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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?)

blue_egg
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2 Answers2

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

Fei Cao
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  • Thanks! For my own reference, you're saying $X= B_{t+\epsilon}$ has greater variance than $Z = B_{\epsilon}$, and adding a constant $c \sim \mathcal{N}(0, t-\epsilon)$ to both, the covariance of the former still has greater covariance than the latter... since $Cov(X,X+c) = E[X(X+c)] - E[X]E[X+c] = E[X^2] - E[X]^2 = Var(X) > Var(Z) = Cov(Z,Z+c)$ – blue_egg Mar 08 '21 at 00:40
  • You are on a good track! In my example, $c$ is some absolute constant, but you can think of $c$ as having some fixed distribution as well. – Fei Cao Mar 08 '21 at 00:58
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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 $B_\epsilon$.

UserA
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