Canonical Representations and Endpoint Conventions

Why Endpoints Matter

A simple predictable process is constant in time between selected partition points, with a random coefficient chosen using information available at the start of each interval. To state that construction without ambiguity, one must also specify which interval contains a partition point. We use the convention $(t_i,t_{i+1}]$: the right endpoint belongs to the interval that ends there, and the left endpoint does not. This assigns a definite value at every time, including partition points.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. The filtration is increasing, so $\mathcal{F}_s\subseteq\mathcal{F}_t$ when $s\le t$. The arguments below concern the values of processes, not stochastic integrals; no integral is defined here.

Definition: Fix $T>0$ and a finite partition $0=t_0<t_1<\cdots<t_n=T$. A simple predictable process on $[0,T]$, in the convention used here, has the form $$ H_t=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(t),\qquad 0\le t\le T, $$ where each coefficient $\xi_i$ is $\mathcal{F}_{t_i}$-measurable and satisfies $\mathbb{E}[\xi_i^2]<\infty$. In particular, $H_0=0$. Its value at $t_{i+1}$ is $\xi_i$, and its value at $t_i$ for $i\ge1$ is the coefficient from the interval ending at $t_i$, namely $\xi_{i-1}$.

The square-integrability condition is useful for the later construction of the Ito integral; it does not affect the endpoint convention. The interval indicators form a disjoint partition of $(0,T]$, so the displayed sum gives exactly one coefficient at every $t\in(0,T]$. At $t=0$, all indicators vanish. If a process is extended beyond $T$, we take it to be zero there unless another definition is stated.

Recovering Coefficients from Endpoint Values

For a fixed partition, the endpoint convention makes the coefficients recoverable directly from the process. The coefficient on $(t_i,t_{i+1}]$ is precisely the value of the process at $t_{i+1}$. This is more than a convenient notation: it gives uniqueness of the coefficients when processes are regarded as pointwise-equal random functions.

Proposition: For a fixed partition $0=t_0<\cdots<t_n=T$, if $H_t=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(t)$, then $\xi_i=H_{t_{i+1}}$ for every $i$. Consequently, if two such representations on this fixed partition agree almost surely at every time, their corresponding coefficients agree almost surely.

Proof. Fix $i\in\{0,\ldots,n-1\}$. At $t=t_{i+1}$, the indicator $\mathbf{1}_{(t_i,t_{i+1}]}(t_{i+1})$ equals $1$. For every $k\ne i$, the point $t_{i+1}$ is not in $(t_k,t_{k+1}]$: intervals with $k<i$ end before it, and the interval with $k>i$ starts at or after it, with its left endpoint excluded. Thus the sum at $t_{i+1}$ has exactly one nonzero term, and $H_{t_{i+1}}=\xi_i$. Applying this identity to each of two representations that agree almost surely at all times gives equality almost surely of their coefficients. $\square$

The phrase “at every time” matters. Equality in the product measure $dt\times d\mathbb{P}$ only says that the processes agree for almost every time, for almost every outcome. Changing a process at a single deterministic time does not change that equivalence class, but it can change its endpoint value and its displayed coefficients. The proposition is about actual endpoint values, not merely time-almost-everywhere values.

Worked Example: Reading Coefficients at Right Endpoints

On $[0,2]$, let the partition be $0<1<2$ and set $\xi_0=2$, $\xi_1=-1$. Compute the process values at the partition points and recover its coefficients.

The representation is $H_t=2\mathbf{1}_{(0,1]}(t)-\mathbf{1}_{(1,2]}(t)$. At $t=0$, neither indicator is one, so $H_0=0$. At $t=1$, the first indicator is one and the second is zero, so $H_1=2$. At $t=2$, the first indicator is zero and the second is one, so $H_2=-1$.

The endpoint recovery rule gives $\xi_0=H_1=2$ and $\xi_1=H_2=-1$, as required. Check: substituting the recovered coefficients into the original expression gives $2$ on $(0,1]$, $-1$ on $(1,2]$, and $0$ at $0$; these are exactly the values just computed, including the values at both partition points.

Refining a Partition Without Changing the Process

A refinement inserts additional times while retaining every old partition point. The new intervals must inherit the coefficient of the old interval containing them. Because that coefficient was measurable at the old interval's left endpoint, it is also measurable at the left endpoint of each smaller interval. This follows from the increasing-filtration property.

Theorem: Let $H_t=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(t)$ be a simple predictable process on $[0,T]$. Let $0=u_0<u_1<\cdots<u_m=T$ be a refinement, meaning every $t_i$ is among the $u_j$. For each $j\in\{0,\ldots,m-1\}$, let $i(j)$ be the unique index such that $t_{i(j)}\le u_j<u_{j+1}\le t_{i(j)+1}$. Then $$ H_t=\sum_{j=0}^{m-1}\xi_{i(j)}\mathbf{1}_{(u_j,u_{j+1}]}(t),\qquad 0\le t\le T, $$ and $\xi_{i(j)}$ is $\mathcal{F}_{u_j}$-measurable.

Proof. Since the $u$-partition contains every $t_i$, each refined interval $(u_j,u_{j+1}]$ lies within exactly one old interval $(t_{i(j)},t_{i(j)+1}]$, apart from a possible shared endpoint. The left endpoint satisfies $t_{i(j)}\le u_j$, so $\mathcal{F}_{t_{i(j)}}\subseteq\mathcal{F}_{u_j}$. The old coefficient $\xi_{i(j)}$, which is $\mathcal{F}_{t_{i(j)}}$-measurable, is therefore $\mathcal{F}_{u_j}$-measurable. It also retains its square-integrability.

To verify equality at every time, first take $t=0$. Every interval indicator on the refined partition is zero, and $H_0=0$. Now take $t\in(0,T]$. There is exactly one refined interval $(u_j,u_{j+1}]$ containing $t$. Its old interval is $(t_{i(j)},t_{i(j)+1}]$, so the old representation assigns the value $\xi_{i(j)}$, and the refined representation assigns that same value. This includes the case $t=u_{j+1}$, since the right endpoint belongs to this interval. Thus the two expressions agree at every $t\in[0,T]$, with the endpoint convention preserved. $\square$

There is another way to identify the refined coefficients: evaluate the unchanged process at the refined right endpoints. For every $j$, the coefficient on $(u_j,u_{j+1}]$ is $H_{u_{j+1}}$. This is the endpoint-recovery proposition applied to the refined partition. In particular, when a new endpoint is inserted inside an old interval, its value is the old interval's coefficient; when it is an old endpoint, its value is the coefficient of the old interval that ends there.

Worked Example: Refining a Two-Interval Process

Let $T=3$, with old partition $0<1<3$. Take $\xi_0=4$, and let $\xi_1=\mathbf{1}_A$, where $A\in\mathcal{F}_1$ and $\mathbb{P}(A)=\tfrac13$. Refine by inserting $2$. Give the new coefficients, check their measurability, and evaluate the process at every new partition point.

In the old representation, $H_t=4$ on $(0,1]$ and $H_t=\mathbf{1}_A$ on $(1,3]$, while $H_0=0$. The refined partition is $0<1<2<3$. The coefficients on $(0,1]$, $(1,2]$, and $(2,3]$ are respectively $4,\mathbf{1}_A,\mathbf{1}_A$. The first is $\mathcal{F}_0$-measurable, as required for its interval. The second and third are both $\mathcal{F}_1$-measurable and hence $\mathcal{F}_1$- and $\mathcal{F}_2$-measurable, respectively. Also $\mathbb{E}[(\mathbf{1}_A)^2]=\mathbb{P}(A)=\tfrac13$, so the random coefficients are square-integrable.

At $0$, $H_0=0$. At $1$, the process still has value $4$, because $1$ is the right endpoint of $(0,1]$. At $2$ and $3$, it has value $\mathbf{1}_A$, since both times belong to the old interval $(1,3]$. The refinement therefore gives exactly the same values. Check: the new representation assigns $4$ on $(0,1]$ and $\mathbf{1}_A$ on $(1,2]\cup(2,3]=(1,3]$; the union identity verifies equality on the entire domain, including $2$ and $3$.

A Common Endpoint Mistake

A tempting alternative is to use left-closed, right-open intervals while keeping the same coefficient timing. That changes which coefficient is assigned at a partition point. The difference is not a matter of notation when the process is being specified pointwise.

Worked Example: The Wrong Half-Open Convention

Keep the process from the first worked example, with partition $0<1<2$ and coefficients $2,-1$. Compare the correct representation with the calculation that incorrectly uses $[t_i,t_{i+1})$ for both intervals and sets the value at $2$ to zero.

Under the stated convention, $H_1=2$, because $1\in(0,1]$ and $1\notin(1,2]$. In the incorrect convention, $1\in[1,2)$, so the value assigned there is $-1$. Thus the incorrect calculation changes the process at the partition point $1$ from $2$ to $-1$. It also excludes $2$ from $[1,2)$; setting $H_2=0$ then changes the correct terminal value $-1$ to $0$.

Correction: use $(0,1]$ and $(1,2]$. The left endpoint $0$ is excluded, the shared endpoint $1$ belongs to the interval ending there, and the terminal endpoint $2$ belongs to the last interval. Direct substitution gives $H_0=0$, $H_1=2$, and $H_2=-1$, consistent with the coefficients' prescribed values.

This pitfall also explains why endpoint conventions should be fixed before comparing representations. If equality is required at every time, a change at a partition point is a genuine change in the process. If one instead works only up to equality for almost every time under $dt\times d\mathbb{P}$, isolated deterministic endpoints do not affect that equivalence class. Those are different notions of equality and should not be interchanged silently.

Practical Checks for a Representation

When writing or refining a simple predictable process, check the following points directly:

The partition is strictly increasing and covers the stated time horizon.
The interval indicators use the declared endpoint convention consistently.
The coefficient on $(t_i,t_{i+1}]$ is measurable with respect to $\mathcal{F}_{t_i}$, not merely with respect to information available at a later time.
The value at $0$ and the value at $T$ follow from the displayed formula rather than from an unstated convention.
Under refinement, each smaller interval receives the coefficient of the old interval containing it, and the resulting coefficients remain measurable at the new left endpoints.

These checks keep the process itself fixed while its partition description changes. In particular, they establish pointwise equality of the refined and original step processes; they do not require, or make a claim about, an integral representation.

Check Your Understanding

For each question, use the right-closed convention $(t_i,t_{i+1}]$ and distinguish endpoint values from values on the open interiors.

For $H_t=\xi_0\mathbf{1}_{(0,1]}(t)+\xi_1\mathbf{1}_{(1,4]}(t)$, write $H_0,H_1,H_4$ in terms of the coefficients.
If $\xi_i$ is $\mathcal{F}_{t_i}$-measurable and $u_j\ge t_i$, explain why it is also $\mathcal{F}_{u_j}$-measurable.
Refine the partition $0<2<5$ by inserting $3$. What coefficients belong on $(0,2]$, $(2,3]$, and $(3,5]$?
Why can equality almost everywhere in time fail to determine the value of a process at a specified partition endpoint?
For a fixed partition, recover each coefficient from the process values at the right endpoints and state why no other interval contributes at that endpoint.