Geometric Brownian motion: why prices come out lognormal

If a price moves by small random percentage changes, what does its distribution ten years from now look like?

Under geometric Brownian motion, Itô's lemma solves the equation for the price exactly: the log of the price is normal, so the price itself is lognormal. Stage 1 asks you to compare two summaries of that future price before the page draws anything.

Before you start · 3 quick checks on Itô's lemma, normal variables and continuous compounding, about 3 min

Setup: the price model

Brownian motion. \(W_t\), \(t \ge 0\), is a standard Brownian motion: \(W_0 = 0\), its paths are continuous, and for \(s \lt t\) the increment \(W_t - W_s\) is normal with mean 0 and variance \(t - s\), independent of the path up to time \(s\) (step 9).
Parameters. \(S_0 \gt 0\) is today's price, \(\mu\) is a real number, the drift, and \(\sigma \gt 0\) is the volatility. Time \(t\) is in years. \(\mu\) is a rate per year; \(\sigma\) is an annualized volatility, measured per square root of a year, so that \(\sigma\sqrt t\) is a standard deviation over \(t\) years.
The equation. The price \(S_t\) solves the stochastic differential equation \[ dS_t = \mu S_t\,dt + \sigma S_t\,dW_t, \] which is shorthand for \(S_t = S_0 + \int_0^t \mu S_u\,du + \int_0^t \sigma S_u\,dW_u\) for every \(t \ge 0\), with the second integral an Itô integral (step 12).

Over a short step \(\Delta t\), the relative change \(\Delta S/S\) is then close to \(\mu\,\Delta t + \sigma\,\Delta W\): a drift proportional to the step plus a normal shock with standard deviation \(\sigma\sqrt{\Delta t}\). Changes are proportional to the current price, which is what "geometric" refers to. As a model of a stock price this assumes constant \(\mu\) and \(\sigma\), no jumps and no dividends; the fine print below lists the rest.

Lognormal. A positive random variable \(Y\) is lognormal when \(\ln Y\) is normal. Its median is the number \(m\) with \(P(Y \le m) = \tfrac12\).

What the model assumes about prices

Using this equation for a stock price assumes that \(\mu\) and \(\sigma\) stay constant, that the price moves continuously with no jumps, that the stock pays no dividends over the horizon, and that log returns over non-overlapping periods are independent and normal. Probabilities on this page are real-world probabilities, the ones that describe how the price actually behaves; pricing an option uses a different probability, the risk-neutral measure (step 17).

Real prices depart from the model: large moves are more frequent than a normal distribution allows, and volatility changes over time. The model is still the base case for option pricing and for simulating long-run wealth.

Paths on this page come from a seeded random number generator, so every visit shows the same paths. They are drawn exactly from the solution proved in stage 3, not by an approximate scheme: the paths of stage 1 at monthly times, and the separate set in stage 2 at yearly times.

1 See

The mean and the median ten years out

Start at \(S_0 = 100\). Choose a drift and a volatility. The page will simulate 2000 price paths over \(T = 10\) years with your choice.

At \(T = 10\)	Exact	2000 paths	Std. error
Mean \(\mathbb E[S_{10}]\)
Median of \(S_{10}\)			—

The curves are not paths

The mean curve \(t \mapsto \mathbb E[S_t]\) and the median curve are computed one time at a time, across all paths: they summarize the distribution of the price at each time, and neither is a realized path. A single path can stay above, below or between them, and it is the only thing an investor holding the stock ever sees. The mean is pulled up by the few paths that end very high: a price can lose at most all of its value, but it can gain without limit.

Key point For every drift and every \(\sigma \gt 0\), the mean price \(S_0e^{\mu T}\) lies above the median price \(S_0e^{(\mu - \frac12\sigma^2)T}\), by the factor \(e^{\frac12\sigma^2T}\). The two curves summarize the whole bundle; no path has to follow either one.

2 Try

Long horizons and the drift of the log

Below, \(\Phi\) is the standard normal CDF, \(\Phi(z) = P(Z \le z)\) for a standard normal \(Z\); the Black–Scholes lesson calls the same function \(N\).

Hit a target

Set the drift and the volatility so that the price at \(T = 10\) meets each goal. The figure shows the mean and the median of \(S_t\) for \(0 \le t \le 10\), with \(S_0 = 100\).

3 Prove

Solving the equation with Itô's lemma

Theorem (the solution of geometric Brownian motion)

Assumptions.

\(W\) is a standard Brownian motion with \(W_0 = 0\). \(S_0 \gt 0\), \(\mu\) is real and \(\sigma \gt 0\).
Itô's lemma (step 13): if \(dX_t = a_t\,dt + b_t\,dW_t\), with adapted \(a\) and \(b\) for which the integrals exist, and \(f\) is twice continuously differentiable on an open interval that contains every value of \(X\), then \(df(X_t) = f'(X_t)\,dX_t + \tfrac12 f''(X_t)\,(dX_t)^2\), with \((dX_t)^2 = b_t^2\,dt\).
Uniqueness, stated without proof: the coefficients \(\mu x\) and \(\sigma x\) are Lipschitz in \(x\), so the equation has at most one solution with continuous paths.

Conclusion. The solution is

\[ S_t = S_0\,e^{(\mu - \frac12\sigma^2)t + \sigma W_t}, \qquad t \ge 0. \]

For every \(t \gt 0\), \(\ln S_t\) is normal with mean \(\ln S_0 + (\mu - \tfrac12\sigma^2)t\) and variance \(\sigma^2t\), so \(S_t\) is lognormal. Its median is \(S_0e^{(\mu - \frac12\sigma^2)t}\) and its mean is \(\mathbb E[S_t] = S_0e^{\mu t}\).

In the proof, \(t \gt 0\) is a fixed time. Lines 1 and 2 find the formula by applying Itô's lemma to \(\ln S_t\), which needs a solution that stays positive. Line 3 then checks the formula directly, so the conclusion does not rest on that guess.

Scratch work, not part of the proof.

The factor \(e^{-\frac12\sigma^2 t}\) is the same one that makes \(e^{\sigma W_t - \frac12\sigma^2t}\) a martingale (step 11). So \(e^{-\mu t}S_t = S_0e^{\sigma W_t - \frac12\sigma^2t}\) is a martingale: the price discounted at its own drift is a fair game.

Key point Itô's lemma turns the price equation into \(d(\ln S_t) = (\mu - \tfrac12\sigma^2)\,dt + \sigma\,dW_t\), so the log price is a Brownian motion with drift. The \(-\tfrac12\sigma^2\) comes from the second-order term that the ordinary chain rule drops, and it produces the factor \(e^{\frac12\sigma^2t}\) between the mean and the median.

4 Apply

Problems

In every problem the price follows geometric Brownian motion, \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t\), with a constant drift \(\mu\) per year, a constant annualized volatility \(\sigma \gt 0\), no dividends, and time in years.