Ensemble Behaviour in Population Processes with Applications to Ecological Systems

Abstract

We study Markovian models for population processes in continuous time, addressing questions concerning the behaviour of ensembles of individuals (equilibrium, quasi-equilibrium and time-dependent behaviour) and, in particular, what can be deduced from models for individual behaviour.

Appendix

This section is necessarily technical. It contains proofs of our major results.

Proof of Theorem 1

The ensemble model is density dependent in the sense of Kurtz [13]: There is an open subset of ${\mathbb R}^S$, namely E = (0,1)^S, and functions $f:E\times {\mathbb Z}^S \to{\mathbb R}$ with the property that $q(\boldsymbol{n},\boldsymbol{n}+\boldsymbol{l}) = n f\left(\boldsymbol{n}/n,\boldsymbol{l}\right)$, $\boldsymbol{n},\boldsymbol{n}+\boldsymbol{l} \in {\mathbb Z}^S$. Clearly, $f(\boldsymbol{x},\boldsymbol{e}_j-\boldsymbol{e}_i)= x_i q_{ij}$, j ≠ i. Our theorem is then proved by applying Theorem 3.1 of Kurtz’s paper. We first define $F:E\to {\mathbb R}^S$ by $F(\boldsymbol{x}) = \sum_{\boldsymbol{l}} \boldsymbol{l} f(\boldsymbol{x},\boldsymbol{l})$. This function is required to be Lipschitz continuous on E (clearly true because F has bounded first partial derivatives on E), thus guaranteeing a solution $\boldsymbol{p}(t)$ to $\boldsymbol{p}^{\,\prime}(t) = F(\boldsymbol{p}(t))$, $\boldsymbol{p}(t)\in E$, 0 ≤ t ≤ u, satisfying $\boldsymbol{p}(0)=\boldsymbol{a}$. Next, two technical conditions must be fulfilled: (1) $\sup_{\boldsymbol{x}\in E} \sum_{\boldsymbol{l}\neq \boldsymbol{0}} |\boldsymbol{l}| f(\boldsymbol{x},\boldsymbol{l}) <\infty$ and (2) $\lim_{d\to\infty } \sup_{\boldsymbol{x}\in E} \sum_{|\boldsymbol{l}|>d} |\boldsymbol{l}| f(\boldsymbol{x},\boldsymbol{l}) = 0$. They are trivially satisfied because S is a finite set.

Kurtz’s theorem states that $\boldsymbol{X}^{{(n)}}_{\phantom{k}}(t)$ converges uniformly in probability over finite time intervals to $\boldsymbol{p}(t)$ as n→ ∞, so it remains for us to show that $F(\boldsymbol{x})=\boldsymbol{x} Q$. This is straightforward:

$$\begin{array}{rcl} F(\boldsymbol{x}) &=& \sum\limits_{i\in S} \sum\limits_{j\neq i} (\boldsymbol{e}_j-\boldsymbol{e}_i) f(\boldsymbol{x},\boldsymbol{e}_j-\boldsymbol{e}_i) \\ &=& \sum\limits_{i\in S} x_i \sum\limits_{j\neq i} (\boldsymbol{e}_j-\boldsymbol{e}_i) q_{ij} \\ &= & \sum\limits_{i\in S} x_i \left( \sum\limits_{j\neq i} q_{ij} \boldsymbol{e}_j - q_i \boldsymbol{e}_i \right) = \sum\limits_{k\in S} x_k \sum\limits_{i\in S} q_{ki} \boldsymbol{e}_i. \end{array}$$

Hence, $F(\boldsymbol{x})=\boldsymbol{x} Q$, because (elementwise) $F_i(\boldsymbol{x}) = \sum_{k\in S} x_k q_{ki}$ (i ∈ S).□

Proof of Theorem 2

In addition to $f(\boldsymbol{x},\boldsymbol{e}_j-\boldsymbol{e}_i)= x_i q_{ij}$, j ≠ i, we have $f(\boldsymbol{x},-\boldsymbol{e}_i)= x_i q_{i0}$ and $f(\boldsymbol{x},\boldsymbol{e}_j)= v_{j}$ (asymptotically). The model is not strictly density dependent, but asymptotically density dependent in the sense of Pollett [16], and the present result follows from Theorem 3.1 of Pollett’s paper, which shows that $\boldsymbol{X}^{{(\nu)}}_{\phantom{j}}(t)$ converges uniformly in probability over finite time intervals to $\boldsymbol{r}(t)$ as ν→ ∞, the unique solution to $\boldsymbol{r}^{\,\prime}(t) = F(\boldsymbol{r}(t))$, $\boldsymbol{r}(t)\in E$, 0 ≤ t ≤ u. We may evaluate $F(\boldsymbol{x})$ as follows:

$$\begin{array}{rcl} F(\boldsymbol{x}) &=&\sum\limits_{i\in C} \sum\limits_{j\in C,\,j\neq i} (\boldsymbol{e}_j-\boldsymbol{e}_i) f(\boldsymbol{x},\boldsymbol{e}_j-\boldsymbol{e}_i)\\ &&-\sum\limits_{i\in C} \boldsymbol{e}_i f(\boldsymbol{x},-\boldsymbol{e}_i) +\sum\limits_{j\in C} \boldsymbol{e}_j f(\boldsymbol{x},\boldsymbol{e}_j)\\ &=& \sum\limits_{i\in C} x_i \sum\limits_{j\in C,\,j\neq i} (\boldsymbol{e}_j-\boldsymbol{e}_i) q_{ij} -\sum\limits_{i\in C} x_i \boldsymbol{e}_i q_{i0} +\sum\limits_{j\in C} \boldsymbol{e}_j v_j\\ &=& \sum\limits_{i\in C} x_i \left( \sum\limits_{j\in C,\,j\neq i} q_{ij} \boldsymbol{e}_j - q_i \boldsymbol{e}_i \right) +\sum\limits_{j\in C} \boldsymbol{e}_j v_j\\ &=& \sum\limits_{k\in C} x_k \sum\limits_{i\in C} q_{ki} \boldsymbol{e}_i +\sum\limits_{i\in C} \boldsymbol{e}_i v_i, \end{array}$$

where recall that q _i  = − q _ii  = ∑  _{j ∈ S, j ≠ i} q _ij . Hence, $F(\boldsymbol{x})=\boldsymbol{v} + \boldsymbol{x} {Q}_{_{^{\!C}}}$, because (elementwise) $F_i(\boldsymbol{x}) = v_i+ \sum_{k\in C} x_k q_{ki}$ (i ∈ C).□

Proof of Theorem 3

Let Q _k be the restriction to E _k of transition rate matrix ${Q}_{_{^{\!E}}}$ of the ensemble process, and let − α _k be the eigenvalue of Q _k with maximum real part (k = 1,...,n). Then, α _k  = k α. To see this, observe that $\alpha_k = \lim_{t\to \infty} -(1/t) \log \Pr(T>t)$, where T is the time to first exit of the process from E _k (see Kingman [11]); the limit does not depend on the initial distribution over states. However, T =  min {T ₁,...,T _k }, where T _i is the time it takes individual i to reach 0, and, as the individuals move independently, $\Pr(T>t)=\prod_{i=1}^k \Pr(T_i>t)$. As the individuals move according to Q, $-(1/t) \log \Pr(T_i>t)\to \alpha$ as t→ ∞. Hence, α _k  = k α. It follows immediately that − α is the eigenvalue of ${Q}_{_{^{\!E}}}$ with maximum real part, and, moreover, that its algebraic, and hence geometric, multiplicity is equal to 1. We may therefore appeal directly to Theorem 5 of Van Doorn and Pollett [21], which implies that the LCD of the ensemble process exists provided that the initial distribution assigns mass to at least one of E ₁,...,E _n (we assume that all n individuals are present initially, and so, all this mass is assigned to E _n ). Furthermore, the LCD is the unique non-negative solution ${\boldsymbol{u}}_{_{^{\!E}}}=(u(\boldsymbol{m}), \boldsymbol{m}\in {C}_{_{^{\!E}}})$ to ${\boldsymbol{u}}_{_{^{\!E}}} {Q}_{_{^{\!E}}} = -\alpha{\boldsymbol{u}}_{_{^{\!E}}}$ with ${\boldsymbol{u}}_{_{^{\!E}}} {\bf 1}^{\!\top} =1$. Therefore, it remains to show that the given ${\boldsymbol{u}}_{_{^{\!E}}}$ satisfies these equations: $u(\boldsymbol{m})=0$ unless $\boldsymbol{m}\in E_1$, in which case $u((n-1,\boldsymbol{e}_j))=\pi_j$ (j ∈ C), where $\boldsymbol{\pi}=(\pi_j, j\in C)$ is the LCD of the individual process, that is, the unique (strictly positive) solution to $\boldsymbol{\pi} Q = -\alpha\boldsymbol{\pi}$ with $\boldsymbol{\pi} {\bf 1}^{\!\top} =1$. Writing out the latter, we get

$$ \sum\limits_{i\in C,\, i\neq j} \pi_i q_{ij} = (q_j -\alpha) \pi_j \qquad (j\in C), $$

where q _j  = ∑  _{k ∈ S, k ≠ j} q _jk , while the former eigenvector equation may be written as

$$ \sum\limits_{\boldsymbol{m}\in E_n,\, \boldsymbol{m} \neq \boldsymbol{n}} u(\boldsymbol{m}) q(\boldsymbol{m},\boldsymbol{n}) = (q(\boldsymbol{n}) -\alpha) u(\boldsymbol{n}), \label{eq:pkp1} $$

(2)

$\boldsymbol{n}\in E_n$, and, for k = 1,..., n − 1,

$$\begin{array}{rcl} \sum\limits_{\boldsymbol{m}\in E_{k+1}} &&u(\boldsymbol{m}) q(\boldsymbol{m},\boldsymbol{n}) + \sum\limits_{\boldsymbol{m}\in E_k,\, \boldsymbol{m} \neq \boldsymbol{n}} u(\boldsymbol{m}) q(\boldsymbol{m},\boldsymbol{n}) \\ &=& (q(\boldsymbol{n}) -\alpha) u(\boldsymbol{n}) \qquad (\boldsymbol{n}\in E_k). \end{array}$$

(3)

Clearly Eq. 2, and Eq. 3 for k = 2,...,n, are satisfied because $u(\boldsymbol{m})=0$ when $\boldsymbol{m}\in \cup_{k=2}^n E_k$, and the remaining equation (k = 1) will hold if and only if

$$\begin{array}{rcl} &&\sum\limits_{j\in C} \sum\limits_{i\in C,\, i\neq j} u(\boldsymbol{n}+\boldsymbol{e}_j-\boldsymbol{e}_i) (n_j+1) q_{ji} \\ &&\quad= \left(\, \sum\limits_{k\in C} n_k q_{k} -\alpha\right) u(\boldsymbol{n}) \qquad (\boldsymbol{n}\in E_1), \end{array}$$

(4)

remembering that the total rate out of state $\boldsymbol{n} \in E_k$ is $q(\boldsymbol{n}) = \sum_{i\in C} n_i q_{i}$. But, $\boldsymbol{m}\in E_1$ if and only if $\boldsymbol{m}= (n-1,\boldsymbol{e}_i)$ for some i ∈ C, and so we require

$$ \sum\limits_{i\in C,\,i\neq j} u((n-1,\boldsymbol{e}_i)) q_{ij} = \left( q_j -\alpha\right) u((n-1,\boldsymbol{e}_j)), $$

for j ∈ C. On substituting $u((n-1,\boldsymbol{e}_i))=\pi_i$, we find that

$$ \sum\limits_{i\in C,\,i\neq j} \pi_j q_{ij} = \left( q_j -\alpha\right) \pi_j \qquad (j\in C), $$

which of course is the individual process eigenvector equation.□

Proof of Theorem 4

This is similar to the proof of Theorem 3, but now, we appeal to Theorem 3 of Van Doorn and Pollett [21] for important information about accessibility; the initial distribution must assign all mass to $\cup_{k\geq {\kappa}_{_{^{\!E}}}} E_k$, where ${\kappa}_{_{^{\!E}}}$ is the smallest index k of those α _k that share the smallest value among α ₁,..., α _n . However, α _k  = kα, implying that ${\kappa}_{_{^{\!E}}}=1$, and so, the accessibility condition is satisfied automatically (we have assumed that the ensemble process starts in E _n ). Evaluating the LCD as directed by Theorem 3 of Van Doorn and Pollett [21] amounts to calculating ${\boldsymbol{u}}_{_{^{\!E}}}$ as in the proof of our Theorem 3, but now the individual process is a simple death process. Therefore, α = q, and the unique (strictly positive) solution $\boldsymbol{\pi}=(\pi_j, j\in C)$ to $\boldsymbol{\pi} Q = -\alpha\boldsymbol{\pi}$ with $\boldsymbol{\pi} {\bf 1}^{\!\top} =1$ is given in Theorem 7 of Van Doorn and Pollett [21], and written out in the statement of our theorem.□