Geometric Brownian Moments From the Lognormal Distribution

From the Lognormal Law to Moments

Knowing the distribution of geometric Brownian motion makes its moments explicit. The calculation also reveals that the mean, variance, and higher moments can have different time trends: a process whose mean decreases may still have a higher moment that increases. The key is to keep the Gaussian variability in the logarithm when taking expectations.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Consider geometric Brownian motion with constant drift $\mu\in\mathbb{R}$, constant diffusion coefficient $\sigma\in\mathbb{R}$, and initial value $s_0>0$: $dS_t=\mu S_t\,dt+\sigma S_t\,dB_t$. The drift is $\mu S_t$, and the diffusion coefficient is $\sigma S_t$. The earlier result on the lognormal law gives, for $t\ge0$,

\log S_t=\log s_0+\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma B_t.

In particular, $\log S_t$ is normally distributed with mean $\log s_0+(\mu-\sigma^2/2)t$ and variance $\sigma^2t$. For finite $T$, the stochastic integral in the SDE is an Ito integral: its integrand $H_s=\sigma S_s$ is progressively measurable and belongs to $\mathcal{L}^2$, the class of progressively measurable processes satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Indeed, the second moment computed below is finite and bounded on every finite time interval, so $\mathbb{E}\int_0^T\sigma^2S_s^2\,ds=\int_0^T\sigma^2\mathbb{E}[S_s^2]\,ds<\infty$. If $s_0=0$, the solution is identically zero and all its positive moments and its variance are zero; below we treat the nondegenerate case $s_0>0$.

Definition: For a positive integer $n$, the $n$th raw moment of $S_t$ is $\mathbb{E}[S_t^n]$. Its mean is the first raw moment, and its variance is $\operatorname{Var}(S_t)=\mathbb{E}[S_t^2]-(\mathbb{E}[S_t])^2$.

The Moment Formula

Theorem: For $s_0>0$, $t\ge0$, and every positive integer $n$, $$ \mathbb{E}[S_t^n]=s_0^n\exp\left(n\mu t+\frac{n(n-1)}{2}\sigma^2t\right). $$ Consequently, $$ \mathbb{E}[S_t]=s_0e^{\mu t},\qquad \operatorname{Var}(S_t)=s_0^2e^{2\mu t}\left(e^{\sigma^2t}-1\right). $$

Proof. The moment-generating function of a normal random variable $Z$ with mean $m$ and variance $v$ is $\mathbb{E}[e^{aZ}]=e^{am+a^2v/2}$ for real $a$. Apply this with $Z=\log S_t$, $a=n$, $m=\log s_0+(\mu-\sigma^2/2)t$, and $v=\sigma^2t$. Since $S_t^n=e^{n\log S_t}$,

\begin{aligned} \mathbb{E}[S_t^n] &=\exp\left(n\left[\log s_0+\left(\mu-\frac{\sigma^2}{2}\right)t\right] +\frac{n^2\sigma^2t}{2}\right)\\ &=s_0^n\exp\left(n\mu t+\frac{n(n-1)}{2}\sigma^2t\right). \end{aligned}

This calculation also proves that every positive integer moment is finite, including when $t=0$ or $\sigma=0$. Substituting $n=1$ gives $\mathbb{E}[S_t]=s_0e^{\mu t}$, and substituting $n=2$ gives $\mathbb{E}[S_t^2]=s_0^2e^{(2\mu+\sigma^2)t}$. Subtracting the square of the mean yields $\operatorname{Var}(S_t)=s_0^2e^{2\mu t}(e^{\sigma^2t}-1)$, as claimed. $\square$

The factor $n(n-1)\sigma^2/2$ is the contribution of variability in the logarithm. It vanishes for the mean ($n=1$) but not for higher moments. Thus the mean depends on the drift $\mu$ alone, while the second moment and variance also depend on volatility.

Worked Example: Mean, Variance, and a Third Moment

Let $s_0=2$, $\mu=\tfrac14$, $\sigma=\tfrac12$, and $t=2$. Compute the mean, variance, and third moment.

First, $\mu t=\tfrac14\cdot2=\tfrac12$, and $\sigma^2t=(\tfrac12)^2\cdot2=\tfrac12$. The mean is

\mathbb{E}[S_2]=2e^{\mu t}=2e^{1/2}.

For the second moment, $2\mu t=1$, so

\mathbb{E}[S_2^2]=2^2e^{2\mu t+\sigma^2t}=4e^{1+1/2}=4e^{3/2}.

The square of the mean is $(2e^{1/2})^2=4e$. Therefore

\operatorname{Var}(S_2)=4e^{3/2}-4e=4e\left(e^{1/2}-1\right).

For $n=3$, the drift contribution is $3\mu t=3/2$, and the volatility contribution is $\tfrac{3\cdot2}{2}\sigma^2t=3/2$. Hence

\mathbb{E}[S_2^3]=2^3e^{3/2+3/2}=8e^3.

Check: Using the lognormal parameters directly, $\log S_2$ has mean $\log 2+(\tfrac14-\tfrac18)2=\log 2+\tfrac14$ and variance $\tfrac12$. The normal moment-generating function gives $\mathbb{E}[S_2^3]=\exp(3(\log 2+\tfrac14)+9(\tfrac12)/2)=8e^{3/4+9/4}=8e^3$, agreeing with the formula. Also, the variance computed as the second moment minus the squared mean is $4e^{3/2}-4e$, the same expression obtained above.

Exponential Growth Rates

For fixed $n$, the $n$th moment is a constant $s_0^n$ times an exponential in time. Its exponential rate is $\kappa_n=n\mu+\tfrac12n(n-1)\sigma^2$. A positive rate means the moment increases exponentially, a negative rate means it decreases exponentially, and a zero rate means it is constant. This statement concerns the moment as a function of time; it does not assert that individual sample paths are monotone.

Comparing adjacent moments makes volatility’s role visible:

\kappa_{n+1}-\kappa_n=\mu+n\sigma^2.

For example, if $\sigma\ne0$, this difference becomes positive for sufficiently large $n$, whatever the value of $\mu$. High-order moments can therefore grow even when the mean declines. This reflects their greater sensitivity to large values of a lognormal random variable; it is not a contradiction with a decreasing mean.

Worked Example: A Decreasing Mean and a Constant Third Moment

Take $s_0=3$, $\mu=-\tfrac14$, $\sigma=\tfrac12$, and $t=2$. Here $\mu t=-\tfrac12$ and $\sigma^2t=\tfrac12$. The mean is

\mathbb{E}[S_2]=3e^{-1/2}.

The second moment and variance are

\begin{aligned} \mathbb{E}[S_2^2] &=9e^{2(-1/2)+1/2}=9e^{-1/2},\\ \operatorname{Var}(S_2) &=9e^{-1/2}-(3e^{-1/2})^2 =9\left(e^{-1/2}-e^{-1}\right). \end{aligned}

For the third moment, the exponent is $3\mu t+\tfrac{3\cdot2}{2}\sigma^2t=3(-\tfrac12)+3(\tfrac12)=0$. Thus $\mathbb{E}[S_2^3]=3^3e^0=27$. In fact, its rate is $\kappa_3=3(-\tfrac14)+3(\tfrac14)=0$, so the third moment stays equal to $27$ at every time.

Check: The logarithm has mean $\log3+(-\tfrac14-\tfrac18)2=\log3-\tfrac34$ and variance $\tfrac12$. The third-moment calculation from this normal law is $\exp(3\log3-\tfrac94+\tfrac92\cdot\tfrac12)=\exp(3\log3)=27$. For the variance, subtracting the squared mean from the second moment gives $9e^{-1/2}-9e^{-1}=9(e^{-1/2}-e^{-1})$, matching the result.

Variance and Relative Dispersion

The variance formula separates into the squared mean and a volatility factor:

\operatorname{Var}(S_t) =\big(\mathbb{E}[S_t]\big)^2\left(e^{\sigma^2t}-1\right).

When $\sigma\ne0$, the variance divided by the squared mean is $e^{\sigma^2t}-1$, which increases strictly with $t$. This ratio measures relative dispersion: it can keep increasing even if the variance itself eventually declines because the overall level of $S_t$ is falling.

The variance need not increase at every time. For $\sigma\ne0$, differentiating its explicit formula gives

\frac{d}{dt}\operatorname{Var}(S_t) =s_0^2e^{2\mu t}\left[2\mu\left(e^{\sigma^2t}-1\right)+\sigma^2e^{\sigma^2t}\right].

If $\mu\ge0$, the bracket is positive for $t\ge0$, so variance increases. If $-\sigma^2/2\le\mu<0$, it is also positive: rewrite it as $(2\mu+\sigma^2)e^{\sigma^2t}-2\mu$, whose terms are nonnegative and whose second term is positive. If $\mu<-\sigma^2/2$, the bracket is positive at $t=0$, but eventually negative; it changes sign once, at $t=\sigma^{-2}\log\!\left(2\mu/(2\mu+\sigma^2)\right)>0$. Thus variance first increases and then decreases. If $\sigma=0$, the process is deterministic and its variance is zero for all $t$.

Worked Example: Variance That Rises and Then Falls

Set $s_0=1$, $\mu=-\tfrac12$, and $\sigma=1$. The moment rates are $\kappa_n=-\tfrac12n+\tfrac12n(n-1)=\tfrac12n(n-2)$. In particular, the mean is $e^{-t/2}$, the second moment is $1$, and

\operatorname{Var}(S_t)=1-e^{-t}.

At $t=\log2$, the mean is $e^{-(\log2)/2}=1/\sqrt2$, the second moment is $e^0=1$, and the variance is $1-(1/2)=1/2$. The third-moment rate is $\kappa_3=\tfrac12\cdot3\cdot1=\tfrac32$, so $\mathbb{E}[S_{\log2}^3]=e^{(3/2)\log2}=2^{3/2}$.

Check: The variance derivative is $e^{-t}>0$, so for these particular parameters the variance increases rather than turns down. This verifies that the variance trend depends on the drift-volatility combination, not just on a negative drift. The third moment also agrees with the rate formula, since $2^{3/2}$ is the exponential of $\tfrac32\log2$.

Common Mistake: Dropping the Gaussian Contribution

A common incorrect calculation is to take the exponential of the mean of $\log S_t$ and call it the mean of $S_t$. For $s_0=1$, $\mu=0$, $\sigma=1$, and $t=1$, that shortcut gives $e^{-1/2}$, because $\mathbb{E}[\log S_1]=-\tfrac12$. But the correct calculation includes the normal moment-generating-function term: $\mathbb{E}[S_1]=\mathbb{E}[e^{\log S_1}]=e^{-1/2+1/2}=1$. The mean of an exponential is not generally the exponential of the mean. The same missing contribution would give an incorrect second moment and variance.

The moment formula is useful because it distinguishes typical growth, measured by the mean, from sensitivity to large outcomes, measured by higher moments. It also provides direct checks for later calculations: any proposed moment of geometric Brownian motion must agree with its lognormal distribution and remain finite at finite times.

Check Your Understanding

Use the lognormal moment formula, showing the exponent you substitute in each case.

For $s_0=1$, $\mu=\tfrac12$, $\sigma=1$, and $t=1$, compute the mean and variance.
For general parameters, find the exponential rate of the fourth moment.
When $\sigma\ne0$, show that the variance divided by the squared mean increases with time.
For $s_0=2$, $\mu=-1$, and $\sigma=1$, determine whether the third moment increases, decreases, or stays constant.
Explain why a decreasing mean does not imply that every higher moment decreases.