Proving Levy's Characterization With Exponential Martingales

From Quadratic Variation to Increment Laws

Lévy’s characterization turns a pathwise stochastic-calculus condition into a distributional statement. Suppose a continuous process is a local martingale, starts at zero, and accumulates quadratic variation at exactly one unit per unit of time. The key step is to apply Itô’s formula to a complex exponential. Its drift cancels, and its martingale property yields the conditional characteristic function of every increment. That function identifies a centered Gaussian law and shows that each increment is independent of the past.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), with the filtration satisfying the usual conditions. No Brownian motion is assumed at the outset: the process being tested is \(M=(M_t)_{t\ge0}\). Assume \(M\) is an adapted, continuous local martingale with \(M_0=0\) almost surely and quadratic variation \([M]_t=t\) almost surely for all \(t\ge0\). Here, a local martingale is a process that becomes a martingale after stopping at a sequence of stopping times increasing to infinity.

Theorem (Lévy’s characterization, exponential-martingale proof): Under these assumptions, \(M\) is a standard Brownian motion relative to \((\mathcal{F}_t)\). In particular, for \(0\le s<t\), \(M_t-M_s\) has distribution \(N(0,t-s)\) and is independent of \(\mathcal{F}_s\).

The Brownian motion \(B_t\) used in the course notation is not an input to this theorem; the conclusion is that \(M\) itself has the defining Brownian properties relative to the given filtration. For an integral against a Brownian motion \(B\), the usual square-integrable integrands are progressively measurable processes \(H\) with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The integral in the proof below is instead against the continuous local martingale \(M\): its integrand is progressively measurable and square-integrable with respect to the quadratic-variation measure, meaning \(\mathbb{E}\int_0^T |H_s|^2\,d[M]_s<\infty\). This condition will hold for the bounded integrands we use.

The Complex Exponential

Fix \(u\in\mathbb{R}\) and define \(Z_t=\exp(iuM_t+\tfrac12u^2t)\), where \(i^2=-1\). Complex Itô formula is applied to the real and imaginary parts, or equivalently to the complex-valued function \(f(t,x)=\exp(iux+\tfrac12u^2t)\). Its derivatives are \(f_t=\tfrac12u^2f\), \(f_x=iuf\), and \(f_{xx}=-u^2f\). The Itô correction is \(\tfrac12 f_{xx}\,d[M]_t\). Thus, using \(d[M]_t=dt\),

dZ_t=f_t\,dt+f_x\,dM_t+\tfrac12 f_{xx}\,d[M]_t =\tfrac12u^2Z_t\,dt+iuZ_t\,dM_t-\tfrac12u^2Z_t\,dt =iuZ_t\,dM_t.

The drift terms cancel. In differential multiplication notation, this cancellation uses the quadratic-variation contribution corresponding to \(dM_t\,dM_t=d[M]_t=dt\); the finite-variation cross terms satisfy \(dM_t\,dt=0\) and \(dt\,dt=0\). The process \(Z\) is therefore a complex local martingale. This alone does not make it a true martingale: a local martingale need not satisfy the martingale conditional-expectation identity.

Proposition: For each fixed \(u\), \(Z_t=\exp(iuM_t+\tfrac12u^2t)\) is a true martingale on every finite time interval.

Proof. Fix \(T<\infty\). Since \(M_t\) is real, \(|\exp(iuM_t)|=1\), so \(|Z_t|=\exp(\tfrac12u^2t)\le\exp(\tfrac12u^2T)\) for \(0\le t\le T\). The process is a local martingale by the Itô computation. Let \((\tau_n)\) be a localizing sequence, so each stopped process \(Z_{t\wedge\tau_n}\) is a martingale. For \(0\le s\le t\le T\), its martingale identity is \(\mathbb{E}[Z_{t\wedge\tau_n}\mid\mathcal{F}_s]=Z_{s\wedge\tau_n}\). Both sides are dominated in absolute value by the deterministic bound \(\exp(\tfrac12u^2T)\); continuity and \(\tau_n\to\infty\) almost surely imply convergence to \(Z_t\) and \(Z_s\). Conditional dominated convergence gives \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\). Also \(Z_t\) is integrable by the same bound. These are precisely the true-martingale requirements on \([0,T]\). Since \(T\) was arbitrary, the claim follows. \(\square\)

Worked Example: Checking the Cancellation at a Fixed Frequency

Take \(u=2\), so \(Z_t=\exp(2iM_t+2t)\). For \(f(t,x)=\exp(2ix+2t)\), the derivatives are \(f_t=2f\), \(f_x=2if\), and \(f_{xx}=-4f\). The second-order Itô correction is \(\tfrac12 f_{xx}\,d[M]_t=-2f\,dt\), because \(d[M]_t=dt\). Therefore

dZ_t=2Z_t\,dt+2iZ_t\,dM_t-2Z_t\,dt=2iZ_t\,dM_t.

On \(0\le t\le1\), \(|Z_t|=e^{2t}\le e^2\), so the bounded-local-martingale argument proves that \(Z\) is a true martingale there. Check: the two finite-variation terms are \(2Z_t\,dt\) and \(-2Z_t\,dt\), whose sum is zero; the remaining term is a stochastic integral with bounded integrand on this interval and \(\mathbb{E}\int_0^1|2iZ_s|^2\,d[M]_s\le4e^4<\infty\).

Conditional Characteristic Functions

Because \(Z_s=\exp(iuM_s+\tfrac12u^2s)\) is never zero and \(Z_s^{-1}\) is bounded on every finite interval, the martingale identity can be multiplied by the \(\mathcal{F}_s\)-measurable factor \(Z_s^{-1}\). For \(0\le s<t\), this gives

\mathbb{E}\left[\exp\big(iu(M_t-M_s)\big)\mid\mathcal{F}_s\right] =\exp\left(-\tfrac12u^2(t-s)\right).

This holds for every real \(u\). The right side is the characteristic function of \(N(0,t-s)\), and crucially it is deterministic: it does not depend on the information in \(\mathcal{F}_s\).

Worked Example: The Conditional Characteristic Function on an Interval

Set \(s=1\), \(t=3\), and \(u=\tfrac12\). The interval length is \(t-s=2\), so the formula gives

\mathbb{E}\left[\exp\left(\tfrac{i}{2}(M_3-M_1)\right)\mid\mathcal{F}_1\right] =\exp\left(-\tfrac12\left(\tfrac12\right)^2(2)\right) =e^{-1/4}.

The characteristic function of \(N(0,2)\) at \(u=\tfrac12\) is \(\exp(-\tfrac12\cdot2\cdot(\tfrac12)^2)=e^{-1/4}\), the same value. Check: setting \(u=0\) gives \(1\), as any characteristic function must, and the calculation depends on the interval length \(2\), not on the absolute time \(1\) at which the interval starts.

For completeness, the deterministic conditional characteristic function implies both the increment law and independence from the past. Fix \(A\in\mathcal{F}_s\), and define a finite measure on the real line by \(\mu_A(C)=\mathbb{P}(A\cap\{M_t-M_s\in C\})\) for Borel sets \(C\). Its characteristic function is, by the conditional-expectation identity,

\int_{\mathbb{R}}e^{iux}\,\mu_A(dx) =\mathbb{E}\left[\mathbf{1}_A e^{iu(M_t-M_s)}\right] =\mathbb{P}(A)e^{-\frac12u^2(t-s)}.

For \(\mathbb{P}(A)>0\), the measure on the right is the characteristic function of \(\mathbb{P}(A)N(0,t-s)\); for \(\mathbb{P}(A)=0\), both sides are zero and the same measure identity holds. By the standard uniqueness theorem for characteristic functions of finite measures, \(\mu_A=\mathbb{P}(A)N(0,t-s)\). This theorem is a standard result of probability theory; its proof is not needed here. Consequently, the increment has law \(N(0,t-s)\), and for every Borel set \(C\),

\mathbb{P}\big(A\cap\{M_t-M_s\in C\}\big) =\mathbb{P}(A)\,\mathbb{P}(M_t-M_s\in C).

This is the definition of independence of \(M_t-M_s\) from \(\mathcal{F}_s\). The process is adapted and continuous by assumption, starts at zero, and has independent Gaussian increments with the required variances. These properties establish that \(M\) is Brownian relative to the filtration. The reverse direction is the standard quadratic-variation property of Brownian motion: a Brownian motion is a continuous martingale starting at zero and has quadratic variation \(t\).

Worked Example: Independence Tested Against a Past Event

Let \(A\in\mathcal{F}_s\) have probability \(p\), where \(0\le p\le1\), and take \(t-s=4\). The increment law is \(N(0,4)\), whose characteristic function at \(u=1\) is \(e^{-2}\). The finite measure \(C\mapsto\mathbb{P}(A\cap\{M_t-M_s\in C\})\) therefore has characteristic function \(p e^{-2u^2}\), since \(-\tfrac12u^2\cdot4=-2u^2\). The measure \(pN(0,4)\) has exactly that characteristic function, so uniqueness gives \(\mathbb{P}(A\cap\{M_t-M_s\in C\})=p\,N(0,4)(C)\) for every Borel \(C\).

When \(p=0\), both sides are zero; when \(p=1\), this says the increment has its full \(N(0,4)\) law. For intermediate \(p\), it verifies independence from the particular past event \(A\). Check: taking \(C=\mathbb{R}\) gives \(p=p\cdot1\), as required by total probability.

Why the Sign and the Martingale Check Matter

The positive sign in \(+\tfrac12u^2t\) is essential. A common incorrect calculation uses \(Y_t=\exp(iuM_t-\tfrac12u^2t)\), borrowing the sign from a real stochastic exponential. For \(g(t,x)=\exp(iux-\tfrac12u^2t)\), the derivatives are \(g_t=-\tfrac12u^2g\), \(g_x=iug\), and \(g_{xx}=-u^2g\). Its Itô drift is therefore

g_t\,dt+\tfrac12g_{xx}\,d[M]_t =-\tfrac12u^2g\,dt-\tfrac12u^2g\,dt =-u^2g\,dt,

not zero. That expression does not yield the needed local martingale or the conditional characteristic-function identity. The correct positive sign makes the time derivative cancel the quadratic-variation correction.

A second pitfall is to stop at “local martingale.” The conclusion \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\) is what permits the conditional characteristic function argument, and it requires a true martingale. Here the bounded modulus on each finite time interval supplies that upgrade. A general local martingale, without an integrability or boundedness argument, need not have the true-martingale property.

Takeaway: The condition \([M]_t=t\) fixes the Itô correction for the complex exponential. The resulting bounded local martingale is a true martingale, and its conditional characteristic functions force exactly the Gaussian increment laws and independence that define Brownian motion.

Check Your Understanding

Use the exponential-martingale argument and distinguish its local and true martingale steps.

For \(u=3\), write the three derivatives of \(f(t,x)=\exp(iux+\tfrac12u^2t)\) and show explicitly that the Itô drift vanishes.
For \(t-s=\tfrac12\) and \(u=2\), compute the conditional characteristic function of \(M_t-M_s\).
Explain why \(|Z_t|\le\exp(\tfrac12u^2T)\) on \([0,T]\) is sufficient to pass from the localized martingale identities to the true-martingale identity.
For \(A\in\mathcal{F}_s\) with \(\mathbb{P}(A)=0\), explain why the finite-measure argument still gives the independence identity.
Compute the Itô drift for \(\exp(iuM_t-\tfrac12u^2t)\) and identify the sign error it reveals.