Brownian Covariance and Increment Calculation Pitfalls

Brownian Covariance Is About Shared Time

Brownian covariance calculations are short, but the signs and time endpoints make them easy to misread. The key distinction is between a Brownian value, which accumulates variation from time zero, and an increment, which records change over a particular interval. Two values at later times share much of their history; two increments share only the part of their intervals that overlaps. Keeping these two pictures separate prevents many errors.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A standard Brownian motion \((B_t)_{t\ge0}\) starts at \(B_0=0\), has continuous paths, and has independent increments: increments over disjoint time intervals are independent. For \(0\le s<t\), \(B_t-B_s\) is normally distributed with mean \(0\) and variance \(t-s\). Brownian motion is a Gaussian process, meaning that every finite vector of its values has a joint Gaussian distribution. The covariance of random variables \(U,V\) with finite second moments is \(\operatorname{Cov}(U,V)=\mathbb{E}[(U-\mathbb{E}U)(V-\mathbb{E}V)]\).

In particular, \(\mathbb{E}B_t=0\) and \(\operatorname{Var}(B_t)=t\). The covariance formula for two Brownian values is

Key formula: For \(s,t\ge0\), \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\). The smaller time is the length of the time history shared by the two values.

This formula is not a claim that Brownian values at different times are independent. When \(s<t\), the value \(B_t\) includes the increment \(B_s\), so dependence is expected. The formula instead quantifies that dependence. For increments, the relevant question is not which endpoint is smaller, but how much the two increment intervals overlap.

From Values to Increments

For \(0\le a\le b\), write \(\Delta_{a,b}B=B_b-B_a\). The same increment is sometimes denoted \(B_b-B_a\) directly. For two intervals \((a,b]\) and \((c,d]\), with \(a\le b\) and \(c\le d\), expanding covariance gives

\operatorname{Cov}(B_b-B_a,B_d-B_c) =\min(b,d)-\min(b,c)-\min(a,d)+\min(a,c).

A useful geometric form of this identity is that the covariance equals the length of the overlap of the two intervals. Endpoints do not affect length, so using \([a,b]\) instead of \((a,b]\) gives the same answer.

Theorem (Covariance of Brownian increments): For \(0\le a\le b\) and \(0\le c\le d\), \(\operatorname{Cov}(B_b-B_a,B_d-B_c)\) is the length of \((a,b]\cap(c,d]\). In particular, it is nonnegative, vanishes for disjoint intervals, and equals the length of either interval if that interval is contained in the other.

Proof. We first verify the overlap identity, including intervals of length zero. For \(x,y\ge0\), \(\min(x,y)\) is the length of \((0,x]\cap(0,y]\). Hence the displayed four-term expression is the length obtained by taking the intersection of the two intervals and applying inclusion and exclusion to their endpoints. More explicitly, for each \(u>0\), the contribution of \(u\) to that expression is

\big(\mathbf{1}_{\{u\le b\}}-\mathbf{1}_{\{u\le a\}}\big) \big(\mathbf{1}_{\{u\le d\}}-\mathbf{1}_{\{u\le c\}}\big).

The first factor is \(1\) exactly when \(a<u\le b\), and otherwise is \(0\); the second is \(1\) exactly when \(c<u\le d\), and otherwise is \(0\). Their product is therefore \(1\) exactly on \((a,b]\cap(c,d]\). Integrating over \(u>0\) gives the length of that intersection. The four-term expression is the covariance by expansion and the formula \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\), so the theorem follows. If either interval has length zero, its increment is identically zero and the intersection has length zero as well. \(\square\)

Two increments over disjoint intervals therefore have covariance zero. Since Brownian motion is Gaussian, any finite collection of its increments is jointly Gaussian. In that setting, zero covariance does imply independence. It is important to retain the Gaussian condition: outside the jointly Gaussian setting, uncorrelated random variables need not be independent.

Result (Independence for disjoint Brownian increments): Increments of Brownian motion over disjoint time intervals are independent. Equivalently, for any finite collection of pairwise disjoint intervals, the corresponding increments are mutually independent. This is part of the independent-increments property in the definition of Brownian motion; for a pair, it also follows from joint Gaussianity and the covariance theorem above.

For completeness, the pairwise Gaussian implication can be checked directly. If two jointly Gaussian variables have positive variances \(v_1,v_2\) and covariance zero, their joint Gaussian density has covariance matrix \(\begin{bmatrix}v_1&0\\0&v_2\end{bmatrix}\), so its density is the product of the two one-dimensional Gaussian densities. That factorization is independence. If either variance is zero, that variable equals its mean almost surely and is independent of every random variable. For a finite jointly Gaussian vector, zero covariances between all distinct coordinates make its Gaussian characteristic function factor into the product of the one-dimensional characteristic functions; uniqueness of characteristic functions then gives mutual independence. Thus, when using zero covariance to conclude independence, first establish joint Gaussianity (or invoke Brownian independent increments directly).

Worked Calculations

Worked Example: Covariance of Two Brownian Values

Compute \(\operatorname{Cov}(B_2,B_5)\), and decide whether the two values are independent.

Both times are nonnegative, so the value formula applies. The smaller time is \(2\), giving \(\operatorname{Cov}(B_2,B_5)=\min(2,5)=2\). Their variances are \(\operatorname{Var}(B_2)=2\) and \(\operatorname{Var}(B_5)=5\), so neither is degenerate. Because the covariance is nonzero, they are not independent: independent variables with finite second moments have covariance zero.

Check: Write \(B_5=B_2+(B_5-B_2)\). The two terms on the right are independent, and the increment has mean zero. Thus \(\operatorname{Cov}(B_2,B_5)=\operatorname{Cov}(B_2,B_2)+\operatorname{Cov}(B_2,B_5-B_2)=2+0=2\), agreeing with the minimum-time calculation.

Worked Example: Overlapping Increments

Compute the covariance of \(B_4-B_1\) and \(B_5-B_3\). Are these increments independent?

The intervals are \((1,4]\) and \((3,5]\). Their overlap is \((3,4]\), of length \(4-3=1\), so the covariance theorem gives covariance \(1\). The same result follows by expanding all four terms:

\begin{aligned} \operatorname{Cov}(B_4-B_1,B_5-B_3) &=\min(4,5)-\min(4,3)-\min(1,5)+\min(1,3)\\ &=4-3-1+1\\ &=1. \end{aligned}

The variances are \(4-1=3\) and \(5-3=2\), respectively. The covariance is nonzero, so these jointly Gaussian increments are not independent. As a further check, their correlation is \(1/\sqrt{3\cdot2}=1/\sqrt{6}\), which is positive and nonzero, consistent with the shared interval.

Check: The first increment is \((B_3-B_1)+(B_4-B_3)\), and the second is \((B_4-B_3)+(B_5-B_4)\). The only shared increment is \(B_4-B_3\), whose variance is \(4-3=1\). All other pieces are over disjoint intervals, so their cross-covariances are zero. This gives covariance \(1\) again.

Worked Example: Touching Endpoints and Independence

Let \(U=B_3-B_1\) and \(V=B_5-B_3\). Compute their covariance and determine whether they are independent.

The intervals \((1,3]\) and \((3,5]\) do not overlap: the first contains times up to and including \(3\), while the second contains times strictly greater than \(3\). The overlap has length zero. Direct substitution confirms this:

\begin{aligned} \operatorname{Cov}(U,V) &=\min(3,5)-\min(3,3)-\min(1,5)+\min(1,3)\\ &=3-3-1+1\\ &=0. \end{aligned}

Here \(U\) and \(V\) are Brownian increments over disjoint intervals, so they are independent. Their variances are \(3-1=2\) and \(5-3=2\). The endpoints meeting at time \(3\) do not create a shared positive-length increment; in particular, this is not an example of overlapping increments.

Check: The independent-increments property applies because the intervals have disjoint interiors and the increments are taken over \((1,3]\) and \((3,5]\). The computed covariance is zero, in agreement with independence.

Worked Example: Zero Covariance Without Independence

To see why the Gaussian condition matters, let \(Z\) be a standard normal random variable and set \(U=Z\), \(V=Z^2\). Show that the covariance is zero but the variables are dependent.

The means are \(\mathbb{E}U=0\) and \(\mathbb{E}V=1\). Since the standard normal distribution is symmetric about zero, \(\mathbb{E}Z^3=0\). Therefore

\operatorname{Cov}(U,V) =\mathbb{E}[Z^3]-\mathbb{E}[Z]\mathbb{E}[Z^2] =0-(0)(1)=0.

Nevertheless, the variables are dependent because \(V\) is determined by \(U\). For a direct event check, take \(A=\{|Z|\le1\}\) and \(C=\{Z^2\le1\}\). These are the same event, so \(\mathbb{P}(A\cap C)=\mathbb{P}(A)\). If \(A\) and \(C\) were independent, this probability would equal \(\mathbb{P}(A)^2\). But \(p=\mathbb{P}(|Z|\le1)\) satisfies \(0<p<1\), so \(p\ne p^2\). The events, and hence the variables, are not independent.

Check: This pair is not jointly Gaussian: \(V=U^2\) is constrained to be nonnegative and is a function of \(U\). Thus the Gaussian implication used for Brownian increments does not apply. The zero covariance calculation is correct, but it does not establish independence.

Common Calculation Pitfalls

Pitfall 1: Treating Brownian values as disjoint increments. A common incorrect calculation is to say \(B_s\) and \(B_t\) are independent whenever \(s\ne t\), and therefore assign covariance zero. For \(0<s<t\), this is wrong: \(B_t=B_s+(B_t-B_s)\), and the first term is shared. The correct calculation is \(\operatorname{Cov}(B_s,B_t)=\min(s,t)=s\), which is positive.

Pitfall 2: Losing a minus sign when expanding increments. In \(\operatorname{Cov}(B_b-B_a,B_d-B_c)\), the terms involving one subtracted endpoint have negative signs, while the term involving both subtracted endpoints has a positive sign. Dropping the last term or changing one sign can give a covariance that disagrees with the actual interval overlap. Write all four terms before simplifying.

Pitfall 3: Confusing touching intervals with overlapping intervals. Increments over \((a,b]\) and \((b,d]\) have covariance zero, even though their intervals share the named endpoint \(b\) if closed intervals are drawn. A single endpoint has length zero and contributes no Brownian increment. In contrast, intervals with positive-length overlap have positive covariance.

Pitfall 4: Concluding independence from zero covariance without checking the setting. For Brownian increments, joint Gaussianity makes the conclusion valid, and independent increments are also part of the Brownian definition. For arbitrary random variables, zero covariance alone is not enough, as the \(Z,Z^2\) example shows.

A reliable workflow is to write each increment as a difference of Brownian values, expand covariance with the signs intact, and then check the result against the interval picture. The answer must be nonnegative and cannot exceed the length of either interval. Disjoint intervals must give zero; identical intervals must give their length, which is also the variance of that increment.

Check Your Understanding

For each calculation, identify the relevant times or intervals before using a formula.

Compute \(\operatorname{Cov}(B_2,B_7)\). Are the two Brownian values independent?
Find \(\operatorname{Cov}(B_6-B_2,B_5-B_4)\) by identifying the overlap, then verify it using the four-term minimum formula.
Are \(B_4-B_1\) and \(B_7-B_4\) independent? State both the covariance and the reason independence follows or does not follow.
Give a reason that zero covariance is sufficient for independence for a pair of Brownian increments but not for every pair of random variables.
If two Brownian increments use the same interval \((a,b]\), what is their covariance? Check your answer by calculating the variance of the increment.