The Integral Process for Partition-Step Integrands

From a Single Interval to a Process

For an integrand that is constant on a time interval, its Ito integral accumulates the Brownian increment over the part of that interval already elapsed. A partition-step integrand therefore gives more than one terminal random variable: it gives a process indexed by every time $t$. At a partition time, all intervals before that time contribute their completed increments, while the next interval contributes zero.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to $(\mathcal{F}_t)$. Fix $T>0$ and a deterministic partition $0=t_0<t_1<\cdots<t_n=T$. For each $i$, let $\xi_i$ be an $\mathcal{F}_{t_{i-1}}$-measurable, square-integrable random variable. Define $H_0=0$ and $H_s=\xi_i$ for $t_{i-1}<s\le t_i$. This is a simple predictable process: its coefficient on each interval is known at that interval's left endpoint. It is progressively measurable and satisfies $\mathbb{E}\int_0^T H_s^2\,ds=\sum_{i=1}^n(t_i-t_{i-1})\mathbb{E}[\xi_i^2]<\infty$, so $H$ belongs to $\mathcal{L}^2$.

Definition: For this partition-step integrand, the time-indexed Ito integral is the process $I_t=\int_0^t H_s\,dB_s$, $0\le t\le T$, defined by $$ I_t=\sum_{i=1}^n \xi_i\big(B_{t\wedge t_i}-B_{t\wedge t_{i-1}}\big). $$ Here $t\wedge u=\min(t,u)$. Each summand is zero before its interval begins, follows the current Brownian increment while $t$ lies in the interval, and remains at the completed increment after the interval ends.

This definition uses the single-interval elementary integral: for an $\mathcal{F}_a$-measurable square-integrable coefficient $\xi$, the integral of $\xi$ over $(a,b]$, stopped at time $t$, is $\xi(B_{t\wedge b}-B_{t\wedge a})$. The displayed finite sum combines these interval contributions. The choice $H_0=0$ and the values at partition endpoints do not change the integral, since changing an integrand at finitely many times does not change its time integral or its Ito integral.

The Formula Inside Intervals and at Endpoints

If $t_{k-1}\le t\le t_k$, the definition becomes $$ I_t=\sum_{i=1}^{k-1}\xi_i(B_{t_i}-B_{t_{i-1}})+\xi_k(B_t-B_{t_{k-1}}). $$ The first $k-1$ terms are completed intervals. The last term is the partial increment on the interval currently in progress. At the left endpoint $t=t_{k-1}$, that partial increment is zero. At the right endpoint $t=t_k$, it is the full increment of interval $k$. In particular, for $j=0,\ldots,n$, $$ I_{t_j}=\sum_{i=1}^j\xi_i(B_{t_i}-B_{t_{i-1}}), $$ where the sum for $j=0$ is defined to be zero.

Theorem: The displayed finite-sum definition gives an adapted process with continuous paths. For $0\le s\le t\le T$, its increment is $$ I_t-I_s=\sum_{i=1}^n \xi_i \begin{cases} B_{r_i}-B_{\ell_i},&\ell_i<r_i,\\ 0,&\ell_i\ge r_i, \end{cases} \qquad \ell_i=\max(s,t_{i-1}),\quad r_i=\min(t,t_i). $$ Thus only the portions of partition intervals overlapping $[s,t]$ contribute.

Proof. Write $J_i(t)=\xi_i(B_{t\wedge t_i}-B_{t\wedge t_{i-1}})$, so $I_t=\sum_iJ_i(t)$. Before $t_{i-1}$, $J_i(t)=0$. For $t\ge t_{i-1}$, $\xi_i$ is $\mathcal{F}_{t_{i-1}}$-measurable and hence $\mathcal{F}_t$-measurable, and the Brownian terms in $J_i(t)$ are also $\mathcal{F}_t$-measurable. Therefore each $J_i(t)$, and hence $I_t$, is $\mathcal{F}_t$-measurable. For a fixed continuous Brownian path, $J_i$ is zero up to $t_{i-1}$, equals $\xi_i(B_t-B_{t_{i-1}})$ on $[t_{i-1},t_i]$, and is constant after $t_i$. These pieces meet continuously at both endpoints. Each $J_i$ is continuous, and their finite sum $I$ is continuous. To calculate the increment, subtract $J_i(s)$ from $J_i(t)$. If $[s,t]$ does not overlap $[t_{i-1},t_i]$ in an interval of positive length, the difference is zero. Otherwise the Brownian increment in the difference is from $\ell_i=\max(s,t_{i-1})$ to $r_i=\min(t,t_i)$, so $J_i(t)-J_i(s)=\xi_i(B_{r_i}-B_{\ell_i})$. Summing these differences proves the formula, including the case $\ell_i=r_i$, where the contribution is zero. $\square$

The theorem is a pathwise description of this finite sum; it does not assert a martingale property. Martingale properties of simple stochastic integral processes are a separate question. Here the purpose is to specify precisely what the integral process is at all times and how its value changes across the partition.

Worked Calculations

Worked Example: Two Deterministic Steps

On $[0,2]$, use the partition $0<1<2$ and the deterministic coefficients $\xi_1=2$, $\xi_2=-1$. The integrand is $2$ on $(0,1]$ and $-1$ on $(1,2]$. It is in $\mathcal{L}^2$, since $$ \mathbb{E}\int_0^2H_s^2\,ds=2^2(1-0)+(-1)^2(2-1)=4+1=5. $$

On the first interval, $I_t=2(B_t-B_0)=2B_t$. On the second, the first interval is completed and the second contributes its partial increment: $$ I_t=2(B_1-B_0)-(B_t-B_1)=2B_1-(B_t-B_1)=3B_1-B_t,\qquad 1\le t\le2. $$ At the partition time $1$, the value is $I_1=2B_1$; the second-interval increment is zero there.

For a path with $B_1=\tfrac13$, $B_{3/2}=\tfrac12$, and $B_2=0$, the values are $I_1=2\cdot\tfrac13=\tfrac23$, $I_{3/2}=\tfrac23-(\tfrac12-\tfrac13)=\tfrac23-\tfrac16=\tfrac12$, and $I_2=\tfrac23-(0-\tfrac13)=1$. Check: at $t=2$, summing the two completed contributions directly gives $2(\tfrac13-0)-1(0-\tfrac13)=\tfrac23+\tfrac13=1$, the same endpoint value.

Worked Example: A Random Coefficient Known at the Left Endpoint

On $[0,1]$, take the partition $0<\tfrac12<1$, set $\xi_1=1$, and set $\xi_2=B_{1/2}$. The second coefficient is known at time $\tfrac12$, so it is admissible on $(\tfrac12,1]$. The process is in $\mathcal{L}^2$: using $\mathbb{E}[B_{1/2}^2]=\tfrac12$, $$ \mathbb{E}\int_0^1H_s^2\,ds =\tfrac12\mathbb{E}[1^2]+\tfrac12\mathbb{E}[B_{1/2}^2] =\tfrac12+\tfrac14=\tfrac34. $$

For $0\le t\le\tfrac12$, $I_t=B_t$. For $\tfrac12\le t\le1$, the first interval is complete and the second contributes its current partial increment: $$ I_t=B_{1/2}+B_{1/2}(B_t-B_{1/2}). $$ For a path with $B_{1/2}=2$, $B_{3/4}=\tfrac52$, and $B_1=3$, this gives $I_{1/2}=2$, $I_{3/4}=2+2(\tfrac52-2)=3$, and $I_1=2+2(3-2)=4$.

Check: at $t=\tfrac12$, the partial second-interval increment is $B_{1/2}-B_{1/2}=0$, so the formula returns the completed first-interval value $2$. At $t=1$, its two contributions are $2$ and $2\cdot1$, giving $4$, as does the terminal sum $\xi_1(B_{1/2}-B_0)+\xi_2(B_1-B_{1/2})$.

Worked Example: A Zero Step and Both Endpoints

Use the partition $0<1<2<3$ with deterministic coefficients $\xi_1=1$, $\xi_2=0$, and $\xi_3=-2$. This integrand is in $\mathcal{L}^2$, with $\mathbb{E}\int_0^3H_s^2\,ds=1^2+0^2+(-2)^2=5$. Its integral process is $$ I_t= \begin{cases} B_t,&0\le t\le1,\\ B_1,&1\le t\le2,\\ B_1-2(B_t-B_2),&2\le t\le3. \end{cases} $$

Consider a path with $B_1=-1$, $B_2=2$, $B_{5/2}=\tfrac32$, and $B_3=1$. Then $I_1=-1$, $I_2=-1$, $I_{5/2}=-1-2(\tfrac32-2)=-1+1=0$, and $I_3=-1-2(1-2)=-1+2=1$.

Check: the terminal sum is $1(-1-0)+0(2-(-1))+(-2)(1-2)=-1+0+2=1$, matching $I_3$. The zero coefficient makes the process constant on $[1,2]$, and the next interval's contribution is also zero at its left endpoint $t=2$.

Endpoint Conventions and a Common Error

The process is continuous even though the coefficient used on one interval can differ from the coefficient on the next. Continuity follows because the new interval's Brownian increment starts at zero. At $t_j$, the completed-sum formula includes precisely the intervals $1,\ldots,j$; terms from intervals $j+1,\ldots,n$ are zero. This convention also gives $I_0=0$ and the full terminal sum $I_T=\sum_{i=1}^n\xi_i(B_{t_i}-B_{t_{i-1}})$.

Common mistake: using only the partial-increment term for the interval to the right of a partition point. In the first example, applying that mistake at $t=1$ gives $-1(B_1-B_1)=0$, omitting the completed first interval. The correct endpoint value is $I_1=2(B_1-B_0)=2B_1$, which is $\tfrac23$ on the displayed path. The partial-increment formula is valid inside an interval only when the already completed terms are included as well.

The time-indexed construction is useful because later arguments about stochastic integrals compare values at different times, not just a single terminal sum. Its formula identifies exactly which Brownian increments have accumulated by time $t$, and its endpoint convention prevents a partition boundary from being counted twice or not at all. The coefficients remain attached to the intervals on which they were specified; a new coefficient does not alter increments already accumulated.

Check Your Understanding

Use the definition and the completed-increment convention to answer these questions.

For a partition $0=t_0<\cdots<t_n=T$, write $I_{t_j}$ explicitly and explain the value of the empty sum when $j=0$.
Suppose $H_s=\xi$ on $(a,b]$, where $\xi$ is $\mathcal{F}_a$-measurable and square-integrable. What is its contribution to the integral at a time $t$ with $a<t<b$?
For two intervals with coefficients $\xi_1,\xi_2$, state the integral value at the shared endpoint $t_1$. Why is the second interval's contribution zero there?
What integrability condition on the coefficients ensures that a finite partition-step process belongs to $\mathcal{L}^2$?
For $s\le t$, which partition intervals contribute to $I_t-I_s$, and what Brownian increment is used on each such interval?

The Integral Process for Partition-Step Integrands

What You'll Learn