Adapted Step-Process Integrals and Brownian Increment Sums

From Step Processes to Brownian Increment Sums

An Ito integral of an adapted step process is computed by a finite sum. On each time interval, the process has a coefficient already measurable at the interval’s left endpoint; the corresponding term in the integral is that coefficient multiplied by the Brownian increment across the interval. The adaptedness requirement is essential: it excludes coefficients that use information from the future end of the interval.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ with respect to this filtration. In particular, for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$, has mean zero, and has variance $t-s$. Fix $T>0$. A process is in the class $\mathcal{L}^2$ on $[0,T]$ when it is progressively measurable and $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Its Ito integral is denoted $\int_0^T H_s\,dB_s$.

Definition: Choose a deterministic partition $0=t_0<t_1<\cdots<t_n=T$. An adapted step process on this partition has the form $H_s=\xi_i$ for $t_i<s\le t_{i+1}$, where each coefficient $\xi_i$ is $\mathcal{F}_{t_i}$-measurable and square-integrable. Its Ito integral is $$ \int_0^T H_s\,dB_s=\sum_{i=0}^{n-1}\xi_i(B_{t_{i+1}}-B_{t_i}). $$

The coefficient on an interval is fixed at its left endpoint, but it may be random and may depend on Brownian values observed up to that endpoint. Each such step process is progressively measurable. Its square-integrability condition is $\mathbb{E}\int_0^T H_s^2\,ds=\sum_{i=0}^{n-1}(t_{i+1}-t_i)\mathbb{E}[\xi_i^2]<\infty$. Values at the finitely many partition points do not affect this time integral.

Why the Sum Is the Ito Integral

The finite-sum formula is the definition of the Ito integral for adapted step processes. It also agrees with the construction by approximating integrands in $\mathcal{L}^2$. The following result verifies directly how the sum behaves in square mean and why the left-endpoint condition matters to the calculation.

Theorem: For an adapted step process $H_s=\xi_i$ on $t_i<s\le t_{i+1}$, with $\xi_i$ square-integrable and $\mathcal{F}_{t_i}$-measurable, $$ \mathbb{E}\left[\left(\int_0^T H_s\,dB_s\right)^2\right] =\sum_{i=0}^{n-1}(t_{i+1}-t_i)\mathbb{E}[\xi_i^2] =\mathbb{E}\int_0^T H_s^2\,ds. $$

Proof. Write $\Delta B_i=B_{t_{i+1}}-B_{t_i}$ and $\Delta t_i=t_{i+1}-t_i$. Expanding the square of the defining sum gives diagonal terms $\xi_i^2(\Delta B_i)^2$ and cross terms $\xi_i\Delta B_i\,\xi_j\Delta B_j$, $i\ne j$. Since $\Delta B_i$ is independent of $\mathcal{F}_{t_i}$, has mean zero and variance $\Delta t_i$, we have $\mathbb{E}[\xi_i^2(\Delta B_i)^2]=\Delta t_i\mathbb{E}[\xi_i^2]$. For $i<j$, the product $\xi_i\Delta B_i\,\xi_j$ is $\mathcal{F}_{t_j}$-measurable. The increment $\Delta B_j$ is independent of $\mathcal{F}_{t_j}$ with mean zero, so $\mathbb{E}[\xi_i\Delta B_i\,\xi_j\Delta B_j]=0$. The reversed cross term has the same expectation. Thus only the diagonal terms remain, giving the first equality. On each interval $H_s^2=\xi_i^2$, so integrating in time and taking expectation gives the second equality. All terms are integrable: each $\xi_i$ is square-integrable, and $\Delta B_i$ has finite second moment; the cross products are integrable by Cauchy–Schwarz. $\square$

There is also no ambiguity if an interval is subdivided while its coefficient remains unchanged. If $a<c<b$, then $\xi(B_b-B_a)=\xi(B_c-B_a)+\xi(B_b-B_c)$. The coefficient $\xi$, measurable at time $a$, is also measurable at time $c$. Any two finite step representations can be compared on their common refinement, so the formula gives the same integral. This is the finite-sum reason that the definition is consistent.

Worked Sums with Random Coefficients

Worked Example: A Coefficient Frozen at the Midpoint

On $[0,1]$, let $H_s=1$ for $0<s\le\tfrac12$ and $H_s=B_{1/2}$ for $\tfrac12<s\le1$. Compute its integral and verify that $H\in\mathcal{L}^2$.

The coefficient on the first interval is $1$, which is known at time $0$. On the second interval, $B_{1/2}$ is $\mathcal{F}_{1/2}$-measurable, exactly the information available at its left endpoint. The defining sum is therefore $$ \int_0^1 H_s\,dB_s =1(B_{1/2}-B_0)+B_{1/2}(B_1-B_{1/2}) =B_{1/2}+B_{1/2}(B_1-B_{1/2}). $$

Worked Example: Checking Energy and a Pathwise Substitution

For the same process, the expected energy is $$ \mathbb{E}\int_0^1 H_s^2\,ds =\frac12\mathbb{E}[1^2]+\frac12\mathbb{E}[B_{1/2}^2] =\frac12+\frac12\cdot\frac12 =\frac34. $$ Here $\mathbb{E}[B_{1/2}^2]=1/2$, since $B_{1/2}$ has mean zero and variance $1/2$. In particular, the energy is finite and $H\in\mathcal{L}^2$.

To check the sum algebraically, substitute $B_{1/2}=2$ and $B_1-B_{1/2}=-1$ into the displayed expression: it gives $2+2(-1)=0$. The same result follows interval by interval: $1\cdot2+2\cdot(-1)=0$. This is a check of the finite-sum formula for these substituted increment values, not a claim about the probability of that particular path.

Worked Example: Three Successive Brownian Values

Divide $[0,1]$ into three equal intervals and take coefficients $1$, $B_{1/3}$, and $B_{2/3}$, respectively. These coefficients are known at times $0$, $1/3$, and $2/3$, so the process is adapted. Its integral is $$ \int_0^1 H_s\,dB_s =(B_{1/3}-B_0) +B_{1/3}(B_{2/3}-B_{1/3}) +B_{2/3}(B_1-B_{2/3}). $$

For a direct substitution check, set the three successive increments to $1$, $2$, and $-1$. Then $B_{1/3}=1$ and $B_{2/3}=1+2=3$, so the three terms are $1\cdot1=1$, $1\cdot2=2$, and $3\cdot(-1)=-3$. Their sum is $1+2-3=0$. Computing with the cumulative Brownian values in the formula gives the same result: $1+1(3-1)+3(2-3)=1+2-3=0$.

Its expected energy is finite as well: $$ \mathbb{E}\int_0^1 H_s^2\,ds =\frac13\mathbb{E}[1^2] +\frac13\mathbb{E}[B_{1/3}^2] +\frac13\mathbb{E}[B_{2/3}^2] =\frac13+\frac19+\frac29 =\frac23. $$ The Brownian second moments used here are $1/3$ and $2/3$, respectively. This verifies the $\mathcal{L}^2$ condition for this example.

A Linear Combination and Its Check

Worked Example: Two Different Random Coefficients

On $[0,2]$, let $H_s=2$ for $0<s\le1$, and $H_s=1+B_1$ for $1<s\le2$. The second coefficient is $\mathcal{F}_1$-measurable. The integral is $$ \int_0^2 H_s\,dB_s =2(B_1-B_0)+(1+B_1)(B_2-B_1) =2B_1+(1+B_1)(B_2-B_1). $$

The process is in $\mathcal{L}^2$, because $$ \mathbb{E}\int_0^2 H_s^2\,ds =1\cdot 2^2+1\cdot\mathbb{E}[(1+B_1)^2] =4+\bigl(1+2\mathbb{E}[B_1]+\mathbb{E}[B_1^2]\bigr) =4+(1+0+1) =6. $$ Here $B_1$ has mean zero and second moment one.

As a substitution check, take $B_1=1$ and $B_2-B_1=3$. The interval terms are $2\cdot1=2$ and $(1+1)\cdot3=6$, for a total of $8$. The combined expression gives $2(1)+(1+1)3=2+6=8$ as well.

Why the Left Endpoint Cannot Be Replaced

A tempting but invalid change is to use the Brownian value at the right endpoint as the coefficient on the interval leading up to it. For example, on $(0,1]$, the process $H_s=B_1$ is generally not adapted: at time $s<1$, the value $B_1$ is future information. The formal product $B_1(B_1-B_0)=B_1^2$ is therefore not the Ito integral of an adapted step process of this form. It does not become a valid Ito step integral merely because it is an algebraically defined random variable.

In contrast, $H_s=B_0=0$ on $(0,1]$ is adapted and its integral is $B_0(B_1-B_0)=0$. These are different integrands. The issue is not whether a coefficient happens to be a finite random variable; it is whether it is measurable using information available at the interval’s left endpoint.

Takeaway: On a partition interval $(t_i,t_{i+1}]$, use the increment $B_{t_{i+1}}-B_{t_i}$ and a coefficient $\xi_i$ measurable with respect to $\mathcal{F}_{t_i}$. Then sum the products. Check square-integrability by summing $(t_{i+1}-t_i)\mathbb{E}[\xi_i^2]$.

Check Your Understanding

For each proposed process, identify the left-endpoint coefficient and write its integral as a Brownian increment sum.

On $[0,2]$, take coefficient $3$ on $(0,1]$ and coefficient $B_1$ on $(1,2]$. Is the process adapted?
For the process in Question 1, calculate $\mathbb{E}\int_0^2 H_s^2\,ds$.
On $[0,1]$, take coefficient $B_{1/2}$ on $(0,1/2]$ and coefficient $1$ on $(1/2,1]$. Which coefficient fails the left-endpoint measurability requirement?
Prove directly from the finite-sum definition that subdividing one interval without changing its coefficient leaves the integral unchanged.
For a partition with intervals of lengths $1/4$ and $3/4$, and coefficients $2$ and $B_{1/4}$, calculate the expected energy and state why it is finite.