I recently noticed something about the covariance function of a Brownian motion that I don't quite understand, and I was wondering if anyone could help me.

Suppose $W$ is a Brownian motion, and we have regularly-sampled times $\{t_k = T / k, 1 \leq k \leq n \}$. Then the covariance of the vector $(W_{t_1}, \dots, W_{t_n})$ is

$\Sigma = \begin{bmatrix}\min(t_1, t_1) &\dots & \min(t_1, t_n) \\ \vdots & \ddots & \vdots \\ \min(t_n, t_1) & \dots & \min(t_n, t_n) \end{bmatrix}$.

Now, what really surprised me is that the *inverse* of sigma is
proportional to the second-difference operator

$\Sigma^{-1} \propto\begin{bmatrix} 2 & -1 & & & &\\ -1 & 2 & -1 \\ & & \ddots &\\ && -1 & 2 & -1 \\ \\& & & & -1 & 2 \end{bmatrix}$.

Presumably as I take a limit in $n$ I get convergence in some sense to the function $k(s, t) = \min(s, t)$ and to the Laplacian operator $\nabla$. Now, the laplacian is closely related to Brownian motion via the Fokker-Planck equation.

My question is, is there a deeper reason for this than the fact that $\nabla_x \min(x, y) = \delta_y?$ More generally is there a relationship between the infinitesimal generator of a stochastic process and its covariance? What's the connection here?