From an Indicator to a Brownian Increment
For a deterministic integrand that is constant on a time interval, the Itô integral is already visible in the definition of the integral: the constant is multiplied by the Brownian increment across that interval. Indicator functions make this link especially direct. They also provide a simple setting in which the joint law of two Wiener integrals can be read from the geometry of their intervals.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to \((\mathcal{F}_t)\). For a fixed finite \(T\), an Itô integral \(\int_0^T H_s\,dB_s\) is defined here for progressively measurable integrands \(H\) in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The integrands used below are deterministic indicator functions, so they are progressively measurable.
The indicator belongs to \(\mathcal{L}^2\), since it is deterministic and \(\mathbb{E}\int_0^T\mathbf{1}_{[a,b]}(s)^2\,ds=b-a<\infty\). In the construction of the Itô integral for simple integrands, the process \(\mathbf{1}_{(a,b]}\) has integral \(B_b-B_a\): it is zero before \(a\), and its value \(1\) on the interval contributes the increment over that interval. Since \([a,b]\) and \((a,b]\) differ at most at a time-measure-zero set, the same holds for \(\mathbf{1}_{[a,b]}\).
Proof. By the defining formula for the Itô integral of a simple integrand, \(\int_0^T\mathbf{1}_{(a,b]}(s)\,dB_s=B_b-B_a\). The squared \(\mathcal{L}^2\) distance between \(\mathbf{1}_{[a,b]}\) and \(\mathbf{1}_{(a,b]}\) is \(\mathbb{E}\int_0^T|\mathbf{1}_{[a,b]}(s)-\mathbf{1}_{(a,b]}(s)|^2\,ds=0\). The Itô isometry therefore gives equality of their integrals in \(L^2(\mathbb{P})\), hence almost surely. Brownian increments have mean \(0\) and variance \(b-a\), including \(a=b\), when the increment is zero. \(\square\)
This identity makes the one-interval law immediate: \(I_{a,b}\) is normal with mean \(0\) and variance \(b-a\). For two intervals, the individual lengths still give the two variances, but the covariance depends on how much the intervals overlap.
Joint Laws and Overlap Length
Proof. Collect the distinct endpoints among \(a,b,c,d\), put them in increasing order, and denote the resulting list by \(r_0<r_1<\cdots<r_n\). The Brownian increments \(B_{r_i}-B_{r_{i-1}}\) are independent centered normal random variables with variances \(r_i-r_{i-1}\). Each interval increment is the sum of exactly those elementary increments whose subinterval lies inside its interval. Therefore \((I_{a,b},I_{c,d})\) is a linear transformation of a finite vector of independent normal random variables, so it is jointly Gaussian. Its first variance is the sum of the lengths of the elementary subintervals in \([a,b]\), which is \(b-a\); its second variance is \(d-c\). In the covariance sum, an elementary increment contributes its variance precisely when its subinterval lies in both intervals. The sum of those lengths is the length of the intersection, namely \(m\). If the intervals do not overlap in positive length, the sum is zero. These computations give the stated covariance matrix and prove the result. \(\square\)
The covariance matrix is allowed to be singular. For example, equal intervals give identical integrals, so the two coordinates are equal rather than having a two-dimensional density. When the determinant \((b-a)(d-c)-m^2\) is positive, the joint law has the usual nondegenerate bivariate normal density with this covariance matrix. When it is zero, the Gaussian law remains valid but is concentrated on a line or, if both intervals have length zero, at the origin.
Proof. If the intersection has length zero, the theorem gives zero covariance. For a jointly Gaussian vector, zero covariance between its coordinates implies independence: its joint characteristic function is \(\exp\bigl(-\tfrac12(u^2(b-a)+v^2(d-c)+2uvm)\bigr)\), which when \(m=0\) factors into the product of the two marginal characteristic functions. Conversely, independent variables with finite second moments have covariance zero, and the theorem says that covariance is the nonnegative intersection length \(m\). Thus independence requires \(m=0\) as well. This includes zero-length intervals, whose integrals are constant zero. \(\square\)
Worked Interval Calculations
Worked Example: Two Overlapping Intervals
Take \(I_{1,4}=B_4-B_1\) and \(I_{3,6}=B_6-B_3\). Find their joint law, and compute the variance of their sum and difference.
The interval lengths are \(4-1=3\) and \(6-3=3\). Their intersection is \([3,4]\), of length \(4-3=1\). The theorem gives mean vector \((0,0)\) and covariance matrix $$ \begin{pmatrix}3&1\\1&3\end{pmatrix}. $$ For the sum, its variance is \(3+3+2(1)=8\). For the difference, its variance is \(3+3-2(1)=4\). Thus the sum is centered normal with variance \(8\), and the difference is centered normal with variance \(4\).
Check: Decompose the increments into disjoint pieces. The first is \((B_3-B_1)+(B_4-B_3)\), and the second is \((B_4-B_3)+(B_6-B_4)\). The three pieces are independent centered normals with variances \(2,1,2\). Each interval increment consequently has variance \(2+1=3\), their covariance is the shared-piece variance \(1\), and their difference is \((B_3-B_1)-(B_6-B_4)\), with variance \(2+2=4\), as computed above.
Worked Example: Disjoint Intervals That Share an Endpoint
Consider \(I_{0,2}=B_2-B_0\) and \(I_{2,5}=B_5-B_2\). Compute their joint law and decide whether they are independent.
The lengths are \(2-0=2\) and \(5-2=3\). Their intervals share only the endpoint \(2\), so their intersection has length zero. The mean vector is \((0,0)\), and the covariance matrix is $$ \begin{pmatrix}2&0\\0&3\end{pmatrix}. $$ The variables are jointly Gaussian, so the zero covariance implies independence. Their sum is \(B_5-B_0\), which has variance \(5-0=5\).
Check: Brownian increments on the disjoint time intervals \((0,2]\) and \((2,5]\) are independent. Their variances are \(2-0=2\) and \(5-2=3\); the sum of these variances is \(2+3=5\), agreeing with the variance of \(B_5-B_0\). The common endpoint does not create an interval of positive length in the covariance calculation.
Worked Example: A Nested Pair
Let \(I_{1,5}=B_5-B_1\) and \(I_{2,4}=B_4-B_2\). Find the covariance matrix and the variance of \(I_{1,5}-I_{2,4}\).
The interval lengths are \(5-1=4\) and \(4-2=2\). The smaller interval lies inside the larger one, so the intersection length is \(4-2=2\). The covariance matrix is therefore $$ \begin{pmatrix}4&2\\2&2\end{pmatrix}. $$ The difference has variance \(4+2-2(2)=2\).
Check: Write \(A=B_2-B_1\), \(C=B_4-B_2\), and \(D=B_5-B_4\). These increments are independent and have variances \(2-1=1\), \(4-2=2\), and \(5-4=1\). The larger integral is \(A+C+D\), while the smaller one is \(C\). Their variances are \(1+2+1=4\) and \(2\), and their covariance is \(\operatorname{Var}(C)=2\). Their difference is \(A+D\), with variance \(1+1=2\), matching the formula.
Common Mistake: Using the Shorter Length
A common incorrect rule is to set the covariance equal to the shorter of the two interval lengths. That works for nested intervals, but not for arbitrary overlap. For \(I_{1,4}\) and \(I_{3,6}\), this mistake gives covariance \(\min(3,3)=3\). The intervals overlap only on \([3,4]\), whose length is \(1\), so the correct covariance is \(1\). Substituting the incorrect value would give variance \(3+3-2(3)=0\) for the difference, falsely suggesting that the two increments are equal. In fact their difference has variance \(4\), as the disjoint-increment calculation verifies.
The useful rule is to compute the intersection itself, not just compare the interval lengths. In general, its length is \(\max(0,\min(b,d)-\max(a,c))\). This also explains why intervals separated in time yield independent integrals and why endpoint contact alone has no effect.
| Interval relation | Covariance | Consequence for the joint Gaussian law |
|---|---|---|
| Disjoint except possibly for an endpoint | Zero | The integrals are independent |
| Positive-length overlap | Length of the overlap | The integrals are dependent |
| One interval contained in the other | Length of the shorter interval | The shorter increment is part of the longer one |
| Zero-length interval | Zero | That integral is the constant zero |
Check Your Understanding
Use the increment identity and the intersection-length formula in each question.
- Write \(\int_0^T\mathbf{1}_{[2,5]}(s)\,dB_s\) as a Brownian increment, assuming \(T\ge5\), and state its variance.
- Find the covariance of \(I_{0,4}\) and \(I_{2,6}\), and state whether the two integrals are independent.
- Find the covariance matrix for \(I_{1,3}\) and \(I_{3,7}\). Does sharing one endpoint prevent independence?
- For intervals of lengths \(4\) and \(3\) with overlap length \(2\), compute the variance of the difference of their integrals.
- Explain why the joint Gaussian law can be degenerate even though both interval integrals are nonconstant.