Two Tests, Not One
A continuous process can have Brownian-looking features and still fail to be Brownian motion. It might be a martingale but accumulate quadratic variation at the wrong rate, or have the right quadratic variation while carrying a drift that prevents it from being a martingale. The useful diagnostic is to check both properties, with respect to the filtration under consideration.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to this filtration. A process \(M_t\) is a martingale if it is adapted, \(\mathbb{E}|M_t|<\infty\) for every \(t\), and \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) whenever \(s\le t\). A local martingale satisfies the martingale property after stopping at a suitable sequence of stopping times increasing to infinity; without additional integrability, it need not be a true martingale.
The quadratic variation \([X]_t\) measures the limiting accumulated squared increments of a continuous process \(X\). For Brownian motion, \([B]_t=t\). In Ito calculations, the corresponding multiplication rules are \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\); these encode quadratic-variation limits, not ordinary algebraic identities for finite increments.
In applications, it is often convenient to prove the stronger martingale property rather than only local martingality. The quadratic-variation requirement remains essential: a continuous martingale with \([X]_t=ct\) for \(c\ne1\) is not standard Brownian motion on the given time scale. Conversely, quadratic variation alone does not remove a drift.
Why \(B_t^2-t\) Is a Martingale but Not Brownian Motion
Worked Example: The Squared-Brownian Martingale
Set \(M_t=B_t^2-t\). Derive its stochastic differential, verify the martingale property, and test whether it is Brownian motion.
Apply Ito's formula to \(f(x)=x^2\). The derivatives are \(f'(x)=2x\) and \(f''(x)=2\), so the second-derivative correction is \(\tfrac12 f''(B_t)\,dt=\tfrac12(2)\,dt=dt\). Therefore
The Ito integral here is defined on the stated filtered space with integrand \(H_s=2B_s\). Brownian paths are continuous, so this integrand is progressively measurable; moreover, for each finite \(T\),
Thus \(2B\) belongs to \(\mathcal{L}^2\) on \([0,T]\), and the Ito integral is a square-integrable martingale. In particular, it is integrable and its conditional expectation satisfies \(\mathbb{E}[\int_0^t2B_s\,dB_s\mid\mathcal{F}_u]=\int_0^u2B_s\,dB_s\) for \(u\le t\). Since \(M_0=0\), the differential identity gives \(M_t=\int_0^t2B_s\,dB_s\), proving the martingale property.
However, the quadratic variation of this stochastic integral is
which is not the deterministic function \(t\). For an explicit check at \(t=1\), \(\mathbb{E}[M_1]=\mathbb{E}[B_1^2]-1=1-1=0\), as required for a martingale starting at zero, while \(\mathbb{E}[[M]_1]=4\int_0^1s\,ds=2\ne1\). A process with this quadratic variation cannot pass the Brownian test. Check: the martingale identity and the quadratic-variation condition answer different questions; passing the first does not imply passing the second.
The general integral fact used here is that if \(H\) is progressively measurable and \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\), then \(\int_0^tH_s\,dB_s\) is a square-integrable martingale and its quadratic variation is \(\int_0^tH_s^2\,ds\). The first assertion includes integrability and the conditional-expectation identity; the second identifies the pathwise quadratic variation.
Quadratic Variation Does Not Detect Drift
Worked Example: A Candidate with the Right Quadratic Variation but the Wrong Drift
Let \(Y_t=B_t+t\). Its finite-variation term \(t\) does not contribute to quadratic variation, so \([Y]_t=[B]_t=t\). Nevertheless, \(Y\) is not a martingale.
For \(0\le s<t\), Brownian independent increments give \(\mathbb{E}[B_t\mid\mathcal{F}_s]=B_s\), and hence
All variables are integrable because \(B_t\) has a finite first moment and \(t\) is deterministic. The conditional-expectation identity fails whenever \(s<t\). For example, with \(s=0\) and \(t=1\), \(\mathbb{E}[Y_1\mid\mathcal{F}_0]=1\), whereas \(Y_0=0\). Thus the candidate has the correct quadratic variation but fails the martingale test. Check: its mean is \(\mathbb{E}[Y_t]=t\), also inconsistent with the constant mean of a martingale starting at zero.
More generally, for \(X_t=B_t+A_t\), where \(A_t\) is a continuous finite-variation process, the finite-variation part contributes zero quadratic variation under the usual continuous semimartingale rules, so \([X]_t=[B]_t=t\). This fact does not establish that \(X\) is a martingale: the conditional expectations must still be checked. A deterministic nonconstant drift such as \(A_t=t\) already shows why.
Scaling and Deterministic Time Changes
A deterministic change of clock changes the rate of quadratic-variation accumulation. Let \(g:[0,\infty)\to[0,\infty)\) be continuous, nondecreasing, and satisfy \(g(0)=0\). If \(c\) is a constant and \(X_t=cB_{g(t)}\), the natural filtration for this candidate is \(\mathcal{G}_t=\mathcal{F}_{g(t)}\). Relative to this filtration, \(B_{g(t)}\) is a martingale: its increment from \(g(s)\) to \(g(t)\) is independent of \(\mathcal{F}_{g(s)}\) and has mean zero.
Its quadratic variation is \([X]_t=c^2g(t)\). Indeed, for a partition \(0=t_0<\cdots<t_n=t\), the squared-increment sum for \(X\) is \(c^2\sum_i(B_{g(t_i)}-B_{g(t_{i-1})})^2\). By continuity of \(g\), these are Brownian quadratic-variation sums over partitions of \([0,g(t)]\), whose mesh tends to zero as the original partition mesh tends to zero. Their limit is \(c^2g(t)\). Thus matching quadratic variation requires \(c^2g(t)=t\) for every \(t\); the martingale property and Lévy characterization then give the Brownian test.
Worked Example: Rescaling a Brownian Clock
Consider \(X_t=2B_{t/4}\) with filtration \(\mathcal{G}_t=\mathcal{F}_{t/4}\). Check whether \(X\) is standard Brownian motion relative to \((\mathcal{G}_t)\).
Here \(c=2\) and \(g(t)=t/4\), so \([X]_t=c^2g(t)=4(t/4)=t\). Also, \(X_0=2B_0=0\). For \(s<t\), \(X_t-X_s=2(B_{t/4}-B_{s/4})\) is independent of \(\mathcal{G}_s=\mathcal{F}_{s/4}\), is normally distributed, and has mean zero. Its variance is
Thus the increments are independent of the past and have distribution \(N(0,t-s)\); the paths are continuous because the paths of \(B\) and the clock \(t/4\) are continuous. These properties verify directly that \(X\) is Brownian motion relative to \((\mathcal{G}_t)\). Check: its variance at time \(t\) is \(4\operatorname{Var}(B_{t/4})=4(t/4)=t\), in agreement with its quadratic variation.
Worked Example: A Time-Changed Martingale That Fails the Brownian Test
Now let \(Z_t=2B_{t/2}\) with filtration \(\mathcal{H}_t=\mathcal{F}_{t/2}\). It is a continuous martingale starting at zero by the same independent-increment and integrability checks: for \(s<t\), its increment has mean zero and variance \(4(t/2-s/2)=2(t-s)\). Its quadratic variation is \([Z]_t=4(t/2)=2t\), not \(t\).
For instance, at \(t=1\), \([Z]_1=2\) and \(\mathbb{E}[Z_1^2]=4\mathbb{E}[B_{1/2}^2]=4(1/2)=2\), rather than \(1\). Therefore this is a Brownian path run at the wrong scale, not standard Brownian motion on the displayed time scale. It is important that the filtration is stated: martingale and Brownian properties are always relative to a filtration, not just to the paths viewed in isolation.
A Practical Candidate-Testing Procedure
Verify \(X_0=0\) and that \(X\) has continuous paths. A nonzero starting value or discontinuity rules out standard Brownian motion in the stated form.
Check adaptedness and integrability, then compute \(\mathbb{E}[X_t\mid\mathcal{F}_s]\). Do not substitute a local-martingale claim for a true martingale claim without the required integrability.
Use Ito's formula, the quadratic-variation rules, or a time-change calculation. The required value for standard Brownian motion is \([X]_t=t\) almost surely for every \(t\).
If \(X\) is continuous, starts at zero, is a local martingale, and has quadratic variation \(t\), Lévy characterization gives Brownian motion relative to the specified filtration.
Common mistake. Seeing \([X]_t=t\) and immediately declaring \(X\) Brownian omits the martingale requirement. The concrete wrong argument “\([B_t+t]_t=t\), therefore \(B_t+t\) is Brownian” fails because \(\mathbb{E}[B_t+t\mid\mathcal{F}_s]=B_s+t\ne B_s+s\). The correction is to check both martingale behavior and quadratic variation; neither test replaces the other.
A second pitfall is treating a local martingale as though it were automatically a true martingale. A local martingale may fail to be integrable or may fail the conditional-expectation identity without stopping. In the examples above, the relevant Ito integrals are square-integrable, so their martingale property follows with the integrability and conditional-expectation checks explicitly in place.
Check Your Understanding
For each candidate, specify the filtration and say which Brownian test succeeds or fails.
- For \(M_t=B_t^2-t\), write the Ito differential and its quadratic variation. Which one of these tests prevents \(M\) from being Brownian motion?
- For \(Y_t=B_t+3t\), compute \(\mathbb{E}[Y_t\mid\mathcal{F}_s]\) and compare it with \(Y_s\) when \(s<t\).
- For \(X_t=3B_{t/9}\) with filtration \(\mathcal{G}_t=\mathcal{F}_{t/9}\), compute its quadratic variation and the variance of \(X_t-X_s\). Does it pass the Brownian test?
- For \(Z_t=3B_{t/3}\) with filtration \(\mathcal{H}_t=\mathcal{F}_{t/3}\), compute its quadratic variation. Which requirement fails?
- State the hypotheses of Lévy characterization that turn a continuous martingale candidate into Brownian motion relative to a filtration.