Partition Refinement and Integral Invariance

Why Refinement Should Not Change the Integral

A time partition is a bookkeeping device for a simple stochastic integral. If an integrand has a constant coefficient throughout one time interval, dividing that interval into smaller pieces should not alter its contribution: the smaller Brownian increments add to the original increment. The important qualification is that the coefficient must be carried unchanged to every new piece. Refinement changes the representation, not the integrand.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to this filtration. Fix \(T>0\). For every stochastic integral below, the integrand is adapted and progressively measurable, and belongs to \(\mathcal{L}^2\) on \([0,T]\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). A simple predictable process on a partition \(0=t_0<t_1<\cdots<t_n=T\) has the form \(H_s=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{[t_i,t_{i+1})}(s)\), where each \(\xi_i\) is \(\mathcal{F}_{t_i}\)-measurable and square-integrable. The Ito integral at \(T\) is defined by the finite sum \(\int_0^T H_s\,dB_s=\sum_{i=0}^{n-1}\xi_i(B_{t_{i+1}}-B_{t_i})\).

This carrying-forward rule is admissible: if a refined interval starts at \(s_j\ge t_i\), then \(\xi_i\), which is \(\mathcal{F}_{t_i}\)-measurable, is also \(\mathcal{F}_{s_j}\)-measurable because the filtration is increasing. Square integrability is preserved as well. In fact, the refined process is the same process \(H\), so its integral of \(H_s^2\) over time is unchanged.

One Split: The Basic Identity

Proof. For every outcome for which the Brownian increments are defined, the increment identity \(B_b-B_a=(B_c-B_a)+(B_b-B_c)\) holds. Multiplying this identity by the same coefficient \(\xi\) gives \(\xi(B_b-B_a)=\xi(B_c-B_a)+\xi(B_b-B_c)\). The coefficient is admissible on both new intervals: it is measurable at time \(a\), and hence at time \(c\). Contributions from all other intervals are unaltered, so the full finite sums agree. \(\square\)

The argument is algebraic. It does not require an independence calculation, an expectation, or a limiting procedure. In particular, it applies to a random coefficient as well as to a deterministic one. It proves equality of the finite-sum Ito integrals, not merely equality in expectation.

Any Finite Refinement

Proof. Within each original interval \([t_i,t_{i+1}]\), list its refined points as \(r_{i,0}=t_i<r_{i,1}<\cdots<r_{i,k_i}=t_{i+1}\). The refined contribution from that interval is

The middle sum telescopes: each intermediate value \(B_{r_{i,j}}\) appears once with a plus sign and once with a minus sign. The final expression is exactly the original contribution from \([t_i,t_{i+1}]\). Summing this equality over \(i=0,\ldots,n-1\) proves equality of the full finite sums. The refined coefficients remain admissible because \(\mathcal{F}_{t_i}\subseteq\mathcal{F}_{r_{i,j}}\), and the refined process has the same finite \(\mathcal{L}^2\) norm as \(H\). Hence both sums are valid simple Ito integrals, and their equality holds almost surely. \(\square\)

The conclusion also applies to the integral process at any time \(t\le T\). To compute the integral up to \(t\), include \(t\) as a partition point, or truncate the interval containing \(t\). On each completed original interval the telescoping argument above applies; on the last, possibly partial, interval both representations end at the same time \(t\) and telescope to the same increment. Thus refinement does not change the simple integral process, not just its terminal value.

What the Invariance Does—and Does Not—Say

The result matters because a simple integrand can have many partition representations. One representation may list only the intervals on which its coefficient changes; another may include extra points. The theorem ensures that inserting such extra points cannot change the integral, provided the integrand itself is unchanged. It is a finite algebraic consistency property of the simple Ito integral and does not rely on an approximation limit.

Refinement is not the same as choosing new coefficients. The coefficient on a refined subinterval must be the coefficient already assigned to its original interval. For example, on \([0,1)\) with \(H_s=2\), suppose a record has \(B_{1/2}-B_0=\tfrac25\) and \(B_1-B_{1/2}=-\tfrac1{10}\). The correct refined sum is \(2\cdot\tfrac25+2\cdot(-\tfrac1{10})=\tfrac45-\tfrac15=\tfrac35\). The unsplit increment is \(\tfrac25-\tfrac1{10}=\tfrac3{10}\), so its contribution is \(2\cdot\tfrac3{10}=\tfrac35\). A wrong calculation that changes the second coefficient to \(3\) gives \(2\cdot\tfrac25+3\cdot(-\tfrac1{10})=\tfrac45-\tfrac3{10}=\tfrac12\), not \(\tfrac35\). That calculation has changed the integrand; it is not a refinement of the same representation.

This identity concerns Ito sums for simple predictable processes. It does not assert that arbitrary integrands can already be integrated by finite sums, nor does it prove a limit theorem for approximating integrands. Those require additional construction. Here the key point is exact: splitting a Brownian increment into smaller increments and multiplying each by the same admissible coefficient leaves the simple Ito integral unchanged.