Brownian Motion for Stock Modelling With a Discrete Walk Example

Standard Brownian motion

W_t

starts at zero, has continuous paths, and has independent increments with

W_t-W_s\sim N(0,t-s)

for

t>s

. In the geometric Brownian motion stock model, the Brownian increment drives random changes in the log price. Its standard deviation over a time interval

\Delta t

\sqrt{\Delta t}

Imagine taking ever smaller, independent random steps. If the variance of each step equals the time it takes, a year of accumulated steps still has variance one. A shorter interval contains less uncertainty, with a four-times shorter interval carrying half the standard deviation. This scaling helps explain why paths can be continuous and still wiggle sharply at small time scales.

In geometric Brownian motion, the stock changes proportionally to its current level. A $1\%$ move at $200$ is twice the dollar move at $100$ . Putting the Brownian term inside an exponential gives a positive price at any finite time, assuming a positive initial price.

From Brownian shocks to a positive stock price

Under a constant-drift, constant-volatility model,

dS_t=\mu S_t\,dt+\sigma S_t\,dW_t,\qquad S_0>0.

Applying Itô's formula to

\log S_t

subtracts half the instantaneous variance from the drift of the log price.

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

Integrating and exponentiating gives

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

The log return over time

t

is normal with mean

(\mu-\sigma^2/2)t

and variance

\sigma^2t

. The expected price is

S_0e^{\mu t}

, using

\mathbb E[e^{\sigma W_t}]=e^{\sigma^2t/2}

. For risk-neutral option pricing in this constant-rate model, use drift

r-q

, where

r

is the risk-free rate and

q

a continuous dividend yield. An empirical price simulation under the real-world measure uses an estimate or scenario for

\mu

instead.

The distinction between

\mu

and

\mu-\sigma^2/2

matters. In the model,

\mu

is the drift of the price relative to its current level. The logarithm curves down, so Itô's second derivative term reduces the drift of the log price. When you exponentiate a normal log return, its variance raises the mean price by the factor

e^{\sigma^2t/2}

. That factor cancels the drift correction in the expectation. It does not affect the median, because exponentiation preserves the order of outcomes.

Take $S_0=100$ , $\mu=0.06$ , and $\sigma=0.20$ , with time measured in years. After one year the log return has mean $0.04$ and standard deviation $0.20$ . The expected price is about $106.18$ , while the median price is about $104.08$ . The difference comes from the right-skewed distribution created by exponentiation.

From a discrete walk to a Brownian-scale path

import numpy as np

rng = np.random.default_rng(17)
steps, paths = 252, 5_000
dt = 1 / steps
mu, sigma, initial = 0.06, 0.20, 100
increments = rng.normal(0, np.sqrt(dt), (paths, steps))
log_returns = (mu - 0.5 * sigma**2) * dt + sigma * increments
prices = initial * np.exp(np.cumsum(log_returns, axis=1))
print(prices[:, -1].mean(), initial * np.exp(mu))

The Gaussian sampling gives the exact distribution at the grid dates for constant parameters. The finite sample still has Monte Carlo error.

The scaling in the discrete walk is deliberate. Divide a year into $n$ intervals of length $h=1/n$ and give each fair step magnitude $\sqrt h$ . One step has mean zero and variance $h$ . The sum over $k$ steps has variance $kh$ , the elapsed time. Giving each step magnitude $h$ instead would make the annual variance $nh^2=h$ , which vanishes on refinement. Holding the step magnitude fixed would make the variance diverge. That calculation explains the units of the Brownian term in the price model and why a day's volatility is not the annual volatility divided by 252.

Brownian Motion for Stock Modelling With a Discrete Walk Example

From Brownian shocks to a positive stock price

From a discrete walk to a Brownian-scale path

Explore a Brownian path at different resolutions

References and next steps