The fitness of the strongest individual in the subcritical GMS model

We derive the strongest individual fitness distribution on a variation for a species survival model proposed by Guiol, Machado and Schinazi \cite{GMS11}. We point out to the fact that this distribution relies on the Gauss hypergeometric function and when $p=\frac{1}{2}$ on the Hypergeometric function type I distribution


Introduction
We consider a discrete time model beginning from an empty set. At each time n ≥ 1, a new species is born with probability p or there is a death (if the system is not empty) with probability q = 1 − p. Let X n be the total number of species at time n. X n is a random walk on Z + that jumps to right with probability p and jumps to left with probability q. When X n is at 0 the process jumps to 1 with probability p or stays at 0 with probability 1 − p. We assign a random number to each new species. This number has a uniform distribution on [0, 1].
We think of this number as a fitness associated to each species. These random numbers are independent to each other. When a death occurs, the individual with lowest fitness dies. This model, latter denominated GMS model, was first proposed and studied in Guiol et al [5]. Some interesting variations were further studied in Guiol et al [6], Ben Ari et al [2] and Skevi and Volkov [10].
In Guiol et al [5] it is shown that there is a sharp phase transition for p > 1/2. For R n , the set of species with fitness higher than f c = 1−p p at time n approachs an uniform distribution in the following sense. For On the other hand every specie born with fitness less than f c disappear after a finite (random) time. The set of species present in the system whose fitness is smaller than f c becomes empty infinitely many times.
Here we focus on the case p ≤ 1/2 in order to understand better the dynamics of this model. In this case, the process X n is recurrent and the system becomes empty infinitely many times. Therefore it is not interesting to study the distribution of the fitness of the species which are alive on the system in the long run. An interesting point is to study the distribution of the fitness of the strongest individual on each excursion between the epochs when the system becomes empty. We propose a variation for the GMS model by considering that each time the system becomes empty, a set of m individuals are introduced with independent set of fitness. This variation is meant to reinforce competition among species before the system becomes empty again.

Results
We deduce explicitly the distribution of the fitness of the strongest individual on excursions between the epochs when the system becomes empty. The last individual to die before the system becomes empty is the strongest on that excursion because the first ones to die are those individuals with the smallest fitness.
Observe that some excursions may have length 2. When this happens, the individual who is born, dies right away without competing with any other individual. To ensure that each excursion has competition among individuals in a sort of natural selection process, we introduce a change-over on the model: Each time after the system becomes empty, m independent new species are placed on the system Competition introduced in GMS(10) avoids that.
The next result computes the fitness distribution of the strongest individual to die right before the system becomes empty on GMS(m) model. It is shown in terms of the hypergeometric function of Gauss (see Luke [8]). This function is denoted by 2 F 1 (a, b; c; z), namely, where a, b, c, are real numbers with c = 0, −1, −2, . . . , and (a) k is the coefficient Pochhammer, namely, (a) k = a(a + 1) · · · (a + k − 1) (a) 0 = 1.
we have that where the last line has been obtained by using Γ(3/2) = √ π/2. Now, using the duplication formula, namely,
In words τ n is the length of a excursion from 0 to 0. As the process X n is homogeneous, the distribution of τ n does not depend on n so we consider the random variable τ := τ 0 . Besides, as p ≤ 1/2 we have that where where the last line has been obtained by using (a) 2k = a In this situation where the last line has been obtained by using (1)