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Deriving Itô’s Lemma From the Second-Order Taylor Term

Original question

Can you derive Itô's lemma step by step and explain why the second-order derivative term appears?

The second-order term comes from squaring the Brownian part of an increment. If XX follows dXt=at dt+bt dWtdX_t=a_t\,dt+b_t\,dW_t and ff is once continuously differentiable in time and twice in space, then

df(t,Xt)=(ft+atfx+12bt2fxx)dt+btfx dWt,df(t,X_t)=\left(f_t+a_t f_x+\frac12b_t^2f_{xx}\right)dt+b_tf_x\,dW_t,
where the derivatives are evaluated at (t,Xt)(t,X_t). The ordinary chain rule accounts for the first two drift terms. Brownian quadratic variation supplies the third.
Over a small interval of length hh, Brownian motion moves on a scale of h\sqrt h. Squaring that movement produces something on the scale of hh, 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 ff 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 hh be the step length and ΔW=Wt+h−Wt\Delta W=W_{t+h}-W_t. For now take X=WX=W, and evaluate all derivatives at the left endpoint (t,Wt)(t,W_t). In the expansion of f(t+h,Wt+ΔW)−f(t,Wt)f(t+h,W_t+\Delta W)-f(t,W_t), keep

fth+fxΔW+12fxx(ΔW)2.f_t h+f_x\Delta W+\frac12 f_{xx}(\Delta W)^2.
The term fthf_t h measures the change in the function if you advance the clock and keep its spatial argument fixed. The term fxΔWf_x\Delta W is the local linear response to the shock. The factor 1/21/2 comes from the Taylor coefficient on the second spatial derivative. Since a Brownian step has standard deviation h\sqrt h, its square is on the scale of hh. 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 ThT\sqrt h on a uniform grid. A bounded third derivative gives this simple remainder estimate. For the usual C1,2C^{1,2} formula, localisation and continuity of the second derivative control the second-order remainder.

Add up the squared increments

On a partition 0=t0<⋯<tn=T0=t_0<\cdots<t_n=T, write hi=ti+1−tih_i=t_{i+1}-t_i, ΔWi=Wti+1−Wti\Delta W_i=W_{t_{i+1}}-W_{t_i}, and ci=fxx(ti,Wti)c_i=f_{xx}(t_i,W_{t_i}). Split the quadratic contribution into its predictable part and a centred error.

∑ici(ΔWi)2=∑icihi+∑ici((ΔWi)2−hi).\sum_i c_i(\Delta W_i)^2 =\sum_i c_i h_i+\sum_i c_i\big((\Delta W_i)^2-h_i\big).
Why does the second sum disappear? The coefficient cic_i is known before the new increment arrives. Conditional on that past, the squared increment has mean hih_i and variance 2hi22h_i^2. 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 cic_i, the remaining variance is at most 2sup⁡i∣ci∣2∑ihi22\sup_i|c_i|^2\sum_i h_i^2, which tends to zero with the longest step. The first sum converges to the time integral of fxxf_{xx}. 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 dXt=at dt+bt dWtdX_t=a_t\,dt+b_t\,dW_t, the linear term contributes atfx dt+btfx dWta_tf_x\,dt+b_tf_x\,dW_t. The squared noise term now has coefficient bt2b_t^2, so d[X]t=bt2 dtd[X]_t=b_t^2\,dt. 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 f(x)=xf(x)=x. The curvature term vanishes and the equation for XX comes back unchanged.

Apply the formula to the square of a scaled Brownian motion

Take Xt=2WtX_t=2W_t and f(x)=x2f(x)=x^2. The formula gives d(Xt2)=4Xt dWt+4 dtd(X_t^2)=4X_t\,dW_t+4\,dt, so Xt2−4tX_t^2-4t is an Itô integral. Over a half-year its compensator is 22, which agrees with E[X0.52]=4Var⁡(W0.5)=2\mathbb E[X_{0.5}^2]=4\operatorname{Var}(W_{0.5})=2. The stochastic integral has mean zero because its integrand is square integrable on every finite horizon. Starting from a nonzero deterministic X0X_0 adds X02X_0^2 to the mean.

Interactive example

Explore a Brownian path at different resolutions

A seeded path is sampled at nested time grids. Compare the sum of squared increments with the theoretical quadratic variation of 1 at time 1.

64 increments · realised quadratic variation 0.7726 · theoretical 1.0000
A fixed Brownian path sampled on the selected time grid-2-1000.250.50.751TimeBrownian value
  • Sampled path

References and next steps