Calculating the Mean and Variance of Geometric Brownian Motion

For $dX_t=\mu X_t\,dt+\sigma X_t\,dW_t$ with constant coefficients and deterministic $X_0=x$ ,

\mathbb E[X_t]=xe^{\mu t},\qquad \operatorname{Var}(X_t)=x^2e^{2\mu t}(e^{\sigma^2t}-1).

You can derive these by writing down the lognormal solution or by applying Itô's lemma to

X_t^2

and solving two ordinary differential equations.

Find the first and second moments separately

Let $m_1(t)=\mathbb E[X_t]$ . Taking expectations in the integral form of the SDE gives $m_1'(t)=\mu m_1(t)$ and $m_1(0)=x$ . The stochastic integral has mean zero, since $X_t$ has finite second moments on finite intervals. This follows directly from the explicit exponential solution, which has finite moments of every positive order. Hence $m_1(t)=xe^{\mu t}$ .

Now set

m_2(t)=\mathbb E[X_t^2]

. The Itô product rule gives

d(X_t^2)=(2\mu+\sigma^2)X_t^2\,dt+2\sigma X_t^2\,dW_t.

The stochastic term is square integrable on finite horizons by finiteness of the fourth moment. This gives

m_2'(t)=(2\mu+\sigma^2)m_2(t)

with

m_2(0)=x^2

, so

m_2(t)=x^2e^{(2\mu+\sigma^2)t}

. Subtract

m_1(t)^2

to get the variance. As an algebra check,

t=0

\sigma=0

makes the variance vanish. If

\mu=0

, the expected level stays

x

while variance grows as

x^2(e^{\sigma^2t}-1)

The extra $\sigma^2$ in the second moment equation comes from $d[X]_t$ , which appears when you differentiate $X_t^2$ using Itô's rule. In the first moment equation there is no curvature term because the function of $X_t$ is linear. Once you have $\mathbb E[X_t^2]$ , subtract the square of the mean. The two terms both contain $x^2e^{2\mu t}$ , and their difference leaves $e^{\sigma^2t}-1$ . Setting $\sigma=0$ makes the variance vanish and recovers deterministic growth.

For an original exercise, take $x=80$ , $\mu=0.06$ and $\sigma=0.30$ . At $t=2$ , the expected level is $80e^{0.12}\approx90.20$ . The variance is

80^2 e^{0.24}(e^{0.18}-1)\approx1604.56,

so the standard deviation is about

40.06

. The log-price variance over this horizon is

0.18

, a dimensionless quantity. The price-level variance has units of price squared. If you want variance of the simple return

X_2/80-1

, divide the price variance by

80^2

to obtain about

0.251

For a forecast made at time $s$ , replace $x$ by the observed $X_s$ and $t$ by the horizon $t-s$ . The future Brownian increment is independent of $\mathcal F_s$ , so the same two formulas give conditional moments. With a positive observed level, doubling the horizon changes the accumulated variance inside the exponential and also changes the expected price around which dispersion is measured.

The closed-form solution gives a second route to the same result. To get the second raw moment, square the entire exponential, including its Brownian coefficient. If $W_t\sim N(0,t)$ , then $E[e^{2\sigma W_t}]=e^{2\sigma^2t}$ . The deterministic part of the squared solution contributes $x^2 e^{(2\mu-\sigma^2)t}$ . Multiplying the two factors gives $E[X_t^2]=x^2 e^{(2\mu+\sigma^2)t}$ . This check catches the common algebraic slip of doubling the variance adjustment when the coefficient inside a Gaussian exponential is doubled.

Calculating the Mean and Variance of Geometric Brownian Motion

Find the first and second moments separately

Change the drift and volatility of geometric Brownian motion

References and next steps