proof that the covariance function for a fractional Brownian motion / fractional Gaussian free field is well defined

Question

Given $0 < t_1 < \dots < t_n$, we can show that the matrix $\Omega$ whose entries are defined by $M_{i,j} = min(t_i,t_j)$ is symmetric definite positive. The proof is immediate once one recognizes that this is the covariance matrix of $(W_{t_i})$ where $W$ is a Brownian motion. One would write that: $$M = t_1 E_1 E_1^T + (t_2-t_1) E_2 E_2^T + \dots + (t_n-t_{n-1}) E_n E_n^T$$ where $E_i$ is a vector of 0s except for the elements starting at position $i$.

One can note that we can also write $M_{i,j} = min(t_i, t_j) = t_i + t_j -|t_i-t_j|$.

I would like to prove that the matrix whose entries are $M_{i,j} = t_i^{2h}+t_j^{2h}-|t_i-t_j|^{2h}$ is also positive. This is the covariance matrix of the fractional brownian motion, but I would like to prove the positivity without relying on this fact. I am merely mentioning it to give context and intuition.

Extra points if the proof applies for a (potentially fractional) Gaussian free field.

Answer 1

Note that the covariance function of fBm with Hurst parameter $H \in (0,1)$ can be written as an integral\begin{align*} C(s,t) &= \frac{1}{2} \left( t^{2 H} + s^{2 H} - |t - s|^{2 H} \right) \\ &= \frac{1}{c_H^2} \int_{\mathbb{R}} \left[ ((t-u)^+)^{H-1/2} - ((-u)^+)^{H-1/2} \right] \left[ ((s-u)^+)^{H-1/2} - ((-u)^+)^{H-1/2} \right] du \end{align*} where $a^+ = \max(a,0)$ and $ c_H = \sqrt{ \frac{1}{2H} + \int_0^{\infty} ((1+u)^{H-1/2} - u ^{H-1/2})^2 du } < \infty \;. $ See, e.g., Proposition 2.3 of

Nourdin, Ivan, Selected aspects of fractional Brownian motion., Bocconi & Springer Series 4. Milano: Springer (ISBN 978-88-470-2822-7/hbk). x, 122 p. (2012). ZBL1274.60006.

Applying this integral form of the covariance function \begin{align*} \frac{1}{2} &\sum_{j=1}^n \sum_{i=1}^n x_i x_j M_{i,j} = \sum_{j=1}^n \sum_{i=1}^n x_i x_j C(t_i, t_j) \\ &= \frac{1}{c_H^2} \int_{\mathbb{R}} \sum_{j=1}^n \sum_{i=1}^n x_i x_j \left[ ((t_i-u)^+)^{H-1/2} - ((-u)^+)^{H-1/2} \right] \left[ ((t_j-u)^+)^{H-1/2} - ((-u)^+)^{H-1/2} \right] du \\ &= \frac{1}{c_H^2} \int_{\mathbb{R}} \left[ \sum_{i=1}^n x_i \left[ ((t_i-u)^+)^{H-1/2} - ((-u)^+)^{H-1/2} \right] \right]^2 du \ge 0 \quad \text{as required.} \end{align*} This proves that the $n \times n$ matrix $M$ is positive semidefinite, which implies that $M$ is the covariance matrix of some Gaussian random $n$-vector.

Answer 2

There is another proof using Fourier transform. Here is a sketch that could be made more rigorous with some work.

Take $H\neq 1/2$. In the sense of tempered distributions, the Fourier transform of $f(x)=|x|^{2H}$ is given by $\hat f(\xi) = c_H |\xi|^{-1-2H}$ for some universal constant $c_H>0$. This may be proved by a scaling argument.

Thus if $\phi$ is any compactly supported nonzero function on $\Bbb R$ then by Plancherel (and the fact that convolution becomes multiplication under Fourier transform) we have that $$\int_\Bbb R |t-s|^{2H} \phi'(s)\phi'(t)dsdt = c_H \int_\Bbb R |\hat \phi(\xi)|^2 |\xi|^{1-2H} d\xi>0,$$ where $\phi'$ is the derivative. Now fix $n\in \Bbb N$ and real numbers $a_1,...,a_n$. Take a sequence of smooth functions $\phi_k\to \sum_{j=1}^n a_j 1_{[0,t_j]}$, so that $\phi_k' \to \sum_{j=1}^n a_j (\delta_{t_j}-\delta_0)$. On the left side you get $\sum_{i,j=1}^n a_ia_j M_{ij}$ where $M$ is your matrix, and the right side is clearly non-negative.