# Determinant of some covariance matrix (Gaussian kernel process)

Let $x_1,\dots,x_p$ be $p$ points in $\mathbb{R}^n$ ($n\geq 2$) with $x_1=0$. Consider the symmetric matrix $M(x)=(m_{ij}(x))_{1\leq i,j\leq p}$ where $m_{ij}(x) = \exp(-\frac{1}{2}\Vert x_i - x_j\Vert^2)$. The norm is the standard Euclidean one. I would like to prove that $\det(M(x))^{-\frac{1}{2}}$ is locally integrable on $(\mathbb{R}^n)^{p-1}$.

I already know that $M(x)$ is positive definite if the $(x_i)_{1\leq i\leq p}$ are pairwise distinct. More precisely, if $x_1,\dots,x_p$ take exactly $q$ different values, then $M(x)$ has rank $q$.

I also managed to prove that, if $\lambda_1(x)$ is the smallest eigenvalue of $M(x)$ then there exists a constant $K_{n,p}>0$ depending only on $n$ and $p$ such that: $$\frac{1}{\lambda_1(x)} \leq K_{n,p}\sum_{i=1}^p \frac{1}{\prod_{j\neq i}\Vert x_i-x_j\Vert^2} \exp(\Vert x_i\Vert^2) \prod_{j\neq i} (1+\Vert x_j\Vert^2).$$ This is enough to prove that $\frac{1}{\sqrt{\lambda_1}}$ is locally integrable but not enough to get the integrability of $\det(M(x))^{-\frac{1}{2}}$.

Does anyone know how to prove a similar inequality for $\frac{1}{\det(M(x))}$ instead of $\frac{1}{\lambda_1(x)}$? Alternatively, does anyone know of a suitable lower bound for $\det(M(x))$ in terms of the $(\Vert x_i-x_j\Vert)_{i\neq j}$? I know that this matrix appears in statistics, but I couldn't find anything concerning its determinant in the literature so far.

Welcome to MO, Thomas.

I am afraid what you would like to prove is false. Namely, $det(M(x))^{-\frac 1 2}$ is locally integrable if and only if $n$ is large enough depending on $p$.

More precisely there is a function $p_0 \colon \mathbf{N} \to \mathbf{N}$ such that $det(M(x))^{-\frac 1 2}$ if and only if $p \leq p_0(n)$.

I do not know the precise value of $p_0(n)$, but I can give the inequalities $n-1\leq p_0(n) < (n+1) n^{n+1}$ (I do not guarantee that these inequalities are correct).

First remark that by multiplying by the diagonal matrix with diagonal entries $\exp({\frac 1 2 \|x_i\|^2})$, the problem is the same as whether $det(N(x))^{- \frac 1 2}$ if locally integrable, where $N_{i,j}(x) = \exp(\langle x_i,x_j\rangle)$.

We can write $N(x) = \sum_{k\geq 0} \frac{1}{k!}\widetilde N_k(x)$ where $\widetilde N_k(x)$ if the positive matrix with entries $\langle x_i,x_j\rangle^k$. If $N_k(x) \in M_{p-1}(\mathbf R)$ is the matrix obtained by removing the first (vanishing if $k \geq 1$) lign and column of $\widetilde N_k(x)$, we get that $det(N(x))=det(\sum_{k \geq 1} \frac{1}{k!}N_k(x))$, and we deduce that $det(N(x)) \geq det(N_1(x)) = det(\langle x_i,x_j\rangle)_{2\leq i,j\leq p}$. A rapid computation on a piece of paper indicates that $det(N_1(x))^{-\frac 1 2}$ is integrable if $n \geq p-1$, which would give the first inequality.

For the other inequality, remark that $N_k(x)$ has rank at most $n^k$. This says that $\sum_{k \geq 1} \frac{1}{k!}N_k(x)$ can be decomposed as the sum from $1$ to $K-1$, which has rank less than $n+\dots+n^{K-1} < n^{K}$ plus the rest, which is a matrix with entries of order $O(\max_i \|x_i\|^{2K})$. If $p \geq n^K$, we therefore get that at least $p-n^K$ eigenvalues of $\sum_{k \geq 1} \frac{1}{k!}N_k(x)$ are less than $\max_i \|x_i\|^{2K}$, whereas the other are locally bounded. This tells that $det(N(x))^{- \frac 1 2} \geq C (\max_i \|x_i\|)^{-K(p-n^K)}$ locally. This last quantity is not locally integrable if $K(p-n^K)\geq n(p-1)$. Taking $K=n+1$, we get that $det(N(x))^{- \frac 1 2}$ is not locally integrable if $p \geq (n+1)n^{n+1}$.