An Evolutionary Model that Satisfies Detailed Balance

Abstract

We propose a class of evolution models that involves an arbitrary exchangeable process as the breeding process and different selection schemes. In those models, a new genome is born according to the breeding process, and after that a genome is removed according to the selection scheme that involves fitness. Thus, the population size remains constant. The process evolves according to a Markov chain, and, unlike in many other existing models, the stationary distribution – so called mutation-selection equilibrium – can easily found and studied. As a special case our model contains a (sub) class of Moran models. The behaviour of the stationary distribution when the population size increases is our main object of interest. Several phase-transition theorems are proved.

Appendix

Proof of claim 4.1

First note that

$$\sup_{q\in \mathcal{P}}|\langle w_{n},q\rangle-\langle w,q \rangle|\leq \|w_{n}-w\|\|q\|\leq \|w_{n}-w\| \to 0.$$

Now use the fact that if f_n, f, g_n, g are nonnegative functions such that $\sup _{x} |f_{n}(x)-f(x)|\to 0$, $\sup _{x} |g_{n}(x)-g(x)|\to 0$, $\inf _{x} g(x)=g_{*}>0$ and f, g are bounded above, then with $g^{*}=\sup _{x} g(x)$ and $f^{*}=\sup _{x} f(x)$

$$ \begin{array}{@{}rcl@{}} \sup_{x} \big|{f_{n}(x)\over f(x)}-{g_{n}(x)\over g(x)}\big|&=&\sup_{x} \big|{f_{n}(x)g(x)-g_{n}(x)f(x)\over g_{n}(x)g(x)}\big|\\ &\leq& \sup_{x} \big|{(f_{n}(x)-f(x))g(x)\over g_{n}(x)g(x)}\big|+\sup_{x} \big|{(g_{n}(x)-g(x))f(x)\over g_{n}(x)g(x)}\big|\\ &\leq& \sup_{x} \big|{(f_{n}(x)-f(x))g^{*}\over g_{n}(x)g(x)}\big|+\sup_{x} \big|{(g_{n}(x)-g(x))f^{*}\over g_{n}(x)g(x)}\big|\to 0, \end{array} $$

because for n big enough $g_{n}(x)g(x)\geq {g^{2}_{*}\over 2}$ for every x. Take x = q, f_n(q) = w_n(k)q(k), f(q) = w(k)q(k), g_n(q) = 〈w_n, q〉 and g(q) = 〈w, q〉. Then g_∗ = w(K) > 0, f^∗ = g^∗ = w(1) and so Eq. 4.8 follows. □

Proof of Proposition 5.1

Recall that f_n and f are continuous and bounded functions on $\mathcal {P}$ so that $\|f_{n}\|_{\infty }<\infty$ and $\|f\|_{\infty }<\infty$. By assumption, π is a finite measure. Since f_n converges to f uniformly, it follows that $\|f_{n}-f\|_{\infty }\to 0$ and so $\| f_{n}\|_{\infty }\to \| f\|_{\infty }.$ For every m,

$$|\|f_{n}\|_{m}-\|f\|_{m} |\leq \|f_{n}-f\|_{m}\leq \pi(\mathcal{P})^{1\over m}\|f_{n}-f\|_{\infty}\to 0.$$

Since $\|f\|_{m_{n}}\to \|f\|_{\infty }$, we have

$$ \begin{array}{@{}rcl@{}} \big|\|f_{n}\|_{m_{n}}-\|f\|_{\infty}\big|&\leq& \big|\|f_{n}\|_{m_{n}}-\|f\|_{m_{n}}\big|+\big|\|f\|_{m_{n}}-|f\|_{\infty}\big|\\ &\leq& \|f_{n}-f\|_{m_{n}}+\big|\|f\|_{m_{n}}-|f\|_{\infty}\big|\leq \pi(\mathcal{P})^{1\over m_{n}}\|f_{n}-f\|_{\infty}+\big|\|f\|_{m_{n}}\\ &&-|f\|_{\infty}\big|\to 0. \end{array} $$

Now fix δ > 0. Since ${\mathcal {P}^{*}_{\delta }}:=\{q: f(q) > \|f\|_{\infty }-\delta \}$, we have ${\mathcal {P}-\mathcal {P}^{*}_{\delta }}=\{q: f(q)\leq \|f\|_{\infty }-\delta \}.$ Define $\delta ^{\prime }:=\delta /\|f\|_{\infty }$. Then

$$ \begin{array}{@{}rcl@{}} \sup_{q\in {\mathcal{P}-\mathcal{P}^{*}_{\delta}}} {f_{n}(q)\over \|f_{n}\|_{m_{n}}}\!\!&=&\!\!\underset{q\in {\mathcal{P}-\mathcal{P}^{*}_{\delta}}}{\sup} {f(q) + (f_{n}(q) - f(q))\over \|f_{n}\|_{m_{n}}} = \underset{q\in {\mathcal{P}-\mathcal{P}^{*}_{\delta}}}{\sup} {f(q) + (f_{n}(q) - f(q))\over \|f\|_{\infty}}{\|f\|_{\infty}\over \|f_{n}\|_{m_{n}}}\\ &\leq&\!\! \underset{q\in {\mathcal{P}-\mathcal{P}^{*}_{\delta}}}{\sup} {f(q)\over \|f\|_{\infty}}{\|f\|_{\infty}\over \|f_{n}\|_{m_{n}}} + {\|f_{n}-f\|_{\infty}\over \|f_{n}\|_{m_{n}}}\!\leq\! 1 - {\delta^{\prime}\over 2},\end{array} $$

provided n is big enough. Thus,

$$ \sup_{q\in \mathcal{P}-\mathcal{P}^{*}_{\delta}}h_{n}(q)\leq \left( 1-{\delta^{\prime}\over 2}\right)^{m_{n}}\to 0,$$

so that $\nu _{n}(\mathcal {P}^{*}_{\delta })\to 1$. We now argue that when $\mathcal {P}^{*}=\{q^{*}\}$, then for any 𝜖 > 0 there exists δ > 0 so that

$$ \mathcal{P}^{*}_{\delta} \subset B(q^{*},\epsilon), $$

(A.1)

where B(q^∗, 𝜖) is a ball in Euclidean sense. If, for an 𝜖 > 0, such a δ > 0 would not exists, then there would exist a sequence q_n → q such that f(q_n) ↗ f(q), but ∥q_n − q∥≥ 𝜖. Since $\mathcal {P}$ is compact, along a subsequence $q_{n^{\prime }}\to q$ and by continuity $f(q_{n^{\prime }})\to f(q)$. On the other hand ∥q − q^∗∥ > 𝜖 and that would contradict the uniqueness of q^∗. Therefore Eq. A.1 holds and so for any 𝜖 > 0, it holds that ν_n(B(q^∗, 𝜖)) → 1. From the definition of the weak convergence, it now follows that $\nu _{n}\Rightarrow \delta _{q^{*}}.$□

Proof of Lemma 6.1

1):

To find

$$ q^{*}=\arg\max_{q\in \mathcal{P}} \left[ \ln \langle e^{-\phi} ,q \rangle + \sum\limits_{k}\alpha_{k}\ln q(k)\right], $$

(A.2)

we define Lagrangian

$$L(q,\beta)=\ln \langle e^{-\phi} ,q \rangle + \sum\limits_{k}\alpha_{k}\ln q(k)-\beta \left( \sum\limits_{k} q(k)-1\right)$$

(here β is a scalar) and maximize L(q, β) over q > 0 (all entries are positive). Taking partial derivatives with respect to q(k), we have

$${e^{-\phi(k)}\over \langle e^{-\phi} ,q \rangle}+{\alpha_{k}\over q(k)}=\beta,\quad \Rightarrow \quad {e^{-\phi(k)}q(k)\over \langle e^{-\phi} ,q \rangle}+{\alpha_{k}}=q(k) \beta \quad \forall k.$$

With $|\alpha |={\sum }_{k} \alpha _{k},$ we have thus β = 1 + |α| and so the solution q^∗ satisfies the set of equalities

$$ q^{*}(k)={1\over 1+|\alpha|}\left( {e^{-\phi(k)}q^{*}_{k}\over \langle e^{-\phi} ,q^{*} \rangle}+\alpha_{k}\right),\quad \forall k. $$

(A.3)

Now with w(k) = e^−ϕ(k) define parameter θ := 〈w, q^∗〉 and rewrite (A.3) as follows

$$ q^{*}(k)={\alpha_{k} \over (1+|\alpha|)-{w(k)\over \theta}} \quad k=1,\ldots,K. $$

(A.4)

We see that amongst the probability vectors satisfying 〈w, q^∗〉 = θ, the solution is unique. Since α_k > 0 for every k, it is easy to see that there is only one parameter θ such that the right hand side of Eq. A.4 would be a probability measure: if $\theta ^{\prime }>\theta$, then for every k, we have

$${\alpha_{k} \over (1+|\alpha|)-{w(k)\over \theta}}>{\alpha_{k} \over (1+|\alpha|)-{w(k)\over \theta^{\prime}}}.$$

Therefore a solution of Eq. A.2 is unique vector q^∗ given by Eq. A.4, where θ = 〈w, q^∗〉.

2):

To find

$$ q^{*}=\arg\max_{q\in \mathcal{P}} \left[-\langle \phi,q\rangle + \sum\limits_{k}\alpha_{k} \ln q(k)\right], $$

(A.5)

we define Lagrangian

$$L(q,\beta)= -\langle \phi,q\rangle + \sum\limits_{k}\alpha_{k}\ln q(k)-\beta \left( \sum\limits_{k} q(k)-1\right).$$

Partial derivatives with respect to q(k) give us the equalities

$$ -\phi(k)+{\alpha_{k}\over q(k)}=\beta \quad \forall k\quad \Rightarrow \quad -\langle \phi,q\rangle + |\alpha| =\beta. $$

Therefore, the inequalities for q^∗(k) are

$$ q^{*}(k)={\phi(k)q^{*}(k) - \alpha_{k} \over \langle \phi,q^{*}\rangle -|\alpha|}={\phi(k)q^{*}(k) - \alpha_{k} \over \theta -|\alpha|},\quad \theta:=\langle \phi,q^{*}\rangle. $$

(A.6)

After rewriting (A.6), we obtain

$$ q^{*}(k)={ \alpha_{k} \over \phi(k)+|\alpha|-\theta},\quad k=1,\ldots,K, $$

Thus, there cannot be two solutions having the same θ. As in the case 1), it is easy to see that when α_k > 0 there is only one θ so that q^∗ in Eq. A.2 sums up to one. Therefore, the solution to the problem (A.5) is unique. Note that the solution is independent of λ.

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