Brownian Motion as a Martingale Criterion
Brownian motion is usually introduced through its continuous paths and independent Gaussian increments. Lévy’s characterization gives a different test: under the right hypotheses, those distributional and independence properties follow from a martingale condition and a specified quadratic variation. This makes the result useful when a process is presented as a stochastic integral or an Ito-process transform, where its increments may be difficult to analyze directly.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), carrying a Brownian motion \(B_t\) relative to that filtration. A process is a martingale if it is adapted, integrable at each time, and satisfies \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for \(s\le t\). A local martingale has this property after stopping at a sequence of stopping times increasing to infinity; it need not be a true martingale without further conditions. For a continuous local martingale \(M\), its quadratic variation is denoted by \([M]_t\).
The filtration matters: the conclusion is Brownian motion relative to the specified \((\mathcal{F}_t)\), not merely a process with Brownian-looking one-time distributions. The initial value, continuity, local martingale property, and exact bracket are all required. A matching bracket alone does not replace the martingale condition.
How to Apply the Characterization
For a stochastic integral, first check that the integrand is progressively measurable and belongs to \(\mathcal{L}^2\) on every finite interval: \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Its Ito integral is then a square-integrable martingale, and its quadratic variation is \(\int_0^t H_s^2\,ds\). The martingale property follows from the Ito integral theorem: the integral is integrable by the Ito isometry, and its future increment has conditional expectation zero given the past. The bracket formula follows from the quadratic-variation rule for Ito integrals.
Thus, a stochastic integral \(\int_0^t H_s\,dB_s\) is a Brownian motion by Lévy’s characterization if it starts at zero and \(\int_0^t H_s^2\,ds=t\) almost surely for all \(t\). In particular, an integrand that takes values \(1\) and \(-1\) has \(H_s^2=1\), so its integral has the required bracket. By contrast, an integrand with a different squared magnitude usually produces a different bracket.
Worked Example: A Sign-Changing Integrand
Let \(H_s=1\) when \(B_s\ge0\) and \(H_s=-1\) when \(B_s<0\), and set \(M_t=\int_0^t H_s\,dB_s\). Determine whether \(M\) is Brownian motion.
The process \(H\) is progressively measurable because it is a Borel function of the adapted continuous process \(B\). It is bounded by \(1\), so for every finite \(T\), \(\mathbb{E}\int_0^T H_s^2\,ds=T<\infty\). The Ito integral is therefore a square-integrable martingale: \(\mathbb{E}[M_t^2]=\mathbb{E}\int_0^t H_s^2\,ds=t\), and its increments have conditional expectation zero given \(\mathcal{F}_s\). It is continuous, and \(M_0=0\).
At every \(s\), \(H_s^2=1\), including when \(B_s=0\), by the definition chosen there. Hence
Lévy’s characterization implies that \(M\) is Brownian motion relative to \((\mathcal{F}_t)\). Check: the Ito isometry gives \(\mathbb{E}[M_t^2]=t\), the variance of a Brownian motion at time \(t\). This moment check is consistent with the conclusion, although the bracket criterion—not the second moment alone—supplies the characterization.
When the Bracket Does Not Match
Worked Example: The Centered Square of Brownian Motion
Let \(M_t=B_t^2-t\). It is a continuous martingale starting at zero, but is it Brownian motion?
Apply Ito’s formula to \(f(x)=x^2\). Its derivatives are \(f'(x)=2x\) and \(f''(x)=2\); the second-derivative correction is \(\tfrac12 f''(B_t)\,dt=\tfrac12\cdot2\,dt=dt\). Therefore
The integrand \(2B_s\) is progressively measurable, and \(\mathbb{E}\int_0^T(2B_s)^2\,ds=4\int_0^T s\,ds=2T^2<\infty\). Thus \(M_t=\int_0^t2B_s\,dB_s\) is a square-integrable martingale, with conditional mean-zero future increments. Its bracket is
This is not \(t\) almost surely. For instance, at \(t=1\), \(\mathbb{E}[[M]_1]=4\int_0^1s\,ds=2\), whereas the expectation of the constant \(1\) is \(1\). So the bracket condition fails and Lévy’s characterization does not identify \(M\) as Brownian motion. Check: the same Ito calculation gives \(\mathbb{E}[M_t]=0\), consistent with the martingale representation. The zero mean does not repair the incorrect bracket.
Worked Example: Brownian Motion Stopped at a Fixed Time
Let \(M_t=B_{t\wedge1}\). Check whether this process is Brownian motion for all \(t\ge0\).
Equivalently, \(M_t=\int_0^t\mathbf{1}_{\{s\le1\}}\,dB_s\); changing the integrand at the single time \(s=1\) does not change the integral. This integrand is progressively measurable and satisfies \(\mathbb{E}\int_0^T\mathbf{1}_{\{s\le1\}}^2\,ds=\min(T,1)<\infty\). The Ito integral is a square-integrable martingale, since its increments have conditional expectation zero, and it is continuous with \(M_0=0\).
Its bracket is
For \(t>1\), this bracket is \(1\), not \(t\), so the process is not Brownian motion on the full time interval. A direct check at \(t=2\) gives \(\mathbb{E}[M_2^2]=\mathbb{E}[B_1^2]=1\), whereas a Brownian motion at time \(2\) has second moment \(2\). The failure is also visible in the bracket: after time \(1\), the stopped process accumulates no further quadratic variation.
The Martingale Condition Is Essential
Worked Example: Correct Bracket, Nonzero Drift
Let \(X_t=B_t+t\). Its paths are continuous, \(X_0=0\), and its quadratic variation is \(t\), because the finite-variation term \(t\) contributes no quadratic variation. Does the characterization apply?
No: \(X\) is not a martingale. For \(s<t\), independence and the zero mean of Brownian increments give
Thus the martingale hypothesis fails, even though continuity, the initial value, and the bracket match. This example also illustrates why drift and diffusion must not be confused: adding a deterministic drift changes the conditional mean of future increments without changing the quadratic variation.
Why the Criterion Matters
Lévy’s characterization lets us establish a process’s Brownian nature without separately deriving Gaussian increment distributions and independence from the past. The most efficient test is to check the conditions in order:
- Is the process adapted, continuous, and zero at time zero?
- Is it a local martingale (or a true martingale)? A drift term can invalidate this step.
- Is its quadratic variation exactly \(t\) for every \(t\), almost surely?
For a Brownian motion, \([B]_t=t\); this is the standard quadratic-variation result. The notation \(dB_t\,dB_t=dt\) records the same quadratic-variation contribution in Ito calculus. It is not an ordinary chain rule and does not say that an individual squared increment equals the corresponding time increment. In applications, a bracket of \(c^2t\) can be normalized as in the corollary, but a random bracket or a bracket that stops growing does not meet the theorem’s exact hypothesis.
Common pitfall. A process with the right quadratic variation is not automatically Brownian. The incorrect calculation “\([B_t+t]_t=t\), therefore \(B_t+t\) is Brownian” checks only one hypothesis. The correction is to test the martingale property as well: \(\mathbb{E}[B_t+t\mid\mathcal{F}_s]=B_s+t\), which differs from \(B_s+s\) whenever \(s<t\). The theorem therefore does not apply.
Check Your Understanding
For each candidate, identify which hypotheses of Lévy’s characterization hold and which fail.
- If \(M_t=\int_0^t H_s\,dB_s\) with progressively measurable \(H\in\mathcal{L}^2\) and \(H_s^2=1\) almost everywhere, what is \([M]_t\), and what follows?
- Why does \(B_t^2-t\) satisfy the martingale condition but fail the bracket condition?
- For \(M_t=3B_t\), what is its bracket, and what rescaling produces a Brownian motion?
- Which hypothesis fails for \(X_t=B_t+t\), despite \([X]_t=t\)?
- What happens to the bracket of \(B_{t\wedge a}\) after a fixed time \(a>0\)?