A Finite-Partition Proof for the Square Function

From a Finite Sum to the Square-Function Identity

For a differentiable path \(x(t)\), the ordinary chain rule gives \(x(T)^2-x(0)^2=2\int_0^T x(t)\,dx(t)\). Brownian paths are not differentiable, and their squared increments do not disappear as a partition is refined. The finite-partition calculation for the square function makes the resulting correction precise: the difference between the Brownian square and the left-point sum is exactly a sum of squared increments.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) adapted to the filtration. In particular, its increments are independent of the past and \(B_t-B_s\) has distribution \(N(0,t-s)\) for \(s<t\). Fix \(T>0\) and a deterministic partition \(\Pi=\{0=t_0<t_1<\cdots<t_n=T\}\). Write \(\Delta t_i=t_i-t_{i-1}\), \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\), and \(|\Pi|=\max_i\Delta t_i\).

The Ito integral \(\int_0^T H_s\,dB_s\) below is defined for progressively measurable integrands \(H\) in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The finite sum we need is represented by the left-point process \(H^\Pi_s=B_{t_{i-1}}\) for \(t_{i-1}\le s<t_i\), with \(H^\Pi_T=B_{t_{n-1}}\). This is an adapted step process and is progressively measurable; the value at the terminal endpoint is chosen from the past. Its square-integrability follows from \(\mathbb{E}\int_0^T(H^\Pi_s)^2\,ds=\sum_i\Delta t_i\,t_{i-1}<\infty\). Thus it is an admissible Ito integrand.

The Exact Calculation on Each Interval

On a single interval, use \(B_{t_i}=B_{t_{i-1}}+\Delta B_i\). Squaring and subtracting the left-endpoint square gives

Summing over \(i\) makes the left side telescope: every intermediate square occurs once with a plus sign and once with a minus sign. Therefore the finite-partition identity above holds without approximation or probability assumptions. The probability enters only when we identify the limits of the two sums.

Why the Squared-Increment Sum Becomes Time

Proof. Each increment has distribution \(N(0,\Delta t_i)\), so \(\mathbb{E}[(\Delta B_i)^2]=\Delta t_i\) and \(\mathbb{E}[(\Delta B_i)^4]=3(\Delta t_i)^2\). Thus \(\mathbb{E}[Q_\Pi]=\sum_i\Delta t_i=T\), and the variance of each squared increment is \(3(\Delta t_i)^2-(\Delta t_i)^2=2(\Delta t_i)^2\). The Brownian increments are independent, so their squares are independent and \(\operatorname{Var}(Q_\Pi)=2\sum_i(\Delta t_i)^2\). Since \(\Delta t_i\le|\Pi|\), we have \(\sum_i(\Delta t_i)^2\le|\Pi|\sum_i\Delta t_i=T|\Pi|\). Consequently, \(\mathbb{E}[(Q_\Pi-T)^2]=\operatorname{Var}(Q_\Pi)\le2T|\Pi|\), which tends to zero as the mesh tends to zero. \(\square\)

Why the Left-Point Sum Becomes an Ito Integral

For each partition, the definition of the Ito integral for a step process gives \(\int_0^T H^\Pi_s\,dB_s=\sum_i B_{t_{i-1}}\Delta B_i\). To pass to the integral with integrand \(B_s\), first check its admissibility: \(B\) is progressively measurable and \(\mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=T^2/2<\infty\). The Ito isometry then bounds the approximation error.

Proof. On each interval \([t_{i-1},t_i)\), \(H^\Pi_s-B_s=B_{t_{i-1}}-B_s\). By the Ito isometry and the Brownian increment variance,

The right side tends to zero with the mesh, proving the asserted \(L^2\) convergence. The endpoint value at \(T\) does not affect the time integral; here it was chosen as an adapted left-point value, so the process remains an admissible integrand. \(\square\)

The Finite-Partition Proof of Ito's Identity

Proof. For each partition, the exact algebraic identity is \(B_T^2-B_0^2=2\sum_i B_{t_{i-1}}\Delta B_i+Q_\Pi\). Along any sequence of deterministic partitions whose mesh tends to zero, the left-point sums converge in \(L^2\) to \(\int_0^T B_s\,dB_s\), and \(Q_\Pi\) converges in \(L^2\) to \(T\). Therefore the right side converges in \(L^2\) to \(2\int_0^T B_s\,dB_s+T\). Since every finite-partition right side equals the same random variable \(B_T^2-B_0^2\), that random variable equals the \(L^2\) limit almost surely. This proves the identity. \(\square\)

For \(f(x)=x^2\), the derivatives are \(f'(x)=2x\) and \(f''(x)=2\). The Ito second-order correction is \(\tfrac12 f''(B_s)\,ds=\tfrac12\cdot2\,ds=ds\), so the theorem has the differential form \(d(B_t^2)=2B_t\,dB_t+dt\). The finite-partition argument explains why the \(dt\) term appears: it is the limit of the squared-increment sum. It is not obtained by applying the ordinary chain rule to a Brownian path.

As a check using expectations, the left side has expectation \(\mathbb{E}[B_T^2]=T\), since \(B_0=0\). The stochastic integral has mean zero: its left-point sums have mean zero because each \(B_{t_{i-1}}\) is measurable with respect to the past and \(\Delta B_i\) is an independent, mean-zero increment; their \(L^2\) limit therefore also has mean zero. Thus the right side has expectation \(0+T\), in agreement with the left.

Common Pitfalls

Dropping the squared increments. A tempting but incorrect calculation is to write \(B_T^2-B_0^2=2\sum_iB_{t_{i-1}}\Delta B_i\), as if the ordinary chain rule applied to each increment. In the two-increment calculation above, this would give \(0=-\tfrac12\), because the left-point term doubled is \(-\tfrac12\), while the actual squared-increment sum is \(\tfrac12\). The missing term is exactly what restores the finite-partition identity.

Choosing inadmissible endpoint values. Changing a process at finitely many times does not alter its time integral when the modification remains a valid progressively measurable integrand. But an arbitrary endpoint choice need not preserve that property. For example, setting the value at an interior partition time \(t_i<T\) equal to \(B_T\) generally uses information unavailable at time \(t_i\), so the modified process may fail to be adapted. The left-point convention used here chooses values measurable with respect to the filtration at the corresponding times and avoids this issue.

The finite identity is pathwise and exact; its limiting interpretation uses two distinct convergence facts. The squared-increment sum converges to elapsed time, while the left-point sums converge to the Ito integral. Keeping those roles separate is the central discipline in this calculation.