Geometric Brownian motion (GBM) is the workhorse model of mathematical
finance — the price process at the heart of Black–Scholes. It is the
stochastic differential equation
dS_t = \mu S_t\,dt + \sigma S_t\,dW_t, \qquad S_0 > 0,
read as: over an instant, the price S_t drifts up at rate
\mu and is kicked by noise of size \sigma —
both proportional to the current price. That proportionality is the modelling insight:
a \$200 stock and a \$2 stock should move by similar percentages, not similar dollar
amounts. Dividing through, the SDE says the return
dS_t / S_t = \mu\,dt + \sigma\,dW_t has constant drift and volatility.
Here \mu is the expected return and
\sigma the volatility.
Unlike arithmetic Brownian motion, this SDE has state-dependent coefficients, so we cannot
integrate it directly. We solve it with the log-transform — and the answer will be clean,
explicit, and forever positive.
Solving GBM, line by line
We follow the
transform-and-integrate
recipe: apply Itô's lemma
to Y = \ln S, which turns the multiplicative noise into additive
noise we can integrate.
Step 1 — choose the transform and its derivatives. Let
f(s) = \ln s, so Y_t = f(S_t) with
f'(s) = \frac{1}{s}, \qquad f''(s) = -\frac{1}{s^2}.
Step 2 — write Itô's lemma. For a function of a single Itô process,
dY = f'(S)\,dS + \tfrac12 f''(S)\,(dS)^2. Substitute the derivatives:
dY = \frac{1}{S}\,dS - \frac{1}{2}\,\frac{1}{S^2}\,(dS)^2.
Step 3 — substitute dS = \mu S\,dt + \sigma S\,dW.
Into the first term:
\frac{1}{S}\,dS = \frac{1}{S}\big(\mu S\,dt + \sigma S\,dW\big) = \mu\,dt + \sigma\,dW.
Step 4 — compute (dS)^2 by the box
algebra. Squaring dS = \mu S\,dt + \sigma S\,dW and keeping
only the surviving term ((dW)^2 = dt, while
(dt)^2 and dt\,dW are higher order):
(dS)^2 = \sigma^2 S^2\,(dW)^2 = \sigma^2 S^2\,dt.
Step 5 — substitute that second-order term. The
S^2 cancels the 1/S^2:
-\frac{1}{2}\,\frac{1}{S^2}\,(dS)^2 = -\frac{1}{2}\,\frac{1}{S^2}\,\sigma^2 S^2\,dt = -\frac{1}{2}\sigma^2\,dt.
Step 6 — collect. Adding the first-order and second-order pieces, the price
S has disappeared entirely — constant coefficients:
dY = \Big(\mu - \tfrac12\sigma^2\Big)\,dt + \sigma\,dW.
Step 7 — integrate from 0 to
t. Both coefficients are constant, so the integrals are elementary —
the drift integral is (\mu - \tfrac12\sigma^2)t and the Itô integral
of a constant is \sigma(W_t - W_0) = \sigma W_t:
Y_t - Y_0 = \Big(\mu - \tfrac12\sigma^2\Big)t + \sigma W_t, \qquad\text{i.e.}\qquad \ln S_t = \ln S_0 + \Big(\mu - \tfrac12\sigma^2\Big)t + \sigma W_t.
Step 8 — exponentiate. Undo the logarithm to recover
S_t = e^{Y_t}:
S_t = S_0 \exp\!\Big(\big(\mu - \tfrac12\sigma^2\big)t + \sigma W_t\Big).
There it is — the closed-form solution. Three features jump out, and the theorem records them.
First, S_t is an exponential, so it is strictly
positive for all time, no matter what W_t does. Second, the
exponent is \ln S_0 + (\mu - \tfrac12\sigma^2)t + \sigma W_t, an affine
function of the normal variable W_t \sim N(0, t) — so
\ln S_t is normal, which makes S_t itself
lognormal. Third, the drift on the log scale is the curious
\mu - \tfrac12\sigma^2, not \mu — the
volatility drag, examined below.
The SDE dS_t = \mu S_t\,dt + \sigma S_t\,dW_t with
S_0 > 0 has the unique solution
S_t = S_0 \exp\!\Big(\big(\mu - \tfrac12\sigma^2\big)t + \sigma W_t\Big),
which has the following properties:
-
S_t > 0 for all t — a price can never
go negative.
-
\ln S_t \sim N\!\big(\ln S_0 + (\mu - \tfrac12\sigma^2)t,\; \sigma^2 t\big),
so S_t is
lognormal.
-
The mean grows at the full rate \mu:
\mathbb{E}[S_t] = S_0\,e^{\mu t}.
The -\tfrac12\sigma^2 in the exponent looks like it should drag the
mean down — but it is exactly cancelled when we take the expectation, and the cancellation is
not a coincidence. Write the exponent as m + \sigma W_t with
m = \ln S_0 + (\mu - \tfrac12\sigma^2)t. Then
\ln S_t \sim N(m,\, \sigma^2 t), so
S_t is lognormal, and the
lognormal mean
formula \mathbb{E}[e^{N(a, b^2)}] = e^{a + b^2/2} gives
\mathbb{E}[S_t] = \exp\!\Big(m + \tfrac12\sigma^2 t\Big) = \exp\!\Big(\ln S_0 + \big(\mu - \tfrac12\sigma^2\big)t + \tfrac12\sigma^2 t\Big).
The -\tfrac12\sigma^2 t from the GBM solution and the
+\tfrac12\sigma^2 t from the lognormal mean annihilate, leaving
\mathbb{E}[S_t] = S_0\,e^{\mu t}.
So \mu really is the expected growth rate of the price, even though
the log-price grows at the slower rate \mu - \tfrac12\sigma^2.
(Equivalently, S_t\,e^{-\mu t} is a martingale — the discounted price
is a fair game, the fact risk-neutral pricing is built on.)
Two growth rates live inside GBM, and confusing them is a classic, costly error. The
mean grows at \mu, but the
median grows at the slower \mu - \tfrac12\sigma^2,
since the median of S_t is
S_0\,e^{(\mu - \frac12\sigma^2)t} (the median of a lognormal is the
exponential of the normal's mean — no \tfrac12\sigma^2 lift). The gap
\underbrace{\mu}_{\text{mean rate}} \;-\; \underbrace{\big(\mu - \tfrac12\sigma^2\big)}_{\text{median rate}} \;=\; \tfrac12\sigma^2
is the volatility drag. The mean is inflated by a handful of very lucky paths
out in the lognormal's long right tail; the typical path — the median — grows more
slowly, and the more volatile the asset, the further the typical investor's outcome lags the
advertised average. High \sigma widens that gap quadratically. This
is the precise sense in which volatility is a cost, and it feeds directly into the
Black–Scholes model of option
pricing.