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This question is a repost from math.se, where I didn't receive an answer.


Are there simple conditions on an $d \times d$ matrix B under which $$ f(t, s) = \begin{cases} \exp(-B |t - s|^\alpha), & t > s, \\[7pt] \exp(-B^\top |t - s|^\alpha), & t < s \end{cases} $$ with $\alpha \in (0,2)$ and $t, \, s \in \Bbb R$ is positive definite as a matrix-valued function (see def below)?

I know that it's true if $B$ is nonnegative definite (hence symmetric). But can these conditions be relaxed? Is it possible to show this for $B$ satisfying $B + B^\top \geq 0$ (eigenvalues have nonnegative real parts) without $B$ being symmetric (in particular, $B$ needs not be diagonalizable)?


Example. Consider a standard normal vector $\mathbf{N}$ and an antisymmetric matrix $V$. Define a Gaussian process $\mathbf{X} (t) = \exp(t V) \, \mathbf{N}$. Then $$ \mathbb{E} \mathbf{X} (t) \, \mathbf{X}^\top (s) = \exp(V (t — s)) = \begin{cases} \exp(V |t — s|), & t > s, \\[7pt] \exp(V^\top |t — s|), & t < s \end{cases} $$ hence it is positive definite.


Is it true in general that if $r(t, s)$ is a positive definite matrix-valued function, then so is $\exp(r(t, s))$? This is obviously true if $r$ is scalar valued.


The reason for this particular question is that I am interested in the corresponding class of multivariate fractional Ornstein-Uhlenbeck processes.


P.S. By positive definiteness I mean $$ \sum_{i, j} \mathbf{x}_i^\top f (t_i, t_j) \, \mathbf{x}_j \geq 0 \qquad \forall \, t_i \in \mathbb{R}, \quad \mathbf{x}_i \in \mathbb{R}^d. $$

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First a remark: the usual and widely used definition for positive definiteness is what you wrote at the very end of your question $\underline{\rm and}$ the fact that the matrix $F(t)=(f(t_i,t_j)_{1\le i, j\le n}$ is non-singular. In other words, a matrix $F=(f_{i,j})_{1\le i, j\le n}$ is positive definite if $$ \forall x\in \mathbb R^n, \quad x\not=0\Longrightarrow\langle F x, x\rangle>0. \tag{1}$$ Take now a $n\times n$ real-valued matrix satisfying (1); defining $$ F_0=\frac12(F+F^T)\ \text{(a symmetric matrix)}, \quad F_1=\frac12(F-F^T) \ \text{(a skew-symmetric matrix)}. $$ Condition (1) means only that $F_0$ is a symmetric positive definite matrix and your question is in fact the following: let $A$ be a symmetric positive definite matrix, is it true that for any real-valued skew-symmetric matrix $S$, we have $ e^{A+S} $ is positive definite?

The answer is negative: take for instance $$ A=I_2=\begin{pmatrix}1&0\\0&1\end{pmatrix}, \quad S=π\begin{pmatrix}0&-1\\1&0\end{pmatrix}. $$ Since the matrices $A,S$ are commuting, we have $$ e^{A+S}=e^Ae^S=\begin{pmatrix}e&0\\0&e\end{pmatrix} \begin{pmatrix}\cosπ&-\sin π\\\sin π&\cos π\end{pmatrix} =\begin{pmatrix}-e&0\\0&-e\end{pmatrix}, $$ which is a negative-definite matrix.

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  • $\begingroup$ Okay, I see what you mean... Thanks for helping me formulate the question more clearly. $\endgroup$
    – tsnao
    Commented Mar 19, 2023 at 18:11
  • $\begingroup$ Also thanks for the counterexample, it is nice and useful $\endgroup$
    – tsnao
    Commented Mar 19, 2023 at 18:12
  • $\begingroup$ I don't see why this is true: "your question is in fact the following: let $A$ be a symmetric positive definite matrix, is it true that for any real-valued skew-symmetric matrix $S$, we have $ e^{A+S} $ is positive definite?" $\endgroup$ Commented Mar 19, 2023 at 18:21
  • $\begingroup$ @Iosif Pinelis The last question (about $r(t,s)$) is asking about $\exp r$ being positive-definite when $r$ is positive definite. Since $r$ can be written as the sum of a symmetric matrix and a skew-symmetric matrix, and moreover since the positiveness assumption does not concern the skew-symmetric part, you arrive at the question formulated in my answer ($A,S$ business). $\endgroup$
    – Bazin
    Commented Mar 20, 2023 at 9:55
  • $\begingroup$ @Bazin, positive definiteness of $r(t, s)$ as a matrix-valued function includes symmetricity of in the sense that $r^\top(t,s) = r(s, t)$. Consider the example I added. Although admittedly I seriously doubt that $\exp(r(t,s))$ is positive definite, but for different reasons to what you've suggested. It doesn't even seem possible (at least to me) to show that the square $r^2(t, s)$ is positive definite. $\endgroup$
    – tsnao
    Commented Mar 20, 2023 at 11:18

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