The maximum entropy principle to predict forager spatial distributions: an alternate perspective for movement ecology

Abstract

The Maximum Entropy (MaxEnt) principle is a powerful inference principle which enables the determination of the distribution of a system, on the basis of the information available, usually in the form of averages of observables (random variables), and the assumption of maximal ignorance (maximum entropy) beyond the stated prior information. In this work we focus on the use of MaxEnt in the context of spatial ecology for building theory to predict equilibrium foraging distributions. Our work represents a new application of MaxEnt and a novel approach to compute spatial foraging distributions based on the use of energy and entropy arguments. We show the capability of the framework to incorporate mechanisms such as resource depletion, optimality of the foraging strategy, travel costs, information uncertainty, and inter-specific and intra-specific competition into the statistical inference. Our model recovers the predictions of several existing models under various parameter limits, including the uniform distribution, the ideal free distribution, and the distribution that maximizes average suitability, and gives a simple quantitative way to connect these different foraging behaviours. Overall, we show the power of MaxEnt to build theory and to relate existing models in spatial ecology by use of a universal principle. We further discuss the potential applicability of MaxEnt to build theory in other contexts of ecology, such as to formulate population dynamics models, and the potential use of the dynamic form of MaxEnt, i.e. the Maximum Caliber Principle, to develop further theory on dynamical systems in spatial ecology.

Appendix: Large-\(\beta\) limit of Boltzmann and depleted Boltzmann distributions

We can handle the necessary theory for both the Boltzmann and Depleted Boltzmann distributions using the following framework. Let I be a finite set and $(\kern0.1500emf_i)_{i \in I}$ a collection of continuous functions with each $f_i:\mathcal {R}_+\rightarrow \mathcal {R}$. Given population size N, we seek $n:I\rightarrow \mathcal {R}_+$ satisfying $\sum _i n_i=N$ and solving either of the following two problems:

• 1. The maximin problem $\arg \max _n F(n)$ where $F(n):=\text {min}\{\kern0.1500emf_i(n_i):i \in supp(n)\}$ and $supp(n)=\{i:n_i \ne 0\}$.

• 2. The Boltzmann-type equation

  $$\begin{aligned} n_i = N \exp (\beta f_i(n_i))/Z_\beta \end{aligned}$$

  (6.1)

  where $Z_\beta = \sum _i \exp (\beta f_i(n_i))$ is a normalizing constant.

When $(\kern0.1500emf_i)_{i\in I}$ describe a suitability function, solutions of the maximin problem are optimal foraging distributions (Pyke 1984). The Boltzmann-type equation corresponds to the Boltzmann distribution (Eq. (2.12)) if $f_i=\epsilon _i$ and to the Depleted Boltzmann (Eq. (2.14)) if $f_i = \partial (n_i \epsilon _i) / \partial n_i$, and the respective large-$\beta$ limits, the ideal free and max energy distributions, are the maximin solutions with the given $f_i$ (see Theorem 2). To solve both of the above problems it is helpful to make the following assumption:

$$\mathrm {(A1)} \quad \forall i, \ f_i\ \ \text {is decreasing and has} \ \ f_i(x)\rightarrow -\infty \ \text {as} \ x\rightarrow \infty .$$

The first part is a typical assumption in optimal foraging theory (Pyke 1984) while the second is a technical point. Subject to this we have the following results.

Theorem 1

Assume (A1) above. For each $N>0$ the maximin problem has a unique solution, and for each $N,\beta >0$ the Boltzmann-type problem has a unique solution.

Theorem 2

Assume Eq. (6.1) above. For each $N>0$ the solutions $(n(\beta ))_{\beta >0}$ of the Boltzmann-type problem tend to the solution of the maximin problem as $\beta \rightarrow \infty$.

Proof (Proof of Theorem 1)

We first solve the maximin problem. To do so we first find all $n:I\rightarrow \mathcal {R}_+$ satisfying

$$\begin{aligned} \forall i\in supp(n), \ f_i(n_i) = \max _j f_j(n_j). \end{aligned}$$

(6.2)

Let c denote this maximal value. If $f_i(0)>c$ let $n_i$ be the unique positive value satisfying $f_i(n_i)=c$ and otherwise let $n_i=0$. If $c<c_*:=\max _i f_i(0)$ the result is the unique $n:I\rightarrow \mathcal {R}_+$ with the above property and maximal value c, that we denote n(c). The assumptions on $f_i$ mean that for each i, the function $c\mapsto n_i(c)$ is continuous, has $n_i(c)=0$ for $c\ge f_i(0)$, is decreasing on $(-\infty ,f_i(0)]$ and has $n_i(c)\rightarrow \infty$ as $c\rightarrow -\infty$. In turn, $N(c):=\sum _i n_i(c)$ is continuous, is decreasing on $(-\infty ,c_*)$ and has $N(c)\rightarrow 0$ as $c\rightarrow c_*$ and $N(c)\rightarrow \infty$ as $c\rightarrow -\infty$, and so the inverse function c(N) is defined on $(0,\infty )$, is continuous, decreasing, and has $c(N)\rightarrow c_*$ as $N\rightarrow 0$ and $c(N)\rightarrow -\infty$ as $N\rightarrow \infty$.

We claim n(c(N)) is the unique solution to the maximin problem with population size N. To see this, suppose $n:I\rightarrow \mathcal {R}_+$ has $\sum _i n_i=N$ and $n\ne n(c(N))$, then $n_i > n_i(c(N))$ for some $i \in supp(n)$ so $F(n) \le f_i(n_i) < f_i(n_i(c(N)) \le c(N)=F(n_i(c(N)))$, completing the demonstration.

Next we solve the Boltzmann-type equation. We work in a similar way, this time by first solving the related equation

$$\begin{aligned} \forall i \in I, \ n_i = c \exp (\beta f_i(n_i)) \end{aligned}$$

(6.3)

for fixed $c,\beta >0$. Write this as $g_i(n_i,c)=0$ for each i, where $g_i(n_i,c):= c \exp (\beta f_i(n_i)) - n_i$. Then $g_i(0,c)=c \exp (\beta f_i(0))>0$ and by assumption on $f_i$, $n_i \mapsto g_i(n_i,c)$ is continuous, decreasing and $g_i(n_i,c) \rightarrow -\infty$ as $n_i\rightarrow \infty$. Let $n_i(c)$ denote the unique solution. Since $c\mapsto g_i(n_i,c)$ is increasing, if $c_1<c_2$ then $g_i(n_i(c_1),c_2) > g_i(n_i(c_1),c_1)=0$ so $n_i(c_2)>n_i(c_1)$, i.e. $c\mapsto n_i(c)$ is increasing. Since $g_i(n_i,c) \le c\exp (\beta f_i(0))-n_i$, $n_i(c)\le c\exp (\beta f_i(0))$ and in particular $n_i(c)\rightarrow 0$ as $c\rightarrow 0^+$. In addition, for any $n_i>0$, $g_i(n_i,c)\ge 0$ if $c\ge n_i/\exp (\beta f_i(n_i))$, so $n_i(c)\rightarrow \infty$ as $c\rightarrow \infty$. $c\mapsto n_i(c)$ is also continuous: this holds at c if $n_i(c)=n_i(c^-)$, i.e. if for any increasing $(c_k)$ with limit c, $n_i(c)=\lim _{k\rightarrow \infty } n_i(c_k)$, or equivalently $g(c,\lim _{k\rightarrow \infty } n_i(c_k))=0$. Since g is continuous, $g(c,\lim _{k\rightarrow \infty } n_i(c_k)) = \lim _{k\rightarrow \infty } g(c,n_i(c_k))$. Moreover,

$$|g(c,n_i(c_k))| = |g(c,n_i(c_k))-g(c_k,n_i(c_k))| = (c-c_k)\exp (\beta f_i(n_i(c_k))) \le (c-c_k)\exp (\beta f_i(0)).$$

In particular, $g(c,n_i(c_k)) \rightarrow 0$ as $k\rightarrow \infty$, which proves the result. It follows that $c\mapsto N(c)$ is continuous, increasing, has $N(c)\rightarrow 0$ as $c\rightarrow 0^+$ and $N(c)\rightarrow \infty$ as $c\rightarrow \infty$. So, the inverse function c(N) has the same domain and the same properties, and $\sum _I n_i(c(N))=N$ for each $N>0$.

We claim that n(c(N)) is the unique solution to Eq. (6.1) with population size N. Since n(c(N)) solves Eq. (6.3) with $c=c(N)$ and has $\sum _i n_i(c(N))=N$ it solves Eq. (6.1) with $Z_\beta = N/c(N)$. On the other hand if n solves Eq. (6.1) then with $c=N/Z_\beta$, n solves Eq. (6.3), so $n=n(c)$. To finish the proof we must show that $c=c(N)$. However, c(N) is the only value of c for which $\sum _i n_i(c)=N$.

Proof (Proof of Theorem 2)

For any $N>0$ the set of $n:I\rightarrow \mathcal {R}_+$ with $\sum _i n_i=N$ is compact, so the limit set $\lim _{\beta \rightarrow \infty } n(\beta )$ is non-empty. To prove convergence, it suffices to show every limit point has the maximizing support property. First notice that

$$\begin{aligned} n_i(\beta )/n_j(\beta ) = \exp (\beta ( f_i(n_i(\beta ))-f_j(n_j(\beta )))). \end{aligned}$$

Let $n_i$ be the solution of the minimax problem and c the constant value of $f_i(n_i), i \in supp(n)$. It’s enough to show that $\limsup _{\beta \rightarrow \infty } n_i(\beta ) \le n_i$ for each i. If not then for some i, j and some $\delta >0$, $n_i(\beta )\ge n_i+\delta$ and $n_j(\beta ) \le n_j-\delta$ infinitely often as $\beta \rightarrow \infty$. But then for such $\beta$,

$$\begin{aligned} f_j(n_j(\beta )) \ge f_j(n_j-\delta )> f_j(n_j)=f_i(n_i) > f_i(n_i+\delta ) \ge f_i(n_i(\beta )) \end{aligned}$$

or in other words

$$\begin{aligned} f_j(n_j(\beta )) - f_i(n_i(\beta )) \ge \alpha := f_j(n_j-\delta ) - f_i(n_i+\delta ) >0. \end{aligned}$$

It follows then that infinitely often,

$$\begin{aligned} n_i(\beta )/n_j(\beta ) \le e^{-\beta \alpha }. \end{aligned}$$

However, $n_i(\beta )/n_j(\beta ) \ge (n_i+\delta )/(n_j-\delta )>0$ by assumption, a contradiction.