2. Main results

For our first result, we show that the replicator coalescent comes down from infinity in a relatively specific sense. We study the time $\gamma_m = \inf\{t> 0 \colon \sigma(t) \leq m\}$ , $m\in\mathbb{N}$ , that the coalescent first reaches a state with m blocks in total, which can be bounded in probability for large N according to the following result.

As it comes down from infinity, the standard Kingman coalescent with merger rate $c>0$ has block count $(\nu(t), t\geq 0)$ , which is approximately described by the ordinary differential equation (ODE)

(1) \begin{equation} \dot \nu(t)=-c\nu(t)^2/2.\end{equation}

It turns out that the corresponding ODE for our coalescent is already known in the evolutionary game theory literature as the replicator equation.

Replicator equations are used to describe a population of k types, with the proportion of the total population of type $i\in\{1,\ldots, k\}$ at time $t\geq 0$ denoted $x_i(t)\in[0,1]$ . These values sum to one, so $\boldsymbol{x}(t) \;:\!=\; (x_1(t),\ldots,x_k(t))$ lives in $\mathcal{S}^k$ . The replicator equations are then written as $\dot x_i(t)=x_i(t)(f_i(\boldsymbol{x}(t))-\overline{f}(\boldsymbol{x}(t)))$ , $i = 1,\ldots,k$ , $t\geq 0$ , where $f_i\colon\mathcal{S}^k\mapsto \mathbb{R}$ describes the ‘fitness’ of type i as a function of the current population density, and $\overline{f}(\boldsymbol{x}(t))=\sum_{i=1}^nx_if_i(\boldsymbol{x}(t))$ is the average population fitness.

Fitness is often assumed to depend linearly upon the population distribution, with coefficients organised in the ‘payoff matrix’ A. Specifically, let $A_{i,j}$ denote the payoff for a player of type i facing an opponent of type j. Then $f_i(\boldsymbol{x})=\sum_{j=1}^n A_{i,j}x_j$ . This replicator equation, henceforth referred to as the $\boldsymbol{A}$ -replicator equation, may be written as

(2) \begin{equation} \dot x_i(t)=x_i(t)\big([{\boldsymbol{A}}{\boldsymbol{x}}(t)]_i - {\boldsymbol{x}}(t)^\top{\boldsymbol{A}}{\boldsymbol{x}}(t)\big), \quad i = 1,\ldots,k, \, t\geq 0.\end{equation}

If the system in (2) admits a fixed point in the simplex, i.e. $x_i(t) = x^*_i$ , $i = 1,\ldots,k$ , for some vector ${\boldsymbol{x}}^* = (x^*_1, \ldots, x^*_k)\in\mathcal{S}^k$ , so that $\text{d}\big(\sum_{i=1}^k x_i(t)\big)/\text{d} t=0$ , then we see that, necessarily, $[\boldsymbol{A}\boldsymbol{x}^*]_i=(\boldsymbol{x}^*)^\top\boldsymbol{A}\boldsymbol{x}^*$ , $i =1, \ldots, k$ . In turn, this implies that there is a constant $c>0$ such that ${\boldsymbol{A}}{\boldsymbol{x}}^*=c\textbf{1}$ , so ${\boldsymbol{x}}^*=c\boldsymbol{A}^{-1}\textbf{1}$ , where $\boldsymbol{1}$ is the vector in $\mathbb{R}^k$ with unit entries. Since ${\boldsymbol{x}}^*\in\mathcal{S}^k$ , it follows that $\textbf{1}^\top{\boldsymbol{x}}^*=1$ , and hence $c=(\textbf{1}^\top A^{-1}\textbf{1})^{-1}$ , thus (2) is solved by ${\boldsymbol{x}}^*={A^{-1}\textbf{1}}/{\textbf{1}^\top A^{-1}\textbf{1}}$ .

If ${\boldsymbol{x}}^*$ satisfies the relation $({\boldsymbol{x}}^*)^\top{\boldsymbol{A}}{\boldsymbol{x}} > {\boldsymbol{x}}^\top{\boldsymbol{A}}{\boldsymbol{x}}$ for all ${\boldsymbol{x}}\neq\boldsymbol{x}^*$ in a neighbourhood of $\boldsymbol{x}^*$ , then it is called an evolutionary stable state (ESS). Theorem 7.2.4 of [Reference Hofbauer and Sigmund3] states that if ${\boldsymbol{x}}^*$ is an ESS, then

(3) \begin{equation} \lim_{t\to\infty}{\boldsymbol{x}}(t) = {\boldsymbol{x}}^*.\end{equation}

The following results give us a remarkable connection between the theory of replicator equations and our coalescent model. To this end we define our A matrix by

(4) \begin{equation} A_{i,j} = -\bigg(C_{j,i}1_{j\neq i} + \frac{1}{2}C_{i,i}1_{i=j}\bigg).\end{equation}

For the remainder of the paper we will assume that the rates C are such that (3) holds.

Let us now state the connection between the notion of coming down from infinity for the replicator coalescent and the corresponding replicator equations.

Theorem 1. Suppose that $\boldsymbol{A}$ is such that (3) holds, and that $(\boldsymbol{\eta}^N, N\geq 1)$ tends to $(\boldsymbol{r}_0,0)$ . Then, for all $T>0$ , $\lim_{N\to\infty}\mathbb{E}_{{\boldsymbol{\eta}}^N}\big[\sup_{t\leq T}\|\boldsymbol{R}(t)-\boldsymbol{x}(t)\|_1\big]=0$ , $i =1,\ldots,k$ , where $\boldsymbol{R}(t) = \boldsymbol{r}(\tau(t))$ , $t\geq 0$ , $\boldsymbol{x}(t) = (x_1(t),\ldots,x_k(t))$ solves the $\boldsymbol{A}$ -replicator equation with initial condition $\boldsymbol{x}(0) = \boldsymbol{r}_0$ , and $\tau(t) = \inf\big\{s>0\colon\int_0^s\sigma(u)\,\text{d}u>t\}$ , $t\geq 0$ . In particular,

\[ \lim_{t\uparrow\infty}\lim_{N\to\infty}\mathbb{E}_{{\boldsymbol{\eta}}^N}[\|\boldsymbol{R}(t)-\boldsymbol{x}^*\|_1]=0, \quad i =1,\ldots, k. \]

In words, Theorem 1 says that by dilating time arbitrarily close to zero, its process $\boldsymbol{r}$ in the simplex behaves deterministically like the solution to an $\boldsymbol{A}$ -replicator equation. For a special choice of the matrix $C_{i,j}$ , Figure 2 shows simulations of the process $(\boldsymbol{R}(t), t\geq 0)$ that resonate with the statement of Theorem 1. As part of the proof of Lemma 1, we will see that under $\mathbb{P}_{\boldsymbol{\eta}^N}$ , the process $\sigma$ is comparable to a Kingman coalescent with some collision rate, say $c>0$ . Noting that $\int_0^{\tau(t)}\sigma(u)\,\text{d} u = t$ , which implies $\dot \tau(t) = 1/\sigma(\tau(t))$ , if, in heuristic terms, we treat $\sigma$ as a solution to (1) with $\sigma(0) = N$ , so that $\sigma(t)^{-1} = N^{-1}+ ct/2$ , then $\tau(t)\approx 2(\exp\{ct/2\}-1)/cN\approx t/N$ .

Figure 2. Simulations of a replicator coalescent with $k = 3$ initiated from a variety of initial states with an initial number of blocks $\sigma(0)=10^{15}$ . Each path represents a simulation from a different initial state, presented in barycentric coordinates in the 3-simplex and a logarithmic axis for the total number of blocks. The matrix C has entries $C_{i,i}=C_{i,i+1}=1$ and other entries zero. The reader will note that this case in particular demonstrates that we clearly do not need to enforce $C_{i,j}>0$ for all i, j.

For a Kingman coalescent, say $(\nu(t), t\geq 0)$ with merger rate c, classical reasoning tells us that $(\nu(t/N)/N, t\geq 0)$ converges in an appropriate sense to the solution to the ODE (1) as $\nu(0) = N\to\infty$ . In the spirit of these arguments, we would therefore expect that we can similarly control $(\sigma(t/N)/N, t\geq 0)$ as well as $(\boldsymbol{n}(t/N)/N, t\geq 0)$ as $N\to\infty$ . In turn, since we can write

\[ \boldsymbol{R}(t)\approx \boldsymbol{r}(t/N) = \frac{\boldsymbol{n}(t/N)/N}{\sigma(t/N)/N},\quad t\geq 0,\]

we can therefore think of Theorem 1 as a version of the aforementioned functional scaling result for Kingman’s coalescent.

The remainder of this paper is structured as follows. In the next section we discuss how we can compare the process $(\sigma(t), t\geq 0)$ with Kingman’s coalescent on the same probability space when it is issued from a finite number of blocks. This comparison is used frequently in several of our proofs. In Section 4 we treat the Markov process $\boldsymbol{n}$ as a semimartingale and study its decomposition as the sum of a martingale and a bounded variation compensator. This provides the basis for the proof of Theorem 1, given in Section 5. Finally, in Section 6 we conclude with some technical remarks and two conjectures concerning further behaviour of the entrance law.

3. Stochastic comparison with Kingman’s coalescent

As previously alluded to, there are various points in our reasoning where we will compare the number of blocks in a replicator coalescent with the number of blocks in an appropriately formulated Kingman coalescent on the same probability space. The first such result gives us the proof that the replicator coalescent comes down from infinity.

Proof of Lemma 1. The process $\sigma$ decreases by one with each block merger, analogously to the block-counting process of a standard Kingman coalescent. It is therefore sufficient to prove that $\sigma$ decreases at least as slowly, or at least as fast, as a Kingman block-counting process with comparable rates.

Formally, from finite starting states we want to construct such a Kingman coalescent $\nu^+$ on the same space as $\sigma$ with $\sigma\leq \nu^{+}$ by considering the minimal rate of $\sigma$ .

The total rate of mergers in state $\boldsymbol{n}$ is given by

\begin{equation*} \rho(\boldsymbol{n}) = \sum_{i=1}^k\Bigg(\sum_{j\neq i}C_{i,j}n_{j}n_i +C_{i,i} \left(\substack{n_i\\2}\right)\Bigg), \end{equation*}

which depends not just on the total number of blocks, but also on the distribution of block types. However, since $C_{i,j}\geq 0$ for all i, j, we can choose $C\gt \max_{i,j}C_{i,j}$ such that

(5) \begin{equation} \rho(\boldsymbol{n}) < \sum_{i=1}^k\frac{1}{2}Cn_i\Bigg(\sum_{j\neq i}n_{j} + n_i-1\Bigg) = \sum_{i=1}^k\frac{1}{2}Cn_i(\|\boldsymbol{n}\|_1-1) = {C}\left(\substack{\|{\boldsymbol{n}}\|_1\\2}\right). \end{equation}

Appealing to the skip-free property, it follows that we can stochastically couple a Kingman coalescent death chain $(\nu^{-}(t), t\geq 0)$ with collision rate C and the process $({\boldsymbol{n}}(t), t\geq 0)$ on the same space such that, with ${\boldsymbol{\eta}}=(\arg({\boldsymbol{n}}(0)),\sigma(0)^{-1})\in\mathcal{S}^k\times\mathbb{N}^{-1}$ and $\nu^-(0) = \sigma(0)$ , $\sigma(t)\geq \nu^-(t)$ , $t\geq 0$ .

Conversely, when $C_{i,j}>0$ for all i, j, writing $\underline{C} = \min_{i,j}C_{i,j}$ , we get

\[ \rho(\boldsymbol{n}) \geq \sum_{i=1}^k\frac{1}{2}\underline{C}n_i\Bigg(\sum_{j\neq i}n_{j} + n_i-1\Bigg) = \underline{C}\left(\substack{\|{\boldsymbol{n}}\|_1\\2}\right). \]

In the same spirit, it follows that we can stochastically embed another Kingman death chain $(\nu^+(t), t\geq 0)$ and the process $({\boldsymbol{n}}(t), t\geq 0)$ on the same space such that, with ${\boldsymbol{\eta}}=(\arg({\boldsymbol{n}}(0))$ , $\sigma(0)^{-1})\in\mathcal{S}^k\times\mathbb{N}^{-1}$ and $\nu^-(0) = \sigma(0)$ , $\sigma(t)\leq \nu^+(t)$ , $t\geq 0$ .

In particular, as $\nu^+$ comes down from infinity, since $\gamma_m \leq \beta^+_{m} \;:\!=\; \inf\{t>0\colon\nu^+(t)=m\}$ , it follows that, for each $\varepsilon>0$ ,

\[ \lim_{m\to\infty}\lim_{N\to\infty}\mathbb{P}_{{\boldsymbol{\eta}}^N}[\gamma_m<\varepsilon ] \geq \lim_{m\to\infty}\lim_{N\to\infty}\mathbb{P}[\beta^+_m<\varepsilon\mid\nu^+(0)=N] = 1, \]

thus completing the proof.

4. Semimartingale representation

We would like to treat the replicator coalescent $(\boldsymbol{n}(t), t\geq 0)$ as a semimartingale. It turns out to be more convenient to consider instead the vectorial process

\[ \boldsymbol{y}(t) = \begin{bmatrix} \boldsymbol{r}(t\wedge\gamma_1) \\[1mm] {1}/{\sigma(t\wedge \gamma_1)} \end{bmatrix}, \quad t\geq 0,\]

where $\gamma_1 = \inf\{t>0\colon \sigma(t) = 1\}$ . Naturally, by expressing the evolution of $(\boldsymbol{y}(t), t\geq 0)$ as that of a semimartingale, our interest is predominantly in the process $(\boldsymbol{r}(t), t\geq 0)$ in order to make a link with the replicator equations in (2).

Lemma 2. For each $\boldsymbol{\eta}\in\mathbb{N}^k_*$ , the process $\boldsymbol{y}$ under $\mathbb{P}_{\boldsymbol{\eta}}$ has a semimartingale decomposition $\boldsymbol{y}(t) = \boldsymbol{y}(0)+ \boldsymbol{m}(t) + \boldsymbol{\alpha}(t)$ , $t\geq 0$ , where $(\boldsymbol{m}(t), t\geq 0)$ is a martingale taking the form $\boldsymbol{m}(t) = \sum_{s\leq t\wedge \gamma_1}\Delta \boldsymbol{y}(s) - \boldsymbol{\alpha}(t)$ , $t\geq 0$ , such that $\Delta\boldsymbol{y}(t) =\boldsymbol{y}(t)- \boldsymbol{y}(t{-})$ and $(\boldsymbol{\alpha}(t), t\geq 0)$ is a compensator taking the form

\begin{equation*} \boldsymbol{\alpha}(t) = \int_{0}^{t\wedge\gamma_1}\frac{\sigma(s)}{\sigma(s)-1}\sum_{i = 1}^k \begin{bmatrix} \sigma(s)({\boldsymbol{r}} (s)-\boldsymbol{e}_{\boldsymbol{i}}) \\[1mm] 1 \end{bmatrix} r_i(s)[\sigma(s)^{-1}{\rm diag}(\boldsymbol{A})\textbf{1}-\boldsymbol{A}\boldsymbol{r}(s)]_i\,\text{d}s. \end{equation*}

Proof. A standard computation using the compensation formula tells us that $\boldsymbol{m}$ is a martingale provided that $\sum_{s\leq t}\|{\Delta\boldsymbol{m}(s)}\|_1=\sum_{s\leq t}\|{\Delta\boldsymbol{y}(s)}\|_1$ has finite expectation for each $t\geq 0$ , which is equivalent to the existence of the compensator $\boldsymbol{\alpha}(t)$ . The latter is given by the rates that define the replicator coalescent. More precisely, recalling (4),

\begin{align*} \boldsymbol{\alpha}(t) & = \int_{0}^{t\wedge\gamma_1}\sum_{i=1}^k \begin{bmatrix} \dfrac{\boldsymbol{n}(s)-\boldsymbol{e}_{\boldsymbol{i}}}{\sigma(s)-1}-\dfrac{\boldsymbol{n}(s)}{\sigma(s)} \\[3mm] \dfrac{1}{\sigma(s)-1}-\dfrac{1}{\sigma(s)} \end{bmatrix} \Bigg[\sum_{j=1,\,j\neq i}^kn_j(s)n_i(s)C_{ji}+\frac{1}{2}n_i(s)(n_i(s)-1)C_{i,i}\Bigg]\,\text{d}s \\ & = \int_{0}^{t\wedge\gamma_1}\sum_{i=1}^k \begin{bmatrix} \dfrac{\boldsymbol{n}(s)-\boldsymbol{e}_{\boldsymbol{i}}\sigma(s)}{(\sigma(s)-1)\sigma(s)} \\[3mm] \dfrac{1}{(\sigma(s)-1)\sigma(s)} \end{bmatrix} \Bigg[n_i(s)A_{i,i}-\sum_{j=1}^kn_j(s)n_i(s)A_{i,j}\Bigg]\,\text{d}s \\ & = \int_{0}^{t\wedge\gamma_1}\dfrac{\sigma(s)}{\sigma(s)-1}\sum_{i = 1}^k \begin{bmatrix} \sigma(s)(\boldsymbol{r} (s)-\boldsymbol{e}_{\boldsymbol{i}}) \\[1mm] 1 \end{bmatrix} r_i(s)[\sigma(s)^{-1}{\rm diag}(\boldsymbol{A})\textbf{1}-\boldsymbol{A}\boldsymbol{r}(s)]_i\,\text{d}s, \end{align*}

as required. Note that if we identify the compensator via the density $(\boldsymbol\lambda(t), t\geq 0)$ , where

(6) \begin{equation} \boldsymbol{\alpha}(t) =\!: \int_0^{t\wedge\gamma_1}\boldsymbol\lambda(s)\,\text{d}s \end{equation}

then, in the above representation of $\boldsymbol{\alpha}$ , the largest term is of order $\sigma(s)$ , from which, because the process $\boldsymbol{n}$ is non-increasing, we can easily conclude that, for all $\boldsymbol{\eta}\in\mathbb{N}^k_*$ and any time $t\geq 0$ ,

(7) \begin{equation} \mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\sum_{s\leq t\wedge\gamma_1}\|{\Delta\boldsymbol{y}(s)}\|_1\Bigg] \leq \mathbb{E}_{\boldsymbol{\eta}}\bigg[\int_0^{t\wedge\gamma_1}\|{\boldsymbol\lambda(s)}\|_1\,\text{d}s\bigg] \leq C\|{\boldsymbol{\eta}}\|_1\mathbb{E}_{\boldsymbol{\eta}^N}[t\wedge\gamma_1]\leq C\|{\boldsymbol{\eta}}\|_2t \end{equation}

for an unimportant constant $C>0$ . This ensures that $\boldsymbol{m}$ is a martingale and that $\boldsymbol{\alpha}$ is well defined.

For the proof of Theorem 1, we are interested in the behaviour of the process $\boldsymbol{r}$ under $\mathbb{P}_{\boldsymbol{\eta}}$ for any $\boldsymbol{\eta}$ such that $\|{\boldsymbol{\eta}}\|_1\to\infty$ and $\arg(\boldsymbol{\eta})\to\boldsymbol{r}$ for some $\boldsymbol{r}\in\mathcal{S}^k_+$ . Heuristically speaking, the term $\sigma(s)({\boldsymbol{r}} (s)-\boldsymbol{e}_{\boldsymbol{i}})$ in the expression for $\boldsymbol{\alpha}$ suggests that $\alpha(t)$ explodes as $t\to 0$ . The undesirable factor $\sigma(s)$ can be removed, however, by an appropriate time change; in doing so, we begin to see where the relationship with the replicator equations emerges.

Lemma 3. Suppose we define the sequence of stopping times $(\tau(t),t\geq 0)$ , which are defined by the right inverse, $\tau(t) = \inf\{u>0\colon\int_0^u\sigma(s)\,\text{d}s>t\}$ , $t\geq 0$ . Then $\boldsymbol{y}^\tau \;:\!=\; \boldsymbol{y}\circ\tau$ has the semimartingale decomposition $\boldsymbol{y}^\tau = \boldsymbol{m}^\tau + \boldsymbol{\alpha}^\tau$ , where $\boldsymbol{m}^\tau \;:\!=\; \boldsymbol{m} \circ \tau$ is a martingale and, for $t\geq 0$ ,

(8) \begin{align} \boldsymbol{\alpha}^\tau(t) & = \int_{0}^{t\wedge\tau^{-1}(\gamma_1)}\dfrac{\sigma(\tau(s))}{\sigma(\tau(s))-1} \notag \\ &\quad \times \sum_{i=1}^k \begin{bmatrix} \boldsymbol{r}(\tau(s))-\boldsymbol{e}_{\boldsymbol{i}} \\[1mm] \dfrac{1}{\sigma(\tau(s))} \end{bmatrix} r_i(\tau(s))[\sigma(\tau(s))^{-1}{\rm diag}(\boldsymbol{A})\textbf{1} - \boldsymbol{A}\boldsymbol{r}(\tau(s))]_i\,\text{d}s. \end{align}

Proof. We use basic Stieltjes calculus to tell us that $\text{d}\boldsymbol{\alpha}^\tau(t) = \text{d}\boldsymbol{\alpha}(s)|_{s=\tau(t)}\,\text{d}\tau(t)$ . Moreover,

(9) \begin{equation} \int_0^{\tau(t)}\sigma(s)\,\text{d}s = t, \quad \text{and hence } \sigma(\tau(t))\,\text{d}\tau(t) = \text{d}t. \end{equation}

Combining these observations with the conclusion of Lemma 2, the result follows. In particular, from (6),

(10) \begin{equation} \int_0^{\tau(t)\wedge\gamma_1}\boldsymbol\lambda(s)\,\text{d}s = \int_0^{t\wedge\tau^{-1}(\gamma_1)}\frac{\boldsymbol\lambda(\tau(u))}{\sigma(\tau(u))}\,\text{d}u. \end{equation}

Technically, we need to verify that $\boldsymbol{m}^\tau$ is a martingale rather than a local martingale; however, a computation similar to (7) taking advantage of (8) is easily executed, affirming the required martingale status.

Next we look at how to control the second moment of the martingale $\boldsymbol{m}^\tau$ in the semimartingale decomposition of $\boldsymbol{y}^\tau$ .

Lemma 4. For each $\boldsymbol{\eta}\in\mathbb{N}^k_*$ , the martingale $\boldsymbol{m}^\tau$ under $\mathbb{P}_{\boldsymbol{\eta}}$ satisfies

\[ \mathbb{E}_{\boldsymbol{\eta}^N}\big[\|{\boldsymbol{m}^\tau(t)}\|^2_2\big] \leq C\mathbb{E}_{\boldsymbol{\eta}}\bigg[\int_0^{\tau(t)\wedge\gamma_1}\frac{\sigma(s)^2}{(\sigma(s)-1)^2}\, \text{d}s\bigg], \quad t\geq 0. \]

Proof. Steiltjes calculus, or equivalently general semi-martingale calculus (see, for example, [Reference Protter8, Theorem II.33]), tells us that, since $\boldsymbol{m}$ has bounded variation,

\begin{align*} \|{\boldsymbol{m}(t)}\|^2_2 & = 2\int_0^{t\wedge\gamma_1}\boldsymbol{m}(s{-})\cdot\text{d}\boldsymbol{m}(s) + \sum_{0<s\leq t\wedge\gamma_1}\big\{\Delta\|{\boldsymbol{m}(s)}\|_2^2 - 2\boldsymbol{m}(s{-})\cdot\Delta\boldsymbol{m}(s)\big\} \\ & = 2\int_0^{t\wedge\gamma_1}\boldsymbol{m}(s{-})\cdot\text{d}\boldsymbol{m}(s) + \sum_{0<s\leq t\wedge\gamma_1}(\Delta\boldsymbol{m}(s))^2. \end{align*}

As all vector entries are bounded, it is easy to show that $\int_0^{t\wedge\gamma_1}\boldsymbol{m}(s{-})\cdot\text{d}\boldsymbol{m}(s)$ , $t\geq 0$ , is a martingale.

Next, we identify the adapted increasing bounded variation process, say $\boldsymbol\beta$ , that is the compensator of $\sum_{0<s\leq t\wedge\gamma_1}(\Delta\boldsymbol{m}(s))^2$ , $t\geq 0$ , so that $\|{\boldsymbol{m}(t)}\|_2^2 - \boldsymbol\beta(t)$ , $t\geq 0$ , is a martingale with mean 0. To this end, note that $\Delta\boldsymbol{m}(t) = \Delta\boldsymbol{y}(t)$ . Hence, we have, on the event that t is a time at which the number of blocks of type i decreases, that $(\Delta\boldsymbol{m}(t))^2$ is given by

\begin{equation*} \chi_i(t) \;:\!=\; \begin{bmatrix} \dfrac{\boldsymbol{n}(t)-\sigma(t)\boldsymbol{e}_{\boldsymbol{i}}}{(\sigma(t)-1)\sigma(t)} \\[3mm] \dfrac{1}{(\sigma(t)-1)\sigma(t)} \end{bmatrix} \cdot \begin{bmatrix} \dfrac{\boldsymbol{n}(t)-\sigma(t)\boldsymbol{e}_{\boldsymbol{i}}}{(\sigma(t)-1)\sigma(t)} \\[3mm] \dfrac{1}{(\sigma(t)-1)\sigma(t)} \end{bmatrix} = \frac{\boldsymbol{n}(t)\cdot\boldsymbol{n}(t)-2\sigma(t)n_i(t)+\sigma(t)^2+1}{\sigma(t)^2(\sigma(t)-1)^2}. \end{equation*}

It is now straightforward to see that there exists a $C>0$ such that

\begin{align*} \boldsymbol{\beta}(t) & = \int_0^{t\wedge\gamma_1}\sum_{i=1}^k\chi_i(s)\Bigg[\sum_{{j=1,\,j\neq i}}^kn_j(s)n_i(s)C_{ji} + \frac{1}{2}n_i(s)(n_i(s)-1)C_{i,i}\Bigg]\,\text{d}s \\ & \leq C\int_0^{t\wedge\gamma_1}\frac{\sigma(s)^2}{(\sigma(s)-1)^2}\,\text{d}s, \end{align*}

where we have used that $\sigma(t)^2 = (n_1(t)+\cdots+n_k(t))^2 \geq \boldsymbol{n}(t)\cdot\boldsymbol{n}(t)$ . Replacing t by $\tau(t)$ and taking expectation gives the desired inequality.

As the reader may now expect, our ultimate objective is to show that for any sequence of starting initial configurations $\boldsymbol{\eta}^N\in \mathbb{N}^k_*$ such that $\|{\boldsymbol{\eta}^N}\|_1\to\infty$ as $N\to\infty$ and $\arg(\boldsymbol{\eta}^N)\to \boldsymbol{r}\in \mathcal{S}^k_+$ , the martingale component $\boldsymbol{m}^\tau$ disappears. This tells us that the behaviour of $(\boldsymbol{r}(t), t\geq 0)$ behaves increasingly like the compensator term, which is a further key to controlling its behaviour. To this end, we conclude this section with two more results that provide us with the desired control of the aforesaid martingale component.

Proof. Recall from the proof of Lemma 1, and specifically (5), that there is a death chain $(\nu(t), t\geq 0)$ representing the number of blocks in a Kingman coalescent with merger rate C such that, for any $\boldsymbol{\eta}\in\mathbb{N}^k_*$ , on the same probability space, we can stochastically bound $\sigma(t)\geq \nu(t)$ , $t\geq 0$ .

We now note that for any large $M>0$ there exists a constant $C>0$ (not necessarily the same as before) such that, for any m sufficiently large,

(11) \begin{align} \lim_{N\to\infty}\mathbb{P}_{{\boldsymbol{\eta}}^N}\bigg(\int_0^{\gamma_m} \sigma(s)\,\text{d}s > M\bigg) & \geq \lim_{N\to\infty}\mathbb{P}_{{\boldsymbol{\eta}}^N}\bigg(\int_0^{\beta_m}\nu(s)\,\text{d}s > M\bigg)\notag \\ & = \Pr\Bigg(\sum_{n=m+1}^\infty\frac{n}{C\left(\substack{n\\2}\right)}\textbf{e}^{(n)}_{1} > M\Bigg) \notag \\ & = \Pr\Bigg(\sum_{n=m+1}^\infty\frac{1}{n-1}\textbf{e}^{(n)}_{1} > \frac{CM}{2}\Bigg), \end{align}

where $\beta_m = \inf\{t>0\colon \nu(t) = m\}$ and $(\textbf{e}^{n}_1, n\geq 1)$ is a sequence of independent and identically distributed unit-mean exponentially distributed random variables. If we write $(N(t), t\geq 0)$ for a Poisson process with unit rate, then almost surely we have

\[ \sum_{n=m+1}^\infty\frac{1}{n-1}\textbf{e}^{(n)}_{1} = \int_0^\infty\frac{1}{N(s)+m}\,\text{d}s = \int_0^\infty\frac{s}{N(s)+m}\,\frac{\text{d}s}{s} = \infty, \]

where the final equality follows by the strong law of large numbers for Poisson processes. As such, the right-hand side of (11) is equal to 1.

Since M and m can be arbitrarily large, this shows the first claim as soon as we note that $\tau^{-1}(t) = \int_0^t\sigma(u)\,\text{d}u$ , which is an easy consequence of (9). On the other hand, note that since $\int_0^{\tau(t)}\sigma(s)\,\text{d}s = t$ when $\sigma(0)={\boldsymbol{\eta}}^N \rightarrow\infty$ , the above comparison with Kingman’s coalescent shows almost surely that, since $\int_0^u\sigma(s)\,\text{d}s$ converges weakly to infinity for all $u>0$ , then $\tau(t)$ converges weakly to 0. Indeed, if with positive probability $\tau(t)>\varepsilon$ in the limit as $N\to\infty$ , then, on that event, $\int_0^{\tau(t)}\sigma(s)\,\text{d}s \geq \int_0^{\varepsilon}\sigma(s)\,\text{d}s$ , which explodes in distribution. This in turn contradicts the definition of $\tau(t)$ . This proves the second and third statements of the lemma.

Since the second moment of the martingale $\boldsymbol{m}^\tau$ can be controlled by its associated time change, we also get a helpful $L^2$ corollary from Lemma 5.

Proof. From Lemma 4 and a change of variable similar to (10),

\begin{align*} \mathbb{E}_{\boldsymbol{\eta}^N}\big[\|{\boldsymbol{m}^\tau(t)}\|^2_2\big] & \leq C\mathbb{E}_{\boldsymbol{\eta}^N}\bigg[\int_0^{\tau(t)\wedge\gamma_1}\frac{\sigma(s)^2}{(\sigma(s)-1)^2}\, \text{d}s\bigg] \\ & = C\mathbb{E}_{\boldsymbol{\eta}^N}\bigg[\int_0^{t\wedge\tau^{-1}(\gamma_1)} \frac{\sigma(\tau(u))^2}{(\sigma(\tau(u))-1)^2}\frac{1}{\sigma(\tau(u))}\,\text{d}u\bigg] \leq Ct\mathbb{E}_{\boldsymbol{\eta}^N}\bigg[\frac{1}{\sigma(\tau(t))}\bigg]. \end{align*}

We can choose N sufficiently large that $\tau(t)$ is less than $\delta$ with probability at least $1-\varepsilon$ , so

\begin{equation*} \mathbb{E}_{\boldsymbol{\eta}^N}\bigg[\frac{1}{\sigma(\tau(t))}\bigg] \leq \mathbb{E}_{\boldsymbol{\eta}^N}\bigg[\frac{1}{\sigma(\delta)};\;\tau(t)<\delta\bigg] + \mathbb{P}_{\boldsymbol{\eta}^N}(\tau(t)\geq\delta) \leq \delta\mathbb{E}_{\boldsymbol{\eta}^N}\bigg[\frac{1}{\delta\nu(\delta)}\bigg] + \varepsilon, \end{equation*}

where we have again compared with a lower bounding Kingman coalescent $(\nu(t), t\geq 0)$ on the same space, as in Lemma 5. Recall the classical result for Kingman’s coalescent coming down from infinity that, when the collision rate is $C>0$ , $\delta\nu(\delta)\to 2/C$ almost surely as $\delta\to 0$ [Reference Berestycki2]. We can now easily conclude with the help of dominated convergence that $\lim_{N\to\infty}\mathbb{E}_{\boldsymbol{\eta}^N}\big[\|{\boldsymbol{m}^\tau(t)}\|^2_2\big] = 0$ , and this concludes the proof once we invoke Doob’s submartingale inequality.

5. Proof of Theorem 1

Recall that we have required from the $\boldsymbol{A}$ -replicator equation that $\boldsymbol{x}(t)\to \boldsymbol{x}^*$ holds. Reinterpreting (2) in its integral form, this tells us that

(12) \begin{equation} x_i(t) = x_i(0) +\int_0^t x_i(s)([{\boldsymbol{A}}{\boldsymbol{x}}(s)]_i - {\boldsymbol{x}}(s)^\top{\boldsymbol{A}}{\boldsymbol{x}}(s))\,\text{d}s, \quad t\geq 0.\end{equation}

This representation makes it easier to give the heuristic basis of the proof of Theorem 1.

Following our earlier heuristic reasoning, we can now see that, under $\mathbb{P}_{\boldsymbol{\eta}}$ as $\|{\boldsymbol{\eta}}\|_1\to\infty$ the integrand in the expression for $\boldsymbol{\alpha}^\tau$ appears to have a similar structure to the replicator equations (2). That is, under $\mathbb{P}_{\boldsymbol{\eta}}$ as $\|{\boldsymbol{\eta}}\|_1\to\infty$ and as $t\to 0$ ,

\[ \frac{\text{d}\boldsymbol{\alpha}^\tau(t)}{\text{d}t} \approx \begin{bmatrix} \boldsymbol\theta(t) \\[1mm] 0 \end{bmatrix},\]

where $\boldsymbol\theta_i(t) = {r}_i(\tau(t))\big([\boldsymbol{A}\boldsymbol{r}(\tau(t))]_i -\boldsymbol{r}(\tau(t))^\top\boldsymbol{A}\boldsymbol{r}(\tau(t))\big)$ . On the other hand, if we can show that $\boldsymbol{m}^\tau\to0$ as $\|{\boldsymbol{\eta}}\|_1\to\infty$ , then, since $\boldsymbol{y}^\tau = \boldsymbol{m}^\tau + \boldsymbol{\alpha}^\tau$ , reading off the first component of $\boldsymbol{y}^\tau$ , i.e. $\boldsymbol{R}(t) \;:\!=\; \boldsymbol{r}(\tau(t))$ , roughly speaking we see that

\[ R_i(t) \approx r_i(0) + \int_0^t R_i(s)\big([\boldsymbol{A}\boldsymbol{R}(s)]_i - \boldsymbol{R}(s)^\top\boldsymbol{A}\boldsymbol{R}(s)\big)\,\text{d}s\]

as $\|{\boldsymbol{\eta}}\|_1\to\infty$ , given that Corollary 1 shows the martingale component is negligible. In other words, the process $(\boldsymbol{R} (t), t\geq 0)$ , begins to resemble the replicator equation in its integral form (12). It now looks like a reasonable conjecture that $\boldsymbol{R}(t)\to \boldsymbol{x}^*$ , just as the solution to the replicator equation does.

Let us thus move to the proof of Theorem 1, which, as alluded to earlier, boils down to the control we have on the martingale $\boldsymbol{m}^\tau$ under $\mathbb{P}_{\boldsymbol{\eta}}$ as $\|{\boldsymbol{\eta}}\|_1\to\infty$ , thanks to Corollary 1.

Proof of Theorem 1. Write $\boldsymbol{R}_i(t) = \boldsymbol{r}(\tau(t))$ on the event $A_t \;:\!=\; \{t<\tau^{-1}(\gamma_1)\}$ , $t\geq 0$ , and note from Lemma 5 and the proof of Lemma 1 that $\lim_{N\to\infty}\mathbb{P}_{\boldsymbol{\eta}^N}(A_t)=1$ . We have, for each $T>0$ ,

\begin{align*} & \mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\sup\limits_{t\leq T}\Bigg\|{\boldsymbol{R}}(t) - {\boldsymbol{r}}(0) - \int_0^t\Bigg(\sum_{i=1}^k{\boldsymbol{e}}_iR_i(s)[\boldsymbol{A}\boldsymbol{R}(s)]_i - (\boldsymbol{R}(s)^\top\boldsymbol{A}\boldsymbol{R}(s))\boldsymbol{R}(s)\Bigg)\,\text{d}s \Bigg\|_11_{A_t}\Bigg] \\ & \leq \mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\sup\limits_{t\leq T}\Bigg\|{\boldsymbol{R}}(t) - {\boldsymbol{r}}(0) \\ & \qquad\qquad\quad - \int_0^t\frac{\sigma(\tau(s))}{\sigma(\tau(s))-1} \sum_{i = 1}^k(\boldsymbol{R}(s)-\boldsymbol{e}_{\boldsymbol{i}})R_i(s) [\sigma(\tau(s))^{-1}{\rm diag}(\boldsymbol{A})\textbf{1}-\boldsymbol{A}\boldsymbol{R}(s)]_i\, \text{d}s\Bigg\|_11_{A_t}\Bigg] \\ & \quad + \mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\sup\limits_{t\leq T}\Bigg\| \int_0^t\frac{1}{\sigma(\tau(s))-1}\Bigg(\sum_{i=1}^k{\boldsymbol{e}}_iR_i(s) [\boldsymbol{A}\boldsymbol{R}(s)]_i - (\boldsymbol{R}(s)^\top\boldsymbol{A}\boldsymbol{R}(s)) \boldsymbol{R}(s)\Bigg)\,\text{d}s\Bigg\|_11_{A_t}\Bigg] \\ & \quad + \mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\sup\limits_{t\leq T}\Bigg\|\int_0^t\!\!\!\frac{1}{\sigma(\tau(s))-1} \Bigg(({\boldsymbol{R}}(s)^\top{\rm diag}(\boldsymbol{A})\boldsymbol{1})\boldsymbol{R}(s) - \sum_{i=1}^k{\boldsymbol{e}}_iR_i(s)[{\rm diag}(\boldsymbol{A})\boldsymbol{1}]_i\Bigg)\, \text{d}s\Bigg\|_11_{A_t}\Bigg]. \end{align*}

From Corollary 1, and the fact that $\|{\boldsymbol{y}}\|_1\leq\sqrt{k}\|{\boldsymbol{y}}\|_2$ for $\boldsymbol{y}\in \mathbb{R}^k$ , the first term after the inequality tends to zero as $N\to\infty$ . Up to a multiplicative constant, the second and third terms after the inequality can be bounded by

\[ T\mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\frac{1}{(\sigma(\tau(T)) -1)}\wedge1\Bigg], \]

where we have used the monotonicity of $\tau(\cdot)$ and $\sigma(\cdot)$ . As noted in the proof of Corollary 1, the latter tends to zero as $N\to\infty$ .

It now follows that

\begin{equation*} \lim_{N\to\infty}\mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\sup_{t\leq T}\Bigg\| {\boldsymbol{R}}(t) - {\boldsymbol{r}}(0) - \int_0^t\sum_{i=1}^k {\boldsymbol{e}}_iR_i(s) [\boldsymbol{A}\boldsymbol{R}(s)]_i - (\boldsymbol{R}(s)^\top\boldsymbol{A}\boldsymbol{R}(s)) \boldsymbol{R}(s)\,\text{d}s\Bigg\|_11_{A_t}\Bigg] = 0. \end{equation*}

As all of the vectorial and matrix terms in this equation are bounded, it is also easy to see, with the help of Lemma 5, that

\begin{align*} & \lim_{N\to\infty}\mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\sup_{t\leq T}\Bigg\| {\boldsymbol{R}}(t) - {\boldsymbol{r}}(0) - \int_0^t\sum_{i=1}^k{\boldsymbol{e}}_iR_i(s) [\boldsymbol{A}\boldsymbol{R}(s)]_i - (\boldsymbol{R}(s)^\top\boldsymbol{A}\boldsymbol{R}(s)) \boldsymbol{R}(s)\,\text{d}s\Bigg\|_11_{A_t^c}\Bigg] \\ & \leq \lim_{N\to\infty}CT\mathbb{P}_{\boldsymbol{\eta}^N}(A_T^{\text{c}}) = 0 \end{align*}

for some constant $C>0$ , which gives us

(13) \begin{equation} \lim_{N\to\infty}\mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\sup\limits_{t\leq T}\Bigg\| {\boldsymbol{R}}(t) - {\boldsymbol{r}}(0) - \int_0^t\sum_{i=1}^k{\boldsymbol{e}}_iR_i(s) [\boldsymbol{A}\boldsymbol{R}(s)]_i - (\boldsymbol{R}(s)^\top\boldsymbol{A}\boldsymbol{R}(s)) \boldsymbol{R}(s)\,\text{d}s\Bigg\|_1\Bigg] = 0. \end{equation}

Consider the replicator equation initiated from any $\boldsymbol{r}(0)\in \mathcal{S}^k_+$ . Similarly to (12), albeit in vectorial form, we can write the solution to (2) when issued from $\boldsymbol{r}(0)$ as

\begin{equation*} \boldsymbol{x}(t) - \boldsymbol{r}(0) - \int_0^t\sum_{i=1}^k{\boldsymbol{e}}_i x_i(s) [{\boldsymbol{A}}{\boldsymbol{x}}(s)]_i - ({\boldsymbol{x}}(s)^\top{\boldsymbol{A}}{\boldsymbol{x}}(s)) \boldsymbol{x}(s)\,\text{d}s = 0. \end{equation*}

Subtracting this from (13), we see that, for each $T>0$ ,

\begin{align*} & \mathbb{E}_{\boldsymbol{\eta}^N}\big[\sup_{t\leq T}\|{\boldsymbol{R}(t)-\boldsymbol{x}(t)}\|_1\big] \\ & \qquad \leq \mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\sup_{t\leq T}\Bigg\|{\boldsymbol{R}}(t) - {\boldsymbol{r}}(0) - \int_0^t\sum_{i=1}^k\boldsymbol{e}_iR_i(s)([\boldsymbol{A}\boldsymbol{R}(s)]_i - \boldsymbol{R}(s)^\top\boldsymbol{A}\boldsymbol{R}(s))\,\text{d}s\Bigg\|_1\Bigg] \\ & \qquad\quad + \int_0^T\sum_{i=1}^k\mathbb{E}_{\boldsymbol{\eta}^N}\big[|R_i(s)-x_i(s)|\, |[\boldsymbol{A}\boldsymbol{R}(s)]_i - \boldsymbol{R}(s)^\top\boldsymbol{A}\boldsymbol{R}(s)|\big]\, \text{d}s \\ & \qquad\quad + \int_0^T\sum_{i=1}^k x_i(s)\mathbb{E}_{\boldsymbol{\eta}^N}\big[|[\boldsymbol{A}(\boldsymbol{R}(s) - \boldsymbol{x}(s))]_i - \boldsymbol{R}(s)^\top\boldsymbol{A}(\boldsymbol{R}(s)-\boldsymbol{x}(s))|\big]\, \text{d}s \\ & \qquad\quad + \int_0^T\sum_{i=1}^k x_i(s)\mathbb{E}_{\boldsymbol{\eta}^N}[|(\boldsymbol{R}(s) - \boldsymbol{x}(s))^\top\boldsymbol{A}\boldsymbol{x}(s)|]\,\text{d}s. \end{align*}

Noting that $0\leq R_i(s), x_i(s) \leq 1$ for all $i =1,\ldots, k$ , $s\geq 0$ , we have

\begin{align*} & \mathbb{E}_{\boldsymbol{\eta}^N}[\sup_{t\leq T}\|{\boldsymbol{R}(t) - \boldsymbol{x}(t)}\|_1] \\ & \qquad \leq \mathbb{E}_{\boldsymbol{\eta}^N}\Bigg[\sup_{t\leq T}\Bigg\|{\boldsymbol{R}}(t) - {\boldsymbol{r}}(0) - \int_0^t\sum_{i=1}^k\boldsymbol{e}_iR_i(s)\big([\boldsymbol{A}\boldsymbol{R}(s)]_i - \boldsymbol{R}(s)^\top\boldsymbol{A}\boldsymbol{R}(s)\big)\,\text{d}s\Bigg\|_1\Bigg] \\ & \qquad\quad + C\int_0^T\mathbb{E}_{\boldsymbol{\eta}^N}[\sup_{u\leq s}\|{\boldsymbol{R}(u) - \boldsymbol{x}(u)}\|_1]\,\text{d}s, \end{align*}

where $C>0$ is an unimportant constant. Hence, using (13), the monotonicity of norms, and dominated convergence,

\begin{align*} u(T) & \;:\!=\; \lim_{N\to\infty}\sup_{N'\geq N}\mathbb{E}_{\boldsymbol{\eta}^N} \big[\sup_{t\leq T}\|{\boldsymbol{R}(t) - \boldsymbol{x}(t)}\|_1\big] \\ & \leq C\int_0^T\lim_{N\to\infty}\sup_{N'\geq N}\mathbb{E}_{\boldsymbol{\eta}^N} \big[\sup_{u\leq s}\|{\boldsymbol{R}(u)-\boldsymbol{x}(u)}\|_1\big]\,\text{d}s = C\int_0^T u(s)\,\text{d}s, \end{align*}

where $C>0$ is an unimportant constant. Grönwall’s lemma now tells us that

\begin{equation*} \lim_{N\to\infty}\mathbb{E}_{\boldsymbol{\eta}^N}\big[\sup_{t\leq T}\|{\boldsymbol{R}(t)-\boldsymbol{x}(t)}\|_1\big] = 0. \end{equation*}

By taking $t\to\infty$ , we easily deduce that

\[ \lim_{t\to\infty}\lim_{N\to\infty}\mathbb{E}_{\boldsymbol{\eta}^N}[\|{\boldsymbol{R}(t)-\boldsymbol{x}^*}\|_1] \leq \lim_{t\to\infty}\lim_{N\to\infty}\mathbb{E}_{\boldsymbol{\eta}^N}[\|{\boldsymbol{R}(t)-\boldsymbol{x}(t)}\|_1] + \lim_{t\to\infty}\|{\boldsymbol{x}(t)-\boldsymbol{x}^*}\|_1 = 0. \]

This completes the proof of the theorem.

6. Concluding remarks

For the purpose of the following discussion, we can assume that $C_{i,j}>0$ for all $i,j\in\{1,\ldots,k\}$ . In further work, it is possible pursue the issue of Skorokhod continuity with respect to any ‘entrance laws at infinity’ that the process can come down from.

Suppose $\mathbb{D}$ is the space of càdlàg paths from $[0,\infty)$ to $\mathbb{N}^k_*$ , with $\mathcal{D}$ as the Borel sigma algebra on $\mathbb{D}$ generated from the usual Skorokhod metric. A Markovian definition of coming down from infinity would require the existence of a law $\mathbb{P}^\infty$ on $(\mathbb{D}, \mathcal{D})$ that is consistent with $\mathbb{P}$ in the sense that

\[ \mathbb{P}^\infty(\boldsymbol{n}(t+s) = \boldsymbol{n}) = \sum_{\substack{\boldsymbol{n}'\in\mathbb{N}^k_*\\\|{\boldsymbol{n}'}\|_1\geq\|{\boldsymbol{n}}\|_1}} \mathbb{P}^\infty(\boldsymbol{n}(t) = \boldsymbol{n}') \mathbb{P}_{\boldsymbol{n}'}(\boldsymbol{n}(s) = \boldsymbol{n}), \quad s, t>0,\,\boldsymbol{n}\in \mathbb{N}^k_*,\]

with $\mathbb{P}^\infty(\sigma(t)<\infty)=1$ for all $t>0$ , and $\mathbb{P}^\infty(\lim_{t\downarrow0}\sigma(t) = \infty)=1$ .

As there is no single point on the boundary of $\mathbb{N}^k_*$ that represents an appropriate ‘infinity’ to come down from, one of the associated issues is whether a unique entrance law exists or whether, e.g., there is an entrance law of $( \boldsymbol{r}, \sigma^{-1})$ for each ‘infinity’ of the form $(\boldsymbol{r}_0, 0)$ , where $\boldsymbol{r}_0\in\mathcal{S}^k_+$ . We believe the latter to hold.

There is also the question of how we can see these different entrance laws in terms of the behaviour of the process at arbitrarily small times. The following conjecture suggests that looking backwards in time, it will be difficult to differentiate between the different entrance laws proposed in Conjecture 1.

Conjecture 2. Suppose $(\boldsymbol{\eta}^N, N\geq 1)$ tends to $(\boldsymbol{r}_0, 0)$ for some $\boldsymbol{r}_0\in\mathcal{S}^k_+$ . Then

\[ \lim_{m\to\infty}\lim_{N\to\infty}\mathbb{E}_{{\boldsymbol{\eta}}^N}[\|{\boldsymbol{r}(\gamma_m)-\boldsymbol{x}^*}\|_1]=0, \]

where we recall that $\gamma_m=\inf\{t>0\colon \sigma(t) \leq m\}$ for $m\geq1$ .

In contrast to Theorem 1, Conjecture 2 claims that, moving backwards through time towards the instantaneous event at which the replicator coalescent comes down from infinity at the origin of time, the process $\boldsymbol{r}$ necessarily approaches $\boldsymbol{x}^*$ . As such, working backwards in time, the replicator coalescent never gets to see the ‘initial state’ $(\boldsymbol{r}_0,0)$ from which its entrance law is constructed.

Put together, Theorem 1 and Conjecture 2 are really claiming that $\boldsymbol{x}^*$ is a ‘bottleneck’ for the replicator coalescent. Figure 1 simulates an example where $k=3$ , in which the bottleneck phenomenon can clearly be seen.

Theorem 1 also shows that under $\mathbb{P}_{\boldsymbol{\eta}^N}$ as $\boldsymbol{\eta}^N\to(\boldsymbol{r}_0,0)$ , in an arbitrarily small amount of time (on the natural time scale of the original Markov process), the process $\boldsymbol{r}$ will effectively jump from $\boldsymbol{r}_0$ to $\boldsymbol{x}^*$ . Taking Conjecture 2 into account, it is for this reason we pose our final conjecture.