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?)
Best Answer
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 $\mathrm{Cov}(Z,W)$?