Covariances of Overlapping Brownian Increments

Why Overlap Determines Covariance

Brownian motion has independent increments over disjoint time intervals. When two intervals overlap, however, the corresponding increments share some of the same random variation. The covariance measures that shared variation: it is exactly the length of the overlap. This geometric description works for intervals in any order, including intervals that are disjoint, nested, or only partially overlapping.

Let \((B_t)_{t\ge0}\) be standard Brownian motion on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with state space \(\mathbb{R}\), continuous paths, \(B_0=0\), and centered Gaussian finite-dimensional distributions with covariance \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\). For \(0\le a\le b\), define the increment over \([a,b]\) by \(B_b-B_a\). The increment is zero if \(a=b\). For finite intervals \(I=[a,b]\) and \(J=[c,d]\), write \(|I\cap J|\) for the length of their intersection, taking that length to be zero if they are disjoint or meet only at an endpoint.

The formula is symmetric in the two intervals. It includes zero-length intervals and intervals that touch at one endpoint. A useful way to understand the result is to represent each increment by its interval indicator: times in both intervals contribute to the covariance, while times in only one interval do not.

Proof. By bilinearity of covariance and the Brownian covariance formula,

To identify this expression as an overlap length, use \(\min(x,y)=\int_0^\infty \mathbf{1}_{\{u\le x\}}\mathbf{1}_{\{u\le y\}}\,du\) for nonnegative \(x,y\). Substituting this identity into the four terms gives

The integrand is \(1\) exactly on the intersection of the two half-open intervals, and is \(0\) elsewhere. Changing whether endpoints are included does not change interval length, so the integral is \(|[a,b]\cap[c,d]|\). If the intervals do not overlap in positive length, the integral is zero. Otherwise the intersection begins at \(\max(a,c)\) and ends at \(\min(b,d)\), so its length is \(\min(b,d)-\max(a,c)\). Combining these cases gives the stated maximum formula. If either interval has length zero, its indicator is zero except possibly at a single endpoint, which has length zero, and the covariance is zero as claimed. \(\square\)

Reading the Formula in Three Configurations

The overlap formula removes the need to compare four Brownian values separately each time. First locate the intervals on the time axis; then compute the length of their common part. A disjoint pair has covariance zero. If one interval lies inside the other, the covariance is the length of the smaller interval. If they overlap partially, it is only the length of the shared segment—not the length of either entire interval.

Variance of Linear Combinations of Increments

The same interval viewpoint handles sums and differences of several increments. For a finite collection of intervals \(I_r=[a_r,b_r]\) and real coefficients \(\alpha_r\), define \(Y=\sum_{r=1}^m\alpha_r(B_{b_r}-B_{a_r})\). At each time \(u\), add the coefficients of the intervals that contain \(u\). Squaring this total and integrating accounts for every pairwise overlap, including the signs of the coefficients.

Proof. Each Brownian increment has mean zero, so \(Y\) has mean zero and \(\operatorname{Var}(Y)=\mathbb{E}[Y^2]\). Expanding the finite square and using bilinearity of covariance gives \[ \operatorname{Var}(Y)=\sum_{r=1}^m\sum_{s=1}^m\alpha_r\alpha_s \operatorname{Cov}(B_{b_r}-B_{a_r},B_{b_s}-B_{a_s}). \] By the covariance theorem, each covariance in this sum is the length of \(I_r\cap I_s\), which equals \(\int_0^\infty\mathbf{1}_{[a_r,b_r)}(u)\mathbf{1}_{[a_s,b_s)}(u)\,du\). There are finitely many terms, so we may combine their integrals and factor the resulting finite sum: \[ \operatorname{Var}(Y) =\int_0^\infty\sum_{r=1}^m\sum_{s=1}^m \alpha_r\alpha_s\mathbf{1}_{[a_r,b_r)}(u)\mathbf{1}_{[a_s,b_s)}(u)\,du =\int_0^\infty\left(\sum_{r=1}^m\alpha_r\mathbf{1}_{[a_r,b_r)}(u)\right)^2du. \] This proves the identity, including cases of zero-length intervals or zero coefficients, whose contributions vanish. \(\square\)

Independence, Interpretation, and a Common Mistake

Brownian increments are jointly Gaussian. The earlier material on zero covariance and independence in Gaussian vectors establishes that jointly Gaussian random variables with zero covariance are independent. Consequently, increments over intervals that are disjoint except possibly for endpoints are independent, not merely uncorrelated. This implication depends on joint Gaussianity; zero covariance alone does not imply independence for arbitrary random variables.

The overlap formula also makes the sign structure of a linear combination visible. In the variance proposition, positive and negative coefficients can cancel on a shared part of the intervals. In the difference example, the common segment disappeared because its two coefficients were \(1\) and \(-1\). For a sum with both coefficients positive, that common segment instead contributes to the variance through the square of their sum.

Common mistake: using the shorter interval’s length for every pair. This works when one interval is contained in the other, but not in general. For example, for \([1,5]\) and \([3,7]\), the shorter interval has length \(4\), but the shared interval is only \([3,5]\), of length \(2\). The incorrect calculation \(\operatorname{Cov}(B_5-B_1,B_7-B_3)=4\) treats all of the shorter interval as shared. The correction is to intersect the intervals first: the covariance is \(|[1,5]\cap[3,7]|=|[3,5]|=2\). Equivalently, the endpoint calculation gives \(5-3-1+1=2\).

For any pair of increments, a reliable calculation is therefore: write down both time intervals, determine their intersection, and take its length. The endpoint formula provides an independent check when the intervals are easy to confuse. Zero covariance for disjoint intervals can then be upgraded to independence because the increments belong to a jointly Gaussian family.