Let be a supercritical continuous-time branching process with immigration; our focus is on the large deviation rates of and thus extending the results of the discrete-time Galton–Watson process to the continuous-time case. Firstly, we prove that is a submartingale and converges to a random variable . Then, we study the decay rates of and for and under various moment conditions on and . We conclude that the rates are supergeometric under the assumption of finite moment generation functions.

1. Introduction

Suppose that is a continuous-time branching process with immigration. Its generating matrix is defined as follows:where

Throughout this paper, we always suppose that , , and .

Setwhere and separately represent the total number of the original individual’s offspring and the offspring of the individual from immigrants at the time . If for all and , then the process degenerates to the continuous-time branching process . It is known from Athreya and Ney [1] or Harris [2] that is a martingale and converges to a r.v. W w.p.1 as . Therefore, converges to w.p.1 as for arbitrary fixed , and henceFor , definewhere and .

In this paper, let , we will demonstrate that is a submartingale and converges to a r.v. almost surely since converges to a r.v. a.s. and converges to a r.v. a.s.

The above discussion of the decay rates is not only of its own meaning but also of some significance to the algorithm tree structure in computer science (Karp and Zhang [3] and Miller [4]). Heyde [5] discussed that there exists a constant sequence which makes converge to a random variable W w.p.1. Athreya [6] considered similar decay rates for the Galton–Watson process. Seneta [7] researched the supercritical Galton–Watson process with immigration, and Riotershtein [8] considered multitype branching processes with immigration in a random environment. Liu and Zhang [9] studied the decay rates of for the supercritical branching processes with immigration. Recently, Sun and Zhang [10] considered the convergence rates of harmonic moments for supercritical branching processes with immigration. Chen and He [11] investigated the lower deviation and moderate deviation probabilities for maximum of a branching random walk. Abraham et al. [12] analyzed the stationary continuous state branching processes. The model considered in this paper involves continuous-time and immigration, while the models considered in the above references do not involve continuous-time. Such change in conditions may affect the large deviation rates and thus is mathematically interesting. We need to find a new method to investigate the effect of continuous-time and immigration.

Based on the previous results, it is natural to develop the large deviation rates for the continuous-time branching processes with immigration. In this paper, we are aiming to discuss the decay rates ofunder various moment conditions on and for any and . We conclude that the rates of (6) are always supergeometric under the assumption of finite moment generation functions. The specific conclusions are presented in Section 3.

2. Preliminaries

Before the investigation, we present the generating functions and of the known sequences and as

Clearly, is well defined and finite at least on with , so 1 is the solution of . Then, is the mean birth rate of . Moreover, is similar to .

We present some preliminaries in this section.

Lemma 1. Let and . Ifexists, then for arbitrary ,

Proof. Omitted.

Lemma 2. exists for any , and . Furthermore, satisfies the functional equation:with condition .

Proof. By the Kolmogorov forward equation,For ,That is,Hence,For ,So,By Lemma 1,It follows from (11) thatHence,Finally, we can obtain the following equality from (11):and moreover,Taking the limit as , we haveThe proof is completed.

Proposition 1. Let be the -algebra generated by and . Then, is a submartingale and converges to a r.v. almost surely.

Proof. According to the definition of , we conclude the following inequalities:which implies is a submartingale. We know that if and only if . Hence, . Thus, is an integrable submartingale and converges to a r.v. almost surely.
Define , with initial condition , where is a continuous-time branching processes with immigration with . Denote andWe will study the properties of and in the subsequent contents. Let . Apparently, is strictly increasing with and . Furthermore, we have for since and for since . Therefore, the iterates of are nonincreasing with and nondecreasing with (with respect to ).

Proposition 2. Let . If for some , then for , as andwhere is the unique solution ofsubject to

Proof. Note that for , we know that for , . Hence, is well defined on for and for .
Since , we have for any , which implying as . On the other hand,that is,Denote . Putting yields . Thus, we obtain , i.e., as .
Now, there exists for such thatHence, exists.
Furthermore, satisfies (26) and (27). For ,Letting yieldsFor another,since for . Thus, .
Finally, it is easily seen that the solution of (26) and (27) is unique.

Proposition 3. Let for some . Then,where is the round of .

Proof. For , noting yieldsAssume that inequality (35) holds for , i.e.,Then,holds for all .
Therefore, (35) is proved.

3. Main Results and Proofs

Theorem 1. Assume that and for some . Then, for arbitrary small ,where .

Proof. Note that could be defined aswhere and are independent and identical distributed, are i.i.d. random variables with the same law as . Hence,Indeed, for any fixed ,where and are arbitrary constants. Thus,where .
Since it is assumed that and , we have and it follows that there exist and such thatfor any and . Hence, there exist such that for all . By Markov’s inequality, we can prove that there exist and such that . Therefore, by Lemma 2,The proof is completed.

Theorem 2. Assume that for fixed and , there exist constants and such that and for all . Then, (39) holds.

Proof. Under the above assumptions, there exists another such that for all since is equivalent to as . In general, we denote the positive constants by . Therefore,With a simple modification of the convergence theorem, it can be proved thatsinceFor all ,and henceSince , the proof shall be completed if we provewhere . Indeed, for any fixed , let be the inverse function of at , then we have as . Note that , thus we can writewhereSince , it follows promptly that satisfyingBy hypothesis and hence for any , we can pick an such that for all . Hence, for ,which impliesand henceApplying for together with (57) implies (51).

Corollary 1. Assume that for some and such that . Then, (39) holds.

Proof. By Markov’s inequality, we haveAccording to the assumption,is finite, and then there exists s.t. for all . Applying Theorem 2, the proof is completed.

Theorem 3. Assume that for some and . Then, there exists satisfying

Proof. In general, we suppose that for . Firstly, we prove thatSince is equivalent to , we haveHence, (61) is proved since . By Proposition 3, note that for any ,Indeed, if , by Proposition 3,But, by definition of ,and hence the above inequality becomessince . Hence, as since . Finally, by Proposition 2,which is positive and finite. Therefore, we can find such that

Theorem 4. Assume that for some and . Then, there exist constants and satisfying