Evolutionary Game Dynamics in a Finite Continental Island Population Model and Emergence of Cooperation

Abstract

We consider the continental island model for a finite haploid population with a total number of ${\textit{n}}$ demes consisting of one continent and $n-1$ islands. We assume viability differences in the population captured by a linear game within each deme as a result of pairwise interactions. Assuming weak selection, conservative migration and the limit case of the structured coalescent assumptions, we derive the first-order approximation for the fixation probability of a single mutant, initially introduced in the continent, with respect to the intensity of selection. This result is applied to the case of the iterated Prisoner’s Dilemma, when the resident strategy is always defect and the mutant cooperative strategy is tit-for-tat. In this context, we investigate the condition under which selection favors the emergence of cooperation and we derive an extension of the “one-third law” of evolution. When the continent and the islands are of the same size, we compare the continental island model to its Wright’s island model counterpart. When the islands have the same size, but this size differs from the size of the continent, we investigate how the asymmetry in the deme sizes can better promote the evolution of tit-for-tat compared to its equal deme sizes model counterpart.

Appendices

Appendix A Proof of Proposition 1

Here, we prove Proposition 1. For $t=0, 1,\ldots$, let us define Z(t), the weighted frequency of A, as:

$$\begin{aligned} Z(t)=u_0\,X_0(t)+\cdots +u_{n-1}\,X_{n-1}(t)\,, \end{aligned}$$

where $\mathbf{u}=(u_0,\ldots ,u_{n-1})$ denotes the stationary distribution of migration matrix $\mathbf{M}$. For any given selection intensity $s\ge 0$, the discrete-time stochastic process $(Z(t))_{t\ge 0}$, is a Markov chain on the finite state space:

$$\begin{aligned} \biggl \{\textstyle u_0\,\frac{k_0}{N_0}+\sum _{i=1}^{n-1} u_i\,\frac{k_i}{N_1} \;;\; k_0=0,\ldots , N_0\,,\ k_i=0,\ldots , N_1\,,\ i=1,\ldots , n-1 \biggr \}\,. \end{aligned}$$

Its initial state is $Z(0)=u_0/N_0$, and there are two absorbing states $z=0$ and $z=1$, which correspond to the fixation of B and A, respectively; all the other states being transient.

This process converges in probability to a random variable $Z(\infty )$, which takes the value 1 with probability u(s), and 0 otherwise. In the neutral scenario ($s=0$), this process is a bounded martingale. By the stopping time theorem [14], we find that the fixation probability of A, which occurs when the absorbing state $z=1$ is reached, is equal to:

$$\begin{aligned} u(0)={\mathbb {E}}_0(Z(\infty ))=Z(0)=\frac{u_0}{N_0}\,. \end{aligned}$$

In the general case when selection intensity is s, we can write following [43] that:

$$\begin{aligned} {\mathbb {E}}[Z(\infty )-Z(0)]=\sum _{t\ge 0}{\mathbb {E}}[Z(t+1)-Z(t)]\,, \end{aligned}$$

which is equivalent to:

$$\begin{aligned} u(s)-u(0)=\sum _{t\ge 0}{\mathbb {E}}[Z(t+1)-Z(t)]\,. \end{aligned}$$

After differentiating with respect to s and assuming the interchangeability of summation and derivation (for a formal proof under mild regularity conditions see [26]), we get:

$$\begin{aligned} u'(0) = \sum _{t\ge 0} \frac{{d }}{{{d }} s}{\mathbb {E}}[Z(t+1)-Z(t)]\Big \vert _{s=0}\,. \end{aligned}$$

Conditioning on the values ${\mathbf {x}}=(x_0, x_1,\ldots , x_{n-1})$ taken by ${\mathbf {X}}(t)$, we can write:

$$\begin{aligned} {\mathbb {E}}[Z(t+1)-Z(t)] = \sum _{{\mathbf {x}}} {\mathbb {E}}\bigl [Z(t+1)-Z(t)\big \vert {\mathbf {X}}(t)={\mathbf {x}}\bigr ]\, {\mathbb {P}}({\mathbf {X}}(t)={\mathbf {x}})\,, \end{aligned}$$

which implies that:

$$\begin{aligned}&\frac{{d }}{{d }s}{\mathbb {E}}[Z(t+1)-Z(t)]\Big \vert _{s=0}\\&\quad = \sum _{{\mathbf {x}}}\frac{{d }}{{d }s}{\mathbb {E}}\big [Z(t+1)-Z(t)\big \vert {\mathbf {X}}(t) ={\mathbf {x}}\big ]\Big \vert _{s=0} \;{\mathbb {P}}_0\big ({\mathbf {X}}(t)={\mathbf {x}}\big )\\&\qquad +\,\sum _{{\mathbf {x}}} {\mathbb {E}}_0\Big [Z(t+1)-Z(t)\Big \vert {\mathbf {X}}(t)={\mathbf {x}}\Big ] \,\frac{{d }}{{d }s}\,{\mathbb {P}}\big ({\mathbf {X}}(t)={\mathbf {x}}\big )\big \vert _{s=0} \,. \end{aligned}$$

In the neutral case ($s=0$), the expected change in Z(t) from one generation to the next is zero, that is:

$$\begin{aligned} {\mathbb {E}}_0\bigl [Z(t+1)-Z(t)\big \vert {\mathbf {X}}(t)={\mathbf {x}}\bigr ]=0\,. \end{aligned}$$

This leads to:

$$\begin{aligned} u'(0)= & {} \sum _{t\ge 0} \sum _{{\mathbf {x}}} {\mathbb {P}}_0\big ({\mathbf {X}}(t)=\mathbf{x}\big )\, \frac{{d }}{{d }s} {\mathbb {E}}\big [Z(t+1)-Z(t)\vert {\mathbf {X}}(t)=\mathbf{x}\big ]\Big \vert _{s=0}\,. \end{aligned}$$

(A.1)

Since the change in the weighted frequency of A from one generation to the next has conditional expectation :

$$\begin{aligned} {\mathbb {E}}\big [Z(t+1)-Z(t)\vert {\mathbf {X}}(t)=\mathbf{x}\big ] = \sum _{i=0}^{n-1}u_i\Big ({\mathbb {E}}\big [X_i(t+1)\vert {\mathbf {X}}(t)=\mathbf{x}\big ]-x_i\Big ), \end{aligned}$$

it follows from (3), (4) and (5), that:

$$\begin{aligned}&\frac{{d }}{{d }s}{\mathbb {E}}\big [Z(t+1)-Z(t)\big \vert {\mathbf {X}}(t)=\mathbf{x}\big ]\Big \vert _{s=0} \nonumber \\&\quad = \sum _{i=0}^{n-1} u_i \, {\tilde{x}}_i^2\,(1-{\tilde{x}}_i)\,({\mathbf {p}}_A-{\mathbf {p}}_B)\cdot W\,({\mathbf {p}}_A-{\mathbf {p}}_B) \nonumber \\&\qquad +\, \sum _{i=0}^{n-1}u_i\,{\tilde{x}}_i\,(1-{\tilde{x}}_i)\,({\mathbf {p}}_A-{\mathbf {p}}_B)\cdot W_i\,{\mathbf {p}}_B\,. \end{aligned}$$

(A.2)

If we plug Eqs. (1) and (2) into (A.2), it follows from (A.1) that:

$$\begin{aligned} u'(0)&= ({\mathbf {p}}_A-{\mathbf {p}}_B)\\&\quad \cdot \left\{ u_0 \sum _{t\ge 0}\lambda _0({\mathbf {m}}_0,{\mathbf {X}}(t)) + \sum _{i=1}^{n-1}u_i \sum _{t\ge 0}\lambda _i({\mathbf {m}}_1,{\mathbf {X}}(t)) \right\} W({\mathbf {p}}_A-{\mathbf {p}}_B)\\&\quad + \,({\mathbf {p}}_A-{\mathbf {p}}_B)\\&\quad \cdot \left\{ u_0 \,\sum _{t\ge 0} \delta _0({\mathbf {m}}_0,{\mathbf {X}}(t)) + \sum _{i=1}^{n-1}u_i \,\sum _{t\ge 0}\delta _i({\mathbf {m}}_1,{\mathbf {X}}(t)) \right\} \,W \, {\mathbf {p}}_B\,, \\&= ({\mathbf {p}}_A-{\mathbf {p}}_B) \cdot \kappa _{\lambda }\, W\,({\mathbf {p}}_A-{\mathbf {p}}_B) + ({\mathbf {p}}_A-{\mathbf {p}}_B) \cdot \kappa _{\delta } \, W \, {\mathbf {p}}_B\,, \end{aligned}$$

where $\lambda _0({\mathbf {m}}_0,{\mathbf {X}}(t))$, $\lambda _i({\mathbf {m}}_1,{\mathbf {X}}(t))$, $\delta _0({\mathbf {m}}_0,{\mathbf {X}}(t))$, and $\delta _i({\mathbf {m}}_1,{\mathbf {X}}(t))$ can be explicitly written in terms of coefficients ${\mathbb {E}}_0[X_i(t)X_j(t)(1-X_k(t))]$ and ${\mathbb {E}}_0[X_i(t)(1-X_j(t))]$, with $i,j,k\in \{0,\ldots , n-1\}$. Due to the interchangeability of the islands, those expressions can be reduced to equations involving coefficients ${\mathbb {E}}_0[X_i(t)X_j(t)(1-X_k(t))]$ and ${\mathbb {E}}_0[X_i(t)(1-X_j(t))]$ only for $i,j,k\in \{0,1,2,3\}$, as follows:

$$\begin{aligned} \lambda _0({\mathbf {m}}_0,{\mathbf {X}}(t))&= (1-m_0)^3 \, {\mathbb {E}}_0[X_0(t)^2 \, (1-X_0(t))] \nonumber \\&\quad +\, m_0 \, (1-m_0)^2 \, {\mathbb {E}}_0[X_0^2(t) \, (1-X_1(t))] \nonumber \\&\quad + \,2 \, m_0 \, (1-m_0)^2 \, {\mathbb {E}}_0[X_0(t) \, (1-X_0(t)) \, X_1(t)] \nonumber \\&\quad + \,2 \, (1-m_0) \, m_0^2 \, \frac{n-2}{n-1} \, {\mathbb {E}}_0[X_0(t) \, X_1(t) \, (1-X_2(t))] \nonumber \\&\quad + \,2 \, (1-m_0) \, \frac{m_0^2}{n-1} \, {\mathbb {E}}_0[X_0(t) \, X_1(t) \, (1-X_1(t))] \nonumber \\&\quad + \,(1-m_0) \, \frac{m_0^2}{n-1} \, {\mathbb {E}}_0[X_1^2(t) \, (1-X_0(t))] \nonumber \\&\quad + \,(1-m_0) \, m_0^2 \, \frac{n-2}{n-1} \, {\mathbb {E}}_0[X_1(t) \, X_2(t) \, (1-X_0(t))] \nonumber \\&\quad + \,\frac{m_0^3}{(n-1)^2} \, {\mathbb {E}}_0[X_1^2(t)(1-X_1(t))] \nonumber \\&\quad + \,\frac{m_0^3(n-2)}{(n-1)^2} \, {\mathbb {E}}_0[X_1^2(t)(1-X_2(t))] \nonumber \\&\quad +\,2 \, m_0^3 \, \frac{(n-2)}{(n-1)^2} \, {\mathbb {E}}_0[X_1(t) \, X_2(t) \, (1-X_1(t))] \nonumber \\&\quad + \,m_0^3 \, \frac{(n-2) \, (n-3)}{(n-1)^2} \, {\mathbb {E}}_0[X_1(t) \, X_2(t) \, (1-X_3(t))] \, , \nonumber \\ \delta _0({\mathbf {m}}_0,{\mathbf {X}}(t))&= (1-m_0)^2 \, {\mathbb {E}}_0[X_0(t) \, (1-X_0(t))] \nonumber \\&\quad + \,m_0 \, (1-m_0) \, {\mathbb {E}}_0[X_0(t) \, (1-X_1(t))] \nonumber \\&\quad + \,m_0(1-m_0) \, {\mathbb {E}}_0[X_1(t) \, (1-X_0(t))] \nonumber \\&\quad + \, \frac{m_0^2}{n-1} \, {\mathbb {E}}_0[X_1(t) \, (1-X_1(t))] \nonumber \\&\quad + \, m_0^2 \, \frac{n-2}{n-1} \, {\mathbb {E}}_0[X_1(t) \, (1-X_2(t))] \, , \end{aligned}$$

(A.3)

and for all $i=1,\ldots , n-1$:

$$\begin{aligned} \lambda _i({\mathbf {m}}_1,{\mathbf {X}}(t))&= m_1^3\,{\mathbb {E}}_0[X_0(t)^2\,(1-X_0(t))] \nonumber \\&\quad +\,m_1^2\,(1-m_1)\,{\mathbb {E}}_0[X_0(t)^2\,(1-X_1(t))] \nonumber \\&\quad +\,2\,m_1^2\,(1-m_1)\,{\mathbb {E}}_0[X_0(t)\,(1-X_0(t))\,X_1(t)] \nonumber \\&\quad +\,2\,m_1\,(1-m_1)^2\,{\mathbb {E}}_0[X_0(t)\,X_1(t)\,(1-X_1(t))] \nonumber \\&\quad +\,m_1\,(1-m_1)^2\,{\mathbb {E}}_0[X_1(t)^2\,(1-X_0(t))] \nonumber \\&\quad +\,(1-m_1)^3\,{\mathbb {E}}_0[X_1^2(t)\,(1-X_1(t))]\,, \nonumber \\ \delta _i({\mathbf {m}}_1,{\mathbf {X}}(t))&= m_1^2\,{\mathbb {E}}_0[X_0(t)\,(1-X_0(t))] \nonumber \\&\quad +\,m_1\,(1-m_1)\,{\mathbb {E}}_0[X_0(t)\,(1-X_1(t))] \nonumber \\&\quad +\,m_1\,(1-m_1)\,{\mathbb {E}}_0[X_1(t)\,(1-X_0(t))] \nonumber \\&\quad +\,(1-m_1)^2\,{\mathbb {E}}_0[X_1(t)\,(1-X_1(t))]\,. \end{aligned}$$

(A.4)

Appendix B Proof of Proposition 2

Here, we calculate ${\mathbb {E}}_0\big [X_i(t)\,X_j(t)\,(1-X_k(t))\big ]$ and ${\mathbb {E}}_0\big [X_i(t)\,(1-X_j(t))\big ]$ with $i,j,k\in \{0,1,2,3\}$, in terms of the submatrices L, F, Q of (8) the transition matrix K of the ancestral process $\sigma (t)$, that describes the locations of the ancestors of 3 individuals.

For each $k\in \{1,\ldots N_1\}$ and $i\in \{0,\ldots n-1\}$, let $\xi _{k,i}$ denote the random variable that assigns the value 1 to the k-th individual in deme i if it is of type A and the value 0, otherwise:

$$\begin{aligned} \xi _{k, i}(t) := \left\{ \begin{array}{ll} 1\,, &{} \text {if individual}\ k\ \text {in deme}\ i\ \text {at time}\ t\ \text {is of type}\ A, \\ 0\,, &{} \text {otherwise}. \end{array} \right. \end{aligned}$$

Let us first focus on $\sum _{t\ge 0}{\mathbb {E}}_0\big [X_0^2(t) \, (1-X_0(t))\big ]$. The frequency of A in the continent at time t can be written in terms of the $\xi _{k,0}(t)$ as:

$$\begin{aligned} X_0(t)= \frac{1}{N_0}\sum _{k=1}^{N_0}\xi _{k,0}(t)\,. \end{aligned}$$

Thus, we have:

$$\begin{aligned}&{\mathbb {E}}_0\big [X_0^2(t)(1-X_0(t))\big ] = \frac{1}{N_0^3} {\mathbb {E}}_0\left[ \left( \sum _{k=1}^{N_0}\xi _{k,0}(t)\right) ^2 \left( \sum _{l=1}^{N_0}(1-\xi _{l,0}(t))\right) \right] \nonumber \\&\quad = \frac{1}{N_1^3}\sum _{k=1}^{N_1}\sum _{l=1}^{N_1}\sum _{m=1}^{N_1} {\mathbb {E}}_0\big [ \xi _{k,0}(t)\,\xi _{l,0}(t)\,(1-\xi _{m,0}(t)) \big ] \nonumber \\&\quad = \Bigl (1-\frac{1}{N_0}\Bigr )\, \Bigl (1-\frac{2}{N_0}\Bigr )\, \alpha _0(t) + \frac{1}{N_0}\,\Bigl (1-\frac{1}{N_0}\Bigr )\,\beta _0(t)\,, \end{aligned}$$

(A.5)

where:

$$\begin{aligned} \alpha _0(t)&:= {\mathbb {P}}_0\Bigl ( \xi _{1,0}(t)=\xi _{2,0}(t)=1\,,\ \xi _{3,0}(t)=0\Bigr )\,,\\ \beta _0(t)&:= {\mathbb {P}}_0\Bigl (\xi _{1,0}(t)=1\,,\ \xi _{2,0}(t)=0\Bigr )\,. \end{aligned}$$

Now, let us calculate $\sum _{t\ge 0}\alpha _0(t)$. According to its definition, $\alpha _0(t)$ is the probability that individuals 1, 2 and 3 from deme 0 at time t are of respective types A, A and B. Since a single mutant A was introduced in the continent at time 0, individuals 1, 2 and 3 in deme 0 at time t are of types A, A and B, respectively, if a single coalescence occurs from time t to time 0, this event being a coalescence between the lineages of individuals 1 and 2, and the two distinct ancestors at time 0 of these three individuals are of respective types A (for the common ancestor of individuals 1 and 2) and B (for the ancestor of individual 3). This implies that the state $\sigma (t)$ in S, the ancestors of those three individuals are in t generations back, must either be 00 or 01, while the type of the ancestor common to individuals 1 and 2 is A. Thus, $\alpha _0(t)$ is equal to the probability that the process $\sigma (t)$ is in state 00 or 01 at time t given that $\sigma (0)$ is in state 000 in $S_{1,2,3}$, times $1/N_0$, which is the frequency of A in the continent at time 0. Using the notations in Eq. (8) for the block form of K, the transition matrix of $\sigma (.)$, this translates as follows:

$$\begin{aligned} \alpha _0(t) = \sum _{\tau =1}^t\Big (F^{\tau -1} \, Q \, L^{t-\tau } \, u\Big )_{000} \, , \end{aligned}$$

where u is the column vector $u=(1/N_0,1/N_0,0,0,0)$, and index 000 refers to the vector’s component that corresponds to the chain starting in state 000. As a consequence, since matrices $I-F$ and $I-L$ are invertible, where I refers to an identity matrix of appropriate order, if we sum $\alpha _0(t)$ over $t\ge 0$, we find that:

$$\begin{aligned} \sum _{t\ge 0} \alpha _0(t) = \Big ((I-F)^{-1} \, Q \, (I-L)^{-1} \, u\Big )_{000} \, . \end{aligned}$$

(A.6)

Similarly, we find that:

$$\begin{aligned} \beta _0(t) := {\mathbb {P}}_0\big (\xi _{1,0}(t)=1 \, ,\ \xi _{2,0}(t)=0\big ) = (L^tu)_{00} \, , \end{aligned}$$

leading to:

$$\begin{aligned} \sum _{t\ge 0} \beta _0(t)=\Big ((I-L)^{-1}u\Big )_{00} \, . \end{aligned}$$

(A.7)

Let:

$$\begin{aligned} {\mathcal {U}}&:=(I-L)^{-1} \, u\,,&\text {and}&{\mathcal {V}}&:=(I-F)^{-1}\,Q\,(I-L)^{-1}u\,. \end{aligned}$$

We then derive from (A.5), (A.6) and (A.7) that:

$$\begin{aligned} \sum _{t\ge 0}{\mathbb {E}}_0\bigl (X_0(t)^2\,(1-X_0(t))\bigr )&= \textstyle (1-\frac{1}{N_0})\,(1-\frac{2}{N_0})\,{\mathcal {V}}_{000} + \frac{1}{N_0}\,(1-\frac{1}{N_0})\,{\mathcal {U}}_{00}\,. \end{aligned}$$

(A.8a)

In the same way, we find that

$$\begin{aligned}&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_1(t)^2\,(1-X_1(t))\big ] = \textstyle (1-\frac{1}{N_1})\,(1-\frac{2}{N_1})\,{\mathcal {V}}_{111} +\frac{1}{N_1}\,(1-\frac{1}{N_1})\,{\mathcal {U}}_{11}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_0(t)^2\,(1-X_1(t))\big ] = \textstyle (1-\frac{1}{N_0})\,{\mathcal {V}}_{001}+\frac{1}{N_0}\,{\mathcal {U}}_{01}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_1(t)^2\,(1-X_0(t))\big ] = \textstyle (1-\frac{1}{N_1})\,{\mathcal {V}}_{110}+\frac{1}{N_1}\,{\mathcal {U}}_{10}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_0(t)\,X_1(t)\,(1-X_0(t))\big ] = \textstyle (1-\frac{1}{N_0})\,{\mathcal {V}}_{010}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_0(t)\,X_1(t)\,(1-X_1(t))\big ] = \textstyle (1-\frac{1}{N_1})\,{\mathcal {V}}_{011}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_1(t)^2\,(1-X_2(t))\big ] = \textstyle (1-\frac{1}{N_1})\,{\mathcal {V}}_{112} +\frac{1}{N_1}\,{\mathcal {U}}_{12}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_1(t)\,X_2(t)\,(1-X_1(t))\big ] = \textstyle (1-\frac{1}{N_1})\,{\mathcal {V}}_{112}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_0(t)\,X_1(t)\,(1-X_2(t))\big ] = {\mathcal {V}}_{012}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_1(t)\,X_2(t)\,(1-X_0(t))\big ] = \textstyle {\mathcal {V}}_{120}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_1(t)\,X_2(t)\,(1-X_3(t))\big ] = {\mathcal {V}}_{123}, \end{aligned}$$

(A.8b)

$$\begin{aligned}&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_0(t)\,(1-X_0(t))\big ] = \textstyle \big (1-\frac{1}{N_0}\big )\,{\mathcal {U}}_{00}, \end{aligned}$$

(A.8c)

$$\begin{aligned}&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_1(t)\,(1-X_1(t))\big ] = \textstyle \big (1-\frac{1}{N_1}\big )\,{\mathcal {U}}_{11}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_0(t)\,(1-X_1(t))\big ] = {\mathcal {U}}_{01}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_1(t)\,(1-X_0(t))\big ] = {\mathcal {U}}_{10}, \nonumber \\&\textstyle \sum _{t\ge 0}{\mathbb {E}}_0\big [X_1(t)\,(1-X_2(t))\big ] = {\mathcal {U}}_{12}. \end{aligned}$$

(A.8d)

Appendix C Approximations under the Structured Coalescent Assumptions

Here, we give the formulae for coefficients $\gamma _{i}$ and $\delta _{i}$, $i=0,1$ as functions of the population parameters n, M, P and $\Lambda$ derived using Maple.

$$\begin{aligned} \gamma _{0} =\frac{\gamma _{01}}{3(Mn-M+1-2P+nP)\gamma _{02}} \end{aligned}$$

(A.9)

where

$$\begin{aligned} \gamma _{01}&=16P^5+6M^5Pn-71M^2nP+29M^2n^2P-39M^2n^2P^2+11M^2nP^2 \\&\quad -\,59M^2n^2P^3+22Mn^2P+M^4P^2-13M^3P^2+3M^5n-2M^5P+10M^2P^2 \\ {}&\quad +\,11M^2n^2P^4+2M^2n^2P^5-52Mn^2P^3+4Mn^2P^4+22MnP^4+4Mn^3P^2 \\ {}&\quad +\,72MP^2+56M^3nP^3-28MP^3-12P^3n^2-34M^3n^2P^2+116MnP^3 \\ {}&\quad -\,11M^3Pn+6M^3n^2-10M^4P^3+11M^3P-36nP^2-15M^3n-20MP \\ {}&\quad -\,24P^4M^2 -40M^2P^3+M^5n^3-3M^5n^2-44MP^4+16MP^5+2M^3n^3 \\ {}&\quad -\,4M^3P^4+113M^2nP^3+20M^4nP^3-16P^2-4n^2P^4+12nP-8M^2nP^5 \\ {}&\quad +\,36nP^4+3M^4n-20MP^5n+8Mn^2P^2+6M^2n^3P+10M^2n^3P^2+12n^2P^2 \\ {}&\quad +\,39M^3nP^2+13M^3n^3P+6MP^5n^2+8M^3n^3P^2-32M^3n^2P^3+4M^3n^2P^4 \\ {}&\quad -\,6M^5n^2P+2M^5n^3P-M^5-6M^2+8M^2n^2+9M^4n^3P+2M^4n^3P^2 \\ {}&\quad -\,23M^4n^2P-3M^4n^2P^2-10M^4n^2P^3+19M^4nP+7M^3-M^2n \\ {}&\quad -\,13M^3n^2P-24M^3P^3-5M^4P+6Mn-8MnP-104MnP^2+56P^3-56P^4 \\ {}&\quad +\,36M^2P+8M^2P^5+4M^2nP^4+4n^2P^5-16nP^5+3M^4n^3-6M^4n^2+4nP^3, \end{aligned}$$

and

$$\begin{aligned} \gamma _{02}&=4M+6M^2nP+6M^2n^2P+6M^2n^2P^2-20M^2nP^2+8P+16M^2P^2-20MP^2 \nonumber \\&\quad -\,8M^3nP^3+32MP^3+2M^3n^2P^2-20MnP^3-11M^3Pn+2M^3n^2+4M^3P \nonumber \\&\quad +\,8nP^2-M^3n-12MP+16M^2P^3-16M^2nP^3-24P^2-2M^4n+4Mn^2P^2 \nonumber \\&\quad -\,2M^3nP^2+M^4-4M^2+2M^4n^2P-4M^4nP-M^3+6M^2n+7M^3n^2P \nonumber \\&\quad +\,8M^3P^3+2M^4P+16MnP-4MnP^2+16P^3-16M^2P+M^4n^2-8nP^3. \nonumber \\ \gamma _{1}&=\frac{\gamma _{11}}{3(n-1)(Mn-M+1-2P+nP)\gamma _{12}} \end{aligned}$$

(A.10)

where

$$\begin{aligned} \gamma _{11}&=-16P^5-8M-8M^5Pn+93M^2nP-92M^2n^2P-2M^2n^2P^2+47M^2nP^2 \\ {}&\quad +\,96M^2n^2P^3+36Mn^2P-16P-M^4P^2+29M^3P^2+8M^3n^4P^2+2M^5n^4P \\ {}&\quad -\,4M^5n+2M^5P-26M^2P^2-26M^3n^3P^3+4M^3n^3P^4+9M^3n^4P \\ {}&\quad +\,29M^2n^2P^4-10M^2n^2P^5+62Mn^2P^3 +14Mn^2P^4-38MnP^4+18Mn^3P^2 \\ {}&\quad -\,100MP^2-50M^3nP^3-4M^3nP^4+36MP^3-33M^2n^3P^3-28P^3n^2 \\ {}&\quad +\,101M^3n^2P^2-M^2n^3P^4+2M^2n^3P^5-58MnP^3+42M^3Pn-20Mn^3P^3 \\ {}&\quad +\,2Mn^3P^4-35M^3n^2+10M^4P^3-17M^3P-68nP^2+M^5n^4+34M^3n \\ {}&\quad +\,60MP+32P^4M^2+24M^2P^3-4M^5n^3+6M^5n^2+12MP^4+12M^3n^3+4M^3P^4 \\ {}&\quad -\,87M^2nP^3-10M^4n^3P^3+M^4nP^2-30M^4nP^3+72P^2+20n^2P^4+12nP \\ {}&\quad +\,16M^2nP^5 -76nP^4+3M^4n^4-3M^4n+4MP^5n-110Mn^2P^2+33M^2n^3P \\ {}&\quad -\,25M^2n^3P^2+12n^2P^2-90M^3nP^2-10M^3n^3P-2MP^5n^2-48M^3n^3P^2 \\ {}&\quad +\,64M^3n^2P^3-4M^3n^2P^4+12M^5n^2P -8M^5n^3P+M^5+18M^2+15M^2n^2 \\ {}&\quad -32M^4n^3P-5M^4n^3P^2+42M^4n^2P+3M^4n^2P^2+30M^4n^2P^3-24M^4nP \\ {}&\quad -\,11M^3+6M^2n^4P^2-33M^2n-24M^3n^2P +12M^3P^3+5M^4P+6Mn \\ {}&\quad -\,86MnP+172MnP^2-112P^3+72P^4+9M^4n^4P+2M^4n^4P^2 -34M^2P \\ {}&\quad -\,8M^2P^5-60M^2nP^4-4n^2P^5+16nP^5-9M^4n^3+9M^4n^2+116nP^3 \end{aligned}$$

and

$$\begin{aligned} \gamma _{12}&=4M+6M^2nP+6M^2n^2P+6M^2n^2P^2-20M^2nP^2+8P+16M^2P^2-20MP^2 \nonumber \\&\quad -\,8M^3nP^3 +32MP^3+2M^3n^2P^2-20MnP^3-11M^3Pn+2M^3n^2+4M^3P \nonumber \\&\quad +\,8nP^2-M^3n -12MP+16M^2P^3-16M^2nP^3-24P^2-2M^4n+4Mn^2P^2 \nonumber \\&\quad -\,2M^3nP^2+M^4-4M^2+2M^4n^2P-4M^4nP-M^3+6M^2n+7M^3n^2P \nonumber \\&\quad +\,8M^3P^3+2M^4P+16MnP-4MnP^2+16P^3-16M^2P+M^4n^2-8nP^3.\nonumber \\ \delta _0&= \frac{-P+Mn+n-M}{Mn-M+1-2P+nP} \ ; \ \delta _1=\frac{1-P+Mn-M}{Mn-M+1-2P+nP} \end{aligned}$$

(A.11)

$$\begin{aligned}&(1-u_0)\delta _1+u_0\delta _0=1.\nonumber \\&\Lambda (M,n,P) = (1-u_0)\,\gamma _1+u_0\,\gamma _0 = \rho _{1}/\rho _{2} \end{aligned}$$

(A.12)

with:

$$\begin{aligned} \rho _{1}&= -8M+4Mn^3P^2+41M^2nP^3-12MP^4+8MP^5+M^4n^3+12n^2P^2 \\ {}&\quad -\,50MnP+39M^2nP^2-16P+56P^2+12nP-52nP^2+44nP^3+4nP^4 \\ {}&\quad +24Mn^2P+6Mn+5M^3P^2+2MnP^2+6Mn^2P^5+10M^2 \\ {}&\quad -\,21M^3n^2P-56P^3-10M^3n^2P^3-23M^2n^2P^2-25M^2n^2P^3 \\ {}&\quad +\,8M^2n^2P^4-10M^3nP^2+20M^3nP^3+8M^2n^3P+4M^2n^3P^2-M^4 \\ {}&\quad -\,5M^3P+3M^4n-10M^3P^3-26Mn^2P^3+2M^4n^3P-14Mn^2P^2 \\ {}&\quad -\,2Mn^2P^4-5M^2n^2P-12M^2nP^4-13M^2nP-12P^3n^2+8M^3n^3P \\ {}&\quad -\,16MP^3-2M^4P+10M^2P-7M^3n^2+62MP^3n+4MP^4n \\ {}&\quad -\,12MP^5n-6M^4Pn^2+6M^4Pn+9M^2n^2+5M^3n-19M^2n+36MP \\ {}&\quad -\,12MP^2+5M^3n^2P^2-20M^2P^2-16M^2P^3-M^3+16P^4+4n^2P^5 \\ {}&\quad -\,8nP^5+3M^3n^3-3M^4n^2+4P^4M^2+18M^3nP-4n^2P^4\,,\\ \rho _{2}&= 3(n-1)(4M-16M^2nP^3+16MnP-20M^2nP^2+8P-24P^2 \\ {}&\quad +\,8nP^2-8nP^3-4MnP^2-4M^2+7M^3n^2P+16P^3 \\ {}&\quad +\,6M^2n^2P^2-2M^3nP^2-8M^3nP^3+M^4+4M^3P-2M^4n \\ {}&\quad +\,8M^3P^3+4Mn^2P^2+6M^2n^2P+6M^2nP+32MP^3+2M^4P \\ {}&\quad -\,16M^2P+2M^3n^2-20MP^3n+2M^4Pn^2-4M^4Pn-M^3n \\ {}&\quad +\,6M^2n-12MP-20MP^2+2M^3n^2P^2+16M^2P^2 +16M^2P^3 \\&\quad -\,M^3+M^4n^2-11M^3nP)\,. \end{aligned}$$

Appendix D Transition Probabilities

Matrix K is the probability transition matrix, under neutrality, for the ancestral process $\sigma (t)$ that traces the spatial locations of the ancestors of 3 individuals, labelled individuals 1, 2, 3, over one time step backwards in time. Following Eq. (8),

For the calculation of the first-order expansion of the fixation probability (see Proposition 2), only submatrices F, Q, R, and L are needed (resp. of sizes $15\times 15$, $5\times 15$, $5\times 5$, $5\times 2$), so we do not compute the other submatrices. We now give the detailed expressions for F, Q, R, and L. As the square $15\times 15$ matrix F is cumbersome, it is decomposed into three $5\times 15$ submatrices:

$$\begin{aligned} F = (F^{(1)} \ F^{(2)}\ , F^{(3)})\,. \end{aligned}$$

(A.13)

The computation of these matrices does not present any specific difficulties apart from their size. Before giving their expressions, we shall present how the first line related to transitions from the “000” state (in which the lineages of individuals 1, 2, 3 are not coalesced and on the continent at the given time t) is computed, leaving the reader to detail the other lines. For the “000” state, only the terms in the first row of sub-matrices F and Q need to be detailed.

From the “000 state” going one generation back in time, the possible transitions are broken down as follows:

• 0 coalescence, corresponding to the matrix F, in which case there can either be:

  • 0 migration: the 3 parents remain on the continent $\rightarrow$ 1 possibility: $000\rightarrow 000$

  • 1 migration (and only 1, this is no longer specified in the following): $\rightarrow$ 3 possibilities: $000\rightarrow \{001, 010, 100\}$

  • 2 migrations

    • $*$ to the same island $\rightarrow$ 3 possibilities: $000\rightarrow \{011, 101, 110\}$

    • $*$ to 2 distinct islands $\rightarrow$ 3 possibilities: $000\rightarrow \{012, 102, 120\}$

  • 3 migrations

    • $*$ to the same island $\rightarrow$ 1 possibility: $000\rightarrow 111$

    • $*$ to 2 distinct islands $\rightarrow$ 3 possibilities: $000\rightarrow \{ 112, 121, 211\}$

    • $*$ to 3 distinct islands $\rightarrow$ 1 possibility: $000\rightarrow 123$

• 1 coalescence (only one is possible: the one of the lineages of individuals 1 and 2) corresponding to matrix Q, in which case there can either be:

  • 0 migration $\rightarrow$ 1 possibility: $000\rightarrow 00$

  • 1 migration (the one of the lineage of individual 3) $\rightarrow$ 1 possibility: $000\rightarrow 01$

  • 2 migrations (the one of the lineages of individuals 1 and 2) $\rightarrow$ 1 possibility: $000\rightarrow 10$

  • 3 migrations:

    • $*$ to the same island $\rightarrow$ 1 possibility: $000\rightarrow 11$

    • $*$ to 2 distinct islands $\rightarrow$ 1 possibility: $000\rightarrow 12$

In conclusion, we obtain 20 possible transitions for the 000 state (i.e., transitions to states in $S_{1,2,3}\cup S_{12,3}$), corresponding to the 20 terms of the first line of F and Q. In each case, the computation of the transition probabilities is immediate.

The expressions for submatrices $F^{(1)}$, $F^{(2)}$, $F^{(3)}$, Q, L and R are now given below.