Deriving the Brownian Covariance Function

Why the Earlier Time Determines the Covariance

Brownian motion has independent increments: after time \(s\), the change from \(B_s\) to \(B_t\) is independent of what has happened by time \(s\). Its increment over that interval has mean zero and variance \(t-s\). These facts give a direct way to calculate the covariance of two Brownian values. The key is that the value at the later time contains the value at the earlier time, plus a new increment that contributes no covariance with the earlier value.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((B_t)_{t\ge0}\) be a real-valued Brownian motion on it, with time index \(t\ge0\) and state space \(S=\mathbb{R}\). We use the standard Brownian motion assumptions: \(B_0=0\) almost surely; paths are continuous almost surely; increments over disjoint time intervals are independent; and, for \(0\le u<v\), \(B_v-B_u\) has distribution \(N(0,v-u)\). Write \(\mathcal{F}_t=\sigma(B_u:0\le u\le t)\) for the natural filtration. In particular, the increment \(B_t-B_s\) is independent of \(B_s\) when \(0\le s\le t\): the latter is determined by the process up to time \(s\), and the former is an increment over the disjoint interval \((s,t]\).

For random variables \(X\) and \(Y\) with finite second moments, their covariance is \(\operatorname{Cov}(X,Y)=\mathbb{E}[(X-\mathbb{E}X)(Y-\mathbb{E}Y)]\). Brownian values have finite second moments. Since \(B_t-B_0\) has distribution \(N(0,t)\), \(\mathbb{E}B_t=0\) and \(\operatorname{Var}(B_t)=t\), including at \(t=0\), when \(B_0=0\) almost surely.

Proof. First suppose \(s\le t\). Write \(B_t=B_s+(B_t-B_s)\). By linearity of covariance, \(\operatorname{Cov}(B_s,B_t)=\operatorname{Cov}(B_s,B_s)+\operatorname{Cov}(B_s,B_t-B_s)\). The first term is \(\operatorname{Var}(B_s)=s\). The second is zero: \(B_s\) and \(B_t-B_s\) are independent, and both have finite expectations, so their covariance is the product of their means minus the product of their means. Thus \(\operatorname{Cov}(B_s,B_t)=s\). This also holds when \(s=0\) or \(s=t\). If \(t\le s\), covariance is symmetric, so \(\operatorname{Cov}(B_s,B_t)=\operatorname{Cov}(B_t,B_s)=t\). In either ordering the value is \(\min(s,t)\), as claimed. \(\square\)

The order matters for the derivation, but not for the final formula. When \(s\le t\), the earlier value is the part shared by the two variables. The later value adds an independent, mean-zero increment. Reversing the labels does not change covariance, so the answer for arbitrary times is the smaller of the two time points.

Calculations from the Increment Decomposition

For a calculation with ordered times, the useful identity is \(B_t=B_s+(B_t-B_s)\) whenever \(s\le t\). The variance of the added increment is \(t-s\), but that is not the covariance between \(B_s\) and \(B_t\). Rather, the increment has zero covariance with \(B_s\), leaving the variance of the shared earlier value. The examples below make each part of this reasoning explicit.

Proof. Define \(\Delta_r=B_{t_r}-B_{t_{r-1}}\) for \(r=1,\ldots,m\). Then \(B_{t_i}=\sum_{r=1}^{i}\Delta_r\), where a sum with no terms, as for \(i=0\), is zero. The increments \(\Delta_r\) are independent, have mean zero, and have variance \(t_r-t_{r-1}\). Suppose first that \(i\le j\). By bilinearity of covariance, the covariance of the two sums is the sum of \(\operatorname{Cov}(\Delta_r,\Delta_q)\) over \(1\le r\le i\) and \(1\le q\le j\). Terms with \(r\ne q\) are zero because the increments are independent and centered. The remaining terms are \(\operatorname{Var}(\Delta_r)=t_r-t_{r-1}\) for \(1\le r\le i\). Their sum is \(t_i-t_0=t_i\), since \(t_0=0\). If \(j\le i\), symmetry gives covariance \(t_j\). These cases yield \(t_{\min(i,j)}\). If \(i=0\) or \(j=0\), the corresponding Brownian value is \(B_0=0\), and the covariance is zero, which agrees with the formula. Repeated time points cause zero-length increments, whose variance is zero, so they are covered as well. \(\square\)

What the Formula Says—and What It Does Not Say

The covariance function records how two values of the process vary together. For Brownian motion, the value at time \(t\) includes all increments from time zero to time \(t\). Two values therefore share the increments up to the earlier time; increments after that time occur in only the later value and are independent of the shared part. The variance of the shared part is the earlier time. This interpretation is a useful way to remember the formula, but the proof rests on the stated independence and variance assumptions.

The function \((s,t)\mapsto\min(s,t)\) is symmetric, nonnegative, and equals \(s\) on the diagonal \(s=t\). The diagonal identity recovers \(\operatorname{Var}(B_t)=t\). At time zero it gives \(\operatorname{Cov}(B_0,B_t)=0\), as it must because \(B_0\) is the constant zero. These are useful checks on a proposed covariance calculation.

The covariance formula alone is not a substitute for the Brownian motion assumptions. In deriving it, we used both the variance of each increment and the independence of an increment from the earlier value. Knowing only that increments have mean zero and variance equal to elapsed time does not, by itself, justify setting the cross-covariance to zero. The independence assumption supplies that step here.

Common mistake. For \(s\le t\), a mistaken calculation may identify \(\operatorname{Cov}(B_s,B_t)\) with \(\operatorname{Var}(B_t-B_s)=t-s\), giving \(3\) for \(\operatorname{Cov}(B_2,B_5)\). That variance belongs to the new increment alone. The correct calculation is

The increment variance is indeed \(5-2=3\), but it is not the covariance in question. Another useful check is the equal-time case: setting \(s=t\) must give \(\operatorname{Cov}(B_t,B_t)=\operatorname{Var}(B_t)=t\). The formula \(\min(s,t)\) passes this check; the incorrect expression \(t-s\) would give zero.