From the Ordinary Chain Rule to Partition Sums
For a differentiable path, the ordinary chain rule describes how a function changes along that path. Its partition-sum form makes an important hypothesis visible: when the path has bounded variation, squared increments vanish in the fine-partition limit. Brownian paths do not have this property. Comparing the two settings shows precisely why applying the ordinary first-order chain rule directly to Brownian motion leaves out a contribution.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). The smooth-path calculations below are deterministic and apply to each fixed path \(x:[0,T]\to\mathbb{R}\). A partition is \(0=t_0<t_1<\cdots<t_n=T\); write \(\Delta t_i=t_i-t_{i-1}\), \(\Delta x_i=x(t_i)-x(t_{i-1})\), and \(|\Pi|=\max_i\Delta t_i\). In this tutorial, integrals involving a smooth path are ordinary calculus integrals, not stochastic integrals.
The Partition-Sum Chain Rule
Suppose \(x\) is continuously differentiable on \([0,T]\), and \(f\) is continuously differentiable on an open interval containing the range of \(x\). The ordinary chain rule says that the derivative of \(f(x(t))\) is \(f'(x(t))x'(t)\). Summing over a partition gives the exact telescoping identity \(\sum_i\big(f(x(t_i))-f(x(t_{i-1}))\big)=f(x(T))-f(x(0))\). The chain rule appears when each summand is approximated by its first-order term.
Proof. Let \(K=\sup_{0\le s\le T}|x'(s)|\), which is finite by continuity, and let \(V=\int_0^T|x'(s)|\,ds\). The mean value theorem gives \(|\Delta x_i|\le K\Delta t_i\), so \(\max_i|\Delta x_i|\le K|\Pi|\to0\), and \(\sum_i|\Delta x_i|\le V\). Consequently, $$ \sum_i(\Delta x_i)^2 \le \max_i|\Delta x_i|\sum_i|\Delta x_i| \le K|\Pi|V\longrightarrow0. $$ For the first assertion, let \(R_i=f(x(t_i))-f(x(t_{i-1}))-f'(x(t_{i-1}))\Delta x_i\). Since \(f'\) is uniformly continuous on a compact interval containing the range of \(x\), its modulus of continuity \(\omega(\delta)=\sup\{|f'(u)-f'(v)|:|u-v|\le\delta\}\) tends to zero as \(\delta\to0\). The mean value theorem yields \(|R_i|\le\omega(\max_j|\Delta x_j|)|\Delta x_i|\). Thus \(\sum_i|R_i|\le\omega(\max_j|\Delta x_j|)V\to0\). The exact telescoping sum is \(f(x(T))-f(x(0))\), so the first-order sums have the same limit. Finally, the ordinary chain rule gives \(\int_0^T f'(x(s))x'(s)\,ds=f(x(T))-f(x(0))\). \(\square\)
The proof identifies why this argument works: the total absolute change stays bounded, while the largest individual increment tends to zero. Their product bounds the sum of squared increments. The same estimate does not apply to Brownian paths, which are almost surely not paths of bounded variation.
Worked Example: A Polynomial Path and Its Partition Sums
Take \(x(t)=t^2\) and \(f(u)=u^2\) on \([0,1]\). Compute the endpoint change and the limit of the first-order partition sum on the equal partition \(t_i=i/n\).
Here \(f(x(t))=t^4\), so the endpoint change is \(f(x(1))-f(x(0))=1-0=1\). Also \(f'(u)=2u\), \(x(t_{i-1})=(i-1)^2/n^2\), and \(\Delta x_i=(i^2-(i-1)^2)/n^2=(2i-1)/n^2\). Therefore the first-order sum is $$ \sum_{i=1}^n f'(x(t_{i-1}))\Delta x_i = \frac{2}{n^4}\sum_{i=1}^n(i-1)^2(2i-1). $$
Putting \(j=i-1\), the sum in the numerator is \(\sum_{j=0}^{n-1}(2j^3+j^2)\). Using \(\sum_{j=0}^{n-1}j^3=n^2(n-1)^2/4\) and \(\sum_{j=0}^{n-1}j^2=n(n-1)(2n-1)/6\), the partition sum equals $$ \frac{(n-1)^2}{n^2}+\frac{(n-1)(2n-1)}{3n^3}\longrightarrow1. $$
Check: the limit agrees with the endpoint change \(1\). For \(n=2\), the first-order sum is \(0+\frac{3}{8}=\frac{3}{8}\), while the exact telescoping change is \(1\); the difference is the finite-partition remainder, which vanishes in the limit.
Worked Example: The Logarithm Along a Positive Smooth Path
Let \(x(t)=1+2t\) on \([0,1]\) and \(f(u)=\log u\), defined for \(u>0\). Find the chain-rule integral and express its left-endpoint sums.
The path stays in \([1,3]\), where \(f'(u)=1/u\). Since \(x'(t)=2\), the chain-rule integrand is \(f'(x(t))x'(t)=2/(1+2t)\). Hence $$ \int_0^1\frac{2}{1+2t}\,dt =\log(1+2t)\big|_0^1 =\log3-\log1 =\log3. $$
For the equal partition \(t_i=i/n\), \(\Delta x_i=2/n\) and \(x(t_{i-1})=1+2(i-1)/n\). The first-order sum is therefore $$ \sum_{i=1}^n\frac{2/n}{1+2(i-1)/n}. $$
This is the left-endpoint Riemann sum for the continuous function \(2/(1+2t)\), so it converges to the integral just evaluated. Independently, its exact telescoping counterpart is \(\sum_i\big(\log x(t_i)-\log x(t_{i-1})\big)=\log3\) for every \(n\). The theorem ensures the first-order sum approaches that exact endpoint change.
Check: \(x(0)=1\), \(x(1)=3\), and \(f(x(1))-f(x(0))=\log3-\log1=\log3\), matching the integral.
What Changes for Brownian Paths?
The key smooth-path estimate was \(\sum_i(\Delta x_i)^2\le\max_i|\Delta x_i|\sum_i|\Delta x_i|\). It depends on bounded total variation. Brownian motion has a different squared-increment limit: along equal partitions of \([0,T]\), its quadratic variation sums converge to \(T\) in \(L^2(\mathbb{P})\). This is the standard Brownian quadratic-variation result used in the earlier material on quadratic variation.
Proof. Each increment \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\) is normal with mean zero and variance \(T/n\); the increments are independent. A centered normal variable of variance \(v\) has second moment \(v\) and fourth moment \(3v^2\). Thus \(\mathbb{E}[Q_n]=n(T/n)=T\), and $$ \operatorname{Var}(Q_n) =\sum_{i=1}^n\operatorname{Var}((\Delta B_i)^2) =n\left(3(T/n)^2-(T/n)^2\right) =\frac{2T^2}{n}. $$ Therefore \(\mathbb{E}[(Q_n-T)^2]=\operatorname{Var}(Q_n)=2T^2/n\to0\), which is exactly the claimed \(L^2(\mathbb{P})\) convergence. \(\square\)
Worked Example: The Square Function Exposes the Missing Contribution
Apply the partition identity for \(f(u)=u^2\) to Brownian motion on \([0,T]\), using the equal partitions above. Compare the exact change in \(B^2\) with the sum suggested by keeping only the ordinary first-order term.
For every increment, ordinary algebra gives $$ B_{t_i}^2-B_{t_{i-1}}^2 =2B_{t_{i-1}}\Delta B_i+(\Delta B_i)^2. $$ Summing this identity telescopes on the left and gives $$ B_T^2-B_0^2 =2\sum_{i=1}^nB_{t_{i-1}}\Delta B_i+Q_n. $$ Since \(B_0=0\), the first-order sum is exactly \(\frac12(B_T^2-Q_n)\). By the proposition, it converges in \(L^2(\mathbb{P})\), and hence in probability, to \(\frac12(B_T^2-T)\).
If one applied the smooth-path chain rule and discarded the squared increments, one would instead predict that the first-order sum accounts for all of \(B_T^2\), or that it should equal \(\frac12B_T^2\). The exact identity shows what this omits: the sum \(Q_n\), whose limit is \(T\). The discrepancy between the limiting first-order sum and the naive prediction is \(-T/2\).
Check: \(\mathbb{E}[B_T^2]=T\), so the actual limiting expression has mean \(\frac12(T-T)=0\). This also agrees with the mean-zero property of each finite sum: \(B_{t_{i-1}}\) is determined by the past, and the independent increment \(\Delta B_i\) has mean zero. The naive expression \(\frac12B_T^2\) has mean \(T/2\), so it cannot be the limit of these first-order sums in \(L^2\).
Common Pitfall: Importing Smooth-Path Remainder Estimates
A tempting but invalid calculation is to write \(\Delta(B^2)=2B\,\Delta B+o(|\Delta B|)\) and then assume the remainders sum to zero, as they do for a continuously differentiable path. For the square function, the remainder is not merely an unspecified error: it is exactly \((\Delta B)^2\). Its sum tends to \(T\), not zero. The smooth-path proof cannot be transferred because Brownian paths do not have the bounded variation estimate that made the total first-order remainder vanish.
The ordinary chain rule remains correct for differentiable paths. What fails is using its first-order partition approximation when the squared increments of the path have a nonzero limit. The calculation above isolates that obstruction without requiring a stochastic chain-rule formula: exact algebra already shows the term that a first-order-only argument would lose.
Check Your Understanding
Use the partition identities and hypotheses above to answer these questions.
- For a continuously differentiable path \(x\), which two quantities bound \(\sum_i(\Delta x_i)^2\) in the proof, and why does their product tend to zero?
- For \(x(t)=1+2t\) and \(f(u)=\log u\), write the left-endpoint sum on the equal partition and identify its limit.
- For \(f(u)=u^2\), write the exact increment identity that separates the first-order term from the squared increment.
- What is the \(L^2(\mathbb{P})\) limit of \(\sum_i(\Delta B_i)^2\) along equal partitions of \([0,T]\), and what is the variance of this sum for \(n\) intervals?
- Why does the smooth-path argument that discards squared increments not apply to Brownian motion?