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?

  • $\begingroup$ I get a similar tridiagonal structure if I invert the gram matrix of the covariance of an Ornstein Uhlenbeck process. The plot thickens... $\endgroup$ May 25, 2018 at 22:33
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    $\begingroup$ Are you trying to convince us that the family of $\Sigma$'s is one-dimensional? I'm not buying that much. At best, what you said holds for $t_j=cj$, which is not so surprising., $\endgroup$
    – fedja
    May 26, 2018 at 1:46
  • $\begingroup$ I'm not really trying to convince anyone of anything - I've just seen a pattern and I'm trying to understand it. But you're right, with irregularly spaced $t_j$ the proportionality breaks down. I'll edit to reflect your comment. $\endgroup$ May 26, 2018 at 11:46

2 Answers 2


I'm not sure what "deeper reason" you are aiming at, but for one thing, there is no surprise here. The restriction of $W_n$ of $W$ to $(t_1,\dots,t_n)$ is a Gaussian vector whose density form is given by

$$ (\Sigma^{-1}_n v;v)=v^2_1\cdot t_1+\sum_{i=2}^n (v_i-v_{i-1})^2(t_i-t_{i-1})=(\nabla_n v;\nabla_n v)=(\nabla_n^*\nabla_n v;v), $$ where $(\cdot,\cdot)$ is the usual dot product and $\nabla_n$ and $\nabla^*_n$ are the finite difference operators: $$ (\nabla_n v)_i = (v_i-v_{i-1})(t_i-t_{i-1})^\frac12;\quad (\nabla^*_n v)_i = (v_{i+1}-v_{i})(t_{i+1}-t_{i})^\frac12. $$ The matrices of these operators are bidiagonal, so their product is always tridiagonal. Actually, it is easy to see that for any Gaussian process with Markov property (e. g. the Ornstein-Uhlenbeck), the inverse covariance matrix will be tridiagonal, and so it can be written as a linear combination of finite difference operators up to second order (generally, with variable coefficients).

Generally, the infinitesimal generator of a stochastic process (e. g. Ito diffusion) is not uniquely determined by its mean and covaraince: if $\mathbb{E}X_t\equiv 0$, then $X_t$ and $-X_t$ have the same mean and covariance, but, of course, need not have the same distribution.


there is a way to see how the covariance operator and its inverse are related by a second derivative in the case of a Brownian motion. it is with the properties of the ramp function We have $min(\tau,\tau_2) = \tau - ramp (\tau - \tau_2)$ and the property $$\frac {d^2}{d\tau^2} ramp(\tau -\tau_2) = \delta (\tau - \tau_2)$$ this implies that
$$-\frac {d^2}{d\tau^2} min(\tau,\tau_2) = \delta (\tau - \tau_2)$$ and that up to a multiplicative coefficient $\Sigma = -\frac {d^2}{d\tau^2}$


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