Simple Predictable Processes on Time Partitions

Why Left-Endpoint Information Matters

The Ito integral is built first for integrands that stay constant over each interval of a finite time partition. The constant on an interval may be random, but it must be determined by information available at the interval's left endpoint. This restriction prevents the coefficient from selecting its value after seeing the Brownian increment it multiplies. It is the elementary form of predictability.

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. In particular, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$ whenever $s<t$. The filtration is increasing: if $s\le t$, then $\mathcal{F}_s\subseteq\mathcal{F}_t$. Fix $T>0$ and a finite partition $0=t_0<t_1<\cdots<t_n=T$.

Definition: A simple predictable process on this partition has the form $H_t=\sum_{i=0}^{n-1}H_i\mathbf{1}_{(t_i,t_{i+1}]}(t)$, where each coefficient $H_i$ is an $\mathcal{F}_{t_i}$-measurable random variable. Its value at time $0$ is $0$. For the square-integrable Ito-integrand class, also require $\mathbb{E}[H_i^2]<\infty$ for every $i$.

The interval convention means $H_i$ is used just after $t_i$, including at $t_{i+1}$. The coefficient is known at $t_i$; the Brownian increment it will multiply, $B_{t_{i+1}}-B_{t_i}$, is not yet known. The process is therefore allowed to respond to the past but not to the increment over its current interval.

Adaptedness and Predictability

A process is adapted if its value at each time $t$ is $\mathcal{F}_t$-measurable. For a simple process, left-endpoint measurability gives this property directly. If $t\in(t_i,t_{i+1}]$, then $H_t=H_i$, and $\mathcal{F}_{t_i}\subseteq\mathcal{F}_t$, so $H_t$ is $\mathcal{F}_t$-measurable. At partition points, the interval convention assigns the value from the interval ending there; that coefficient was measurable at an earlier time and is measurable at the partition point as well. At time $0$, the value is the constant zero.

Predictability is a stronger way of expressing that a process's value is available just before the time at which it is used. The predictable sigma-field is generated by adapted left-continuous processes; equivalently, it is generated by sets $A\times(s,t]$, where $A\in\mathcal{F}_s$, together with the time-zero sets from $\mathcal{F}_0$. For each coefficient $H_i$, the process $H_i\mathbf{1}_{(t_i,t_{i+1}]}$ is measurable with respect to this sigma-field: its level sets on the interval are generated by $A\times(t_i,t_{i+1}]$ with $A\in\mathcal{F}_{t_i}$. Thus the finite sum is predictable.

Proposition: A simple process with $\mathcal{F}_{t_i}$-measurable coefficients is adapted and predictable. It is progressively measurable, and it belongs to $\mathcal{L}^2$—the class of progressively measurable processes satisfying $\mathbb{E}\int_0^T H_t^2\,dt<\infty$—if and only if $\sum_{i=0}^{n-1}(t_{i+1}-t_i)\mathbb{E}[H_i^2]<\infty$.

Proof. Adaptedness follows from the increasing filtration and the fact that $H_i$ is used only at times $t\ge t_i$, as shown above. Predictability follows because each term is measurable with respect to the predictable sigma-field generated by $A\times(t_i,t_{i+1}]$, $A\in\mathcal{F}_{t_i}$, and finite sums of predictable processes are predictable. To verify progressive measurability, fix $u\in[0,T]$. On $[0,u]$, each term is a product of a Borel-measurable time indicator and an $\mathcal{F}_u$-measurable random variable: if the term is nonzero anywhere before $u$, then $t_i<u$ or $t_i=u$, and $H_i$ is measurable with respect to $\mathcal{F}_{t_i}\subseteq\mathcal{F}_u$. Thus the restriction to $[0,u]\times\Omega$ is measurable with respect to $\mathcal{B}([0,u])\otimes\mathcal{F}_u$. For the integrability condition, the intervals are disjoint except at endpoints, which have zero Lebesgue measure. Therefore $$ \mathbb{E}\int_0^T H_t^2\,dt =\sum_{i=0}^{n-1}\mathbb{E}\int_{t_i}^{t_{i+1}}H_i^2\,dt =\sum_{i=0}^{n-1}(t_{i+1}-t_i)\mathbb{E}[H_i^2]. $$ This expectation is finite exactly when the displayed finite sum is finite, proving the criterion. $\square$

For simple predictable $H\in\mathcal{L}^2$, the associated elementary Brownian-increment sum is $$ J_T(H)=\sum_{i=0}^{n-1}H_i(B_{t_{i+1}}-B_{t_i}). $$ It is the finite-partition candidate for the Ito integral over $[0,T]$. The point here is the form and admissibility of the integrand; properties of the resulting integral are developed separately.

Worked Examples

Worked Example: Deterministic Coefficients

Take the partition $0<1<2$ and let $H_0=2$, $H_1=-1$. These constant coefficients are measurable at their left endpoints and have finite second moments. The process equals $2$ on $(0,1]$ and $-1$ on $(1,2]$. Its square-integrability calculation is $$ \mathbb{E}\int_0^2 H_t^2\,dt =1\cdot 2^2+1\cdot(-1)^2=4+1=5. $$ Thus it is in $\mathcal{L}^2$.

Its elementary sum is $J_2(H)=2B_1-(B_2-B_1)$. On any path with $B_1=\tfrac35$ and $B_2-B_1=-\tfrac14$, this gives $$ J_2(H)=2\cdot\tfrac35-(-\tfrac14)=\tfrac65+\tfrac14=\tfrac{24}{20}+\tfrac5{20}=\tfrac{29}{20}. $$ Directly multiplying each interval's coefficient by its increment gives the same two terms, $6/5$ and $1/4$, whose sum is $29/20$. The specified path values illustrate the calculation; they are not claims about an event of positive probability.

Worked Example: A Coefficient Observed at the Left Endpoint

On the partition $0<1<2$, let $H_0=1$ and $H_1=B_1$. The first coefficient is known at time $0$, and the second is $\mathcal{F}_1$-measurable, so the process is simple predictable. Its integrability check is $$ \mathbb{E}\int_0^2 H_t^2\,dt =1\cdot\mathbb{E}[1^2]+1\cdot\mathbb{E}[B_1^2] =1+1=2, $$ because $B_1\sim N(0,1)$. The elementary sum is $J_2(H)=B_1+B_1(B_2-B_1)$.

For $B_1=\tfrac12$ and $B_2-B_1=-\tfrac13$, it is $$ J_2(H)=\tfrac12+\tfrac12\left(-\tfrac13\right)=\tfrac12-\tfrac16=\tfrac13. $$ As a check, substituting $B_2=\tfrac12-\tfrac13=\tfrac16$ into the expanded expression $B_1+B_1B_2-B_1^2$ gives $\tfrac12+\tfrac1{12}-\tfrac14=\tfrac13$ as well. The random coefficient adapts to the value at time $1$, but it cannot depend on the next increment.

Worked Example: An Event Known Before the Interval

Use the partition $0<1<3$, with $H_0=2$ and $H_1=3\mathbf{1}_{\{B_1\ge0\}}$. The event $\{B_1\ge0\}$ belongs to $\mathcal{F}_1$, so the second coefficient is permitted. Since $B_1\sim N(0,1)$, symmetry and $\mathbb{P}(B_1=0)=0$ give $\mathbb{P}(B_1\ge0)=1/2$. Consequently, $$ \mathbb{E}\int_0^3H_t^2\,dt =1\cdot 2^2+2\cdot 3^2\mathbb{P}(B_1\ge0) =4+18\cdot\tfrac12=13. $$ The elementary sum is $J_3(H)=2B_1+3\mathbf{1}_{\{B_1\ge0\}}(B_3-B_1)$.

If $B_1=\tfrac12$ and $B_3-B_1=-\tfrac13$, the event indicator is $1$, and the sum is $2\cdot\tfrac12+3\cdot1\cdot(-\tfrac13)=1-1=0$. If instead $B_1=-\tfrac12$ with the same next increment, the indicator is $0$, and the sum is $2\cdot(-\tfrac12)+0=-1$. These substitutions check that the coefficient is selected using the value already observed at time $1$.

What Fails Without the Information Restriction

Consider replacing $H_1=B_1$ in the second example by $G_1=B_2-B_1$. This coefficient uses the very increment on $(1,2]$ that it would multiply. It is not $\mathcal{F}_1$-measurable: the increment $B_2-B_1$ is independent of $\mathcal{F}_1$ and has a nondegenerate normal distribution. For instance, if $A=\{B_2-B_1>0\}$ were in $\mathcal{F}_1$, independence would imply $\mathbb{P}(A)=\mathbb{P}(A)^2$, whereas the normal distribution gives $\mathbb{P}(A)=1/2$. Thus $A$ cannot be in $\mathcal{F}_1$. A coefficient based on this future increment is not an admissible simple predictable coefficient.

Common mistake. It is tempting to write “choose the coefficient to be $1$ when the next Brownian increment is positive and $-1$ otherwise.” This uses $G_1=\mathbf{1}_{\{B_2-B_1>0\}}-\mathbf{1}_{\{B_2-B_1\le0\}}$, which is not $\mathcal{F}_1$-measurable by the same independence argument. Although the resulting finite sum can be computed path by path, that computation does not make the coefficient predictable. The correction is to choose the coefficient from information at time $1$, for example $H_1=\mathbf{1}_{\{B_1\ge0\}}$, and then multiply it by the increment from time $1$ to time $2$.

Adapted coefficients need not be bounded. They may be unbounded as random variables, provided the required second moments are finite after weighting by the interval lengths. For a finite partition, the exact criterion is the proposition's sum; checking measurability alone does not establish membership in $\mathcal{L}^2$. Conversely, deterministic or random coefficients with infinite second moment do not qualify for this square-integrable class, even if they are measurable at the correct times.

Simple predictable processes provide a controlled starting class: their time dependence is explicit, their coefficients use only available information, and their square-integrability can be checked interval by interval. These features make them suitable for defining the Ito integral before extending the construction to more general progressively measurable integrands.

Check Your Understanding

For each question, distinguish measurability at the left endpoint from square-integrability.

On $0<2<5$, take $H_0=1$ and $H_1=B_2$. Compute $\mathbb{E}\int_0^5H_t^2\,dt$.
Explain why $H_1=\mathbf{1}_{\{B_3-B_2>0\}}$ is not an admissible coefficient on $(2,3]$.
For $H_0=0$ and $H_1=2\mathbf{1}_{\{B_1\ge0\}}$ on $0<1<4$, compute $\mathbb{E}\int_0^4H_t^2\,dt$.
State which property is guaranteed by $\mathcal{F}_{t_i}$-measurability of each coefficient: adaptedness, square-integrability, or both.
Write the elementary Brownian-increment sum for a simple process on $0=t_0<\cdots<t_n=T$, and explain what information each coefficient may use.

Simple Predictable Processes on Time Partitions

What You'll Learn