Moments Of Geometric Brownian Motion

What the Moments Measure

The mean of geometric Brownian motion describes its average level, while its second moment and variance quantify the spread around that average. The second moment is not the variance: the variance subtracts the square of the mean. This distinction matters in applications where a model’s expected value and uncertainty enter different calculations.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ with respect to the filtration. In particular, for $u>t$, $B_u-B_t$ is independent of $\mathcal{F}_t$. Fix constants $s_0>0$, $\mu\in\mathbb{R}$, and $\sigma\in\mathbb{R}$. The geometric Brownian motion $S_t$ satisfies the SDE with drift coefficient $\mu S_t$ and diffusion coefficient $\sigma S_t$:

dS_t=\mu S_t\,dt+\sigma S_t\,dB_t,\qquad S_0=s_0.

Here $S$ is adapted and continuous. Its strong solution, meaning a solution constructed on this probability space using this Brownian motion, has the explicit form $S_t=s_0\exp((\mu-\tfrac12\sigma^2)t+\sigma B_t)$. This is the solution obtained in the earlier material on geometric Brownian motion. The coefficients $x\mapsto\mu x$ and $x\mapsto\sigma x$ are globally Lipschitz and have linear growth: their squared differences are bounded by $(\mu^2+\sigma^2)|x-y|^2$, and their squared values sum to $(\mu^2+\sigma^2)|x|^2$. The earlier existence and uniqueness result therefore applies.

Definition: For a random variable $X$ with finite second moment, its mean is $\mathbb{E}[X]$, its second moment is $\mathbb{E}[X^2]$, and its variance is $\operatorname{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X])^2$.

Moments from the Lognormal Law

For $t>0$, $B_t$ has the normal distribution $N(0,t)$, so the explicit solution shows that $S_t$ is lognormal. The same moment formulas hold at $t=0$ by direct substitution. The Gaussian moment-generating function is $\mathbb{E}[e^{aB_t}]=e^{a^2t/2}$ for real $a$; it follows by completing the square in the normal density.

Theorem: For $t\ge0$, geometric Brownian motion has $$ \mathbb{E}[S_t]=s_0e^{\mu t},\qquad \mathbb{E}[S_t^2]=s_0^2e^{(2\mu+\sigma^2)t},\qquad \operatorname{Var}(S_t)=s_0^2e^{2\mu t}\big(e^{\sigma^2t}-1\big). $$

Proof. Put $a=\mu-\tfrac12\sigma^2$. From the explicit solution and the Gaussian moment-generating function,

\mathbb{E}[S_t] =s_0e^{at}\mathbb{E}[e^{\sigma B_t}] =s_0e^{at}e^{\sigma^2t/2} =s_0e^{\mu t}.

Squaring the solution before taking expectations gives

\mathbb{E}[S_t^2] =s_0^2e^{2at}\mathbb{E}[e^{2\sigma B_t}] =s_0^2e^{2at}e^{(2\sigma)^2t/2} =s_0^2e^{(2\mu+\sigma^2)t}.

Subtracting the square of the mean yields $\operatorname{Var}(S_t)=s_0^2e^{(2\mu+\sigma^2)t}-s_0^2e^{2\mu t} =s_0^2e^{2\mu t}(e^{\sigma^2t}-1)$. These expressions are finite for every finite $t$, so the first two moments exist. At $t=0$, they give $s_0$, $s_0^2$, and $0$, respectively, as required for the deterministic initial value. $\square$

Worked Example: Nonzero Drift and Volatility

Let $s_0=2$, $\mu=\tfrac13$, $\sigma=\tfrac12$, and $t=3$. Compute the mean, second moment and variance.

For the mean, substitute $\mu t=\tfrac13\cdot3=1$, so $\mathbb{E}[S_3]=2e$. For the second moment, $2\mu+\sigma^2=\tfrac23+\tfrac14=\tfrac{11}{12}$, and $(2\mu+\sigma^2)t=\tfrac{11}{12}\cdot3=\tfrac{11}{4}$. Thus $\mathbb{E}[S_3^2]=4e^{11/4}$.

The variance is $\operatorname{Var}(S_3)=4e^{2\mu t}(e^{\sigma^2t}-1) =4e^2(e^{(1/4)\cdot3}-1)=4e^2(e^{3/4}-1)$.

Check: Subtract the squared mean from the second moment: $4e^{11/4}-(2e)^2=4e^{11/4}-4e^2=4e^2(e^{3/4}-1)$, since $11/4=2+3/4$. This agrees with the variance formula.

Recovering the Moments from the SDE

The SDE gives another route to the same results. To take expectations of its stochastic integrals, we use the following property of the Itô integral. If $H$ is progressively measurable and belongs to $\mathcal{L}^2$ on $[0,T]$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, then $M_t=\int_0^t H_s\,dB_s$ is a square-integrable martingale and $\mathbb{E}[M_t\mid\mathcal{F}_r]=M_r$ for $r\le t$. In particular, $\mathbb{E}[M_t]=0$ when $M_0=0$. This property follows first for adapted step integrands: each future increment has conditional mean zero because Brownian increments are independent of the past filtration. Approximation in the Itô integral’s $L^2$ norm extends the conditional-mean identity to $\mathcal{L}^2$ integrands.

Theorem: For the geometric Brownian motion above, the SDE implies $\mathbb{E}[S_t]=s_0e^{\mu t}$ and $\mathbb{E}[S_t^2]=s_0^2e^{(2\mu+\sigma^2)t}$. Consequently its variance is $s_0^2e^{2\mu t}(e^{\sigma^2t}-1)$.

Proof. Fix $T>0$. The lognormal moment formulas already established show that $\mathbb{E}[S_s^2]$ and $\mathbb{E}[S_s^4]$ are finite and bounded as $s$ ranges over $[0,T]$. Therefore $\mathbb{E}\int_0^T(\sigma S_s)^2\,ds<\infty$ and $\mathbb{E}\int_0^T(2\sigma S_s^2)^2\,ds<\infty$. These are the integrability conditions for the Itô integrals below. The formulas that follow use Itô’s formula; in each case the second-derivative correction is displayed.

First take $f(x)=x$, with $f'(x)=1$ and $f''(x)=0$. Itô’s formula gives the correction $\tfrac12 f''(S_t)\sigma^2S_t^2\,dt=0$, hence $dS_t=\mu S_t\,dt+\sigma S_t\,dB_t$, as in the SDE. Taking expectations in its integral form is valid because the time integral is integrable and $\sigma S$ is in $\mathcal{L}^2$. The Itô integral has mean zero, so $m(t)=\mathbb{E}[S_t]$ satisfies $m(t)=s_0+\mu\int_0^t m(s)\,ds$. This integral equation gives $m'(t)=\mu m(t)$ and $m(0)=s_0$. Solving yields $m(t)=s_0e^{\mu t}$.

Next take $f(x)=x^2$, with $f'(x)=2x$ and $f''(x)=2$. The full Itô formula is

d(S_t^2)=2S_t\,dS_t+\tfrac12(2)(dS_t)^2 =(2\mu+\sigma^2)S_t^2\,dt+2\sigma S_t^2\,dB_t.

Here $(dS_t)^2=\sigma^2S_t^2\,dt$: the multiplication rules are $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. In particular, the $\tfrac12 f''$ correction contributes $\sigma^2S_t^2\,dt$. Since $2\sigma S^2$ is in $\mathcal{L}^2$, its Itô integral is a square-integrable martingale with zero expectation. Thus $q(t)=\mathbb{E}[S_t^2]$ satisfies $q(t)=s_0^2+(2\mu+\sigma^2)\int_0^t q(s)\,ds$. Solving $q'(t)=(2\mu+\sigma^2)q(t)$, $q(0)=s_0^2$, gives $q(t)=s_0^2e^{(2\mu+\sigma^2)t}$. Subtracting $m(t)^2$ gives the stated variance. $\square$

This calculation uses the explicit law to check the integrability needed for taking expectations, but obtains the moment equations from the SDE. It does not assume that an unverified stochastic integral has mean zero: both integrands have been checked to lie in $\mathcal{L}^2$.

Worked Example: Zero Drift and Unit Volatility

Let $s_0=5$, $\mu=0$, $\sigma=1$, and $t=\ln 2$. Compute all three moments.

The mean is $5e^{0\cdot\ln2}=5$. The second moment is $25e^{(0+1)\ln2}=25\cdot2=50$. The variance is $25e^0(e^{1\cdot\ln2}-1)=25(2-1)=25$.

Check: The variance definition gives $\mathbb{E}[S_t^2]-(\mathbb{E}[S_t])^2=50-5^2=50-25=25$. Also, the lognormal expression has $\log S_t=\log 5-\tfrac12\ln2+B_{\ln2}$, consistent with the specified drift and volatility.

Conditional Moments

The same formulas apply conditionally from a current time $t$ to a later time $u\ge t$. The solution and independent increments give $S_u=S_t\exp((\mu-\tfrac12\sigma^2)(u-t)+\sigma(B_u-B_t))$. Because $S_t$ is $\mathcal{F}_t$-measurable and $B_u-B_t$ is independent of $\mathcal{F}_t$, the Gaussian moment-generating function gives

\mathbb{E}[S_u\mid\mathcal{F}_t]=S_te^{\mu(u-t)},\qquad \mathbb{E}[S_u^2\mid\mathcal{F}_t]=S_t^2e^{(2\mu+\sigma^2)(u-t)}.

These conditional expectations are well-defined: the unconditional first and second moments at $u$ are finite, which implies integrability. Subtracting the square of the first conditional moment from the second gives the conditional variance $\operatorname{Var}(S_u\mid\mathcal{F}_t)=S_t^2e^{2\mu(u-t)}(e^{\sigma^2(u-t)}-1)$. At $u=t$, the formulas reduce to $S_t$, $S_t^2$, and $0$; no positive time increment is required for this endpoint case.

Worked Example: Deterministic Zero-Volatility Case

Let $s_0=4$, $\mu=\tfrac12$, $\sigma=0$, and $t=2$. Compute the moments and compare them with the path itself.

The solution is $S_2=4e^{(1/2)2+0\cdot B_2}=4e$, a deterministic value. The mean formula gives $4e^{(1/2)2}=4e$. The second-moment formula gives $16e^{(2(1/2)+0)2}=16e^2$. The variance formula gives $16e^{2}(e^0-1)=0$.

Check: Since $S_2=4e$ deterministically, $\mathbb{E}[S_2^2]=(4e)^2=16e^2$, and $\mathbb{E}[S_2^2]-(\mathbb{E}[S_2])^2=16e^2-(4e)^2=0$. This agrees with all three formulas.

Common Mistake: Calling the Second Moment the Variance

A common error is to report $\mathbb{E}[S_t^2]$ as $\operatorname{Var}(S_t)$. In the zero-drift example, that mistake gives $50$ for the variance, whereas the correct value is $50-5^2=25$. Always subtract the squared mean. As a further check, when $\sigma=0$, the process is deterministic and its variance must be zero; the formula $s_0^2e^{2\mu t}(e^{\sigma^2t}-1)$ has exactly that property. For $\sigma\ne0$ and $t>0$, it is positive.

If $s_0=0$, which was excluded from the lognormal formulation, the SDE has the identically zero solution and all three moments are zero. Thus the positive-initial-value formulas have a consistent zero limit as $s_0$ tends to zero.

Check Your Understanding

Use the formulas and the SDE calculation to answer the following.

For $s_0=3$, $\mu=\tfrac14$, $\sigma=1$, and $t=2$, calculate the mean, second moment, and variance.
For $\sigma=0$, explain directly from the solution why the variance is zero for every $t\ge0$.
Which integrand must be in $\mathcal{L}^2$ to take expectations of the stochastic integral in the Itô formula for $S_t^2$?
Write the conditional mean and variance of $S_u$ given $\mathcal{F}_t$, for $u\ge t$.
Explain why $\mathbb{E}[S_t^2]$ alone does not determine $\operatorname{Var}(S_t)$.