When Geometric Brownian Motion Is a Martingale

Conditional Expectations Determine the Classification

Geometric Brownian motion is often introduced through its explicit exponential solution. That formula also gives a direct way to decide whether the process is a martingale: condition at time \(s\), separate the known past from the Brownian increment after \(s\), and compute the expectation of the resulting exponential. The coefficient in the SDE drift, not the coefficient appearing in the logarithm, determines the answer.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to this filtration. In particular, \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\) for \(s<t\). Let \(\mu(t)\) and \(\sigma(t)\) be deterministic measurable functions such that \(\int_0^T|\mu(u)|\,du<\infty\) and \(\int_0^T\sigma(u)^2\,du<\infty\) for every finite \(T\). Take a deterministic initial value \(S_0>0\), and define \(A(t)=\int_0^t\mu(u)\,du\) and \(V(t)=\int_0^t\sigma(u)^2\,du\).

The geometric Brownian motion with these coefficients is the positive process

The stochastic integral here is the Ito integral. Its deterministic integrand \(\sigma\) is progressively measurable and belongs to \(\mathcal{L}^2\) on every finite interval, meaning \(\mathbb{E}\int_0^T\sigma(u)^2\,du<\infty\). By Ito's formula, \(S\) solves \(dS_t=\mu(t)S_t\,dt+\sigma(t)S_t\,dB_t\). On each finite interval, \(\sigma(t)S_t\) is progressively measurable and satisfies \(\mathbb{E}\int_0^T\sigma(t)^2S_t^2\,dt<\infty\), so the stochastic integral in this SDE is also defined in the Ito sense. The drift coefficient is \(\mu\); the diffusion coefficient is \(\sigma\).

The Conditional-Expectation Criterion

Proof. Split the stochastic integral at \(s\) and use \(A(t)-A(s)=\int_s^t\mu(u)\,du\) and \(V(t)-V(s)=\int_s^t\sigma(u)^2\,du\). This gives

The last integral is independent of \(\mathcal{F}_s\), is normal with mean zero and variance \(V(t)-V(s)\), and therefore its exponential after subtracting half its variance has expectation \(1\). This follows from the normal moment-generating function \(\mathbb{E}[e^{\lambda Z}]=e^{\lambda^2v/2}\) for \(Z\sim N(0,v)\), including \(v=0\). Hence the conditional expectation is \(S_s e^{A(t)-A(s)}\). The same calculation from time zero shows \(\mathbb{E}[S_t]=S_0e^{A(t)}<\infty\), so the process is integrable at each time.

Since \(S_s>0\) almost surely, comparing the conditional expectation with \(S_s\) shows that the martingale property is equivalent to \(e^{A(t)-A(s)}=1\) for all \(s\le t\), the submartingale property to \(A(t)-A(s)\ge0\), and the supermartingale property to \(A(t)-A(s)\le0\). The exponential is strictly increasing, which justifies these comparisons. Thus the corresponding conditions are that \(A\) is constant, nondecreasing, or nonincreasing. Since \(A(t)=\int_0^t\mu(u)\,du\) is absolutely continuous, these are equivalent to \(\mu=0\), \(\mu\ge0\), or \(\mu\le0\), respectively, almost everywhere. For the converse in this last equivalence, the derivative of an absolutely continuous nondecreasing function is nonnegative almost everywhere, and likewise a nonincreasing function has nonpositive derivative almost everywhere. \(\square\)

This is a statement under \(\mathbb{P}\), the probability measure used to specify the Brownian motion and coefficients above. In a financial model, coefficients and Brownian motion may instead be specified under a risk-neutral measure \(\mathbb{Q}\). The same calculation applies under that measure using its drift coefficient; a drift under \(\mathbb{P}\) should not be substituted for a drift under \(\mathbb{Q}\).

Positive Deterministic Rescalings

Let \(g(t)\) be any finite, strictly positive deterministic function, and put \(Y_t=g(t)S_t\). No differentiability of \(g\) is needed for the conditional-expectation test. Define \(H(t)=g(t)e^{A(t)}\). For each \(s\le t\),

Proof. The displayed conditional expectation and the identity \(Y_s=g(s)S_s>0\) show that the ratio of the conditional expectation to \(Y_s\) is \(H(t)/H(s)\). This ratio equals \(1\) for all \(s\le t\) exactly when \(H\) is constant; it is at least \(1\) for all such pairs exactly when \(H\) is nondecreasing; and it is at most \(1\) exactly when \(H\) is nonincreasing. Integrability follows from \(\mathbb{E}[Y_t]=g(t)S_0e^{A(t)}<\infty\). This proves all three assertions. \(\square\)

In particular, the unique positive rescaling up to a constant multiplier that makes \(S\) a martingale is \(g(t)=ce^{-A(t)}\), where \(c>0\). If \(g\) is absolutely continuous, there is also a differential explanation. The Ito product rule gives, almost everywhere in time,

Here \(dt\,dB_t=0\) and \(dt\,dt=0\); the quadratic-variation rule \(dB_t\,dB_t=dt\) is already accounted for in the Ito solution for \(S\). When \(g'/g=-\mu\), the finite-variation drift in this expression vanishes, so Ito's formula identifies \(Y\) as a local martingale. The conditional-expectation theorem above supplies the further integrability and conditional-expectation check that makes this particular process a true martingale. Vanishing drift by itself is not a general proof that an arbitrary local martingale is a true martingale.

A Common Drift-Correction Pitfall

The explicit solution contains \(A(t)-\tfrac12V(t)\) in its exponent. It is tempting to treat \(\mu(t)-\tfrac12\sigma(t)^2\), the coefficient in the logarithm, as the drift that determines whether \(S\) is a martingale. That misses the expectation of the Brownian exponential. For \(Z\sim N(0,v)\), the factor \(e^{Z-v/2}\) has expectation \(1\), so the \(-V/2\) term is exactly offset when taking conditional expectation.

For a concrete wrong calculation, take constant \(\mu=\tfrac14\) and \(\sigma=1\). The logarithmic exponent has drift \(\mu-\tfrac12\sigma^2=-\tfrac14\). If one incorrectly uses this as the drift to cancel, one might propose \(e^{t/4}S_t\) as a martingale. But

which is strictly greater than \(e^{s/4}S_s\) whenever \(s<t\). The proposed rescaling is not a martingale. The correct factor is \(e^{-t/4}\), obtained from the SDE drift \(\mu=\tfrac14\); it gives \(e^{-t/4}S_se^{(t-s)/4}=e^{-s/4}S_s\). The quadratic-variation correction belongs in the log solution, but the conditional mean is governed by the drift of \(S\) in its SDE.

The same distinction matters when interpreting discounted asset prices. Discounting at a chosen rate makes a geometric Brownian motion a martingale only when that rate matches the drift under the probability measure being used. Under \(\mathbb{Q}\), for example, the relevant drift is the drift specified under \(\mathbb{Q}\), not automatically the drift under \(\mathbb{P}\).