The second-order term comes from squaring the Brownian part of an increment. If follows and is once continuously differentiable in time and twice in space, then
where the derivatives are evaluated at . The ordinary chain rule accounts for the first two drift terms. Brownian quadratic variation supplies the third.
Over a small interval of length , Brownian motion moves on a scale of . Squaring that movement produces something on the scale of , the same scale as elapsed time. For a curved function, positive and negative shocks can have a nonzero average effect even when the shocks average to zero. If is convex near the current state, its positive change in response to an upward move outweighs the magnitude of its negative change in response to an equal downward move.
Start with the Taylor expansion over one step
Let be the step length and . For now take , and evaluate all derivatives at the left endpoint . In the expansion of , keep
The term measures the change in the function if you advance the clock and keep its spatial argument fixed. The term is the local linear response to the shock. The factor comes from the Taylor coefficient on the second spatial derivative. Since a Brownian step has standard deviation , its square is on the scale of . Squared time steps and mixed time-space steps are smaller after summation. For example, the sum of the expected absolute cubic Brownian terms is on the scale of on a uniform grid. A bounded third derivative gives this simple remainder estimate. For the usual formula, localisation and continuity of the second derivative control the second-order remainder.
Add up the squared increments
On a partition , write , , and . Split the quadratic contribution into its predictable part and a centred error.
Why does the second sum disappear? The coefficient is known before the new increment arrives. Conditional on that past, the squared increment has mean and variance . When you square the sum of the errors, each cross term has expectation zero by conditioning on the past before the later increment. With bounded , the remaining variance is at most , which tends to zero with the longest step. The first sum converges to the time integral of . This argument replaces a sum of squared steps by its limit. An individual squared Brownian step is still random.
Put the drift and diffusion back in
For , the linear term contributes . The squared noise term now has coefficient , so . The drift has zero quadratic variation and no covariation with the continuous martingale term. Adding the time derivative gives the full formula above. You can check which terms go where by taking . The curvature term vanishes and the equation for comes back unchanged.
Apply the formula to the square of a scaled Brownian motion
Take and . The formula gives , so is an Itô integral. Over a half-year its compensator is , which agrees with . The stochastic integral has mean zero because its integrand is square integrable on every finite horizon. Starting from a nonzero deterministic adds to the mean.