aa r X i v : . [ m a t h . P R ] J u l A basic model of mutations
Maxime Berger ∗ Rapha¨el Cerf ∗† August 1, 2018
Abstract
We study a basic model for mutations. We derive exact formulaefor the mean time needed to discover the master sequence, the meanreturning time to the initial state, or to any Hamming class. Theselast two formulae are the same than the formulae obtained by MarkKac for the Ehrenfest model.
According to the Darwinian paradigm, the evolution of living creatures isdriven by two main forces: mutations and selection. Mutations create newforms of behaviour or new characters, some less fit to their environment,some more, whereas selection favors the reproduction of fitter individuals.On the one hand, mutations may discover very fit characters but withoutselection, they would be quickly erased by further mutations. On the otherhand, selection alone would result in uniform populations, lacking in ge-netic diversity. The success of an evolutionary process rests on a subtleinteraction between mutations and selection.Let us consider for instance a population of HIV viruses or
Drosophilamelanogaster . The genetic material of one individual, also called its geno-type, is encoded into its DNA, which is a long chain of nucleobases A,T,Gor C. To simplify the analysis, we suppose here that there are only twotypes of nucleobases instead of four, and we denote them by 0 or 1. Selec-tion enters the game through fitness. The fitness describes the adaptationof the individuals to the environnement. The fitness of an individual canbe thought as a function of its genotype. For instance, a possible choice ∗ D´epartement de math´ematiques et applications, Ecole Normale Sup´erieure, CNRS,PSL Research University, 75005 Paris † Laboratoire de Math´ematiques d’Orsay, Universit´e Paris-Sud, CNRS, Universit´eParis–Saclay, 91405 Orsay. · · ·
0, which has a superior fitness. There exist severalmathematical models of evolution combining mutations and selection. Thesimplest one is perhaps the Moran model, whose dynamics is the following.At each time step, one individual dies, while one individual gives birth toa child (in particular, the population size stays constant). All individualsare equally likely to die, but the fitter individuals having genotype 0 · · · · · · · · ·
A model for mutations . We follow the genotypes of the individuals alonga lineage. We fix the probability of mutation p ∈ (0 , X in { , } N , which represents the genotype of the first individual.We denote by X n the genotype of the n –th individual in the lineage. Attime n , for each bit of X n , we toss a coin of parameter p to decide whethera mutation occurs on the bit, in which case it is transformed into the com-plementary digit. All the coins are taken independent, so the probabilityof having no mutations at all at time n is (1 − p ) n . The assumption of inde-pendance simplifies considerably the mathematical analysis, and it is alsobiologically plausible. Indeed the mutations arise because of transcriptionerrors during the replication process, and for long genomes, they are notcorrelated. It seems however that the mutation probability p varies alongthe genome, so the next modelling step would be to incorporate this spatialdependance into the model. We end up with a random walk ( X n ) n ∈ N onthe hypercube { , } N , for which the transition probabilities only dependon the number of differences between two states. Let τ be the hitting timeof the master sequence 0 · · ·
0, i.e., τ = inf (cid:8) n ≥ X n = 0 · · · (cid:9) . The goal is then to compute the expected value of τ . Lumping.
The smart way to analyze this random walk is to lump togetherthe states of the hypercube into Hamming classes. The Hamming classnumber i consists of the points which have i digits equal to 1 and N − i equal to 0. So we define a new process ( Y n ) n ∈ N by setting Y n = number of digits of X n equal to 1 .
2e obtain a Markov chain with state space { , . . . , N } , and we shall pro-vide explicit formulas for its mean passage times. Discovering and recovering time.
For 0 ≤ j ≤ N , we define the hittingtime of the Hamming class j by τ j = inf (cid:8) n ≥ Y n = j (cid:9) . The three theorems below give exact formulas for the expected value of τ j ,when starting from 0, N , or j . These formulas are surprisingly simple andcome out from tricky computations. The discovering time of the mastersequence is bounded from above by the traversal time, which is the timeneeded to reach the class 0 starting from the class N . This correspondsto the situation where we start with a string containing only ones, and wewait until we see a string containing only zeroes. Theorem 1.1
The mean traversal time is given by E (cid:0) τ (cid:12)(cid:12) Y = N (cid:1) = N X k =1 (cid:18) Nk (cid:19) − ( − k − (1 − p ) k . The recovering time of the master sequence corresponds to the returningtime to the class 0 when starting away from a genotype different from0 · · ·
0. This time is bounded from below by the mean return time to 0starting from 0, which we compute next.
Theorem 1.2
The mean returning time to the class 0 is given by E (cid:0) τ (cid:12)(cid:12) Y = 0 (cid:1) = 2 N . Of course, the formula of theorem 1.2 is also a straightforward consequenceof the classical result expressing the invariant probability measure of aMarkov chain in terms of mean recurrence times. However, it does notseem that the other formulas presented in theorems 1.1 or 1.3 are easyconsequences of more general results. We compute next a beautiful formulafor the returning time to the class j when starting away from a genotypeof the same class. Theorem 1.3
For 1 ≤ j ≤ N , the mean returning time to the class j is E (cid:0) τ j (cid:12)(cid:12) Y = j (cid:1) = 2 N (cid:18) Nj (cid:19) . x , E (cid:0) τ (cid:12)(cid:12) Y = 0 (cid:1) ≤ E (cid:0) τ (cid:12)(cid:12) X = x (cid:1) ≤ E (cid:0) τ (cid:12)(cid:12) Y = N (cid:1) . Taking into account theorems 1.1 and 1.2, we conclude that2 N ≤ E (cid:0) τ (cid:12)(cid:12) X = x (cid:1) ≤ N X k =1 (cid:18) Nk (cid:19) − ( − k − (1 − p ) k ≤ N p . These inequalities show that the discovering and the recovering times ofthe master sequence are of order 2 N . It turns out that these results areakin to those related to an old classical model, the Ehrenfest model, thatwe describe briefly. The Ehrenfest model.
Let us consider N balls and two boxes. Initially,all the balls are in the first box. At each time step, one ball is selected atrandom and is moved from its current box to the other box. The centralquestion is then the following:On average, how long will it take to return to the initial state?In 1947, Mark Kac gave a simple answer to this question in a celebratedpaper [ ? ]. He considered the evolution of the number of balls in the firstbox. This process is a Markov chain on { , · · · , N } , which is quite differentfrom our process ( Y n ) n ∈ N . For instance, its increments are either − , N , which is theanalog of theorem 1.2. He showed also that, when starting from the state j , the average time until return to j is equal to 2 N / (cid:0) Nj (cid:1) . Theorem 1.3 givesthe analogous result for our model of mutations. A glimpse of potential theory . We attack here the general case, i.e.,we look for a formula for the mean returning time to the class j , whenthe process starts from the class i . We start from the formula obtained intheorem 1.1, for the specific case where i = N and j = 0. We expand in ageometric series the denominator and we get E (cid:0) τ (cid:12)(cid:12) Y = N (cid:1) = N X k =1 X n ≥ (cid:18) Nk (cid:19)(cid:16) (1 − p ) nk − ( − k (1 − p ) nk (cid:17) . We exchange the order of the summations and, using the formulas (5.1)and (5.2), we obtain E (cid:0) τ (cid:12)(cid:12) Y = N (cid:1) = 2 N X n ≥ (cid:16) P (cid:0) Y n = 0 (cid:1) − P N (cid:0) Y n = 0 (cid:1)(cid:17) . E (cid:0) τ (cid:12)(cid:12) Y = 0 (cid:1) = 2 N . The abovedisplay formula is actually a particular case of a more general identityvalid for a large class of Markov chains, that we state in the next theorem. Theorem 1.4
For any i, j in { , . . . , N } , we have E (cid:0) τ j (cid:12)(cid:12) Y = i (cid:1) = E (cid:0) τ j (cid:12)(cid:12) Y = j (cid:1)(cid:18) X n ≥ (cid:16) P j (cid:0) Y n = j (cid:1) − P i (cid:0) Y n = j (cid:1)(cid:17)(cid:19) . In the case of our specific model, we have further a formula for each termin the sum (see formula (4.1)). This way, we get an exact formula for E (cid:0) τ j (cid:12)(cid:12) Y = i (cid:1) , which we present in the next theorem. Theorem 1.5
For 1 ≤ j ≤ N , the mean returning time to the class j starting from class i is given by E (cid:0) τ j (cid:12)(cid:12) Y = i (cid:1) =1 (cid:18) Nj (cid:19) X n ≥ N X k =0 (cid:18) jk (cid:19)(cid:18) N − jk (cid:19) − (cid:18) ij − k (cid:19)(cid:18) N − ik (cid:19)(cid:16) − (1 − p ) n − p ) n (cid:17) i − j !(cid:16) − (1 − p ) n (cid:17) k (cid:16) − p ) n (cid:17) N − k . (1.6) Strategy of the proof . We will employ the same method that Mark Kacused for the Ehrenfest model. We shall try to compute E (cid:0) τ j (cid:12)(cid:12) Y = i (cid:1) for0 ≤ i, j ≤ N . These computations are tricky. We will in fact compute thegenerating functions of the event { Y n = j } and of the random variable τ j ,and we will relate them through a functional equation that we derive inthe next section. The mean passage times are equal to the left derivativeat 1 of the generating function of τ j . We compute these derivatives byperforming a local expansion of the functions around 1. Finally, in the lastsection, we prove a general formula for E (cid:0) τ j (cid:12)(cid:12) Y = i (cid:1) for 0 ≤ i, j ≤ N ,which is based on the potential theory for Markov chains. Throughout the computations, we shall denote by P i the probability con-ditioned on the event that Y = i . Let us fix 0 ≤ i, j ≤ N . For n ≥
1, wecompute P i ( Y n = j ) by decomposing the event { Y n = j } according to thehitting time τ j , as follows: P i ( Y n = j ) = n X k =1 P i ( Y n = j, τ j = k )5 n X k =1 P i (cid:0) Y = j, . . . , Y k − = j, Y k = j, Y n = j (cid:1) . We perform a conditioning in the probability in the sum and we get P i ( Y n = j ) = n X k =1 P i (cid:0) Y = j, . . . , Y k − = j, Y k = j (cid:1) × P i (cid:0) Y n = j (cid:12)(cid:12) Y = j, . . . , Y k − = j, Y k = j (cid:1) . The constitutive property of a Markov chain is that the past influences thefuture only through the present state. This is rigorously formalized in theMarkov property, which yields that P i (cid:0) Y n = j (cid:12)(cid:12) Y = j, . . . , Y k − = j, Y k = j (cid:1) = P i (cid:0) Y n = j (cid:12)(cid:12) Y k = j (cid:1) . Since in addition the Markov chain is time homogeneous, we have that P i (cid:0) Y n = j (cid:12)(cid:12) Y k = j (cid:1) = P j (cid:0) Y n − k = j (cid:1) . Plugging the last two equalities in the sum, we conclude that P i ( Y n = j ) = n X k =1 P i (cid:0) τ j = k (cid:1) P j (cid:0) Y n − k = j (cid:1) . (2.1)To take advantage of this identity, we introduce the generating functionsof the event { Y n = j } and of the random variable τ j , i.e., we consider theseries F ij ( z ) = X n ≥ P i ( Y n = j ) z n ,G ij ( z ) = X n ≥ P i ( τ j = n ) z n . Their radius of convergence is at least one. In equation (2.1), we isolatethe term corresponding to k = n , and we multiply by z n to get P i ( Y n = j ) z n = P i (cid:0) τ j = n (cid:1) z n + n − X k =1 P i (cid:0) τ j = k (cid:1) z k P j (cid:0) Y n − k = j (cid:1) z n − k . Looking at the sum, we recognize the Cauchy product of the two series.Summing this identity, we conclude that F ij ( z ) = G ij ( z ) + F jj ( z ) G ij ( z ) . (2.2)The strategy is now the following. We try to compute the functions F ij .We use then the above identity to obtain the functions G ij . Finally, themean passage times can be computed from G ij by taking its left derivativeat 1: E (cid:0) τ j (cid:12)(cid:12) Y = i (cid:1) = X n ≥ nP i ( τ j = n ) = G ′ ij (1) . Single nucleotide dynamics
We suppose here that N = 1, i.e., we focus on the dynamics of a singlenucleotide. In this case, the process ( X n ) n ≥ is the Markov chain withstate space { , } and transition matrix M = (cid:18) − p pp − p (cid:19) . The eigenvalues of M are 1 and 1 − p . We compute, for n ≥ M n = 12 (cid:18) − p ) n − (1 − p ) n − (1 − p ) n − p ) n (cid:19) . Here is a simple illuminating way to realize the dynamics and to understandthe expression of the n –th power M n . Let ( ε n ) n ≥ be an i.i.d. sequence ofBernoulli random variables with parameter p . At each time step, we usethe variable ε n to decide whether X n mutates or not. More precisely, weset X n = ( X n − if ε n = 0 , − X n − if ε n = 1 . Now, the event X n = X occurs if and only if the total number of mutationswhich happened until time n is even, i.e., P ( X n = X ) = P (cid:0) ε + · · · + ε n is even (cid:1) . Let us set S n = ε + · · · + ε n . Here is a little trick to compute the probability that S n is even. We computein two different ways the expected value of ( − S n . Indeed, we have E (cid:0) ( − S n (cid:1) = (cid:16) E (cid:0) ( − ε (cid:1)(cid:17) n = (cid:0) − p + 1 − p (cid:1) n = (cid:0) − p (cid:1) n = P (cid:0) S n is even (cid:1) − P (cid:0) S n is odd (cid:1) . Obviously, we have P (cid:0) S n is even (cid:1) + P (cid:0) S n is odd (cid:1) = 1 , therefore we obtain that P ( X n = X ) = P (cid:0) S n is even (cid:1) = 12 (cid:16) − p ) n (cid:17) . This way we recover the expression of the diagonal coefficients of M n . Letus define p n = 12 (cid:16) − p ) n (cid:17) . (3.1)7rom the above computations, we conclude the following. Conditionallyon X = 1, X n is a Bernoulli random variable with parameter p n , i.e., P (cid:0) X n = 1 (cid:12)(cid:12) X = 1 (cid:1) = p n , P (cid:0) X n = 0 (cid:12)(cid:12) X = 1 (cid:1) = 1 − p n . Similarly, conditionally on X = 0, X n is a Bernoulli random variable withparameter 1 − p n . We consider now the case where the number N of nucleotides is largerthan one. In our model, the mutations occur independently at each site.An important consequence of this structural assumption is that the com-ponents of X n , ( X n ( i ) , ≤ i ≤ N ), are themselves Markov chains like theone studied in the previous section, and these Markov chains are more-over independent. This remark, combined with the results of the previoussection, allows to derive explicitly the distribution of Y n . Indeed, supposethat we start from Y = i . This means that i digits in X are equal to 1and N − i to 0. At time n , in X n , the i digits which were initially equalto 1 are distributed according to a Bernoulli law of parameter p n , the oth-ers are distributed according to a Bernoulli law of parameter 1 − p n . Theevolution of the nucleotides being independent, these Bernoulli variablesare independent, so their sum is distributed as the sum of two independentBinomial random variables: Y n ∼ Bin( i, p n ) + Bin( N − i, − p n ) . This yields for instance the following formula: P i ( Y n = j ) = X ≤ k ≤ i ≤ j − k ≤ N − i P (cid:0) Bin( i, p n ) = k (cid:1) P (cid:0) Bin( N − i, − p n ) = j − k (cid:1) = X ≤ k ≤ i ≤ j − k ≤ N − i (cid:18) ik (cid:19)(cid:18) N − ij − k (cid:19) (1 − p n ) i + j − k ( p n ) N − i − j +2 k . (4.1)This formula is quite complicated. Yet it becomes particularly simple inthe cases where i or j is equal to 0 or N . Indeed, we have, for 0 ≤ i ≤ N , P i ( Y n = 0) = (1 − p n ) i ( p n ) N − i ,P i ( Y n = N ) = ( p n ) i (1 − p n ) N − i , and for 0 ≤ j ≤ N , P ( Y n = j ) = (cid:18) Nj (cid:19) (1 − p n ) j ( p n ) N − j , (4.2)8 N ( Y n = j ) = (cid:18) Nj (cid:19) ( p n ) j (1 − p n ) N − j . (4.3)For once, surprisingly enough, these two cases are also the most relevantfor genetic applications, so we treat them first. We will indeed comparethese extreme cases to the general chain and deduce an estimation on thediscovering and returning time. This section is devoted to the completion of the proof of theorems 1.1and 1.2. We shall implement the strategy explained at the end of section 2.Our first goal is to compute the generating function F N ( z ) = X n ≥ P N ( Y n = 0) z n . From formulas (4.3) and (3.1), we have P N ( Y n = 0) = (1 − p n ) N = (cid:16) − (1 − p ) n (cid:17) N . We use the binomial expansion to develop the N –th power in order tocompute the generating function F N as a sum of geometric series: P N ( Y n = 0) = 12 N N X k =0 (cid:18) Nk (cid:19) ( − k (1 − p ) nk . (5.1)Notice that P N (cid:0) Y = 0 (cid:1) = 0. For convenience, we start the sum defining F N at n = 0 and we obtain a finite number of geometric series: F N ( z ) = X n ≥ P N ( Y n = 0) z n = X n ≥ N N X k =0 (cid:18) Nk (cid:19) ( − k (1 − p ) nk z n = 12 N N X k =0 (cid:18) Nk (cid:19) ( − k − (1 − p ) k z . Our next goal is to compute the generating function F ( z ) = X n ≥ P ( Y n = 0) z n . P ( Y n = 0) = ( p n ) N = (cid:16) − p ) n (cid:17) N = 12 N N X k =0 (cid:18) Nk (cid:19) (1 − p ) nk . (5.2)This time, we have P (cid:0) Y = 0 (cid:1) = 1. Adding this term to F , we get againnice geometric series:1 + F ( z ) = X n ≥ P ( Y n = 0) z n = X n ≥ N N X k =0 (cid:18) Nk (cid:19) (1 − p ) nk z n = 12 N N X k =0 (cid:18) Nk (cid:19) − (1 − p ) k z . For 0 ≤ k ≤ N , we introduce the auxiliary functions φ k ( z ) = (cid:18) Nk (cid:19) − (1 − p ) k z , and we rewrite the expressions of F N and 1 + F as F N ( z ) = 12 N N X k =0 ( − k φ k ( z ) , F ( z ) = 12 N N X k =0 φ k ( z ) . (5.3)We have computed F N and 1 + F . From the probabilistic identity (2.2),we obtain G N ( z ) = F N ( z )1 + F ( z ) . Remember that our ultimate goal is to compute the left derivative of G N at 1. The functions φ k are regular around 1, except the first one, φ , indeed, φ ( z ) = 11 − z .
10o get G ′ N (1), we perform a local expansion of G N around 1, as follows: G N ( z ) = 11 − z + N X k =1 ( − k φ k ( z )11 − z + N X k =1 φ k ( z )= 1 + (1 − z ) N X k =1 (cid:0) ( − k − (cid:1) φ k ( z ) + o ( z − . This expansion readily yields the value of the left derivative of G N at 1: G ′ N (1) = N X k =1 (cid:0) − ( − k (cid:1) φ k (1) . Replacing φ k (1) by its value, we obtain the formula stated in theorem 1.1.We proceed similarly to prove theorem 1.2. In fact, we have to compute G ′ (1), and the probabilistic identity (2.2) yields G ( z ) = F ( z )1 + F ( z ) = 1 −
11 + F ( z ) . We have already computed 1 + F ( z ) in formula (5.3). We use this expres-sion and we expand around z = 1: G ( z ) = 1 − NN X k =0 φ k ( z ) = 1 − N (1 − z ) + o (1 − z ) . This expansion shows that G ′ (1) = 2 N . We shall finally prove the analog of Kac theorem on the mean returningtime to the class j , when the process starts from the class j . We write theformula (4.1) with i = j , we reindex the sum by setting ℓ = j − k and weperform the two binomial expansions: P j ( Y n = j ) = X ≤ k ≤ j ≤ j − k ≤ N − j (cid:18) jk (cid:19)(cid:18) N − jj − k (cid:19)(cid:16) − (1 − p ) n (cid:17) j − k (cid:16) − p ) n (cid:17) N − j +2 k j ∧ ( N − j ) X ℓ =0 (cid:18) jℓ (cid:19)(cid:18) N − jℓ (cid:19)(cid:16) − (1 − p ) n (cid:17) ℓ (cid:16) − p ) n (cid:17) N − ℓ = j ∧ ( N − j ) X ℓ =0 N (cid:18) jℓ (cid:19)(cid:18) N − jℓ (cid:19) ℓ X α =0 N − ℓ X β =0 (cid:18) ℓα (cid:19)(cid:18) N − ℓβ (cid:19) ( − α (1 − p ) ( α + β ) n . For n = 0, we have P j ( Y = j ) = 1, therefore, after a geometric summation,we get1 + F jj ( z ) = X n ≥ P j ( Y n = j ) z n = j ∧ ( N − j ) X ℓ =0 N (cid:18) jℓ (cid:19)(cid:18) N − jℓ (cid:19) ℓ X α =0 N − ℓ X β =0 (cid:18) ℓα (cid:19)(cid:18) N − ℓβ (cid:19) ( − α − (1 − p ) α + β z . We expand this function around z = 1 and we get1 + F jj ( z ) = j ∧ ( N − j ) X ℓ =0 N (cid:18) jℓ (cid:19)(cid:18) N − jℓ (cid:19) − z + O (1)= 12 N (cid:18) Nj (cid:19) − z + O (1) , thanks to the combinatorial identity stated in the next lemma. Lemma 6.1
For 0 ≤ j ≤ N , we have j ∧ ( N − j ) X ℓ =0 (cid:18) jℓ (cid:19)(cid:18) N − jℓ (cid:19) = (cid:18) Nj (cid:19) . Proof.
Let us fix j in { , . . . , N } , and let us consider a set E havingcardinality N . We fix also a subset A of E having j elements. We classifythe subsets of E having cardinality j according to the cardinality of theirintersection with A and we readily obtain the formula of the lemma. (cid:3) From the probabilistic identity (2.2), we have G jj ( z ) = 1 −
11 + F jj ( z ) , thus G jj ( z ) admits the following expansion around z = 1: G jj ( z ) = 1 + 2 N (cid:18) Nj (cid:19) ( z −
1) + o ( z − . E (cid:0) τ j (cid:12)(cid:12) Y = j (cid:1) = G ′ jj (1) = 2 N (cid:18) Nj (cid:19) and this concludes the proof of theorem 1.3. It has been observed a long time ago that the equations defining the in-variant measure, or the returning time to a set for a random walk, or moregenerally a Markov chain, are formally equivalent to the equations arisingin potential theory, if one interprets the transition probabilities as con-ductances (see the very nice book [2]). In fact, the formula presented intheorem 1.4 takes its roots in potential theory [4]. Let us denote by P thetransition matrix of the process ( Y n ) n ≥ , defined by ∀ i, j ∈ { , . . . , N } ∀ n ≥ P ( i, j ) = P ( Y n +1 = j | Y n = i ) . The arguments presented below are in fact valid for a general class ofMarkov chains with finite state space. For instance, it suffices that P , orone of its powers, has all its entries positive. In this situation, the classicalergodic theorem for Markov chains ensures the existence and uniqueness ofan invariant probability measure and the following convergence holds: ∀ i, j ∈ { , . . . , N } lim n →∞ P n ( i, j ) = 1 E (cid:0) τ j (cid:12)(cid:12) Y = j (cid:1) . (7.1)From now on, we fix j in { , . . . , N } and we try to compute E (cid:0) τ j (cid:12)(cid:12) Y = i (cid:1) .The idea is to study the behavior of the Markov chain until the time τ j .To do so, we introduce the companion matrix G defined by ∀ i, k ∈ { , . . . , N } G ( i, k ) = E i (cid:16) τ j − X n =0 { Y n = k } (cid:17) . The matrix G is called the potential matrix associated with the restrictionof P to { , . . . , N } \ { j } . The quantity G ( i, k ) represents the averagenumber of visits of the state k before reaching the state j when startingfrom i . We introduce also the matrix H given by ∀ i, k ∈ { , . . . , N } H ( i, k ) = ( k = j k = j . H describes the distribution of the exit point from the set { , . . . , N } \ { j } . In our case, it is necessarily the Dirac mass on j , yetin the general case, the matrix H is more involved! The three matrices P, G, H are linked through a simple identity.
Lemma 7.2
Denoting by I the identity matrix, we have GP = H + G − I .
Proof.
The matrix G encodes the behaviour of the process until ithits j . Multiplying on the right G by the transition matrix P amounts toperform one further step of the process. This step might either stay inside { , . . . , N } \ { j } , in which case we recover the matrix G − I , or it mightland in j , and this is where the matrix H enters the game. Let us makethis argument rigorous. We have to check that ∀ i, k ∈ { , . . . , N } GP ( i, k ) = H ( i, k ) + G ( i, k ) − I ( i, k ) . For i, k ∈ { , . . . , N } , we compute GP ( i, k ) = X ≤ ℓ ≤ N G ( i, ℓ ) P ( ℓ, k )= X ≤ ℓ ≤ N E i (cid:16) X n ≥ { τ j >n } { Y n = ℓ } (cid:17) P ( ℓ, k )= X ≤ ℓ ≤ N X n ≥ P i (cid:16) τ j > n, Y n = ℓ (cid:17) P (cid:0) Y n +1 = k (cid:12)(cid:12) Y n = ℓ (cid:1) = X n ≥ X ≤ ℓ ≤ N P i (cid:16) τ j > n, Y n = ℓ, Y n +1 = k (cid:17) = P ( i, j ) + X n ≥ P i (cid:16) τ j > n, Y n +1 = k (cid:17) . We consider now two cases. If k = j , the formula becomes GP ( i, j ) = X n ≥ P i (cid:0) τ j = n + 1 (cid:1) = 1 = H ( i, j ) + G ( i, j ) − I ( i, j ) . If k = j , the formula becomes GP ( i, k ) = P ( i, k ) + X n ≥ P i (cid:16) τ j > n + 1 , Y n +1 = k (cid:17) = G ( i, k ) − I ( i, k ) . This ends the proof, since H ( i, k ) = 0 in this case. (cid:3)
14e complete finally the proof of theorem 1.4. We multiply the formula oflemma 7.2 by P n and we sum from 0 to m to obtain G − GP m +1 = m X n =0 (cid:0) P n − HP n (cid:1) . We focus on the coefficients ( i, j ) of the matrices and we send m to ∞ :lim m →∞ (cid:0) G ( i, j ) − GP m ( i, j ) (cid:1) = X n ≥ (cid:0) P n ( i, j ) − P n ( j, j ) (cid:1) . Now G ( i, j ) = 0 and from the convergence (7.1), we havelim m →∞ GP m ( i, j ) = (cid:16) N X k =0 G ( i, k ) (cid:17) × E (cid:0) τ j (cid:12)(cid:12) Y = j (cid:1) . Noticing that N X k =0 G ( i, k ) = E (cid:0) τ j (cid:12)(cid:12) Y = i (cid:1) , and putting together the previous identities, we obtain the formula statedin theorem 1.4. It then suffices to replace the probabilities with their ex-pression to get theorem 1.5. References [1] Raphael Cerf. Critical population and error threshold on the shaap peaklandscape for a Moran model.
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Americanmathematical monthly , 54(7):369-391, 1947.[4] Jacques Neveu. Chaˆınes de Markov et th´eorie du potentiel.