Well-Definedness Under Different Step Representations

Why the Step Representation Must Not Matter

A simple predictable process can be written using different time partitions. For example, a process that is constant on \((0,2]\) can also be written as two constant pieces on \((0,1]\) and \((1,2]\). The Ito integral is defined from the coefficients and Brownian increments in such a representation, so its definition is meaningful only if changing the representation leaves the answer unchanged.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\), and fix a finite horizon \(T>0\). A partition is a finite, strictly increasing list \(0=t_0<t_1<\cdots<t_n=T\). We use the endpoint convention \((t_{i-1},t_i]\). A partition-step process is \(H_t=\sum_{i=1}^n \xi_i\mathbf{1}_{(t_{i-1},t_i]}(t)\), where each \(\xi_i\) is \(\mathcal{F}_{t_{i-1}}\)-measurable and square-integrable. Thus the coefficient on an interval is known at its left endpoint. Such a process is progressively measurable and satisfies \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); it belongs to the class \(\mathcal{L}^2\) of progressively measurable integrands with this property.

The displayed finite sum is the construction for simple processes; it is not an appeal to a later extension theorem for general \(\mathcal{L}^2\) integrands. The integrability condition follows directly: \(\mathbb{E}\int_0^T H_s^2\,ds=\sum_i(t_i-t_{i-1})\mathbb{E}[\xi_i^2]<\infty\). If \(T=0\), there are no intervals and the integral is zero. In what follows \(T>0\), so each cell of a partition or refinement has strictly positive length.

Common Refinements Compare the Coefficients

Given two partitions, their common refinement is obtained by listing all their endpoints in increasing order, without repetitions. On each open-left, closed-right cell of this refined partition, both original processes are constant in time (though their coefficients may be random). The key point is that equality of the processes forces equality of those coefficients almost surely on every such cell.

Proof. On the \(j\)th cell, \(H_t-G_t=\eta_j-\zeta_j\). If \(\eta_j=\zeta_j\) almost surely for every \(j\), then, outside the union of finitely many null events, the processes agree on all cells, hence \(dt\,d\mathbb{P}\)-almost everywhere. Conversely, if \(H=G\) for \(dt\,d\mathbb{P}\)-almost every \((t,\omega)\), then on the \(j\)th cell the product-measure of the set where they differ is \((u_j-u_{j-1})\mathbb{P}(\eta_j\ne\zeta_j)\). This must be zero. Since \(u_j-u_{j-1}>0\), it follows that \(\mathbb{P}(\eta_j\ne\zeta_j)=0\). This holds for each of the finitely many cells. \(\square\)

The proposition uses equality of processes up to product-measure null sets, the natural notion for integrands in \(\mathcal{L}^2\). Values assigned at isolated partition endpoints do not affect this equality or the time integral. With the stated \((t_{i-1},t_i]\) convention, the integral formula is nevertheless unambiguous at those endpoints.

Representation Invariance of the Integral

Proof. Take a common refinement \(0=u_0<\cdots<u_m=T\). Let \(\eta_j\) and \(\zeta_j\) be the coefficients of \(H\) and \(G\) on its \(j\)th cell. By the proposition, there is a probability-one event on which \(\eta_j=\zeta_j\) for all \(j\); finiteness of the refinement lets us take one event for every cell at once. Brownian paths are continuous almost surely, and Brownian increments are additive: for any \(t\), \(B_{t\wedge u_j}-B_{t\wedge u_{j-1}}\) is the increment over the part of that cell lying before \(t\). The finite-sum definition, with each original interval split at the refinement points, therefore gives $$ \int_0^t H_s\,dB_s =\sum_{j=1}^m\eta_j(B_{t\wedge u_j}-B_{t\wedge u_{j-1}}), \qquad \int_0^t G_s\,dB_s =\sum_{j=1}^m\zeta_j(B_{t\wedge u_j}-B_{t\wedge u_{j-1}}). $$ On the probability-one event where all coefficients match, the two sums agree for every \(t\). This proves indistinguishability, including at \(t=0\), when both sums are zero. \(\square\)

The proof is deliberately finite-dimensional: it needs only equality of the coefficients on refined cells and the pathwise identity for Brownian increments. It does not require an independence calculation, a conditional-expectation argument, or a martingale claim. In particular, the issue here is not whether either integral process is a martingale; it is whether the definition assigns one value to one integrand.

Check: on a path with \(B_1=1\), \(B_{3/2}=2\), and \(B_2=\tfrac12\), the event indicator is \(1\). The first expression is \(1+(\tfrac12-1)=\tfrac12\); the refined expression is \(1+(2-1)+(\tfrac12-2)=1+1-\tfrac32=\tfrac12\). On a path with \(B_1=-1\), the indicator is \(0\), so both expressions reduce to \(B_1=-1\), regardless of the later values.

Common Mistake: Reusing the Whole Increment After Refinement

A refinement splits an interval's Brownian increment; it does not copy the whole original increment onto every new piece. For \(H_t=1\) on \((0,2]\), the correct refined sum at \(t=2\) is \((B_1-B_0)+(B_2-B_1)=B_2\). A wrong calculation assigns the original increment \(B_2-B_0\) to both refined cells, obtaining \((B_2-B_0)+(B_2-B_0)=2B_2\). This is not the refined integral: it counts the same increment twice. The correction is to use the increment over each cell, whose sum telescopes to \(B_2-B_0\).

A second potential error is to alter a coefficient while splitting an interval. Refinement alone copies the original coefficient to each subinterval. It cannot replace a coefficient known at time \(u\) by one depending on information revealed after \(u\); that would be a different process and could violate predictability. These safeguards make the construction independent of notation while preserving the left-endpoint information rule essential to the Ito integral.