Geometric Brownian Motion From Its SDE

From Multiplicative Noise to a Logarithm

Geometric Brownian motion is the scalar process whose drift and diffusion are both proportional to its current value. Its equation is simple to write, but the solution depends on the quadratic variation of Brownian motion: taking a logarithm introduces a second-derivative term that changes the drift. The calculation also explains why the exponential expression must be checked by Itô’s formula rather than by the ordinary chain rule.

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 this filtration. Throughout, $\mu,\sigma\in\mathbb{R}$ are constants and the initial value $S_0=s_0$ is deterministic. An Itô integrand on $[0,T]$ is progressively measurable and belongs to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The equation is interpreted in the Itô sense:

S_t=s_0+\int_0^t\mu S_s\,ds+\int_0^t\sigma S_s\,dB_s.

Here $\mu S_t$ is the drift coefficient and $\sigma S_t$ is the diffusion coefficient. For each finite $T$, the coefficients $x\mapsto\mu x$ and $x\mapsto\sigma x$ are globally Lipschitz and satisfy linear growth. The global Lipschitz existence and uniqueness theorem therefore gives a unique global strong solution, with finite second moments on finite time intervals. In particular, the stochastic integrand $\sigma S_t$ is in $\mathcal{L}^2$ on every such interval. Strong means the solution is constructed on the given space using the given Brownian motion; pathwise uniqueness means any two such solutions with the same Brownian motion and initial value agree almost surely. These are not the same as uniqueness in law, which concerns equality of distributions. We work under $\mathbb{P}$; no risk-neutral measure $\mathbb{Q}$ is being introduced.

Definition: For $s_0>0$, geometric Brownian motion with parameters $\mu,\sigma$ is the solution of the Itô SDE $dS_t=\mu S_t\,dt+\sigma S_t\,dB_t$, $S_0=s_0$. The logarithmic transform is valid only while $S_t>0$; positivity will be established below.

Deriving the Exponential Formula

Suppose first that $s_0>0$. To apply Itô’s formula without assuming positivity in advance, fix a time interval on which $S$ stays bounded away from zero and apply the formula to $f(x)=\log x$. The derivatives are $f'(x)=1/x$ and $f''(x)=-1/x^2$. Since the drift is $\mu S_t$, the diffusion coefficient is $\sigma S_t$, and $dB_t\,dB_t=dt$, the second-order correction is essential:

d\log S_t =\frac{1}{S_t}\,dS_t+\frac12\left(-\frac{1}{S_t^2}\right)(\sigma S_t)^2\,dt =\left(\mu-\frac{\sigma^2}{2}\right)dt+\sigma\,dB_t.

The $-\sigma^2/2$ term comes from $\tfrac12 f''(S_t)(\sigma S_t)^2dt$; it would be absent under the ordinary chain rule. Integrating on such an interval gives $\log S_t=\log s_0+(\mu-\sigma^2/2)t+\sigma B_t$. Exponentiation suggests the candidate solution. The next result justifies that this calculation applies for every finite time.

Theorem: If $s_0>0$, the unique solution of the geometric Brownian motion SDE is $$ S_t=s_0\exp\left(\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma B_t\right),\qquad t\ge0. $$ In particular, $S_t>0$ for every finite $t$, almost surely.

Proof. Let $S$ be the global continuous solution given by the global Lipschitz existence theorem, and let $\tau_0=\inf\{t\ge0:S_t=0\}$, with $\inf\varnothing=\infty$. For any $t<\tau_0$, continuity and $S_0>0$ imply that $S$ is bounded above and bounded away from zero on $[0,t]$. We may therefore apply Itô’s formula to $\log S$ on that interval. More formally, stop first when $S$ exits $(1/n,n)$; the stopped integrands are bounded and hence square-integrable on finite time intervals. The formula gives $\log S_t=\log s_0+(\mu-\sigma^2/2)t+\sigma B_t$ for all $t<\tau_0$, almost surely. If $\tau_0<\infty$, then as $t$ increases to $\tau_0$, continuity gives $S_t\to0$, so $\log S_t\to-\infty$. But the right-hand side converges to the finite value $\log s_0+(\mu-\sigma^2/2)\tau_0+\sigma B_{\tau_0}$, since Brownian paths are continuous. This contradiction shows $\tau_0=\infty$. Exponentiating the identity now gives the displayed formula for every $t\ge0$. $\square$

Worked Example: Solving with a Positive Initial Value

Take $s_0=2$, $\mu=\tfrac13$, and $\sigma=\tfrac12$. Find the solution and identify the logarithmic drift correction.

First compute $\sigma^2/2=(1/4)/2=1/8$. Thus the drift in the logarithmic equation is $\mu-\sigma^2/2=1/3-1/8=8/24-3/24=5/24$. Integrating the logarithmic equation gives $\log S_t=\log 2+\tfrac{5}{24}t+\tfrac12 B_t$, so

S_t=2\exp\left(\frac{5}{24}t+\frac12B_t\right).

Check: at $t=0$, $B_0=0$, so $S_0=2e^0=2$. The coefficient of $dt$ in the logarithm is also confirmed directly as $\mu-\sigma^2/2=1/3-1/8=5/24$.

Verifying the Candidate in the Original SDE

The logarithm derivation identifies a candidate; a direct Itô calculation verifies that it solves the original equation. For $s_0>0$, define $G_t=s_0\exp((\mu-\sigma^2/2)t+\sigma B_t)$. Apply Itô’s formula to $g(t,x)=s_0\exp((\mu-\sigma^2/2)t+\sigma x)$. Its derivatives are $g_t=(\mu-\sigma^2/2)g$, $g_x=\sigma g$, and $g_{xx}=\sigma^2g$. The full formula, including its correction, is

dG_t=g_t(t,B_t)\,dt+g_x(t,B_t)\,dB_t+\frac12g_{xx}(t,B_t)\,dt =\left(\mu-\frac{\sigma^2}{2}\right)G_t\,dt+\sigma G_t\,dB_t+\frac12\sigma^2G_t\,dt =\mu G_t\,dt+\sigma G_t\,dB_t.

Also $G_0=s_0\exp(0+\sigma B_0)=s_0$, so integrating this differential gives exactly the required integral equation. The integrand is admissible: for finite $T$, the normal moment-generating function gives $\mathbb{E}[G_t^2]=s_0^2\exp((2\mu+\sigma^2)t)$, and hence $\mathbb{E}\int_0^T(\sigma G_t)^2dt\le \sigma^2T s_0^2\exp(|2\mu+\sigma^2|T)<\infty$. Thus $\sigma G\in\mathcal{L}^2$. The global Lipschitz pathwise uniqueness result applies to these coefficients, so the verified candidate is the unique strong solution.

Worked Example: Direct Verification with Different Parameters

Let $s_0=3$, $\mu=1$, and $\sigma=2$. The proposed process is $G_t=3\exp(-t+2B_t)$, because $\mu-\sigma^2/2=1-4/2=-1$. Verify its drift and diffusion coefficients.

Set $g(t,x)=3e^{-t+2x}$. Then $g_t=-g$, $g_x=2g$, and $g_{xx}=4g$. Itô’s formula gives $dG_t=-G_tdt+2G_tdB_t+\tfrac12(4G_t)dt=G_tdt+2G_tdB_t$. This is the required equation with $\mu=1$ and $\sigma=2$.

Check: the drift coefficient recomputes as $-1+\tfrac12\cdot4=1$, while the diffusion coefficient is $2$. At time zero, $G_0=3e^0=3$, the specified initial value.

Degenerate Initial Values and a Martingale Case

The logarithmic argument requires a positive initial value. If $s_0=0$, the process $S_t=0$ satisfies the equation, and pathwise uniqueness for the globally Lipschitz coefficients shows it is the unique solution. If $s_0<0$, apply the positive-initial-value result to $-S_t$, which satisfies the same linear SDE with initial value $-s_0>0$. Therefore the formula $S_t=s_0\exp((\mu-\sigma^2/2)t+\sigma B_t)$ holds for every real $s_0$; it is positive for $s_0>0$, negative for $s_0<0$, and identically zero for $s_0=0$.

The explicit solution is not automatically a martingale. For example, when $\mu=0$, it is $S_t=s_0\exp(\sigma B_t-\sigma^2t/2)$. If $s_0$ is fixed, this is a true martingale: the normal moment-generating function gives $\mathbb{E}|S_t|=|s_0|<\infty$. For $0\le s\le t$, independent Brownian increments give

\mathbb{E}[S_t\mid\mathcal{F}_s] =S_s\,\mathbb{E}\left[\exp\left(\sigma(B_t-B_s)-\frac{\sigma^2}{2}(t-s)\right)\middle|\mathcal{F}_s\right] =S_s.

The last conditional expectation is $1$, since $B_t-B_s$ is independent of $\mathcal{F}_s$, is normal with variance $t-s$, and has exponential moment $\exp(\sigma^2(t-s)/2)$. For general $\mu$, this argument instead gives $\mathbb{E}[S_t\mid\mathcal{F}_s]=e^{\mu(t-s)}S_s$, so it is not a martingale unless $\mu=0$ (apart from the identically zero case). A local martingale is a process that becomes a martingale after stopping at a suitable increasing sequence of stopping times; that weaker property alone does not imply integrability or the conditional-expectation identity above.

Worked Example: The Zero-Drift Martingale

Take $s_0=1$, $\mu=0$, and $\sigma=1$. The solution is $S_t=e^{B_t-t/2}$. Verify the conditional expectation over an interval of length $h>0$.

Since $B_{s+h}-B_s\sim N(0,h)$ independently of $\mathcal{F}_s$, its exponential moment is $\mathbb{E}[e^{B_{s+h}-B_s}]=e^{h/2}$. Therefore $\mathbb{E}[S_{s+h}\mid\mathcal{F}_s]=S_s e^{-h/2}e^{h/2}=S_s$. Its integrability follows from $\mathbb{E}[S_t]=e^{-t/2}e^{t/2}=1$.

Check: the conditional expectation equals the current value, not a fixed constant depending on the future time. This verifies the martingale property; it is stronger than merely observing that the SDE has no $dt$ term.

Common Pitfall: Omitting the Itô Correction

A common incorrect calculation is to write $\log S_t=\log s_0+\mu t+\sigma B_t$, as if the ordinary chain rule applied. For the example $\mu=1,\sigma=2$, this would give the wrong candidate $\widetilde S_t=s_0e^{t+2B_t}$. Applying Itô’s formula with $f(t,x)=s_0e^{t+2x}$, $f_t=f$, $f_x=2f$, and $f_{xx}=4f$, gives $d\widetilde S_t=(1+\tfrac12\cdot4)\widetilde S_tdt+2\widetilde S_tdB_t=3\widetilde S_tdt+2\widetilde S_tdB_t$. Its drift is $3\widetilde S_t$, not the required $\widetilde S_t$. The correction $-\sigma^2/2$ in the exponential precisely cancels the extra $\tfrac12\sigma^2$ contributed when the candidate is differentiated.

This is an Itô SDE, where $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. In Stratonovich notation the same dynamics would be written $dS_t=(\mu-\sigma^2/2)S_tdt+\sigma S_t\circ dB_t$: the Stratonovich-to-Itô correction adds $\tfrac12\sigma^2S_tdt$. The two forms use different drift coefficients; confusing them changes the model.

Check Your Understanding

Use the logarithmic transform and the direct Itô verification to test your reasoning.

For $s_0=4,\mu=\tfrac12,\sigma=1$, calculate the logarithmic drift and write the explicit solution.
For $s_0=2,\mu=0,\sigma=3$, compute the drift of $\log S_t$ and the drift of $S_t$.
Explain why applying $\log$ requires a positivity argument, and identify the contradiction that rules out a finite hitting time of zero.
For which value of $\mu$ is the nonzero geometric Brownian motion a martingale? State the conditional-expectation check needed.
Apply Itô’s formula to $s_0e^{at+\sigma B_t}$ and determine which value of $a$ makes its drift coefficient equal to $\mu S_t$.