Proof of Jump Representation and Positive Semidefiniteness of the Aggregate Fluctuation Covariance

Fix $\Sigma\in\Delta^l$ and $\alpha\in\mathbb{R}^m$, and for each ordered pair $(\sigma,\gamma)\in\{1,\dots,l\}^2$ with $\sigma\neq\gamma$ abbreviate $w_{\sigma\gamma}=\Sigma^\sigma\,\beta(\sigma,\gamma,\Sigma,\alpha)$. By the definition of the probability simplex, $\Sigma^\sigma\ge0$, and by clause 1 (bounds) of the definition of a transition-rate family, $\beta(\sigma,\gamma,\Sigma,\alpha)\ge0$; hence $w_{\sigma\gamma}\ge0$.

Step 1 (Entries of the outer products). Fix an ordered pair $(\sigma,\gamma)$ with $\sigma\neq\gamma$ and set $u=e_\gamma-e_\sigma$, whose component $u^p$ equals $1$ for $p=\gamma$, equals $-1$ for $p=\sigma$, and equals $0$ otherwise. By the definitions of the matrix product and the transpose, $u\,u^{\top}$ is the matrix with $l$ rows and $l$ columns whose entry in row $p$ and column $q$ is $u^p\,u^q$: the row-$p$, column-$q$ entry of the product of the one-column matrix $u$ and the one-row matrix $u^{\top}$ is the product of the sole entry $u^p$ of row $p$ of $u$ and the sole entry $u^q$ of column $q$ of $u^{\top}$.

Step 2 (Jump representation). Write $\Theta'$ for the right-hand side of part 1, so that, sums of matrices being entrywise,

with $u=e_\gamma-e_\sigma$ depending on the pair as in Step 1.

First take $q=p$. The term of the pair $(\sigma,\gamma)$ contributes $w_{\sigma\gamma}(u^p)^2$, which equals $w_{\sigma\gamma}$ when $p\in\{\sigma,\gamma\}$ and $0$ otherwise. The pairs with $\gamma=p$ contribute $\sum_{\sigma:\sigma\neq p}w_{\sigma p}$ and the pairs with $\sigma=p$ contribute $\sum_{\gamma:\gamma\neq p}w_{p\gamma}$, so, renaming the summation index to $\sigma$ in both sums,

which is exactly the diagonal entry $\Theta^{pp}(\Sigma,\alpha)$ prescribed by the definition of the aggregate fluctuation covariance.

Now fix $p\neq q$. The term of the pair $(\sigma,\gamma)$ contributes $w_{\sigma\gamma}\,u^pu^q$, which vanishes unless both $p$ and $q$ lie in $\{\sigma,\gamma\}$; since $p\neq q$ and $\sigma\neq\gamma$, this happens exactly for $(\sigma,\gamma)=(p,q)$ and $(\sigma,\gamma)=(q,p)$. In the first case $u^p\,u^q=(-1)\cdot 1=-1$, and in the second $u^p\,u^q=1\cdot(-1)=-1$. Hence

again by the definition of the aggregate fluctuation covariance. Thus $\Theta'=\Theta(\Sigma,\alpha)$, proving part 1.

Step 3 (Quadratic form). Let $x\in\mathbb{R}^l$. By part 1, entrywise, and interchange of the finite sums,

and $\sum_{p=1}^{l}u^p\,x^p=x^\gamma-x^\sigma$ by the description of $u$ in Step 1. This proves part 2.

Step 4 (Positive semidefiniteness). $\Theta(\Sigma,\alpha)$ is symmetric by its definition. Let $x\in\mathbb{R}^l$. By the definitions of the matrix-vector product and the dot product,

which by part 2 and $w_{\sigma\gamma}\ge0$ is a finite sum of nonnegative terms, hence nonnegative. By the definition of a positive semidefinite matrix, $\Theta(\Sigma,\alpha)$ is positive semidefinite, proving part 3.