Covariance of Brownian motion intuition

Question

So the covariance between two time-instances of Brownian motion is $$\text{Cov}(B_s, B_t) = \min(s,t).$$

This post gives a derivation of this fact, but I'm lacking intuition.

Suppose $\epsilon \ll 1 $ is small and $t$ is larger. Then $$ \text{Cov}(B_\epsilon, B_t) = \epsilon,$$ $$\text{Cov}(B_{t+\epsilon}, B_{2t}) = t+\epsilon.$$

In both cases the time step is $t-\epsilon$, but the covariances are drastically different.

Brownian motion is stationary, so why is this true? (And if not, what am I doing wrong?)

Answer 1

Basically you are saying $X-Y$ and $Z-W$ has the same distribution, why $\mathrm{Cov}(X,Y)$ can be very different from $\mathrm{Cov}(Z,W)$? Think of a baby example: $X$ has some distribution with large variance, take $Y = X+ c$, where $c$ is a constant. Now say $Z$ has some distribution with small variance, take $W = Z+ c$. Can you see that $\mathrm{Cov}(X,Y)$ can be much larger than $\mathrm{Cov}(Z,W)$?

Answer 2

Your computation is correct. However, Brownian motion itself is not a stationary process. It's increments are though. This means that for for any $\epsilon>0$, the process $\{B_{t+\epsilon}-B_t\}_{t\geq 0}$ is stationary, ie has the same distribution. In particular by setting $t=0$ we get that $B_{t+\epsilon}-B_t$ is equal in distribution to $B_\epsilon-B_0$ and hence to $B_\epsilon$.