Analysis of some entrance probabilities for killed birth-death processes

Analysis of some entrance robabilities for killed birth-death rocesses Master s Thesis O.J.G. van der Velde Suervisor: Dr. F.M. Sieksma July 5, 207 Mathematical Institute, Leiden University

Contents Introduction 3 2 The model 4 2. Formulas........................................ 5 2.2 Examles........................................ 7 3 Observations and results for n and x n 9 3. Observations based on Excel............................. 0 3.. Observations of the grah of n........................ 0 3..2 Observations of the grah of x n........................ 2 3..3 Heuristic for x n................................. 2 3.2 Monotonicity and convergence results for the M M c -model, with c N { }. 3 3.3 The sloe of n..................................... 6 3.3. Sloe of the M M -model.......................... 6 3.3.2 Sloe of the M M c -model, with c N................... 8 4 Convergence of n for a general birth-death rocess 20 4. Alication of Theorem 4. art 3.......................... 2 4.2 Proof of arts 2 and 3 of Theorem 4......................... 22 4.3 More monotonicity and convergence results for the M M c -model (with c N). 30 5 Convergence of q n for a general birth-death rocess 35 5. Part 2 of Theorem 5.2................................. 36 5.. Stochastic monotonicity............................ 4 5..2 Alication of stochastic monotonicity.................... 44 5.2 Proof of art 2 of Theorem 5.2............................ 5 6 Summary of the theoretical results in [] 52 6. Maximum over an exonential interval; initial and terminal state given...... 52 6.2 Maximum over a deterministic interval........................ 55 7 Discussion 58

List of Figures The grahs of n for n {0,,..., 25} for M M c, c {,..., 5, }, with µ =, n = 3 4 for all n N 0, and τ = 2............................ 9 2 The grahs of n for n {0,,..., 25} for M M c, c {,..., 5, }, with µ =, n = 3 4 for all n N 0, and τ = 000.......................... 0 3 The grah of n for n {0,,..., 00} with as in Eqn. (4), n = 2 for all n N 0, and τ = 3 2.................................... 4 The grahs of x n for n {0,,..., 25} for M M c, c {,..., 5, }, with µ =, n = 3 4 for all n N 0, and τ = 000.......................... 3 5 The grah of n for n {0,,..., 50} for M M, with µ =, n = 3 4 for all n N 0, and τ = 000.................................. 2 List of Tables Fixed and µ, with various τ............................. 59 2 Fixed µ and τ, with various............................. 59 3 Fixed µ and τ, with various............................. 59 4 Fixed µ and τ, with various............................. 60 5 Fixed µ and τ, with various............................. 60 6 36 situations where, µ and τ can vary........................ 6 2

Introduction The introduction in the aer of Ellens et al., from now on notated as [], describes analytical techniques to analyze metrics that are related to random fluctuations between consecutive observations. In this thesis, we will take a closer look at the robability that the rocess, starting at level n, reaches level n + before an indeendent exonentially distributed time T exires. Also the robability that the rocess, starting at level n, reaches level n before an exonentially distributed time T exires, is examined. These exonentially distributed times T are studied in [] to determine the distribution of the maximum of a birth-death rocess over a deterministic interval with given initial and terminal conditions. The robabilities to reach a higher, resectively lower, state before an exonentially distributed time T exires, both satisfy recursive formulas that look interesting to analyze. The interesting art was that numerical results in Excel seemed to imly that these robabilities do not always converge. Therefore, in this thesis we will analyze their limiting behaviour. The remainder of this thesis is as follows. First we will describe the model of the birth-death rocess that we will use throughout this thesis. This chater will also give the basic formulas we got from [] and we introduce the basic concets and formulas. Also an examle will be given for which we will consider the relation between two sequences of robabilities. After that, we have enough background to analyze the recursive formulas given in Section 2. of []. Chater 3 will give observations of the grah of the recursive formula for the robability that the rocess arrives at a state n + before time T given that the rocess starts in state n, this robability will be denoted by n. The observations will be based on examles in Excel of the M M c -model, with c {, 2,..., 5}, and M M -model. The remainder of the chater will give counterexamles and roofs of observations. Chater 4 and Chater 5 will state the main results of this thesis. These main results are limiting results for the robabilities n and q n, where q n is the robability that the rocess X arrives at state n before time T given that the rocess starts in state n. Chater 5 also introduces the concet of stochastic monotonicity. This will be needed in the roof of the main result of this chater. Chater 6 summarizes the results of Sections 2.2 and 2.3 of [] and will give a motivation to the choice of considering a killed birth-death rocess. This thesis concludes with Chater 7. This chater will discuss what we originally lanned to research in this thesis and what we actually researched. 3

2 The model In this chater we will describe the model that will be used in this thesis. For notational uroses, define N := {, 2, 3,...} and N 0 := {0,, 2, 3,...}. Definition 2.. Let the stochastic rocess X = (X t ) t 0 be a birth-death rocess, where X t N 0 denotes the state of the rocess at time t. The birth rate in state n is n > 0, and the death rate in state n is > 0, for n N 0, where µ 0 = 0. Remark 2.2. The stochastic rocess X is a continuous time Markov chain on a discrete state sace. Since we will rove roerties of these functions in later chaters of this thesis, we will give the stes of the derivation of these recursive formulas in section 2. of [] in this chater. Given an initial state X 0, we consider an indeendent, exonentially distributed time T till exiration. Given a level m N 0, we are interested in the robability that the rocess X equals the level m. The goal of this chater is to give the derivation of the robability that the maximum of the stochastic rocess X is equal to m over the interval [0, T ], of exonential length, given that the rocess starts in state i. Remark 2.3. Since we consider the rocess X on the stochastic interval [0, T ], we call ( ) Xt T t 0 a killed birth-death rocess, where Xt T := X t T. If necessary, we will model the killed birthdeath rocess as a birth-death rocess with an extra state, which will be an absorbing state, i.e., the rocess will never leave this state once it has arrived there. We will use this extra state in Chater 5. Before we can rove results for the rocess X, first some notation and definitions need to be introduced. Notation 2.4.. Use m to denote a level. Then, m is a state in the birth-death rocess (X t ) t 0 about which we want to know several robabilities. (For examle, we could be interested in the robability that the birth-death rocess (X t ) t 0 is in a state larger than m.) 2. Let t 0. Let X t := su s [0,t] X s denote the running time maximum associated with rocess X. 3. Let t 0. Let ˆX t := inf s [0,t] X s denote the running time minimum associated with rocess X. 4. Let T be an exonentially distributed random variable with rate τ > 0, indeendent of rocess X. 4

Definition 2.5. Let n N 0.. Let n := P ( XT n + X 0 = n ) be the robability that the rocess X, starting at level n, reaches level n + before time T. 2. Let n := n = P ( XT = n X 0 = n ) be the robability that the rocess X, starting at level n, does not reach level n + before time T. ( ) 3. Let q n := P ˆXT n X 0 = n be the robability that the rocess X, starting at level n, reaches level n before time T. 4. Let i N 0. Let r m,i := P ( XT = m X 0 = i ) be the robability that the maximum of rocess X equals m over the interval [0, T ] (of random duration), given that it starts in state i. Definition 2.6.. Define + := lim su n n and := lim inf n n. 2. Define q + := lim su n q n and q := lim inf n q n. 3. Define α := lim n n and β := lim n τ, rovided these limits exist. Definition 2.7. Define the robability that the maximum of the birth-death rocess (X t ) t 0 does not exceed level m, given its initial state i and its final value j at a deterministic time t by q m,i,j,t := P ( Xt m X 0 = i, X t = j ). To analyze q m,i,j,t, [] first considers the maximum over an exonential interval given the initial state of the rocess. The authors deduce two recursive formulas describing the robability that the maximum and minimum of the rocess X, starting at level n, reaches level n + and level n, resectively, before an exonential time T exires. These robabilities are given in arts and 3 of Definition 2.5 as ( n ) n N0 and (q n ) n N0. 2. Formulas The rocess X is a birth-death rocess, hence it satisfies the memoryless roerty. For a birthdeath rocess, at rate n a birth occurs when the rocess is in state n, and at rate a death. In this model the rocess is only considered until the exonential time T exires. Therefore, the rocess moves from state n to another state (where reaching time T stos the rocess) at rate n + + τ. We can model this by extending the state sace by an extra state,, to which the rocess jums to with rate τ. However, in this chater we do not yet need to consider the extended state sace. Hence the robability to reach state n + equals µ robability to reach state n equals n n+τ+ (for n N 0 ). n n+τ+ and the Since T d = ex(τ), with τ > 0, it holds that T satisfies the memoryless roerty. This means that after a jum of the birth-death rocess on time t < T, the remaining time T t is again exonentially distributed with rate τ. Using the memoryless roerty of the birth-death rocess and the memoryless roerty of T gives that 0 = 0 0 + τ 5

and for n N it holds that n = Analogously, it holds for all n N 0 that q n = n + n n. n + τ + n + τ + n + q n+ q n. n + τ + n + τ + For all i > m it holds that r m,i = P( X T = m X 0 = i) = 0, because in this case X T = su s [0,T ] i > m. In the next lemma it will be roven that r m,i = ( i... m ) m for all i m. Lemma 2.8. For all i m, it holds that r m,i = ( i... m ) m. () Proof. Throughout this roof, we use the memoryless roerty of the random variable T. Let i = m. Then, r m,i = r m,m = P( X T = m X 0 = m) = m. Let i = m. Then, r m,i = r m,m = P( X T = m X 0 = m ) = P( X T m X 0 = m ) P( X T = m X 0 = m), since to arrive in state m, there must be a birth before time T when the rocess is in state m, and, moreover, no birth may occur in state m before time T. In other words, r m,i = m m. Let i < m. Since we consider a birth-death rocess, to reach state m while starting in state i before time T, the rocess must first reach the states i+,..., m 2 before time T, before being able to jum to state m. Thus, Recursively, we get that r m,i = P( X T = m X 0 = i) = P( X T m X 0 = i) P( X T = m X 0 = m). r m,i = m 2 n=i [ P( XT n X 0 = n) ] m m = ( i... m ) m. Hence, for all i m Eqn. () holds. 6

Lemma 2.9. It holds that n = Proof. For n = 0 it holds that 0 = P( X T X 0 = 0) = Let n. Then, n = which we needed to show. Lemma 2.0. It holds that n n+τ+ + { 0 0+τ, n = 0, n n+τ+( n ), n. (2) 0, because µ 0+t 0 = 0. n+τ+ n n. Rewriting this equation gives ( n + τ + ) n = n + n n ( n + τ + n ) n = n ( n + τ + ( n )) n = n n n = n + τ + ( n ), { 0, n = 0, q n =, n. (3) n+τ+( q n+) Proof. For n = 0 it holds that q 0 = P( ˆX T X 0 = 0) = 0, since X t 0 for all t 0. Let n. It holds that q n = which we needed to show. n+τ+ + n n+τ+ q n+ q n. Rewriting this equation gives ( n + τ + )q n = + n q n+ q n ( n + τ + n q n+ )q n = ( + τ + n ( q n+ ))q n = q n = + τ + n ( q n+ ), 2.2 Examles In this section we consider for two models the relation between the sequence ( n ) n N0 and the sequence (q n ) n N0. Both models have a finite state sace, S = {0,,..., N}. Again, let T be an exonentially distributed random variable with rate τ, indeendent of X. Let the arameters n > 0, for n {0,,..., N }, N = 0, µ 0 = 0 and > 0, for n {, 2,..., N} be given. Consider the birth-death rocess X = (X t ) t 0 on the state sace S with birth- and deatharameters given by n = n > 0, for all n {0,,..., N }, N = 0, µ 0 = 0 and µ n = > 0, for all n {, 2,..., N}. Furthermore, consider the birth-death rocess X 2 = (X 2 t ) t 0 on the state sace S with swaed birth and death arameters given by 2 n = > 0, for all n {0,,..., N }, 2 N = 0, µ2 0 = 0 and µ 2 n = n > 0, for all n {, 2,..., N}. 7

Then it holds that For n {,..., N} it holds that n = Furthermore it holds that 0 = n n + τ and 2 0 = + τ. n n + τ + ( n ) and 2 n = + τ + n ( 2 n ). q N = For n {0,,..., N } it holds that q n = + τ and q2 N = n n + τ. + τ + n ( qn+ ) and n q2 n = n + τ + ( qn+ 2 ). It holds that 0 = qn 2 and that 2 0 = qn. Using induction to n it follows for all n {0,,..., N} that n = qn n 2 and 2 n = qn n. If n =, for all n {0,,..., N }, and =, for all n {, 2,..., N}, then it follows that the rocesses X and X 2 are equal. Hence, n = qn n for all n {0,,..., N}. Note that S = N + is even, if N is off. Vice versa, if S is odd, then N is even. In the latter case it holds that n = qn, n = N 2 If n = µ N n, for all n {0,,..., N }, and N = 0 = µ 0, then it follows that n = q N n for all n {0,,..., N}. 8

3 Observations and results for n and x n When studying the aer [] we became interested in the structural roerties of the formula for the robabilities n, n N 0. Therefore, we decided to numerically analyze six M M c -models, c {,..., 5, }, in Excel to get a better understanding of the behaviour of n as a function of n. For the M M c -model it holds that n = for all n N and { nµ if n < c, = cµ if n c. Notation 3.. For n N 0, c N { }, we use the suerscrit c for the investigated robabilities in the M M c -model. Moreover, we use the suerscrit to indicate the corresonding limits as n tends to. For examle, c n is the robability that the rocess X, starting at level n, reaches level n + before time T in the M M c -model, µ c n is the dearture rate of the M M c -model in state n, and,c := lim n c n. For all the examles we considered, we saw that the grah of n as a function of n is non-increasing and that it has a limiting value. See, for examle, Figures and 2 below. Moreover, we saw that, when lotting the grahs of n for different values of c in one lot, that the grah of n was monotonic. Furthermore, we noticed that the grahs had a short stee art in the grah of n and for the remainder a relatively flat one. If the grah has oints before this stee art of the grah, then the grah changes from relatively flat to stee and then back to relatively flat. Figures and 2 show two examles on realizations of the grahs of n. The lots gave rise to a number of conjectures that we have tried to rove. These conjectures concern the characterization of this stee art of the grah, the limiting value of the grah of n and some monotonic roerties deending on the number of servers c, the birth-rates, the death-rates µ, and the rate τ. Figure : The grahs of n for n {0,,..., 25} for M M c, c {,..., 5, }, with µ =, n = 3 4 for all n N 0, and τ = 2. 9

Figure 2: The grahs of n for n {0,,..., 25} for M M c, c {,..., 5, }, with µ =, n = 3 4 for all n N 0, and τ = 000. 3. Observations based on Excel This section will first focus on observations of the grah of n for the M M c -model, c {, 2,..., 5, } We will describe the examles we considered in Excel and the observations we made using these examles. Then, we focus on the sloe of the grah of n. 3.. Observations of the grah of n In Excel we lotted the grah of n for the M M c -model, c {,..., 5, } for n = 0,,..., 00 and for various values of the arameters, µ or τ. The observations below are based on the grahs in these different Excel sheets. Before resenting the observations, we will first discuss what examles we used to base these observations on. First we fixed and µ to study the effects of τ. We came u with the 7 situations in Table because we wanted to consider the influence of τ on the M M c -model, with c {, 2,..., 5}. Therefore, we chose = 0.75 and µ =, that way we made sure that < cµ and hence the queue length of this queueing system will not exlode (see Chaters 3 and 4 in [4]). Then we wanted to see what would haen if τ became small. Therefore we considered the situations given in Table of the Aendix. After researching the influence of τ on the system, we wanted to see what would haen if we would change the ratio /µ. This made us choose µ = and considered for four situations of τ what would haen if we would decrease (with < µ). The arameter combinations we considered here can be found in in the Tables 2, 3, 4 and 5 of the Aendix. Consider the M M c -model, c {,..., 5, }. examles are stated in Conjecture 3.2. Conjecture 3.2.. the sequence ( c n) n N0 is non-increasing in n. 2. the sequence ( c n) n N0 has a limit. The conjectures we made using the different 0

3. c n = c+ n for all n N 0 with n c and c n c+ n for all n > c. It looks like c n = n for all n N 0 with n c and c n n for all n > c. So, it seems that the grahs of n (for the same n N 0 ) are non-increasing as c is increasing. 4. lim n c n lim n c+ n. This is the limit version of Conjecture 3.2.3. 5. lim c,c = 0. 6. τ n rates. τ2 n for τ, τ 2 > 0 such that τ < τ 2, where τ and τ 2 denote two distinct extinction A counterexamle for Conjecture 3.2. can be given. However, Lemma 3.5 shows the validity of Conjecture 3.2. for the M M c -model, c N 0 { }. Consider a birth-death rocess with n = 0.5 for all n N 0, τ =.5 and 0 if n = 0, = n 2 if n is even, n 0, ln(n) + if n is odd. Then it holds for all n N that > 0 (so n is well-defined for all n N) and it holds that as n. Therefore, α := lim n n = 0 and β τ := lim n = 0. The grah o n shows that it does not hold that the sequence ( n ) n N0 is non-increasing in n. Hence, this is not a generic roerty of birth-death rocesses. The corresonding grah can be found in Figure 3. Moreover, this examle also seems to give a counterexamle of Conjecture 3.2.2. However, as we will see in Chater 4, also for this examle lim n n exists. (4) Figure 3: The grah of n for n {0,,..., 00} with as in Eqn. (4), n = 2 for all n N 0, and τ = 3 2. Moreover, for the M M c -model, c N { }, we can rove that the sequence ( c n) n N0 is non-increasing in n. This will be roven in Lemma 3.5. Conjecture 3.2.2 will be roven under some mild conditions in Chater 4. The other conjectures will be roven for the M M c -model in this chater.

3..2 Observations of the grah of x n Definition 3.3. Define the sloe of the grah as the difference between two consecutive robabilities n, i.e., x n := n n+ for all n N 0. In this subsection we will study the grah of x n. We will study two asects: monotonicity and the behaviour around its maximum. The latter is motivated by our interest in the art of the grah c n, where it is steeest. First, we considered the same examles as in Section 3.. for the sloe of the grah of c n. In this way, we can take into account the observations made in Section 3... However, we also considered an examle with =, µ = 0.5 and τ = 0.0, and an examle with = 0.75, µ = 0.5 and τ = 0.0. This is related to the question of the monotonicity of the grah of x c n. Recall Notation 3.. The conjectures we made using the different examles are given in Conjecture 3.4. Conjecture 3.4.. for the M M c -model, c N { }, the grah of x c n is not always monotone in c. 2. max n N0 x c n is non-decreasing as a function of c. 3. n := arg max n N0 x c n denotes the oint where the grah of c n has the steeest descent, the steeest descent is at c n or c n +. 4. arg max n N0 x c n is non-increasing as c decreases. The goal of these observations is to come u with a heuristic method to evaluate the to of the grah of x c n. The grahs of x c n in Figure 4 are an examle showing that the grahs of x c n can intersect for different values of c. Hence, this examle shows that monotonicity for x c n cannot be roven for the M M c -models, c N { }. 3..3 Heuristic for x n For notational convenience, we leave the suerscrit c out of the notation. In this subsection we will describe our attemts to find a heuristic method to redict n = arg max n N0 x n for M M c -models, c N { }. Excel calculations for the sloe of n for M M c, c N { }. In this subsection we will analyze some Excel sheets we built. These Excel sheets calculate the oint n = arg max n N0 x n. In these Excel sheets multile examles are considered and can be found in Table 6 in the Aendix. Note that these Excel sheets could easily be used to calculate n for other arameters for M M c, with c {, 2, 3, 4, 5}, and M M. This can be done easily by just changing the resective values for, µ or τ. 2

Figure 4: The grahs of x n for n {0,,..., 25} for M M c, c {,..., 5, }, with µ =, n = 3 4 for all n N 0, and τ = 000. First, the values for n are calculated in these Excel files. These values are then used to calculate the values for x n. Finally, n is calculated by a formula that comares the calculated maximum with the Excel cells containing the values of x n and this formula returns how many cells are above the number we sought including the cell that contains this maximum. Since we start counting from n = 0 we have to subtract to get the correct value for n. Excel observations for the sloe of n for M M c, c N { }. This subsection will contain some observations based on the results of the examles.. n is non-decreasing as increases. This observation is based on situations 5. 2. n is non-decreasing as µ increases. This observation is based on situations 6 20. 3. n is non-increasing as τ increases. This observation is based on situations 2 34. 4. Increasing the number of servers c does not always lead to n non-decreasing or nonincreasing. An examle is as follows. Let =, µ = 0.5 and τ = 0.0. For M M 4 and M M it can be calculated that n = 6, while for M M 5, n = 5. The conclusion is that since increasing the number of servers does not always lead to n nondecreasing or non-increasing, this did not succeed. Therefore, this could be a toic of further research. 3.2 Monotonicity and convergence results for the M M c -model, with c N { } Consider the M M c -model, c N { }. Let τ > 0 be given. Assume that µ > 0 and that < cµ. Recall the definition of n, Eqn. (2). Next, we will show that the sequence ( n ) n N is a non-increasing sequence for the M M c -model, c N { }. Then, an easy consequence is that the sequence ( n ) n N has a limit. 3

Lemma 3.5. Let, µ 0 and let τ > 0. For the M M c -model, c N { }, it holds that. n = 0, n N 0, if = 0 and µ 0. Moreover, the limit exists and = lim n n = 0. 2. n = +τ, n N 0, if > 0 and µ = 0. Moreover, the limit exists and = lim n n = 3. n > n+, n N 0, if > 0 and µ > 0. Moreover, = lim n n exists. Proof.. Let = 0 and µ 0. Then, by Eqn. (2), it immediately follows that n = 0 for all n N 0. Hence, the sequence ( n ) n N0 has a limit, namely lim n n = 0. 2. Let > 0 and µ = 0. Then, it follows immediately for all n N 0 that n = +τ. So, then it holds for all n N 0 that n = n+. Thus, the sequence ( n ) n N0 is constant, and hence, it has a limit, namely lim n n = +τ. 3. Let, µ > 0. Note that since > 0 and τ > 0 it holds that 0 = +τ (0, ). It suffices to consider two situations, namely 0 n < c, and n c. Note that for c = only the first case has to be considered. Suose that 0 n < c. It holds that 0 = ( ) > + τ =, where (*) holds since µ > 0 and 0 (0, ). + τ + µ( 0 ) Since µ > 0, it then follows that µ( 0 ) µ( ). Moreover, since 0 and τ > 0, it therefore holds that + τ + µ( 0 ) + τ + µ( ). Hence,. Therefore, we have that +τ+µ( 0) +τ+µ( ) = + τ + µ( 0 ) + τ + µ( ) > + τ + 2µ( ) = 2. So, it holds that > 2. Analogously, n > n+ for all 2 n < c. So, for all 0 n < c we have that n > n+. Suose that n c. Above, we already have roven that c > c. Since µ > 0, it therefore follows that cµ( c ) < cµ( c ). Moreover, since > 0 and τ > 0, we also have that + τ + cµ( c ) < + τ + cµ( c ). Therefore, c = + τ + cµ( c ) > + τ + cµ( c ) = c+. +τ. 4

So, it holds that c > c+. Similarly we get for all n c that n > n+, thus it holds for all n c that n is strictly decreasing. Thus, we have roven for all n N 0 that n > n+. For all n N 0 it holds that n is a robability, hence n [0, ] for all n N 0. Thus the sequence ( n ) n N0 is bounded. Then, by the Monotone Convergence Theorem, it follows that the sequence ( n ) n N0 has a limit. Remark 3.6. In Theorem 4. we will derive an exression for, when, µ > 0. Remark 3.7. If we would have that 0 = µ 0 + and n = 0 for all n N 0, then it would follow analogously that the sequence of ( n ) n N0 is non-increasing. It is even strictly decreasing if > 0 and > 0 for all n N. Remark 3.8. To be able to aly the Monotone Convergence Theorem we need a non-increasing or a non-decreasing sequence ( n ) n N0. For a general birth-death rocess we do not necessarily have this roerty. Therefore we cannot use the Monotone Convergence Theorem to rove in general that the sequence ( n ) n N0 has a limit. Recall Notation 3.. Lemma 3.9. Let, µ, τ > 0. Let c N. Then, c n = c+ n for all n c, c n > c+ n for all n > c. Proof. It holds that µ c n = nµ for all n c and µ c n = cµ for all n c. We use induction on n to rove the statement of the lemma. Let n = 0. Then it holds that c 0 = = c+ 0. Hence, c 0 = +τ = c+ 0. Suose that c n = c+ n, n < c, then c n = + τ + nµ( c n ) = + τ + nµ( c+ n ) = c+ n. Next, let n = c. By assumtion c n = c+ n. It holds that µc n = cµ = µ c+ n. Hence, c n = c+ n. Let n = c +. By assumtion c n = c+ n. It holds that µc n = cµ and µ c+ n = (c + )µ. It holds that cµ < (c + )µ, and thus µ c n < µ c+ n. Then, +τ +µ c n( c n ) = +τ +cµ( n c+ ) < +τ +(c+)µ( c+ n ) = +τ +µc+ n ( c+ n ). Hence, c n = + τ + µ c n( c n ) > + τ + µ c+ n ( c+ n ) = c+ n. 5

Finally, assume that c n > c+ n, with n > c +. It holds that µ c n = cµ < (c + )µ = µ c+ n. Then it follows that µ c n( c n ) < µ c+ n ( c+ n ), since µc n < µ c+ n and since c n < c+ n. This gives that c n = + τ + µ c n( c n ) > + τ + µ c+ n ( c+ n ) = c+ n. So for all n N 0 with n c it holds that c n = c+ n and for all n > c it holds that c n > c+ n. Corollary 3.0. Let c, c 2 N such that c < c 2. Then, c n c n = c2 n for all n c, > c2 n for all n > c. Proof. Using Lemma 3.9 iteratively gives the statement we need to show. Lemma 3.. Let c N. Then, c n = n for all n c, c n > n for all n > c, lim c c n = n. Proof. Let c N. Let n c. Then, c n = c2 n for all c 2 > c. Also, c2 n = n, since the birth- and death-rates are equal for M M c 2 and M M for the states {0,,..., n}. Now, let n > c. Fix c 2 > n. By Corollary 3.0 we get c n case, we get c2 n = n. Then it holds for all n N 0 with n c that c n = c2 > c2 n = n, which concludes the first art of the roof. c n > c2 n, and, similarly to the revious n = n and for all n > c it holds that Since for all n c it holds that c n = n, it immediately follows that lim c c n = n. 3.3 The sloe of n This subsection will again focus on the sloe of the grah of n, in other words, we will focus on the maximum of the grah of x n. The question is, whether this grah has a unique maximum. We will give a result about conditions when the grah of x n is increasing and decreasing. First we will focus on the M M -model. Then we study the M M c -model, c N. From now on, we omit the suerscrit as it will be clear which model will be considered. Recall Definition 3.3, i.e., x n := n n+, n N 0. 3.3. Sloe of the M M -model Consider the M M -model. Let, µ, τ > 0. Then, Eqn. (7) becomes { +τ if n = 0, n = +τ+nµ( n ) if n N. 6

Lemma 3.2. Consider the M M -model. Let, µ, τ > 0. Then, { µ 0 (+τ)(+τ+µ( x n =, n = 0, 0)) n( µ n n)+ n (+τ+nµ(, n. n ))(+τ+(n+)µ( n)) Proof. Let n = 0. Then, Let n N. Then, x 0 = 0 = + τ + τ + µ( 0 ) = + τ + µ( 0) ( + τ) ( + τ)( + τ + µ( 0 )) 0 = µ ( + τ)( + τ + µ( 0 )). (5) x n = n n+ = + τ + nµ( n ) + τ + (n + )µ( n ) = + τ + (n + )µ( n) ( + τ + nµ( n )) ( + τ + nµ( n ))( + τ + (n + )µ( n )) (n + )µ( n ) nµ( n ) = ( + τ + nµ( n ))( + τ + (n + )µ( n )) n( n n ) + n = µ ( + τ + nµ( n ))( + τ + (n + )µ( n )). (6) Lemma 3.3. Consider the M M -model with, µ, τ > 0. Let n N be such that x n n+2 n x n. Then it holds that x n x n+. Proof. Assume that n N is such that x n n+2 n x n. Rewriting Eqn. (6) gives that n( n n ) + n x n = µ ( + τ + nµ( n ))( + τ + (n + )µ( n )) n( n n ) + n µ ( + τ + nµ( n+ ))( + τ + (n + )µ( n )) n( n n ) + n µ ( + τ + (n + 2)µ( n+ ))( + τ + (n + )µ( n )), where the first inequality holds by Lemma 3.5 (i.e., n > n+ ). To rove that x n x n+ it remains to be roven that n( n n ) + n (n + )( n n+ ) + n+. At the contrary, assume that n( n n ) + n < (n + )( n n+ ) + n+. Then, n( n n ) + n (n + )( n n+ ) + n+ < 0. 7

This imlies that n n n n + n n n + n n+ n + n+ + n+ < 0 n n 2(n + ) n + (n + 2) n+ < 0 n( n n ) + (n + 2)( n+ n ) < 0 nx n (n + 2)x n < 0 nx n < (n + 2)x n, which yields a contradiction with the assumtion that x n n+2 n x n. Hence, This gives x n µ n( n n ) + n (n + )( n n+ ) + n+. n( n n ) + n ( + τ + (n + 2)µ( n+ ))( + τ + (n + )µ( n )) (n + )( n n+ ) + n+ = x n+, thus comleting the roof of the lemma. 3.3.2 Sloe of the M M c -model, with c N Consider the M M c -model, with c N. Again, let, µ, τ > 0. Therefore, Eqn. (7) becomes +τ, n = 0, n = +τ+nµ( n ), n c, +τ+cµ(, n > c. n ) Lemma 3.4. For the M M c -model, with c N, it holds that µ 0 (+τ)(+τ+µ(, n = 0, 0)) n( x n = µ n n)+ n (+τ+nµ( n ))(+τ+(n+)µ( n)), n < c, cµ n n (+τ+cµ(, n c. n ))(+τ+cµ( n)) Proof. Consider n = 0. Then, analogously to deriving Eqn. (5), we get that x 0 = µ 0 ( + τ)( + τ + µ( 0 )). Let n N such that n < c. Then, analogously to deriving Eqn. (6), we get that x n = µ n( n n ) + n ( + τ + nµ( n ))( + τ + (n + )µ( n )). 8

Now, let n N such that n c. Then, x n = n n+ = + τ + cµ( n ) + τ + cµ( n ) = + τ + cµ( n) ( + τ + cµ( n )) ( + τ + cµ( n ))( + τ + cµ( n )) n n = cµ ( + τ + cµ( n ))( + τ + cµ( n )). Lemma 3.5. Consider the M M c -model, with c N. Let, µ, τ > 0. If n < c and x n n+2 n x n, then x n x n+. If n c and x n x n, then x n x n+. Proof. Let n < c and x n n+2 n x n, then, analogously to the roof of art of Lemma 3.3, it follows that x n x n+. Now, let n c. Assume that x n x n. Then it holds that n n x n x n+ = cµ ( + τ + cµ( n ))( + τ + cµ( n )) n n+ cµ ( + τ + cµ( n ))( + τ + cµ( n+ )) x n cµ ( + τ + cµ( n+ ))( + τ + cµ( n )) x n cµ ( + τ + cµ( n ))( + τ + cµ( n+ )) 0, where the first inequality holds by Lemma 3.5 (i.e., n > n+ ), and where the second inequality holds by the assumtion that x n x n. This concludes the roof of the lemma. 9

4 Convergence of n for a general birth-death rocess In this chater we will rove that for a birth-death rocess the limit lim n n exists under mild conditions. Before we rove this theorem, we will give an examle to show an alication of the theorem. One of the mild conditions is that the limit of the ratio between the birth and death arameters must exist. The other mild condition is that the limit of the ratio between the exonential rate τ and the death arameters must exist. Theorem 4.. Suose that α = lim n n 0 and β τ = lim n 0 both exist. Let n, n N 0, be given by { 0 n = 0+τ if n = 0, n (7) n+τ+( n ) if n N.. Suose α = 0 and β > 0. Then, lim n n = = 0. 2. Suose α > 0 and β 0. Then, lim n n = exists with the smallest solution of the equation x 2 (α + β + ) x + α = 0, (8) in other words = (α + β + ) (α + β + ) 2 4α. 2 3. Suose α = β = 0. Then, lim n n = = 0. The roof for arts 2 and 3 of the theorem requires more work than the roof for art. Therefore we will first rove art of the theorem. Proof of Theorem 4. art. Let α = 0 and let β > 0. Let n, n N 0 be given by Eqn. (7). Note that > 0 for all n N. Therefore, we can rewrite Eqn. (7), for all n N, by dividing by. This gives Hence, since α = 0 and β > 0. Furthermore, it holds that n = n = lim inf n lim inf n n n + τ + ( n ) n n + τ n n + τ +. (9) + = α α + β = 0, (0) + Hence, n = n n + τ + ( n ) + = lim su n lim su n n n n + τ n n + τ. = α α = 0. () + β Combining Eqns. (0) and () yields lim n n = 0, which comletes the roof of art of Theorem 4.. 20

4. Alication of Theorem 4. art 3 As an illustration of art 3 of Theorem 4., we consider the M M -model. Recall that τ > 0, n = > 0 for all n N 0, µ 0 = 0 and = nµ > 0 for all n N. Then, α = lim n n = 0 and β τ = lim n = 0. An examle of the grah of n is given in Figure 5. Figure 5: The grah of n for n {0,,..., 50} for M M, with µ =, n = 3 4 for all n N 0, and τ = 000. Since the sequence ( ) n N0 has the roerty that it is monotonically increasing to, we can show via a direct roof that lim n n = 0. The roof of Lemma 4.2 was initially based on the roof of Lemma 3 in [2], but later we showed Lemma 3.5 art 3 and the roof below uses this lemma. Lemma 4.2. For the M M model it holds that lim n n = 0. Proof. By Lemma 3.5 we know that the sequence ( n ) n N0 all n N it holds that n 0. Therefore, is a decreasing sequence. Hence, for since n 0. n+ = + τ + (n + )µ( n ) + τ + (n + )µ( 0 ), Now we can take the limit of n to of the right-hand side. This gives lim n + τ + (n + )µ( 0 ) = 0. Thus, = lim n n = 0. 2

4.2 Proof of arts 2 and 3 of Theorem 4. The roof of the second art of Theorem 4. consists of multile stes. These stes are formulated as lemmas that will be roven in this subsection. The final roof of Theorem 4. art 2 can then be found at the end of this subsection and this roof will use all lemmas in this subsection. Throughout this section, we assume that α > 0. Recall Definition 2.6., i.e., + := lim su n n and := lim inf n n. Lemma 4.3. Let α > 0 and β 0. It holds that +, > 0. Proof. Let α > 0 and β 0. Let n, n N 0 be given by Eqn. (7). Analogously to the roof of the first art of Theorem 4. we get again Eqn. (9). Hence, also Eqn. (0) holds, which gives that α α + β + > 0, since α > 0. Because = lim inf n n > 0 it follows immediately that n > 0 for all n N 0. Moreover, since + we also have that + > 0, which concludes the roof. Lemma 4.4. Let α > 0. It holds that + and are both solutions of Eqn. (8). in other words, (α + β + ) (α + β + ) 2 4α (α + β + ) 2 4α +, 2, (α + β + ) + 2. Proof. By virtue of Lemma 4.3 we have that +, > 0. The roof of that lemma even yields that n > 0 for all n N 0. Hence, dividing by +, and n, n N 0, is allowed. Because ( n ) n N0 is a sequence of robabilities, it holds that + and both exist with +, [0, ]. Hence, there exist a subsequence ( nk ) k ( n ) n N0 such that nk + as k and there exists a subsequence ( ml ) l ( n ) n N0 such that ml as l. Recall Eqn. (7), so for all n N 0 that n+ = nk + = n+ n++τ++( n). Therefore, nk + nk + + τ + k +( nk ). Note that lim k nk + exists, since lim k nk Furthermore, it holds that exists. lim n k + lim su k k su j n k + j lim su n = +. n 22

Thus, lim k nk + +, and so So, we have that + Rewriting Eqn. (7) yields Therefore, + nk + lim k nk + + τ + k +( nk ) = lim k nk + k + nk ++τ+k +( nk ) k + α = α + β + lim k nk α = α + β + +. α α +β + +. This can be rewritten as + ( α + β + +) α, ( + ) 2 + (α + β + ) + α 0, ( + ) 2 (α + β + ) + + α 0. (2) n = n( n + τ + ) n n. ( ) nk = nk n k ( nk + τ + k ) nk nk k + τ k + n k k =. (3) k nk nk Because lim k nk = + exists, it follows that lim k nk also exists. It holds that lim n k lim su k k su j n k Since lim k nk = + (α +β +) α, we have that + Therefore, which yields j lim su n = +. n + (α + β + ) α + +. ( + ) 2 + (α + β + ) α, ( + ) 2 + (α + β + ) + α 0. (4) Combining Eqns. (2) and (4) yields that + is a solution of Eqn. (8). Hence, (α + β + ) (α + β + ) 2 4α +, (α + β + ) + (α + β + ) 2 4α 2 2. Analogously (only the sign of the inequalities changes, because the limit inferior is considered instead of the limit suerior), it follows that is also a solution of Eqn. (8). This concludes the roof of the lemma. 23

Remark 4.5. Since it holds that + and both exist, it necessarily must hold that both solutions of the quadratic exression exist. Since β 0 we only need to consider two situations, namely β > 0 and β = 0 in order to be able to rove the theorem. First consider the case where β > 0. If (α + β + ) + (α + β + ) 2 4α then it holds that 2 > (α + β + ) (α + β + ) 2 4α 0, 2 (5) + = = (α + β + ) (α + β + ) 2 4α, 2 since +, [0, ], which is the smallest solution of Eqn. (8). Lemma 4.6. Let α, β > 0. Then Eqn. (5) holds. Proof. First we will rove that (α + β + ) + (α + β + ) 2 4α > (6) 2 holds. To show this, it is sufficient to rove that (α + β + ) + (α + β + ) 2 4α > 2. Suose that is not true. Then, (α + β + ) 2 4α 2 (α + β + ) = α β. This imlies that in other words, ((α + β ) + ) 2 4α ( (α + β )) 2, Rewriting yields (α + β ) 2 + 2 (α + β ) + 4α 2 (α + β ) + (α + β ) 2. 4 (α + β ) 4α 0, so that β 0. This contradicts the fact that β > 0. Hence, Eqn. (6) holds. Next, we will rove that 0 (α + β + ) (α + β + ) 2 4α To show this, it is sufficient to rove that 0 (α + β + ) The fact that α, β > 0, imlies that (α + β + ) 2 (α + β + ) 2 4α 0.. (7) (α + β + ) 2 4α 2. 24

To rove that (α + β + ) (α + β + ) 2 4α 2, we will assume the contrary. In other words, (α + β + ) 2 4α > 2 (α + β + ), or, equivalently, ((α + β ) + ) 2 4α < α + β. Similarly as before, this imlies that β < 0. Thus, we have arrived at a contradiction, so that Eqn. (7) holds. So, if β > 0, then it holds that (α + β + ) + (α + β + ) 2 4α which concludes the roof of the claim. 2 > (α + β + ) (α + β + ) 2 4α 0, 2 Corollary 4.7. Let α > 0. If β > 0, then + = (α + β + ) (α + β + ) 2 4α =, 2 which is the smallest solution of Eqn. (8). Proof. By virtue of Lemma 4.4 and Lemma 4.6, and since +, [0, ], we can conclude that for β > 0 it holds that + = (α + β + ) (α + β + ) 2 4α =, 2 which is the smallest solution of Eqn. (8). Corollary 4.7 concludes the roof of Theorem 4. in case β > 0. Now it remains to consider the case β = 0. Lemma 4.4 yields that + and are both solutions to Eqn. (8), which reduces to x 2 (α + ) x + α = 0, (8) so that (α + ) (α + ) 2 4α +, 2 { (α + ) (α = ) 2 2, (α + ) +, (α + ) + (α ) 2 2 (α + ) 2 4α 2 }. (9) Consider three situations, namely α =, α > and 0 α <. These three situations cover all ossible situations, since α 0. These situations will be considered in Lemma 4.8 and Lemma 4.9, resectively. 25

Lemma 4.8. Let β = 0. Let α. Then it holds that + = =, which is the smallest solution of Eqn. (8). Proof. Let α =. Then is the unique root of Eqn. (8), and so + = =. So, then it follows that + = =, and hence exists and is given by =. Let α >. It holds that and α are the roots of the quadratic Eqn. (8). Since +, [0, ], necessarily + = =. Lemma 4.9. Let β = 0. Let 0 < α <. Then +, {α, }, which are the solutions of the quadratic Eqn. (8). Proof. Since 0 < α <, Eqn. (9) reduces to α + ( α ) 2 +, 2 { α + ( α ) = 2 { 2α = 2, 2 } 2 = {α, }., α + + 2, α + + ( α ) 2 ( α ) 2 However, because 0 < α < this does not give a unique solution. Hence +, {α, }. However, we want to rove that for β = 0 and 0 < α < we have that + = = α <. Therefore we will consider the following lemma. Lemma 4.0. Let β = 0 and let 0 α <. It holds that + = lim su n n <. Proof. It holds that lim n n = α <. Let ɛ > 0 be such that α + ɛ <. Then there exists N N 0, such that for all n N it holds that n α + ɛ <. Define, for all n N 0, and n := n := { n if n < N, (α + ɛ) if n N, n n + τ + ( n ) For all 0 n < N it holds that n = n and thus also that n = n. For all n N, it holds that that n > n. By virtue of Eqn. (7), and n = + τ n + µn n ( n ) n = + τ n + µn n 26 ( n ). }

Let n = N. Since N = N and N > N, it holds that This yields + τ N + µ N ( N ) = + τ N N N = > + τ N + τ N = N. + µ N ( N ) N < + τ N + µ N N ( N ). + µ N ( N N ) + µ N N ( N ) Let n > N and assume that n > n. Then n < n. By assumtion it holds that n > n, hence / n < / n. Therefore, so that + τ n + n ( n ) < + τ n + ( n ) n < + τ n + n ( n ), n = + τ + µn n ( n ) n > + τ n + µn n ( n ) = n. As a conclusion, n n for all n N 0, so that lim su n n lim su n n. Thus, it is sufficient to show that lim su n n <. We will now rove by induction to n that n < for all n N 0. Notice that 0 = 0 0 +τ and n = n n +τ+µn( n ), n N. Since we have for all n N that n = (α + ɛ) = α + ɛ =: c, we can write, for all k N, c k = c + τ µ k +. (20) k It holds that τ > 0, hence 0 <. Then there exists σ (0, ) such that 0 σ <. Without loss of generality we may assume that c < σ, otherwise we could have chosen σ larger. Then, c c c = c + τ µ + 0 c + <, (2) 0 c + σ 27

where the first inequality holds because τ µ 0. Using induction to k, we will show that Eqn. (2) gives which is Eqn. (22) for k =. Therefore, k Using this for Eqn. (20) with k = 2, we get c <. (22) c + σ c c + σ, c c + σ = c + σ c = σ c + σ c + σ. c c c 2 = c + τ µ 2 + c + c + Since c < σ, we get c + σ, hence σ c+ σ σ. Therefore, c 2 which is Eqn. (22) for k = 2. c + c+ σ σ c+ σ σ c+ σ.. Multilying by σ then gives that c c + σ, Assume that Eqn. (22) holds for k = n. We will rove that Eqn. (22) also holds for k = n. Since τ 0, we get that c c n = c + τ + n c + n Similarly to the case k = 2 we get Therefore, This roves Eqn. (22) for k = n. Eqn. (22) imlies that As a consequence, we get that which is what we needed to show. c n c + σ c + σ σ. σ c+ σ c c + c c + σ <. lim su c n n c + σ <. c lim su n n c + σ <, c c+ σ c = c + σ c+ σ. 28

Corollary 4.. For β = 0 and 0 < α < it holds that + = = α, and thus = α. Proof. From Lemma 4.0 it follows that + = lim su n n <. Since + {α, } it follows that + = α <. Since it always holds that +, we also have that = α <. Thus, + = = α <. Therefore, exists and is equal to α <, which is the smallest solution of Eqn. (8). Now we can give the roof the second art of Theorem 4.. Proof of Theorem 4. art 2. Let X = (X t ) t 0 be a birth-death rocess with birth arameters n > 0, n N 0, and death arameters > 0, n N and µ 0 = 0. Let τ > 0 be given. Assume that lim n n = α τ > 0 and lim n = β 0 both exist. Let n, n N, be given by Eqn. (7). To rove that lim n n exists, it is sufficient to rove that + =, i.e., lim su n n = lim inf n n. Lemma 4.4 gives that + and are both solutions of Eqn. (8). For β > 0, Lemma 4.6 gives that Eqn. (5) holds. Then Corollary 4.7 gives that = (α + β + ) (α + β + ) 2 4α, 2 which is the smallest solution of Eqn. (8). Now, consider β = 0. Lemma 4.8 gives for α that =, which is the smallest solution of Eqn. (8). Moreover, Lemma 4.9 gives, for 0 < α <, that +, {α, }, which is not a unique solution yet. However, from Lemma 4.0 it follows that + c c+ σ <. Then with Corollary 4. it follows that = α, which is the smallest solution of Eqn. (8). Thus, for all β 0 and for all α > 0, it holds that exists and it is the smallest solution of Eqn. (8), which is what we needed to show. To finish the roof of Theorem 4., it remains to rove art 3 of this theorem. This roof will now be given. Proof of Theorem 4. art 3. Let X = (X t ) t 0 be a birth-death rocess with birth arameters n > 0, n N 0, and death arameters > 0, n N and µ 0 = 0. Let τ > 0 be given. Assume that lim n n = α τ = 0 and lim n = β = 0. Let n, n N, be given by Eqn. (7). To rove that = lim n n = 0 exists, it is sufficient to rove that + = 0, i.e., lim su n n = 0. Lemma 4.0 gives that + <. So, to rove that + = 0, we will assume the contrary and rove that + =. Therefore, assume that + > 0. Hence, there exists a subsequence ( nk ) k ( n ) n N0 such that 29

nk + as k. Then, Eqn. (3) holds. Then, since α = β = 0 we get ( ) nk lim nk k + τ k + n k k n k = lim k k nk nk = lim + τ + n k k k k k nk = 0 + 0 + 0. Therefore, it holds that + =. However, this is a contradiction to the fact that + <. Therefore, + = 0. Since, + and +, [0, ], we immediately get that = 0. 4.3 More monotonicity and convergence results for the M M c -model (with c N) Let c N and consider the M M c -model. From [4] we know that we have to assume that n = < cµ, for all n N 0, and { nµ for all n < c, = cµ for all n c, where we assume that µ > 0. Recall Notation 3.. This subsection uses the suerscrit index again. From Theorem 4. we know that the sequence ( c n) n N0 has a limit, therefore we introduce some notation for this limit. Notation 4.2. Denote by,c the limiting value of the sequence ( c n) n N0. In this chater we will rove that,c is monotonic in c N and comute the limit lim c,c. For the M M c -model, c N, it holds that α = cµ > 0 and β = τ cµ > 0. Lemma 4.3. Let c N. Then,c is non-increasing as a function of c, i.e.,,c,c+ 0. Proof. By Theorem 4. we get that,c = ( ) 2 cµ + τ cµ + cµ + τ cµ + 4 cµ Multilication of both the denominator and the numerator by cµ yields,c = +τ+cµ (cµ) 2 ( cµ + τ cµ +)2 4cµ 2cµ = +τ+cµ (+τ+cµ) 2 4cµ 2cµ. 30 2.

To rove that,c is non-increasing in c, we will rove that the derivative to c is non-ositive. The derivative of,c to c is given by d dc,c = + τ [ 2c 2 µ 2c 2 µ ( + τ + cµ) 2 4cµ + [ ]] 2cµ d ( + τ + cµ) 2 4cµ dc = + τ 2c 2 µ + 2c 2 µ ( + τ + cµ) 2 4cµ 2cµ 2 ( + τ) + = = ( + τ) (2µ( + τ) + 2cµ 2 4µ ) ( + τ + cµ) 2 4cµ ( + τ + cµ) 2 4cµ 2µ ( + τ + cµ) 2c 2 µ 4cµ ( + τ + cµ) 2 4cµ ( + τ + cµ) 2 4cµ + ( + τ + cµ) 2 4cµ cµ ( + τ + cµ). 2c 2 µ ( + τ + cµ) 2 4cµ To rove that d dc,c 0, we will assume the contrary. Hence, we assume that ( + τ) ( + τ + cµ) 2 4cµ + ( + τ + cµ) 2 4cµ cµ ( + τ + cµ) > 0. Then, ( + τ + cµ) 2 4cµ > ( + τ) ( + τ + cµ) 2 4cµ + cµ ( + τ + cµ), in other words, ( + τ) 2 + 2cµ( + τ) + c 2 µ 2 4cµ > ( + τ) ( + τ + cµ) 2 4cµ + c 2 µ 2 + cµ (τ ). If ( + τ) 2 + cµ(τ ) < 0, then we get a contradiction and hence the roof is comlete. So, assume that ( + τ) 2 + cµ(τ ) 0. Then, ( + τ) 2 + cµ(τ ) > ( + τ) ( + τ + cµ) 2 4cµ. Taking the square on both sides yields, ( + τ) 4 + 2cµ(τ )( + τ) 2 + c 2 µ 2 (τ ) 2 > ( + τ) 2 Rewriting this gives ( ) ( + τ + cµ) 2 4cµ. ( + τ) 4 + 2cµ(τ )( + τ) 2 + c 2 µ 2 (τ ) 2 > ( + τ) 2 (( + τ) 2 + 2cµ( + τ) + c 2 µ 2 4cµ ) 2cµ(τ )( + τ) 2 + c 2 µ 2 (τ ) 2 > 2cµ( + τ) 3 + c 2 µ 2 ( + τ) 2 4cµ( + τ) 2 2( + τ) 2 (τ τ) + cµ(τ ) 2 > cµ( + τ) 2 4( + τ) 2 4( + τ) 2 + cµ(τ ) 2 > cµ( + τ) 2 4( + τ) 2 (τ ) 2 > ( + τ) 2. 3

However, since, τ > 0, (τ ) 2 > ( + τ) 2 gives a contradiction. Hence, the assumtion that ( + τ) ( + τ + cµ) 2 4cµ + ( + τ + cµ) 2 4cµ cµ ( + τ + cµ) > 0 is not correct. Thus, d dc,c 0. Corollary 4.4. Let c, c 2 N. Then it holds that,c,c2. Proof. Aly Lemma 4.3 reeatedly. Lemma 4.5. It holds that,c cµ, hence lim c,c = 0. Proof. Before roving the uer bound for,c, we will first give a lower bound for ( + τ + cµ) 2 4cµ, which is the term under the square root in. For all c N we have that ( + τ + cµ) 2 4cµ = ( + τ) 2 + 2( + τ)cµ + c 2 µ 2 4cµ = 2 + 2τ + τ 2 + 2cµ + 2cµτ + c 2 µ 2 4cµ = 2 + 2τ + τ 2 2cµ + 2cµτ + c 2 µ 2 = 2 2τ + 4τ + τ 2 2cµ + 2cµτ + c 2 µ 2 = (τ ) 2 + 2(τ )cµ + c 2 µ 2 + 4τ = (τ + cµ) 2 + 4τ (τ + cµ) 2 0, where the last inequality holds since, τ 0 and < cµ. It holds that where (*) holds because < cµ.,c = + τ + cµ ( + τ + cµ) 2 4cµ 2cµ = + τ + cµ (τ + cµ) 2 + 4τ 2cµ + τ + cµ (τ + cµ) 2 2cµ ( ) + τ + cµ (τ + cµ) = 2cµ = 2 2cµ = cµ, 32

So, when we take the limit of c we get 0 lim c,c lim c cµ = 0. Notation 4.6.. Let c N. For all n N and for τ > 0 denote by τ n the robability c n in the M M c -model. 2. Denote by,τ the limiting value of the sequence ( τ n) n N0. Lemma 4.7. Let c N. It holds that lim τ 0,τ = cµ. Proof. It holds that lim τ 0,τ = lim τ 0 + τ + cµ ( + τ + cµ)2 4cµ 2cµ = 2cµ. Proosition 4.8. Let c N and let τ, τ 2 > 0 such that τ < τ 2. Then, it holds for all n N 0, that τ n τ2 n > 0. Moreover,,τ,τ2 > 0. Proof. We will give a roof by induction on n. Let n = 0. Then, τ 0 τ2 0 = 0 0 ( ) 0 > 0 = 0, 0 + τ 0 + τ 2 0 + τ 2 0 + τ 2 where (*) holds because τ < τ 2. So, for n = 0 it holds that τ 0 τ2 0 > 0. Now consider n =. Then it holds that τ τ2 = + τ + µ ( τ 0 ) + τ 2 + µ ( τ2 ( ) > ( ) > = 0, + τ + µ ( τ2 0 ) 0 ) + τ 2 + µ ( τ2 0 ) + τ 2 + µ ( τ2 0 ) + τ 2 + µ ( τ2 0 ) where (**) holds because τ 0 > τ2 0, and where (***) holds because τ < τ 2. So, for n = it holds that τ τ2 > 0. Suose that for n it holds that τ n τ2 n n = it follows that τ n τ2 n > 0. Thus, for all n N 0, it holds that τ n τ2 n 0. 33 > 0. Then consider n. Analogously to the case

It holds that,τ = lim n τ n. Theorem 4. states that this limit exists. Then for τ < τ 2 we have that,τ,τ2 So, it holds for τ < τ 2 that,τ,τ2 0. = lim n τ n lim n τ2 n = lim n (τ n τ2 n ) lim (0) n = 0. 34

5 Convergence of q n for a general birth-death rocess Consider a birth-death rocess X = (X t ) t 0 with birth arameters n > 0, n N 0, and death arameters > 0, n N and µ 0 = 0. Let τ > 0 be given. Assume that α, β 0 both exist. Recall from Eqn. (3) that q n = + τ + n ( q n+ ). (23) Lemma 5.. It holds, that q n > 0 for all n N. Moreover, it holds that for all n N. Proof. Let n N. Then, Hence, lim inf n q n lim n q n+ = + τ + n n n q n, (24) q n = + τ + n ( q n+ ) + τ. + n + τ + n = since β 0 and α 0 both exist. Hence, q n > 0 for all n N. + β + α > 0, It holds that q 0 = 0, therefore we consider only n N. Let n N be arbitrary. Recall that n > 0 for all n N 0. Lemma 5. states that q n > 0 for all n N, therefore dividing by q n is allowed. Rewriting Eqn. (23) gives Then, q n ( + τ + n ( q n+ )) =. q n+ = + τ + n n n q n, which is equal to Eqn. (24), and which is what we needed to show. Note that since q 0 = 0, we do not get such a formula for q. Therefore, to rove that the sequence (q n ) n N0 has a limit under mild conditions is harder than for the sequence ( n ) n N0. However, in the remainder of this chater we will rove that for a birth-death rocess the sequence (q n ) n N0 has a limit under mild conditions. 35

Theorem 5.2. Suose that α = lim n n 0 and β τ = lim n 0 both exist. Let q 0 = 0 and let q n, n N, be given by Eqn. (23).. Suose α = 0. Then lim n q n = q exists with q = β +. 2. Suose α > 0. Then lim n q n = q exists with q the smallest solution of the equation α x 2 (α + β + ) x + = 0, (25) in other words q = α + β + (α + β + ) 2 4α 2α. The roof for α > 0 requires more work than the roof for α = 0. Therefore we will first rove art of the theorem. Proof of Theorem 5.2 art. Let α = 0. Let β = lim n τ 0 exist. Note that > 0, n N. Hence, we can rewrite Eqn. (23) by dividing by. This gives Hence, since β 0 and α = 0. Furthermore, it holds that lim inf n q n lim n q n = + τ + n ( q n+ ) + τ. + n + τ + n = q n = + τ + n ( q n+ ) + τ. + β + α = + β, Hence, lim su q n lim n n + τ = + β. Hence, lim n q n = +β, which comletes the roof of art of Theorem 5.2. 5. Part 2 of Theorem 5.2 The roof of the second art of Theorem 5.2 consists of multile stes that will be formulated and roven in this subsection. Similarly to the Theorem 4., the art of Theorem 5.2 concerning β > 0, can be roven directly. For β = 0 and 0 < α, we can also give a direct roof. However, for the case that β = 0 and α >, we will need to introduce the concet of stochastic monotonicity, discussed in Section 5... Stochastic monotonicity will be used to 36