Defining the Integral on a Single Time Interval

One Coefficient, One Brownian Increment

A simple predictable integrand can be constant in time on an interval while its value is random. The essential timing condition is that this coefficient is known at the interval’s left endpoint. Once that condition holds, the Itô integral over the interval is the coefficient multiplied by the Brownian increment across it. At an intermediate time, the increment stops at that time; it does not use information from the future endpoint.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to the filtration: in particular, $B_t$ is adapted, has continuous paths, and $B_t-B_s$ is independent of $\mathcal{F}_s$ with distribution $N(0,t-s)$ for $s<t$. Fix a finite horizon $T$, and let $0\le a<b\le T$. Assume that $\xi$ is $\mathcal{F}_a$-measurable and $\mathbb{E}[\xi^2]<\infty$.

Definition: Set $H_s=\xi\mathbf{1}_{(a,b]}(s)$ for $0\le s\le T$. This is an adapted, progressively measurable integrand, and it belongs to $\mathcal{L}^2$, the class of progressively measurable processes satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Its Itô integral at time $t$ is defined by $$ \int_0^t H_s\,dB_s=\xi\bigl(B_{t\wedge b}-B_{t\wedge a}\bigr),\qquad 0\le t\le T. $$

The notation $dB_s$ here denotes an Itô integral. The indicator is zero at $s=a$ and one at $s=b$, but changing an integrand at a single time does not change this integral. If $a=b$, the interval is empty, $H$ is zero, and the integral is zero; the displayed formula also gives zero.

Reading the Formula at Each Time

The minimums in the definition encode three cases. Before the interval starts, both stopped times equal $t$, so their difference is zero. During the interval, the first stopped time is $t$, while the second is $a$. After the interval ends, they are $b$ and $a$. Thus the integral process has the piecewise form

\int_0^t H_s\,dB_s= \begin{cases} 0, & 0\le t\le a,\\ \xi(B_t-B_a), & a<t\le b,\\ \xi(B_b-B_a), & b<t\le T. \end{cases}

In particular, the value at $t=a$ is zero, and at $t=b$ the integral has accumulated the full interval increment. The coefficient remains $\xi$ throughout; it is not replaced by a value of the process at the intermediate time or at the right endpoint.

Before the interval.
If $t\le a$, then $t\wedge b=t\wedge a=t$, so the integral is zero.

Inside the interval.
If $a<t\le b$, then $t\wedge b=t$ and $t\wedge a=a$, giving $\xi(B_t-B_a)$.

After the interval.
If $t>b$, then the stopped times are $b$ and $a$, so the integral remains $\xi(B_b-B_a)$.

Worked Example: A Deterministic Coefficient

Let $T=4$, $a=1$, $b=3$, and $\xi=3$. Suppose a realized Brownian path has $B_1=-\tfrac14$, $B_{5/2}=\tfrac54$, and $B_3=2$. Compute the integral at $t=\tfrac12$, $t=\tfrac52$, and $t=4$.

Since $\tfrac12\le1$, the integral is zero at $t=\tfrac12$. At $t=\tfrac52$, which lies inside the interval,

\int_0^{5/2}H_s\,dB_s =3\left(B_{5/2}-B_1\right) =3\left(\frac54-\left(-\frac14\right)\right) =3\cdot\frac64 =\frac92.

At $t=4$, the interval has ended, so

\int_0^4H_s\,dB_s =3(B_3-B_1) =3\left(2+\frac14\right) =3\cdot\frac94 =\frac{27}{4}.

Check: the value at $t=4$ is the same as at $t=3$, because $H_s=0$ after $b=3$. The value at $t=1$ is zero, since the Brownian increment $B_1-B_1$ is zero.

Why the Definition Is Square-Integrable

The single-interval formula gives a finite random variable with a finite second moment under the stated assumption on $\xi$. The key fact is that $\xi$ is measurable at time $a$, whereas the increment $B_b-B_a$ is independent of $\mathcal{F}_a$. In particular, the coefficient and the future increment can be separated in the second-moment calculation.

Proposition: For $0\le t\le T$, the process $I_t=\int_0^t H_s\,dB_s$ is adapted and has almost surely continuous paths. Moreover, $$ \mathbb{E}[I_t^2]=\mathbb{E}[\xi^2]\bigl((t\wedge b)-(t\wedge a)\bigr). $$ In particular, $\mathbb{E}[I_T^2]=(b-a)\mathbb{E}[\xi^2]<\infty$, and $H\in\mathcal{L}^2$.

Proof. For $t\le a$, $I_t=0$, which is $\mathcal{F}_t$-measurable. For $t>a$, $\xi$ is $\mathcal{F}_a$-measurable and hence $\mathcal{F}_t$-measurable; also $B_{t\wedge b}$ and $B_{t\wedge a}$ are $\mathcal{F}_t$-measurable. The defining formula therefore makes $I_t$ adapted. The Brownian path is continuous, and multiplying its stopped increment by the finite random variable $\xi$ preserves path continuity. For the second moment, put $u=t\wedge a$ and $v=t\wedge b$. If $t\le a$, then $u=v=t$ and $I_t=0$. If $t>a$, then $u=a$, and $B_v-B_a$ is independent of $\mathcal{F}_a$, has mean zero, and has second moment $v-a$. Since $\xi^2$ is $\mathcal{F}_a$-measurable, $$ \mathbb{E}[I_t^2] =\mathbb{E}\!\left[\xi^2(B_v-B_a)^2\right] =\mathbb{E}[\xi^2]\mathbb{E}[(B_v-B_a)^2] =\mathbb{E}[\xi^2](v-a). $$ This is the claimed formula in both cases. Finally, by integration in time, $$ \mathbb{E}\int_0^T H_s^2\,ds =\mathbb{E}\left[\xi^2\int_0^T\mathbf{1}_{(a,b]}(s)\,ds\right] =(b-a)\mathbb{E}[\xi^2]<\infty. $$ Thus $H\in\mathcal{L}^2$, and the displayed second moments are finite. $\square$

The proposition verifies the integrability required to use this integrand in the Itô integral. It does not claim any martingale property; that requires a separate argument. Nor is the formula an ordinary Riemann–Stieltjes integral along a Brownian path. It is the defining single-interval rule for the Itô integral.

Worked Example: A Coefficient Known at the Left Endpoint

Let $T=2$, $a=1$, $b=2$, and $\xi=B_1$. The coefficient is $\mathcal{F}_1$-measurable. For a realized path with $B_1=2$, $B_{3/2}=\tfrac52$, and $B_2=\tfrac12$, compute the integral at $t=\tfrac32$ and $t=2$.

At $t=\tfrac32$, the coefficient is $B_1=2$, so

I_{3/2}=B_1(B_{3/2}-B_1) =2\left(\frac52-2\right) =2\cdot\frac12 =1.

At $t=2$,

I_2=B_1(B_2-B_1) =2\left(\frac12-2\right) =2\left(-\frac32\right) =-3.

Check: using the interval formula, the change from the intermediate value to the terminal value is $$ I_2-I_{3/2}=B_1(B_2-B_{3/2}) =2\left(\frac12-\frac52\right) =2(-2)=-4. $$ The computed values give $-3-1=-4$, as required.

Worked Example: An Indicator Coefficient

Let $T=3$, $a=1$, $b=3$, and $\xi=\mathbf{1}_{\{B_1>0\}}$. The event uses a strict inequality. Suppose a realized path has $B_1=-2$, $B_2=1$, and $B_3=4$. Compute the integral at $t=2$ and $t=3$.

Because $B_1=-2$ does not satisfy $B_1>0$, the indicator is zero. Therefore

I_2=\mathbf{1}_{\{B_1>0\}}(B_2-B_1) =0\cdot(1-(-2))=0, \qquad I_3=\mathbf{1}_{\{B_1>0\}}(B_3-B_1) =0\cdot(4-(-2))=0.

If instead the realized value were $B_1=1$, the same strict event would hold, and the integral at $t=2$ would be $1\cdot(B_2-1)$. Equality at $B_1=0$ gives indicator zero because the event is $\{B_1>0\}$, not $\{B_1\ge0\}$. Check: for the stated path, the coefficient is zero, so the integral stays zero throughout the interval regardless of the later Brownian values.

How the Rule Fits Simple Stochastic Sums

The definition agrees with the elementary sums used to build simple stochastic integrals. For a fixed $t$ with $a<t\le b$, choose any finite partition $a=s_0<s_1<\cdots<s_n=t$. If the same coefficient $\xi$ is used on each subinterval, the sum is

\sum_{i=1}^n \xi(B_{s_i}-B_{s_{i-1}}) =\xi\sum_{i=1}^n(B_{s_i}-B_{s_{i-1}}) =\xi(B_t-B_a).

The middle sum telescopes exactly; it does not require a limiting argument or a pathwise integration theory. For $t\ge b$, applying the same calculation on $[a,b]$ gives $\xi(B_b-B_a)$, with no contribution from times after $b$. This explains both the interval formula and why the intermediate-time integral accumulates only the Brownian increment seen so far.

Common Pitfall: Using Information from the Right Endpoint

The coefficient must be measurable at time $a$, not merely at time $b$. For example, take $a=1$, $b=2$, and use the valid coefficient $\xi=B_1$. On a path with $B_1=1$ and $B_2=4$, the integral is $B_1(B_2-B_1)=1\cdot3=3$. The tempting calculation $B_2(B_2-B_1)=4\cdot3=12$ replaces the coefficient by a value only known at the right endpoint. That coefficient is not $\mathcal{F}_1$-measurable and does not meet the definition’s hypothesis. The correction is to use the left-endpoint coefficient throughout: $B_1(B_2-B_1)=3$.

This timing condition is the point of the single-interval construction: the integrand’s value is selected using information already available when the Brownian increment begins. The formula then tells exactly how much of that interval’s increment has entered the integral at each time.

Check Your Understanding

Use $I_t=\xi(B_{t\wedge b}-B_{t\wedge a})$ and check which information is available at time $a$.

For $a=1$, $b=3$, and $t=2$, write the integral in terms of $\xi$, $B_1$, and $B_2$.
If $t<a$, what are the two stopped times in the defining formula, and what is the integral?
Suppose $\xi\in\mathcal{F}_a$ and $\mathbb{E}[\xi^2]<\infty$. Compute $\mathbb{E}[I_T^2]$ for $T\ge b$.
For $\xi=\mathbf{1}_{\{B_a\ge0\}}$, what is the coefficient when $B_a=0$? How would that answer differ for $\mathbf{1}_{\{B_a>0\}}$?
Why does choosing $B_b$ as the coefficient generally fail the left-endpoint measurability requirement?