Geometric Brownian Motion

The standard model for asset prices: constant proportional drift and volatility driving log-normal prices.

1. The idea

Geometric Brownian Motion (GBM) models a price $S_{t}$ via the stochastic differential equation $d S_{t} = μ S_{t} d t + σ S_{t} d W_{t}$ , meaning both the drift and the noise scale proportionally with the current price — a natural model for returns rather than absolute price changes.

Applying Itô's lemma to $lo g S_{t}$ removes the proportionality, giving an explicit solution: $S_{t} = S_{0} exp ((μ - \frac{1}{2} σ^{2}) t + σ W_{t})$ , so $lo g S_{t}$ is Normally distributed.

The $- \frac{1}{2} σ^{2} t$ term (the Itô correction) means the expected log-return is less than $μ t$ , even though the expected price itself grows at exactly rate $μ$ : $E [S_{t}] = S_{0} e^{μ t}$ .

2. Why it matters

GBM is the process underlying Black-Scholes option pricing, and recognizing that log-prices are Normal (not prices themselves) is essential for correctly computing probabilities and expectations of terminal stock prices.

3. Core mathematics

SDE

d S_{t} = μ S_{t} d t + σ S_{t} d W_{t}

Closed-form solution

S_{t} = S_{0} exp ((μ - \frac{1}{2} σ^{2}) t + σ W_{t})

Log-price distribution

lo g S_{t} \sim N (lo g S_{0} + (μ - \frac{1}{2} σ^{2}) t, σ^{2} t)

Expected price

E [S_{t}] = S_{0} e^{μ t}

Variance of price

Var (S_{t}) = S_{0}^{2} e^{2 μ t} (e^{σ^{2} t} - 1)

4. How to recognize it

Think of this when…

Asset prices described with constant expected return $μ$ and volatility $σ$
SDEs of the form $d S = μ S d t + σ S d W$
Questions asking for the distribution of $lo g S_{t}$ or the probability the stock exceeds some level at time $T$
Requests for $E [S_{t}]$ that require distinguishing it from the median (which uses no Itô correction)

5. Worked example

Worked example

A stock follows GBM with $S_{0} = 100$ , $μ = 0.08$ , $σ = 0.2$ . Find $E [S_{1}]$ and the median of $S_{1}$ .

1
Expected price uses the simple drift formula, with no volatility correction.
$E [S_{1}] = 100 e^{0.08} \approx 100 \times 1.0833 \approx 108.33$
2
The median of a log-normal equals $exp$ of the mean of the log, i.e. uses the Itô-corrected drift.
$median (S_{1}) = 100 exp (0.08 - \frac{1}{2} (0.2)^{2}) = 100 e^{0.06}$
3
Evaluate numerically.
$100 e^{0.06} \approx 106.18$
Answer: $E [S_{1}] \approx 108.33$ ; median $(S_{1}) \approx 106.18$ — the mean exceeds the median because of log-normal skew.

6. Common mistakes

Using $μ t$ instead of $(μ - σ^{2} /2) t$ when computing the mean of $lo g S_{t}$
Confusing $E [S_{t}] = S_{0} e^{μ t}$ with the median, which uses the Itô-corrected exponent
Treating $S_{t}$ itself as Normal rather than $lo g S_{t}$
Forgetting that $σ^{2} t$ , not $σ^{2}$ , is the variance of $lo g S_{t}$ over horizon $t$

7. Quick check

Under GBM with $μ = 0.1, σ = 0.3$ , what is the mean of $lo g (S_{1} / S_{0})$ ?

Why is $E [S_{t}] = S_{0} e^{μ t}$ even though $lo g S_{t}$ has mean $lo g S_{0} + (μ - σ^{2} /2) t$ ?

8. Practice

3 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Change Variables / Normalize No-Arbitrage Replication

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