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How Ordinary Calculus and Itô Calculus Treat the Chain Rule

Original question

What is the difference between ordinary calculus and Itô calculus, and why does stochastic calculus require different rules?
Ordinary calculus uses the standard chain and product rules for differentiable paths. Itô calculus also handles paths with Brownian noise, whose squared increments accumulate over time. For Yt=f(Xt)Y_t=f(X_t), the Itô chain rule adds 12f′′(Xt) d[X]t\tfrac12 f''(X_t)\,d[X]_t, where [X][X] is quadratic variation. For a differentiable path, that quadratic variation is zero and the two rules coincide.

What becomes of the squared steps

For a C1C^1 path xx with bounded derivative on [0,T][0,T], a partition with mesh ∣π∣|\pi| satisfies

∑i(xti+1−xti)2≤sup⁡t∣xt′∣2 T∣π∣⟶0.\sum_i (x_{t_{i+1}}-x_{t_i})^2\le \sup_t |x'_t|^2\,T|\pi| \longrightarrow 0.
The same vanishing holds for any continuous finite-variation path by bounding the squared sum by the largest absolute increment times the total variation. For Brownian motion, ∑i(ΔWi)2\sum_i(\Delta W_i)^2 converges in mean square to TT, since the expectation is TT and the variance is 2∑i(Δti)22\sum_i(\Delta t_i)^2.
In ordinary calculus, the chain rule discards terms quadratic in the small increment. The Itô rule retains this term for Brownian motion. In the product rule, use the exact discrete identity Δ(XY)=XiΔY+YiΔX+ΔXΔY\Delta(XY)=X_i\Delta Y+Y_i\Delta X+\Delta X\Delta Y. Summing and taking the stochastic limit yields d(XY)=X dY+Y dX+d[X,Y]d(XY)=X\,dY+Y\,dX+d[X,Y]. For smooth paths, the final term vanishes by the same small-increment bound.

Compare xt=tx_t=t and Xt=WtX_t=W_t with f(x)=x2f(x)=x^2. For the smooth path, d(t2)=2t dtd(t^2)=2t\,dt. For the stochastic path, d(Wt2)=2Wt dWt+dtd(W_t^2)=2W_t\,dW_t+dt. Therefore

∫0tWs dWs=12(Wt2−t).\int_0^t W_s\,dW_s=\frac12(W_t^2-t).
At t=1t=1, a Brownian path whose endpoint happens to be near 0.40.4 gives an Itô integral near −0.42-0.42, although 12W12\tfrac12W_1^2 is near 0.080.08. The missing 0.50.5 in an ordinary-calculus calculation is the accumulated quadratic variation correction. The Itô integral has mean zero, consistent with an integrand fixed at the start of each interval.

You can see exactly where the two answers part company by telescoping Xti+12−Xti2=2XtiΔXi+(ΔXi)2X_{t_{i+1}}^2-X_{t_i}^2=2X_{t_i}\Delta X_i+(\Delta X_i)^2. For the smooth path, the second sum vanishes. For Brownian motion, it converges to elapsed time. The left-endpoint sum in the first term is what defines the Itô integral. In the Brownian example, even if the endpoint is close to zero, the integral can be negative because the squared increments accumulated along the whole path.

You can diagnose the rule by asking how the function was sampled. In the Itô integral of WW against itself, each term uses WtiW_{t_i} before the shock. If you instead average the two endpoint values in each term, you add half of each squared increment. The symmetric sum therefore tends to Wt2/2W_t^2/2, while the left-endpoint sum tends to (Wt2−t)/2(W_t^2-t)/2. This explains how an ordinary-looking chain rule can reappear in the symmetric convention. The two limits represent different integrals of the same path.