Combining Integrands and Splitting Time
The extended Ito integral is constructed by approximating an admissible integrand with simple predictable processes and taking a limit in mean square. Two useful operations must be compatible with that limit: combining integrands before integrating, and splitting an integral over adjacent time intervals. These properties let us reorganize integrals without returning to the original approximation at every step.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) adapted to the filtration, with increments independent of the past. Fix \(T>0\). An integrand \(H\) on \([0,T]\) is admissible here if it is progressively measurable and satisfies \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Progressively measurable means that, for every \(t\le T\), the restriction \((s,\omega)\mapsto H_s(\omega)\) to \([0,t]\times\Omega\) is measurable with respect to \(\mathcal{B}([0,t])\otimes\mathcal{F}_t\). The class of such integrands is denoted by \(\mathcal{L}^2\). Every Ito integral below is for an integrand in this class; the integral over a subinterval is defined by restricting an integrand in \(\mathcal{L}^2\).
The construction of the extended integral provides two facts used below. Simple predictable integrands are dense in \(\mathcal{L}^2\), and the extended Ito isometry holds: \(\mathbb{E}[I_T(H)^2]=\mathbb{E}\int_0^T H_s^2\,ds\). Consequently, convergence of integrands in the norm \(\|H\|_2=(\mathbb{E}\int_0^T H_s^2\,ds)^{1/2}\) implies convergence of their integrals in \(L^2(\mathbb{P})\). These are the construction and isometry results for the extended integral; the proofs here use them to establish the compatibility properties.
Linearity for Extended Integrands
Proof. The linear combination is progressively measurable because \(H\) and \(K\) are. The inequality \((x+y)^2\le 2x^2+2y^2\) gives \(\mathbb{E}\int_0^T(\alpha H_s+\beta K_s)^2\,ds\le2\alpha^2\|H\|_2^2+2\beta^2\|K\|_2^2<\infty\), so it is admissible. By density, choose simple predictable \(H_n,K_n\) with \(\|H_n-H\|_2\to0\) and \(\|K_n-K\|_2\to0\). For simple predictable integrands, the integral is a finite sum of coefficient times Brownian increment, so it is linear term by term. Thus \(I_T(\alpha H_n+\beta K_n)=\alpha I_T(H_n)+\beta I_T(K_n)\). The integrands \(\alpha H_n+\beta K_n\) converge to \(\alpha H+\beta K\), since \(\|\alpha(H_n-H)+\beta(K_n-K)\|_2\le|\alpha|\|H_n-H\|_2+|\beta|\|K_n-K\|_2\). By the Ito isometry, the corresponding integrals converge in \(L^2(\mathbb{P})\). Taking limits in the simple-integrand identity proves the stated equality in \(L^2\). Two square-integrable random variables equal in \(L^2\) are equal almost surely, because their difference has squared expectation zero. \(\square\)
The constants in this theorem are deterministic. Linearity does not say that a random variable can be pulled outside an Ito integral. Such an operation needs separate hypotheses and is not part of this result.
Worked Example: A Linear Combination of Simple Integrands
On \([0,2]\), let \(H_s=2\mathbf{1}_{(0,1]}(s)+B_1\mathbf{1}_{(1,2]}(s)\) and \(K_s=\mathbf{1}_{(0,1]}(s)+3\mathbf{1}_{(1,2]}(s)\). Compute the integral of \(2H-K\).
On \((0,1]\), the coefficient of \(2H-K\) is \(2\cdot2-1=3\). On \((1,2]\), it is \(2B_1-3\). The coefficient \(B_1\) on \((1,2]\) is \(\mathcal{F}_1\)-measurable, so it is known at the left endpoint of that interval; both integrands are simple predictable. Directly summing coefficient-times-increment gives \(\int_0^2(2H_s-K_s)\,dB_s=3(B_1-B_0)+(2B_1-3)(B_2-B_1)\). Since \(B_0=0\), the result is \(3B_1+(2B_1-3)(B_2-B_1)\).
Linearity gives the same answer by separate integration: \(\int_0^2 H_s\,dB_s=2B_1+B_1(B_2-B_1)\) and \(\int_0^2 K_s\,dB_s=B_1+3(B_2-B_1)\). Therefore \(2\int_0^2H_s\,dB_s-\int_0^2K_s\,dB_s =4B_1+2B_1(B_2-B_1)-B_1-3(B_2-B_1)\), which simplifies to the direct expression. This checks the identity by computing both sides.
Additivity Across Adjacent Intervals
Proof. Each restricted integrand belongs to \(\mathcal{L}^2\), since multiplication by an indicator preserves progressive measurability and \(\mathbb{E}\int_0^T(\mathbf{1}_{(u,v]}H_s)^2\,ds\le\mathbb{E}\int_0^T H_s^2\,ds\). For \(a\le c\le b\), the indicators satisfy \(\mathbf{1}_{(a,b]}=\mathbf{1}_{(a,c]}+\mathbf{1}_{(c,b]}\) except possibly at endpoints; the two sides are therefore equal for almost every time \(s\). The corresponding restricted integrands are equal in the \(\mathcal{L}^2\) norm, because their squared difference has zero time integral. Apply linearity to these integrands to obtain the asserted identity. If \(a=b\), the indicator \(\mathbf{1}_{(a,a]}\) is zero, so the integral is zero. This also covers \(a=c\) or \(c=b\), when one of the two intervals is empty. \(\square\)
In particular, for \(0\le a\le b\le T\), define \(I_t(H)=\int_0^t H_s\,dB_s\) for each fixed \(t\). The definition and linearity give \(\int_a^b H_s\,dB_s=I_b(H)-I_a(H)\) in \(L^2(\mathbb{P})\). This is an identity at each fixed choice of endpoints. It does not require choosing a simultaneous pathwise version for every time.
Worked Example: Splitting a Three-Interval Integral
On \([0,3]\), set \(H_s=1\) on \((0,1]\), \(H_s=B_1\) on \((1,2]\), and \(H_s=B_2\) on \((2,3]\). These coefficients are measurable at the left endpoints of their respective intervals. The integrand is in \(\mathcal{L}^2\): the expected squared-integrand integral is \(1+\mathbb{E}[B_1^2]+\mathbb{E}[B_2^2]=1+1+2=4\).
Each interval integral is a simple coefficient times a Brownian increment: \(\int_0^1H_s\,dB_s=B_1\), \(\int_1^2H_s\,dB_s=B_1(B_2-B_1)\), and \(\int_2^3H_s\,dB_s=B_2(B_3-B_2)\). Adding gives \(\int_0^3H_s\,dB_s=B_1+B_1(B_2-B_1)+B_2(B_3-B_2)\). The same result follows by writing the simple-process definition directly on all three intervals, which checks additivity.
Worked Example: Restricting an Integrand to a Subinterval
Use the integrand \(H_s=2\) on \((0,1]\) and \(H_s=B_1\) on \((1,2]\). Compute its integral restricted to \((\tfrac12,\tfrac32]\), then split that restricted integral at \(1\).
Restriction multiplies the integrand by \(\mathbf{1}_{(\frac12,\frac32]}\). On \((\tfrac12,1]\) the coefficient remains \(2\), and on \((1,\tfrac32]\) it remains \(B_1\). Hence \(\int_{\frac12}^{\frac32}H_s\,dB_s =2(B_1-B_{\frac12})+B_1(B_{\frac32}-B_1)\). Splitting at \(1\) gives \(\int_{\frac12}^{1}H_s\,dB_s=2(B_1-B_{\frac12})\) and \(\int_1^{\frac32}H_s\,dB_s=B_1(B_{\frac32}-B_1)\). Their sum is exactly the restricted integral just computed. The coefficient \(B_1\) on the second interval is available at its left endpoint, so the restricted process is admissible.
Common Mistake: Confusing an Interval with an Initial-Time Integral
A frequent error is to treat \(\int_a^b H_s\,dB_s\) as though it meant \(\int_0^b H_s\,dB_s\). For example, if \(H_s=1\) on \((0,2]\), then \(\int_0^2H_s\,dB_s=B_2\), while \(\int_1^2H_s\,dB_s=B_2-B_1\). The incorrect calculation \(\int_1^2H_s\,dB_s=B_2\) has omitted the contribution before time \(1\) from the initial-time integral. The correction is to insert the interval indicator: \(\int_1^2H_s\,dB_s=\int_0^2\mathbf{1}_{(1,2]}(s)H_s\,dB_s\). Equivalently, subtract the two initial-time integrals, \(I_2(H)-I_1(H)=B_2-B_1\).
These rules are useful because they allow an integrand to be decomposed into pieces supported on different time intervals, or several integrands to be combined before integration. The results follow from the definition by approximation, rather than from ordinary Riemann-integral intuition: each operation must preserve admissibility, and equalities between extended integrals are understood in \(L^2(\mathbb{P})\), hence almost surely for each fixed set of endpoints.
Check Your Understanding
For each question, assume all stated integrands belong to \(\mathcal{L}^2\), and interpret integral identities in \(L^2(\mathbb{P})\).
- Why does multiplying an admissible integrand by the indicator of a deterministic interval preserve admissibility?
- For deterministic \(\alpha,\beta\), express \(\int_0^T(\alpha H_s+\beta K_s)\,dB_s\) using the separate integrals of \(H\) and \(K\).
- Write \(\int_a^bH_s\,dB_s\) as a difference of initial-time integrals, for \(0\le a\le b\le T\).
- What is \(\int_a^aH_s\,dB_s\), and why?
- For \(H_s=1\) on \((0,2]\), compute the integrals over \([0,2]\) and \([1,2]\) in terms of \(B_1\) and \(B_2\).