Kingman's model with random mutation probabilities: convergence and condensation II

Kingman's model describes the evolution of a one-locus haploid population of infinite size and discrete generations under the competition of selection and mutation. A random generalisation has been made in a previous paper which assumes all mutation probabilities to be i.i.d.. The weak convergence of fitness distributions to a globally stable equilibrium for any initial distribution was proved. A condensation occurs if almost surely a positive proportion of the population travels to and condensates on the largest fitness value due to the dominance of selection over mutation. A criterion of condensation was given which relies on the equilibrium whose explicit expression is however unknown. This paper tackles these problems based on the discovery of a matrix representation of the random model. An explicit expression of the equilibrium is obtained and the key quantity in the condensation criterion can be estimated. Moreover we examine how the design of randomness in Kingman's model affects the fitness level of the equilibrium by comparisons between different models. The discovered facts are conjectured to hold in other more sophisticated models.

effects of different designs of randomness by comparing the moments and condensation sizes of the equilibriums in several models.

Models
This section is mainly a summarisation of Section 2 in [21], in addition to the introduction of a new random model where all mutation probabilities are equal but random.

Two deterministic models
Let M 1 be the space of probability measures on [0, 1] endowed with the topology of weak convergence. Let (b n ) = (b n ) n≥1 be a sequence of numbers in [0, 1), and P 0 , Q ∈ M 1 . Kingman's model with time-varying mutation probabilities or simply the general model has parameters (b n ), Q, P 0 . In this model, (P n ) = (P n ) n≥0 is a (forward) sequence of probability measures in M 1 generated by where denotes Then S u is interpreted as the largest fitness value of a population of distribution u. Let h := S P0 and assume that h ≥ S Q . This assumption is natural because in any case we have S P1 ≥ S Q . We are interested in the convergence of (P n ) to a possible equilibrium, which is however not guaranteed without putting appropriate conditions on (b n ). To avoid triviality, we do not consider Q = δ 0 , the dirac measure on 0.
Kingman's model is simply the model when b n = b for any n with the parameter b ∈ [0, 1). We say a sequence of probability measures (u n ) converges in total variation to u if the total variation u n − u converges to zero. It was shown by Kingman [15] that (P n ) converges to a probability measure, that we denote by K, which depends only on b, Q and h but not on P 0 . Theorem 1 (Kingman's theorem, [15]). If Q(dx) 1−x/h ≥ b −1 , then (P n ) converges in total variation to where θ b , as a function of b, is the unique solution of If Q(dx) 1−x/h < b −1 , then (P n ) converges weakly to We say there is a condensation on {h} in Kingman's model if Q(h) = Q({h}) = 0 but K(h) > 0, which corresponds to the second case above. We call K(h) the condensate size on {h} in Kingman's model when Q(h) = 0. The terminology is due to the fact that if we let additionally P 0 (h) = 0, then any P n has no mass on the extreme point h; however asymptotically a certain amount of mass K(h) will travel to and condensate on h.

Two random models
We recall the notation of weak convergence for random probability measures. Let (µ n ) be random probability measures supported on [0, 1]. The sequence converges weakly to a limit µ if and only if for any continuous function f on [0, 1] we have Next we introduce two random models which generalise Kingman's model. Let β ∈ [0, 1) be a random variable. Let (β n ) be a sequence of i.i.d. random variables sampled from the distribution of β. If b n = β n for any n we call it Kingman's model with random mutation probabilities or simply the first random model. It has been proved in [21] that (P n ) converges weakly to a globally stable equilibrium, that we denote by I whose distribution depends on β, Q, h but not on P 0 .
For comparison we introduce another random model. If b n = β for any n, we call it Kingman's model with the same random mutation probability or the second random model. Conditionally on the value of β, it becomes Kingman's model. So we can think of this model as a compound version of Kingman's model, with b replaced by β. We denote the limit of (P n ) by A which is a compound version of K.
In this paper, we continue to study the equilibrium and the condensation phenomenon in the first random model. By Corollary 4 in [21], if Q(h) = 0, then I(h) > 0 a.s. or I(h) = 0 a.s.. We say there is a condensation on {h} in the first random model if Q(h) = 0 but I(h) > 0 a.s.. We call I(h) the condensate size on {h} if Q(h) = 0. A condensation criterion, which relies on a function of β and I, was established in [21]. As the equilibrium has no explicit expression, the condensation criterion cannot be used in concrete cases. This paper aims to solve these problems based on a matrix representation of the general model which can be inherited to the first random model. The objectives include an explicit expression of I, and finer properties of I on the moments and condensation. The comparison of Kingman's model and the two random models will be performed and to this purpose we assume additionally that The case with b = 0 is excluded for triviality.

Preliminary results
In this section, we again recall some necessary results from [21]. We introduce with β independent of ν. Note that I, the limit of (P n ) in the first random model, is an invariant measure. In the general model a forward sequence (P n ) does not necessarily converge. But the convergence may hold if we investigate the model in a backward way. A finite backward sequence (P n j ) = (P n j ) 0≤j≤n has parameters n, (b j ) 1≤j≤n , Q, P n n , h with h = S P n n and satisfies Consider a particular case with P n n = δ h . Then P n j converges in total variation to a limit, denoted by G j = G j,h (and G = G 0 , G Q = G 0,SQ ), as n goes to infinity with j fixed, such that where Then G 0 can be considered as a convex combination of probability measures {δ h , Q, Q 1 , Q 2 , · · · }. We introduce also G j = G j,h for G j,h for any j and G = G 0 , G Q = G 0,SQ .
The above results hold regardless of the values of (b n ). So they hold also in the other three models. In particular, we replace the symbol G, G by I, I in the first random model, by A, A in the second random model and by K, K in Kingman's model.
For the first random model, (I j ) is stationary ergodic and I is the weak limit of (P n ). Moreover E ln (1−β) yIQ(dy) ∈ [−∞, − ln yQ(dy)] is well defined, whose value does not depend on the joint law of (β, I). This term is the key quantity in the condensation criterion. Note that we neither have an explicit expression of I Q nor an estimation of Theorem 2 (Condensation criterion, Theorem 3 in [21]).

Notations on matrices
The most important tool in this paper is the matrix representation in the general model. We need to firstly introduce some notations and functions related to matrix. One can skip this part at first reading. 1). Define where the 4 terms all belong to (0, ∞]. For any 1 ≤ j ≤ n ≤ ∞ (except j = n = ∞), define and W j,n := W j,n Introduce and W n = W 1,n ; W = W 1,∞ ; W n+1,n = (m 1 ); W m,n = (1), ∀m > n + 1.
2). For a matrix M of size m × n, let r i (M ) be the ith row and c j (M ) be the jth column, for 1 ≤ i ≤ m, 1 ≤ j ≤ n. If the matrix is like Here U r k increases the indices of the first row by k, with r referring to "row", and U to "upgrade". Similarly define which increases the indices of the last column by k, with c referring to "column". In particular we write . Let | · | denote the determinant operator for square matrices. It is easy to see that, if none of γ j , γ j+1 , · · · , γ n is equal to infinity, Define L j,n := |W j+1,n | |W j,n | , R n j,k := |U r k W j,n | |W j,n | , R n j := R n j,1 , ∀ 1 ≤ j ≤ n, k ≥ 1.
Specifically, let L n+1,n = 1 m1 , R n n+1,k = m k+1 m1 . In the above definition, if one or some of γ j , γ j+1 , · · · , γ n are infinite, we consider L j,n , R n j,k as obtained by letting the concerned variables go to infinity. As a convention, we will not mention again the issue of some γ j 's being infinite, when the function can be defined at infinity by limit.
Notice that expanding W j,n along the first column, we have If γ j = ∞, let L j,n = 0, γ j L j,n = 1 R n j+1 . Lemma 1. In the general model, R n j,k increases strictly in n to a limit that we denote by R j,k (and R j = R j,1 ) which satisfies And γ j L j,n decreases strictly in n to a limit that we denote by γ j L j which satisfies Moreover

1). Matrix representation.
We set a convention that for a term, say α j , in the general model, we use α j to denote the corresponding term in the first random model and α j in the second random model, α j in Kingman's model. If the corresponding term does not depend on the index j, we just omit the index.
Consider a finite backward sequence (P n j ) in the general model: The previous sequence used in Section 3.1 starts with P n n = δ h and this one starts with P n n = Q. The advantage of this change is that the latter enjoys a matrix representation, which is the most important tool in this paper. and Letting n go to infinity, we obtain the following.
Theorem 3. For j fixed and n tending to infinity, P n j converges weakly to a limit, and Note that (H j ) is the limit of (P n j ) with P n n = Q, and (G j ) is the limit of (P n j ) with P n n = δ h . When h = S Q , it remains open whether H = G Q or not. But the equality holds in the first random model.
A remarkable application of the matrix representation is that the condensation criterion in Theorem 2 can be written into a simpler and tractable form using matrices.
Note that the key quantity E ln (1−β) yIQ(dy) in Theorem 2 is now rewritten as E ln Γ 1 L 1 . An estimation of it is highly necessary to make the criterion applicable. To achieve this, we introduce the second important tool of this paper in the following lemma, which is interesting by itself.
The estimation of E[ln Γ 1 L 1 ] is given as follows. and

Remark 1. The two inequalities in (24) are not strict in general. Here is an example. By Theorem 1, if
1−x/SQ ≤ β −1 almost surely, then Γ L = 1/S Q almost surely. So taking β and b small enough, the two inequalities in (24) become equalities.
As Kingman's model is a special kind of the first random model, Corollary 2 applies to Kingman's model as well. The second inequality in (24) implies that Kingman's model is easier to have condensation than the first random model in general. This is made more clear in the next Theorem 5.

3). Comparison between the first random model and the other models.
For succinctness, the results that we present in this part are only in the case h = S Q . However all the results can be easily proved for h > S Q , if we do not stick with strict inequalities. The main idea is to take a new mutant distribution (1 − 1 n )Q + 1 n δ h and consider the limits of equilibriums as n tends to infinity.
We consider an equilibrium to be fitter if it has higher moments and bigger condensate size. In the following, we provide three theorems on the comparison of moments and/or condensate sizes.

Theorem 5. Between Kingman's model and the first random model, if
2. in terms of condensate size, if Q(S Q ) = 0 and I Q > 0, a.s., then Theorem 6. Between the two random models, the following inequality holds

Theorem 7. Between Kingman's model and the second random model, it holds that
But there is no one-way inequality between E[ yA Q (dy)] and yK Q (dy).
It turns out that the first random model is completely dominated by Kingman's model in terms of condensate size and moments of all orders of the equilibrium. We conjecture that the first random model is also dominated by the second random model in the same sense, as supported by a different comparison in Theorem 6. The relationship between Kingman's model and the second random model is more subtle.

Perspectives
Recently, the phenomenon of condensation has been studied a lot in the literature. Biaconi et al [2] argued that the phase transition of condensation phenomenon is very close to Bose-Einstein condensation where a large fraction of a dilute gas of bosons cooled to temperatures very close to absolute zero occupy the lowest quantum state. See also [3] for another model which can be mapped into the physics context. Under some assumptions, Dereich and Mörters [9] studied the limit of the scaled shape of the traveling wave of mass towards the condensation point in Kingman's model, and the limit turns out to be of the shape of some gamma function. A series of papers [11,19,10,16,12] were written later on to investigate the shape of traveling wave in other models where condensation appears and have proved that gamma distribution is universal. Park and Krug [17] adapted Kingman's model to a finite population with unbounded fitness distribution and observed in a particular case emergence of Gaussian distribution as the wave travels to infinity.
The first random model, as a natural random variant of Kingman's model, provides an interesting example to study condensation in detail. The matrix representation can be a handy tool to study the shape of the traveling wave to verify if the gamma-shape conjecture holds. On the other hand, we can also ask the question: will the relationships between the three models revealed and conjectured in this paper be applicable to other more sophisticated models under the competition of two forces, particularly to those models on the balance of selection and mutation? It is very tempting to say yes. The verification of the universality constitutes a long term project.

Proof of Lemma 2
Proof of Lemma 2. Note that Assume that for some 0 ≤ j ≤ n − 1, Then The last equality is obtained by expanding W j+1,n x and W j+1,n on the first column. By induction, we prove (17). As a consequence, we also get (18).
Expanding the first row of W j,n x and using the above result, we get Then we plug it in (18), changing j to j + 2.

Proof of Lemma 1
We need to prove first a few more results on monotonicity. The following Hölder's inequality will be heavily used: Lemma 4. For j ≥ 1, n ≥ j − 1, R n j increases strictly in n to R j ∈ (0, 1], as Proof. By Hölder's inequality, for j = n + 1, Consider n ≥ j. Without loss of generality let j = 1. Using (11) The two matrices U r W n , W n differ only on the first row, which is (m 2 , · · · , m n+2 ) for the former, and (m 1 , · · · , m n+1 ) for the latter. Again by Hölder's inequality, we have For the comparison of R n 1 and R n+1 Simply applying the above lemma and (12), we obtain the following Corollary.
Corollary 4. For any j ≥ 1, γ j L j,n decreases strictly in n to γ j L j . Define Then Φ j,l,n = Φ j,l = 0 if γ j+l = ∞, otherwise Φ j,l,n decreases strictly in n to Φ j,l .
Corollary 5. For any j ≥ 1, l ≥ 1, R n j,k increases strictly in n to R j,k .
Proof. The case k = 1 has been proved by Lemma 4. We consider here k ≥ 2. Without loss of generality we let j = 1. The idea is to apply Lemma 8 in the Appendix. Following the notations in Lemma 8 we set and c l = C n−1 0,l , c ′ l = C n 0,l , ∀ 0 ≤ l ≤ n − 1; c n = 0, c ′ n = C n 0,n . Then by the definition of R n 1,k and Lemma 2 So by (26) For any n ≥ 1, by Hölder's inequality Moreover a 0 , · · · , a n , b 0 , · · · , b n are all strictly positive numbers.
Next we consider the c l 's and c ′ l 's. Note that c 0 = c ′ 0 = b 1 . By Corollary 4, for Now we apply Lemma 8 to conclude.
Proof of Lemma 1. As we have already proved Corollary 4 and 5, it remains to tackle (13) and (15). Expanding U r k W j,n and W j,n on the first column, we get Letting n → ∞, we obtain (13).
As R n 2,1 decreases to R 2,1 , we have also R 2,1 < 1 which gives the strict upper bound for R 2,1 . Using (12), the above display yields Since γ 1 L 1,n decreases strictly to γ 1 L 1 , we obtain the following using again (12) .
Then we get R 2,1 > m 1 . Moreover as R 2,1 < 1, So we have found the strict lower and upper bounds for R 2,1 and γ 1 L 1 .

Proofs of Theorem 3 and Corollary 1
For Proof of Theorem 3. Note that Q j ≤ Q j+1 for any j. Then using Corollary 3 and Lemma 1, P n j ≤ P n+1 j . So P n j converges at least weakly to a limit H j . The weak convergence allows to obtain (20) from (4). Expanding (20), we obtain where To prove (21), we firstly use (18) and definition (11) to obtain that A reformulation of the above equality reads Using the convergences as n → ∞, we obtain (21).
Proof of Corollary 1. By (19), H j is equal in distribution for all j's. By (20), H j is an invariant measure on [0, S Q ] with S Hj = S Q a.s.. Recall that I j,SQ is also invariant on [0, S Q ]. Then by Theorem 4 in [21], H j d = I j,SQ . By (5) and (20), for both sequences, the multi-dimensional distributions are determined in the same way by one dimensional distribution. So the two sequences have the same multi-dimensional distributions, and the multi-dimensional distributions are consistent in each sequence. By Kolmogorov's extension theorem (Theorem 5.16, [14]), consistent multi-dimensional distributions determine the distribution of the sequence, which yields the identical distribution for both two sequences.

Proof of Corollary 2
Proof of Corollary 2. Recall that E 1−β yIQ exists and does not depend on the joint law of β, I Q . Using (21) in the first random model, together with Corollary 1, we can rewrite Theorem 2 into Corollary 2.

Proof of Lemma 3
Proof of Lemma 3. Since (ξ 1 , · · · , ξ n ) is exchangeable, we can directly take a symmetric function f and prove the inequality under f x1x2 ≤ 0. For any a > b, we first show that f (a, b, · · · , b n−1 ) + f (b, a, · · · , a n−1 which is proved as follows.
Letting i travel from 1 to n − 1, we prove the lemma.

Proof of Theorem 4
Define Ψ n := n j=1 γ j |W n | , n ≥ 1. Proof. We prove only the case in the first random model. Note that Here we use the fact that Γ j L j,n d = Γ 1 L 1,n−j+1 . Then we apply Lemma 1 to conclude that (24) holds. The rest are direct applications of (21). Lemma 6. ln Ψ n is strictly concave down in every b j , 1 ≤ j ≤ n.
Proof. By basic computations we obtain for b j ∈ (0, 1), By Lemma 11 in the Appendix, ∂ 2 ln Ψn Proof of Theorem 4. To prove (24), we can use Lemma 5 and show instead For any 1 ≤ j < i ≤ n, due to Proposition 1 in the Appendix, Then we apply Lemma 3 to obtain the first inequality of (32). Next we apply Lemma 6 and Janson's inequality for the second inequality of (32). To prove (25), we use (21), and Theorem 1.

Proof of Theorem 5
We need two preparatory results before proving the theorem.
Lemma 7. For any k, n, R n 1,k is strictly concave down in every b i , 1 ≤ i ≤ n.
Let f ′ , f ′′ , g ′ , g ′′ be derivatives with respect to γ i ∈ (0, ∞). Then by Corollary 8 in the Appendix The above statements are not difficult to see if it is clear how f, g can be computed. Or one can refer to Lemma 10 in the Appendix. Then we obtain Moreover, Then where the inequality is due to Lemma 11 in the Appendix.
and if Q(S Q ) = 0, Proof. By (20), we obtain The above display together with (21) lead to (33). If Q(S Q ) = 0, then lim k→∞ S −k Q m k+1 = 0. Using this fact and (18), we obtain Proof of Theorem 5. There are two statements to prove.

By (13)
By Corollary 8 in the Appendix, R 1,k is strictly increasing in γ 1 . Then The above inequality entails that for b 1 ∈ (0, 1) So R 1,k is strictly concave down in b 1 .
In the following display, the first equality is due to (18) and the first inequality is by the above strict concavity. The second equality is due to Lemma 1 and the second inequality is by Lemma 7. The last equality is a consequence of (18) and Corollary 5.

Proof of Theorem 6
Proof of Theorem 6. Note that similarly as in the proof of Lemma 5 For the second random model, similarly By Lemma 13 and (21), and lim n→∞ E ln | W n |β n /n = E ln β We compare next E ln | W n | n j=1 β j and E ln | W n |β n . Note that Then second order partial derivative of ln |W n | n j=1 b i with respect to b s , b t equals ∂ 2 ln |W n | ∂bs∂bt which is, by Lemma 11 in the Appendix, strictly positive for any 1 ≤ s = t ≤ n. Applying Lemma 3, we obtain Then by (35) and (36) we conclude that E ln yI Q (dy) ≤ E ln yA Q (dy) .

Proof of Theorem 7
Proof of Theorem 7. By Theorem 1, So K Q is a concave up function of b, and consequently E[ To show that there is no one-way inequality between E[ yA Q (dy)] and yK Q (dy), we give a concrete example. Let Q(dx) = dx. In this case, We show that d 2 θ b db 2 can be strictly positive and negative for different b ′ s. The above equation can be rewritten as bdx .
with m(t) the numerator and n(t) the denominator. Then As n(t) 2 > 0 and db dt < 0 for any t ∈ (0, 1), we have 6 Appendix 6.1 Appendix A Lemma 8. Let n > 1. Let a 0 , · · · , a n , b 0 , · · · , b n all be strictly positive numbers such that a l b l < a n b n , a l < a n , b l < b n , ∀ 0 ≤ l ≤ n − 1.
Without loss of generality, we consider only l = 0. We have Note that by the assumptions on a l 's and b l 's, That implies (b n − b 0 )A < (a n − a 0 )B which entails ∂f ∂c0 < 0.

Appendix B
Lemma 9. Let X n = (x n 0 , · · · , x n n ) be the unique solution of the equation Then m 1 < x n 0 < x n+1 0 < 1 for any n ≥ 1.
Proof. By Cramer's rule and Lemma 4 For any n ≥ 1, we are going to construct X n+1 from X n and compare x n 0 , x n+1 0 . The main argument is Hölder's inequality (28).
It is clear that if γ i = ∞, then x n,ε i = 0; otherwise x n,ε i is continuous and strictly increasing on ε.
Then X n+1 = Y. To achieve this, let Then the dot product of Y and the second last column of W n+1 gives m n+2 : x n,ε 0 m n+1 + · · · + x n,ε n m 1 − γ n+1 A ε = m n+2 . If A ε ≡ 0, then A ε is continuous and strictly increasing on ε with A 0 = 0. Therefore, in view of (39), there exists a unique ε > 0 such that the dot product of Y and the last column of W n+1 gives m n+3 : x n,ε 0 m n+2 + · · · + x n,ε n m 2 + A ε m 1 = m n+3 . Then together with (40), So X n+1 = Y. As x n,ε 0 is strictly increasing in ε and the ε in the above equality is strictly positive, we obtain that 0 < x n 0 < x n,ε 0 = x n+1 0 < 1.

Appendix C
Proposition 1. For any 1 ≤ j < i ≤ n and b i , b j ∈ (0, 1), Proof. Notice that Dividing both sides by |W n | yields Using the above display By Lemma 1, we can conclude ∂ 2 ln |W n | ∂bi∂bj > 0. Letting n → ∞ we get the following Corollary 7. For any i ≥ 1, γ i L i is strictly decreasing in b i and strictly increasing in b j , ∀j > i. The same result holds for γ i L i,n .
Proof. We shall only consider γ 1 L 1 . The strict monotonicity in b 1 stems from (14). Take j > 1. By (14), the monotonicity of γ 1 L 1 in b j does not depend on b 1 . For convenience let b 1 = c ∈ (0, 1). Then we can study L 1 instead. Note that Notice that the following holds when b 1 = 1, Then by Proposition 1 Then we obtain ∂L1 ∂bj > 0.
Corollary 8. For any k > 1, both R n 1,k and R 1,k strictly decrease in b j , for any j ≥ 1.
Proof. We shall prove only for R 1,k . Without loss of generality, we show that R k+1,k strictly decreases in b m , m ≥ k + 1. Take |W n | |W n | and expand the top W n for the first k elements on the first row. A similar approach was used in obtaining (27) where the expansion was made on the whole first row. Letting n go to infinity we obtain the following, with detailed steps omitted Taking derivative on b m on both sides, and using Corollary 7, the derivative of R k+1,k on b m is strictly negative for b m ∈ (0, 1).

Appendix D
We introduce below a new notation for the special structure of matrix W n . By definition, W n is of type ( * ). To compute the determinant of a matrix of type ( * ), we need some more notations. Define E n k := {e = (e 1 , · · · , e k ) : 1 = e 1 < e 2 < · · · < e k = n + 1}, ∀ 2 ≤ k ≤ n + 1.
So E n k consists of all sequences of length k increasing from 1 to n + 1. Let Proof. By decomposing M along the last row, we can prove it by induction. Details are omitted.
Remark 2. Leibniz formula says that |M | = σ∈sn sgn(σ) n j=1 M j,σ(j) . It is easy to see that the set {σ : σ ∈ s n , n j=1 M j,σ(j) = 0} is in one-to-one correspondence to E n . Moreover sgn(σ) = 1 for any σ in the former set. If we use σ e to denote the corresponding element in s n of an e ∈ E k , k−1 j=1 d(M (e j , e j+1 − 1)) = n j=1 M j,σ e (j) > 0.
Let M be the matrix obtained by deleting the row and column of W n containing γ j . Then The purpose is to compare |W j−1 ||W j+1,n | and |M |. Denote A = {e ∈ E n+1 : j + 1 ∈ e}.
To compute |M |, we also seek to find an expression similar to the above display. Let t(e) be the corresponding location such that e t(e) = j + 1 for any e ∈ A. Denote .

  
There is a clear one-to-one correspondence between A and B. It is easy to verify that Consequently Let e ∈ A ∩ E n+1 k and e ′ its corresponding element in A ′ . Recalling the Definition 1,

Acknowledgment
The author thanks Takis Konstantopoulos, Götz Kersting and Pascal Grange for discussions. The author acknowledges the support of the National Natural Science Foundation of China (Youth Program, Grant: 11801458), and the XJTLU RDF-17-01-39.