Two Algebraic Properties of Simple Ito Integrals
Before the Ito integral is extended by limits to a broader class of integrands, its definition is a finite sum. That finite-sum definition already contains two essential rules: combining integrands combines their integrals linearly, and splitting a time interval splits the integral additively. Both rules follow by organizing the same Brownian increments in different ways. The key is to keep each random coefficient attached to increments that occur after the coefficient is known.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) adapted to the filtration. Fix a finite horizon \(T\). An Ito integrand on \([0,T]\) is taken to be progressively measurable and to satisfy \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); this class is denoted \(\mathcal{L}^2\). Here we focus on simple predictable integrands, which are particular members of this class.
The coefficient on an interval is known at its left endpoint, while the Brownian increment is taken over that interval. Square-integrability of the coefficients ensures \(\mathbb{E}\int_0^T H_s^2\,ds=\sum_i(t_{i+1}-t_i)\mathbb{E}[\xi_i^2]<\infty\). Simple predictable processes are progressively measurable, so these integrals belong to the stated Ito-integrand class. Values at individual endpoints do not affect the time integral; the convention \((t_i,t_{i+1}]\) fixes a convenient representation.
For two integrands given on different partitions, insert all their partition points into one common refinement. On each refined interval, each integrand has a single coefficient, measurable at that interval's left endpoint. The defining sums can then be compared term by term. The well-definedness of the simple integral under different step representations ensures that this refinement does not change the integral.
Linearity from the Partition Sum
Proof. Refine the two partitions together, and write their coefficients on the refined intervals as \(\xi_i\) and \(\eta_i\), respectively. The coefficient \(a\xi_i+b\eta_i\) is measurable at the left endpoint because both \(\xi_i\) and \(\eta_i\) are. Also, \((a\xi_i+b\eta_i)^2\le 2a^2\xi_i^2+2b^2\eta_i^2\), so the refined integrand is square-integrable. Put \(\Delta B_i=B_{t_{i+1}}-B_{t_i}\) on this refinement. Applying the finite-sum definition gives \[ \int_0^T(aH_s+bK_s)\,dB_s =\sum_i(a\xi_i+b\eta_i)\Delta B_i =a\sum_i\xi_i\Delta B_i+b\sum_i\eta_i\Delta B_i. \] The two sums on the right are the defining integrals of \(H\) and \(K\). The equality therefore holds almost surely, with any equality of coefficients understood up to the usual almost-sure identification. \(\square\)
The scalars in this statement are deterministic. This restriction is not a technical decoration: a random multiplier may depend on Brownian information that is not available at the start of an interval. The next examples use coefficients that satisfy the required measurability condition.
Worked Example: Combining Integrands on a Common Partition
On the partition \(0<1<2\), let \(H\) have coefficients \(2\) and \(B_1\), and let \(K\) have coefficients \(-1\) and \(3\). Compute the integrals of \(H\), \(K\), and \(2H-K\) over \([0,2]\), and check the linearity identity.
The first coefficients are known at time \(0\); the second coefficients are known at time \(1\). They are square-integrable: the only random coefficient is \(B_1\), and \(\mathbb{E}[B_1^2]=1\). Write \(\Delta B=B_2-B_1\). The defining sums give \[ \int_0^2H_s\,dB_s=2(B_1-B_0)+B_1(B_2-B_1)=2B_1+B_1\Delta B, \] using \(B_0=0\), and \[ \int_0^2K_s\,dB_s=-1(B_1-B_0)+3(B_2-B_1)=-B_1+3\Delta B. \]
On the first interval, the coefficient of \(2H-K\) is \(2\cdot2-(-1)=5\). On the second, it is \(2B_1-3\). Thus \[ \int_0^2(2H_s-K_s)\,dB_s=5B_1+(2B_1-3)\Delta B. \] The right side of the linearity identity gives \(2(2B_1+B_1\Delta B)-(-B_1+3\Delta B)=5B_1+(2B_1-3)\Delta B\), the same expression. This verifies the result directly from the two partition increments.
Additivity Across Adjacent Intervals
To state additivity precisely, define the integral process for a simple integrand with partition points \(t_i\) by \[ I_t=\int_0^t H_s\,dB_s =\sum_{i=0}^{n-1}\xi_i\big(B_{t\wedge t_{i+1}}-B_{t\wedge t_i}\big),\qquad 0\le t\le T. \] When \(t\) lies before an interval, the two clipped times agree and that term is zero. When it lies inside an interval, the term uses only the Brownian increment up to \(t\). This formula is the finite-sum definition applied up to time \(t\).
Proof. Add \(s\) and \(t\) to the partition for \(H\), obtaining a refinement. On any refined interval \((v_j,v_{j+1}]\), let \(\zeta_j\) denote the coefficient of \(H\). It is measurable at \(v_j\): if the interval lies within an original step, the original coefficient was measurable at that step's left endpoint, which is no later than \(v_j\). Restrict the refined sum to the intervals contained in \((s,t]\). This defines a simple predictable integrand there, with zero coefficients outside that interval if it is viewed on all of \([0,T]\). Its integral is \(\sum_{[v_j,v_{j+1}]\subseteq[s,t]}\zeta_j(B_{v_{j+1}}-B_{v_j})\). The sum defining \(I_t\) contains exactly the original increments up to \(t\), and \(I_s\) contains exactly those up to \(s\). Subtracting cancels every increment up to \(s\), leaving precisely the displayed restricted sum. Hence \(\int_s^tH_u\,dB_u=I_t-I_s\). Applying this equality to \([r,t]\), \([r,s]\), and \([s,t]\) gives \(I_t-I_r=(I_s-I_r)+(I_t-I_s)\), which proves adjacent-interval additivity. If two endpoints coincide, the corresponding sum is empty and equals zero, so the statement also covers degenerate intervals. \(\square\)
Worked Example: Splitting Inside a Step Interval
Let \(H_s=2\) on \((0,\tfrac12]\) and \(H_s=B_{1/2}\) on \((\tfrac12,1]\). Split its integral at \(s=\tfrac14\), and verify that the two pieces add to the integral over \([0,1]\).
The coefficient \(2\) is deterministic, and \(B_{1/2}\) is \(\mathcal{F}_{1/2}\)-measurable with \(\mathbb{E}[B_{1/2}^2]=\tfrac12\). Thus the integrand is simple predictable and square-integrable. The full integral is \[ \int_0^1H_u\,dB_u=2(B_{1/2}-B_0)+B_{1/2}(B_1-B_{1/2}) =2B_{1/2}+B_{1/2}(B_1-B_{1/2}). \]
Splitting the first constant-coefficient interval at \(\tfrac14\) gives \[ \int_0^{1/4}H_u\,dB_u=2(B_{1/4}-B_0)=2B_{1/4}, \] and \[ \int_{1/4}^1H_u\,dB_u =2(B_{1/2}-B_{1/4})+B_{1/2}(B_1-B_{1/2}). \] Adding these expressions cancels \(2B_{1/4}\) against \(-2B_{1/4}\), leaving \(2B_{1/2}+B_{1/2}(B_1-B_{1/2})\), exactly the full integral. The split point lies inside a step interval, but the coefficient there is still the same adapted coefficient \(2\).
Worked Example: Splitting a Deterministic Step Integrand Twice
Let \(H=3\) on \((0,1]\) and \(H=-2\) on \((1,3]\). Compute its integral on \([0,3]\), then split at \(2\) and check additivity.
The full partition sum is \[ \int_0^3H_u\,dB_u=3(B_1-B_0)-2(B_3-B_1)=3B_1-2B_3+2B_1=5B_1-2B_3. \] For the first piece, split \([0,2]\) at \(1\): \[ \int_0^2H_u\,dB_u=3B_1-2(B_2-B_1)=3B_1-2B_2+2B_1=5B_1-2B_2. \] For the second piece, the coefficient is \(-2\) throughout \((2,3]\), so \[ \int_2^3H_u\,dB_u=-2(B_3-B_2)=-2B_3+2B_2. \] Their sum is \(5B_1-2B_2-2B_3+2B_2=5B_1-2B_3\), agreeing with the full-interval calculation. Each coefficient is deterministic and hence known at every required left endpoint.
Common Pitfall: Random Multipliers Can Be Anticipative
A tempting but invalid extension of linearity is to replace a deterministic scalar by any random variable. For example, since \(\int_0^1 1\,dB_s=B_1\), one might write \(B_1\int_0^1 1\,dB_s=\int_0^1B_1\,dB_s\). The left side is the valid random variable \(B_1^2\). But the proposed integrand on the right is not predictable on \((0,1]\): its coefficient \(B_1\) is not known at time \(0\), and in general is not known at the left endpoint \(s<1\). Thus this is not an admissible simple Ito integral under the definition above. The correction is to use deterministic scalars in the linearity theorem, or to use a random coefficient only on an interval beginning after the information it depends on has become available. For instance, \(B_1\) is an admissible coefficient on \((1,2]\), and the defining integral there is \(B_1(B_2-B_1)\).
These algebraic rules are valuable because they are already exact before any limiting construction is used. They let a complicated step integrand be refined into manageable intervals, regrouped, and split at deterministic times without changing its integral. They do not, by themselves, assert a martingale property, an expectation formula, or an isometry; those are separate properties requiring their own arguments.
Check Your Understanding
Use the finite-sum definition in each question, keeping track of when each coefficient is measurable.
- For two simple integrands on different partitions, explain why refining both partitions allows their integrals to be compared term by term.
- If \(H\) has coefficients \(1\) and \(B_1\) on \((0,1]\) and \((1,2]\), respectively, write \(\int_0^2H_s\,dB_s\) and split it at time \(1\).
- For deterministic \(a,b\), identify the coefficient on each interval of a common refinement for \(aH+bK\).
- Why does \(\int_s^tH_u\,dB_u=I_t-I_s\) remain valid when \(s\) lies inside an original step interval?
- Explain why the expression \(\int_0^1B_1\,dB_s\) is not a simple predictable integral with the integrand class defined here.