ a martingale.

$\mathbb{E}[S_t]=S_0e^{\mu t}$ is constant in $t$ only if $\mu=0$ . Under $\mu=0$ , $S_t$ is a martingale: $\mathbb{E}[S_t\mid\mathcal{F}_s]=S_s$ for $s<t$ , since the entire drift has been removed and only the driftless piece $\sigma W_t$ (composed with the exponential's own martingale structure from Chapter 06, Exercise 6.3) remains. This is the driftless case that risk-neutral pricing (Chapter 13) constructs by choice of measure, not by literally setting the real-world $\mu$ to zero.

For what value of $\mu$ is $\mathbb{E}[S_t]$ constant in $t$ ? What extra property does $S_t$ have under that value of $\mu$ ?

EXERCISE 8.3

Median means $P(S_t\leq m)=1/2$ ; use that $\log S_t$ is Gaussian and symmetric about its mean.

Since $\log S_t\sim N(\log S_0+(\mu-\sigma^2/2)t,\sigma^2 t)$ is symmetric about its mean, the median of $\log S_t$ equals that mean, so the median of $S_t$ is $S_0e^{(\mu-\sigma^2/2)t}$ — matching Section 03's claim. As $\sigma\to\infty$ with $\mu,t$ fixed, the median $\to 0$ while $\mathbb{E}[S_t]=S_0e^{\mu t}$ stays fixed: extreme volatility drags almost every path toward zero while a shrinking-probability tail of enormous outcomes keeps the mean unchanged.

Derive the median of $S_t$ from the lognormal distribution in Section 03. What happens to the median as $\sigma\to\infty$ with $\mu$ and $t$ fixed, holding the mean fixed by comparison?

EXERCISE 8.4

GBM's returns over disjoint intervals are functions of disjoint pieces of $W_t$ ; use independence of Brownian increments (Chapter 02).

$S_{t+h}/S_t = \exp[(\mu-\sigma^2/2)h+\sigma(W_{t+h}-W_t)]$ , a function of the increment $W_{t+h}-W_t$ alone — not of $W_t$ itself. Since Brownian increments over disjoint intervals are independent (Chapter 02), $S_{t+h}/S_t$ is independent of $\{S_s : s\leq t\}$ , and its distribution depends only on $h$ . This is the precise sense in which "GBM returns are stationary and scale-free," claimed informally in Section 04.

Show directly from the closed-form solution that the ratio $S_{t+h}/S_t$ is independent of $S_t$ and has a distribution depending only on $h$ .

06 · Chapter Summary

Concept	Meaning
GBM SDE	$dS_t=\mu S_t\,dt+\sigma S_t\,dW_t$
Closed-form solution	$S_t=S_0\exp[(\mu-\sigma^2/2)t+\sigma W_t]$
Distribution	$\log S_t$ Gaussian; $S_t$ lognormal
Mean	$S_0 e^{\mu t}$
Median	$S_0 e^{(\mu-\sigma^2/2)t}$ , below the mean when $\sigma>0$
Why the default	positivity, scale-free returns, uniqueness under those constraints

Next: Chapter 09 — Mean-Reverting SDEs: The Ornstein–Uhlenbeck Process, where multiplicative GBM growth is traded for a pull back toward a fixed level.