Lévy's Characterization of Brownian Motion

Suppose someone hands you a process built from a strange-looking stochastic differential equation and asks: is this secretly just ordinary Brownian motion in disguise? Checking every property of Brownian motion directly — independent increments, Gaussian distribution at every horizon, continuous paths — is painful. Lévy's characterization gives a shortcut: check only two things, both of them properties you can often verify from the model itself rather than from simulated paths.

A continuous process $M_t$ starting at zero is a Brownian motion if and only if it is a martingale (no drift — its expected future value, given everything so far, is its current value) and its quadratic variation equals $t$ itself, written $\langle M \rangle_t = t$. Quadratic variation measures the accumulated "jitteriness" of a path; for Brownian motion this jitter accumulates at exactly rate one per unit time, no faster and no slower. Any continuous martingale with that same jitter rate must, in fact, be indistinguishable in distribution from Brownian motion — a genuinely surprising fact, since a martingale can otherwise look wildly different path by path.

The payoff shows up constantly in derivatives pricing. Take a driftless process built from an Itô integral, $M_t = \int_0^t \sigma_s \, dW_s$, where $\sigma_s$ varies over time. Itô's isometry gives its quadratic variation as $\langle M \rangle_t = \int_0^t \sigma_s^2 \, ds$. If a model is constructed so that $\int_0^t \sigma_s^2 \, ds = t$ exactly — for instance $\sigma_s = 1$ for all $s$ — Lévy's theorem says $M_t$ is a Brownian motion, even though it was written down as an integral against some other Brownian motion $W_t$. This is exactly the trick used to justify time-changed and stochastic-volatility constructions: change the clock so the quadratic variation runs at unit speed, and the result is provably standard Brownian motion again.