Branching Processes with Negative Offspring Distributions

c Birkh¨auser Verlag, Basel, 2003


Annals of Combinatorics 7 (2003) 35-47

Annals of Combinatorics

0218-0006/03/010035-13

Branching Processes with Negative Offspring Distributions Ioana Dumitriu1, Joel Spencer2, and Catherine Yan3∗ 1Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139

USA [email protected] 2Courant Institute of Mathematical Sciences, New York University, New York, NY 10012, USA

[email protected] 3Department of Mathematics, Texas A&M University, College Station, TX 77843, USA

[email protected] Received January 3, 2003 AMS Subject Classification: 05A16, 05D40, 60J80

Abstract. A branching process is a mathematical description of the growth of a population for which the individual produces offsprings according to stochastic laws. Branching processes can be modeled by one-dimensional random walks with non-negative integral step-sizes. In this paper we consider the random walk with a similar algebraic setting but the step-size distribution is allowed to take negative values. We gave the exact formula for the limit probability under which the random walk continues forever. Asymptotic results are also presented. Keywords: branching processes, negative offsprings

1. Introduction A branching process is a mathematical description of the growth of a population for which the individual produces offsprings according to stochastic laws. A typical process is of the following form. Considering a population of individuals developing from a single progenitor — the initial individual. The initial individual produces a random number of offsprings, each of them in turn produces a random number of offsprings; and so the process continues as long as there are live individuals in the population. An interesting question is to find the probability that the population survives, (or, extincts). The branching process was proposed by Galton [5], and the probability of extinction was first obtained by Watson [17] by considering the probability generating function for the number of children in the nth generation. This mathematical model was known as ∗

Research supported by NSF grants DMS-0070574 and a Sloan fellowship. Part of the research was done during the author’s visit to Institute for Advanced Study.

35

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I. Dumitriu, J. Spencer, and C. Yan

the Galton-Watson branching process, and had been studied thoroughly in literature, for example, [2, 6, 7, 10]. Some interesting details on the early history of branching processes can be found in [9]. Another model for the branching processes was based on the interpretation of the random walk Sn − n and the branching processes in terms of queuing theory, which is due to Kendall [8]. Here Sn = Z1 + Z2 + · · · + Zn , where Zi ’s are independent random variables with the identical distribution as the offsprings. In this model, the random family of a branching process is represented by a queue with a single server, with the service time for each customer being a unit period. The queue starts with one customer, corresponding to the initial individual in the branching process. Each individual in the random family represents a different customer, who arrives and waits for service in the queue. A customer j arrives during the service period of i if j is a child of i. Let Yt be the number of customers in the queue at the end of the t-th service period. Then Y0 = 1, and Yi = Yi−1 + Zi − 1 (i ≥ 1), where Zi are independent random variables with the offspring distribution. Note that the total progeny is represented by the length of the first busy period of the server, i.e., smallest t such that Yt = 0. The process continues forever iff Yi > 0 for all i. By the above correspondence, the probabilities of extinction given by the GalsonWatson model and the random walk model are identical. Let Z be the offspring distribution. A basic fact about branching processes states that when E[Z] = c < 1, with probability 1 the process dies out; but when E[Z] = c > 1, there is a positive probability α that the process will continue forever [4]. Let f (x) be the probability generating function for Z. In the case c > 1, α can be computed as 1 − s where s, the probability of extinction, is the unique solution of s = f (s) in the interval [0, 1). In particular, if the offspring distribution is Poisson with mean 1 + ε, then the probability of survival is asymptotically 2ε (c.f., [15]). An important condition in solving such probability in Galton-Watson branching processes is that the possible values of the offspring distribution Z are non-negative integers. In this paper we consider a random walk with a similar algebraic setting as the ones corresponding to branching processes. However, the step-size distribution can take negative values. Explicitly, let Z be a probability distribution with non-negative integral values, and E[Z] = 1 − ε, where 0 < ε < 1. Define a random walk Y = Y0 , Y1 , . . . by the following formula
Y0 = 1, (1.1) Yi = Yi−1 + 1 − Zi, where Z1 , Z2 , . . . are independent random variables each with distribution Z. In particular, we consider the case where Z is distributed according to the Poisson law with mean 1 − ε, denoted by Po(1 − ε). Let T be the least t for which Yt ≤ 0. If no such t exists, we say that T = +∞. Since the expected step-size E[1 − Z] = ε > 0, with probability approaching 1, Yn > 0 when n → ∞. We will compute the probability that Yn > 0 for all n, or equivalently, Pr[T = +∞], in which case we say the random walk Y escapes. We also consider the asymptotics of Pr[T > N]. Note that this problem is essentially different from the one defined by branching processes, since the values of

Branching Processes with Negative Offspring Distributions

37

the step-size random variable 1 − Z are mostly negative. Traditional methods (e.g., [4]) for Galton-Watson branching processes based on the existence of a functional equation for probability generating function are not easily applied. In this paper we find the exact value of the escaping probability for the random walk Y. We present two proofs. The first is a pure probabilistic one, based on the linearity of expectations. The second is more analytic, using a theorem of Otter and Dwass in branching processes, which is equivalent to the Lagrange inversion formula for formal power series. The second proof applies to more general offspring distributions Z, provided that both the expected value and variance of Z exist. It is also used to obtain asymptotics of Pr[T > N]. With suitable assumptions on Z, we prove that Pr[T = ∞|T > N] is asymptotically 1 if N  ε−2 , when ε → 0+ . We also give a formula that generalizes the Otter-Dwass theorem, and use it to prove that if Z = Po(1 − ε), then N  ε−2 is both necessary and sufficient for Pr[T = ∞|T > N] ∼ 1. Generalizations of the Galton-Watson branching processes have been studied in statistical literatures, for example, by Quine and Szczotka [12–14]. It is showed that some standard branching process results hold in more general settings, e.g., the offspring random variables are permitted to take negative integer values, and are not necessarily i.i.d. The statistical work concentrated on the stochastic process {Xn , n = 0, 1, 2, . . .}, where Xn represents the size of the n-th generation in a branching process. Our model is different from theirs. Firstly, we use a different scale. We consider the limit behavior of the random walk as the time evolves, which corresponds to the evolution of the size of a queueing system. There is no natural correspondence of generation in a queueing system. Secondly, we assume that the step-size random variables are i.i.d. with the distribution 1 − Z, where Z is non-negative. These assumptions enable us to get exact solutions instead of general description. Thirdly, we take a different approach. Our proofs are more elementary, which combine the basic probability theory and combinatorial methods, and lead to applications in enumeration. 2. The Escaping Probability Theorem 2.1. Fix ε > 0. For the random walk Y defined by the formula (1.1) with Z = Po(1 − ε), the escaping probability Pr[T = +∞] is e1−ε ε. 2.1. The Probabilistic Proof Let pi = Pr[Z = i] for i ≥ 0. Considering a new random walk Y defined by the same recurrence as (1.1) except that the beginning position is zero, i.e.,
 Y0 = 0, (2.1)  +1−Z , Yi = Yi−1 i where Z1 , Z2 , . . . are independent random variables with the same distribution Z. Let T  be the least t > 0 for which Yt ≤ 0. If no such t exists, we say that T  = +∞. Denote by s the escaping probability Pr[T  = +∞] of Y . It is clear that if s is the escaping probability of the original random walk Y, then s = p0 s and hence s = s /p0 .

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I. Dumitriu, J. Spencer, and C. Yan

Let q = 1 − s be the probability of extinction for Y  . Let Wn be the indicator random variable for the event Yn = 0, i.e.,
1, if Yn = 0, Wn = 0, otherwise,  and let W = ∑∞ i=1 Wn . Then W is the number of t such that Yt = 0. (It is possible that W = ∞. ) A crucial property for the random walk Y is that, in each step, Y can raise at most one unit. Therefore, any random walk with W = k (k is an integer and k ≥ 2) consists of k + 1 part (see Figure 1): The first k parts are random walks that begin at zero, drop to the x-axis, or drop below x-axis and then move back to zero. The last part is a random walk that begins at 0, and remains above the x-axis for all t > 0. It is clear that the last part corresponds to a random walk defined as Y with T  = ∞. This happens with probability s .

Y’

0

t1

...

t2

tk

Figure 1.

In Y , let T be the least t such that t > 0 and Yt = 0. Hence Pr[W = k] =

∑



 s

k

∏ Pr[T = ti ]

t1 ,...,tk >0 i=1

= s



k

∑ Pr[T = t]

t>0

= s (Pr[T < ∞])k .

Branching Processes with Negative Offspring Distributions

39

If T < ∞, then T  < ∞. Conversely, we claim that if T  < ∞, then with probability one T < ∞. To prove the claim, first note that T  < ∞ but T = ∞ implies that there exists N > 0 such that for all n ≥ N, Yn < 0. Let An , n ∈ N be the event that Yn < 0. Yn has the distribution n − Sn where Sn is distributed according to the Poisson law with mean n(1 − ε). For the Poisson distribution, we have the following large deviation result (e.g., see [1, Appendix A.15]), Lemma 2.2. Let P have Poisson distribution with mean µ. For δ > 0, we have Pr[P ≤ µ(1 − δ)] ≤ e−δ

2 µ/2

,

Pr[P ≥ µ(1 + δ)] ≤ [cδ ]µ , where cδ = eδ (1 + δ)−(1+δ) < 1 is a constant depending only on δ. Using the lemma, Pr[An ] = Pr[Yn < 0] = Pr[Sn > n] ≤ (cδ )n(1−ε) where δ = ε/(1 − ε), and cδ is a constant in (0, 1). It follows that ∑∞ n=1 Pr[An ] converges. By Borel-Cantelli Lemma, with probability one Yn < 0 is true only for finitely many n. This proves the claim. Therefore Pr[T < ∞] = Pr[T  < ∞] = q and Pr[W = k] = qk s . It follows that Pr[W ≥ k] = q k . Hence ∞

∑ Pr[W ≥ k] =

E[W ] =

k=1

q 1 − s = . 1 − q s

On the other hand, by the linearity of the expectation, E[W ] = Again, note Therefore

that Yn

∞

∞

n=1

n=1

∑ E[Wn] = ∑ Pr[Yn = 0].

= n − Sn, where Sn has the Poisson distribution with mean n(1 − ε). E[W ] =

∞

∞

e−(1−ε)n (n − nε)n n! n=1

∑ Pr[Sn = n] = ∑

n=1

=

∞

nn 

∑ n!

(1 − ε)e−(1−ε)

n

= te−t R (te−t ),

(2.2)

n=1

where t = 1 − ε and R(x) = (xe−x ) is the inverse function of y = xe−x . (See, e.g., [16, Chap. 5.3].) Let y = te−t . From R(te−t ) = t, differentiating with respect to t, we have R (y)(e−t − te−t ) = 1, that is, R (y) =

et . 1−t

Substituting into (2.2), we have 1−ε 1 − s t = . = E[W ] =  s 1−t ε Therefore s = ε and s = ε/p0 = e1−ε ε.

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I. Dumitriu, J. Spencer, and C. Yan

Remark. Theorem 2.1 asserts that the escaping probability s is ε for the random walk Y . Such a clean answer deserves an easy explanation. The following is an intuitive one: In the infinite random walk Y , for any i > 0, let Wi be the number of t such that Yt = i. For a fixed i, almost surely the random walk will hit the line y = i, i.e., Pr[Wi ≥ 1] = 1. For any k > 1, similar to the argument in the preceding proof, we have Pr[Wi = k] = qk−1 s , and Pr[Wi ≥ k] = qk−1 . Therefore E[Wi ] = 1+q +q2 +· · · = 1/s , consequently E[W1 + W2 + · · · + WNε ] = Nε/s . On the other hand, E[W1 + W2 + · · · + WNε ] =

∞

∑ Pr[1 ≤ Y j ≤ Nε],

j=0

where Y j has distribution j − S j , S j = Z1 + Z2 + · · · + Z j ∼ Po( j(1 − ε)). √ √ Using Lemma 2.2, for N < j < N − K N, (K  1), we have Pr[Y j > Nε] = Pr[S j < j − Nε] ≤ e

ε2 (N− j)2

− 12 j(1−ε)

≤e

ε2 − 12 (1−ε)

K 2√ (1−K/ N)

,

Pr[Y j ≤ 0] = Pr[S j − j ≥ 0] ≤ (cε ) j , where cε < 1 is a constant. Taking K = N 1/3 , we have N−N 5/6

∑√

Pr[1 ≤ Y j ≤ Nε] ∼ N(1 + o(1)).

j= N

√ While for j > N + K N, K = N 1/3 , we have Pr[1 ≤ Y j ≤ Nε] ≤ Pr[Y j ≤ Nε] = Pr[S j − j(1 − ε) ≥ ( j − N)ε] ≤ (cδ ) j(1−ε) , ε where δ = (1−ε) for δ small. So

j−N j .

From Lemma 2.2, cδ is decreasing as δ increases, and cδ ∼ e−δ

2 /2

cδ ≤ cε/(1−ε)N −1/6 ∼ e  j ε2 N −1/3 − 2(1−ε)  Thus Pr[1 ≤ Y j ≤ Nε] ≤ e and

∑

Pr[1

≤ Y j

−

ε2 N −1/3 2(1−ε)2

.

  ε2 N 1/3 − 2(1−ε) ≤ Nε] = O e = o(1).

j=N+N 5/6

Therefore E[W1 + W2 + · · · + WNε ] = N(1 + o(1)) + O(N 1/2) + O(N 5/6 ) + o(1) = N(1 + o(1)). Comparing the two expressions for E[W1 + W2 + · · · + WNε ], we have Nε/s ∼ N when N → ∞. This implies that the escaping probability s equals ε.

Branching Processes with Negative Offspring Distributions

41

2.2. The Analytic Proof We compute the probability Pr[T > N] directly, using the Otter-Dwass theorem in the theory of branching processes. This proof applies to all integral distributions Z, provided that E(Z) and var(Z) exist. In a branching process with offspring distribution Z, let #Fk be the total progeny in the process with k individuals to start with. The following theorem presents a remarkable formula for the distribution of #Fk restricted to positive integers. This formula was discovered by Otter [11] for k = 1, and extended to all k ≥ 1 by Dwass [3]. Theorem 2.3 (Otter and Dwass). For all n, k = 1, 2, . . . , we have Pr[#Fk = n] =

k Pr[Sn = n − k], n

(2.3)

where Sn = Z1 + Z2 + · · · + Zn . To prove Theorem 2.1, as in the first proof, we consider the random walk Y defined by  Y0 = 0, Yi = Yi−1 + 1 − Zi,

where Zi are independent with distribution Z, E(Z) = 1 − ε > 0, and var(Z) = σ. Again let T  be the least t > 0 such that Yt ≤ 0. Lemma 2.4. For 1 ≤ k ≤ n, we have Pr[T  > n, Yn = k] =

k Pr[Sn = n − k]. n

Proof. For any walk {Y0 , Y1 , Y2 , . . . , Yn } satisfying Y0 = 0, Yi > 0 for i = 1, 2, . . . , n − 1  . and Yn = k, define the reversed walk X = {X0 , X1 , . . . , Xn } as follows: X0 = k, Xi = Yn−i   Then Xi = Yn−i = Yn−i−1 + 1 − Zn−i = Xi+1 + 1 − Zn−i . That is, X satisfies the recurrence X0 = k, Xi+1 = Xi + Zi − 1 where Zi are independent random variables with the identical distribution Z. In other words, {X0 , X1 , . . . , Xn } is a random walk corresponding to the Galton-Watson branching process with the offspring distribution Z and k individuals at the zero generation. The conditions Y0 = 0, Yi > 0 for i = 1, 2, . . . , n − 1 imply that Xi > 0 for i = 1, 2, . . . , n − 1, and Xn = 0. So this corresponds to the case that the total progeny in the branching process is exactly n, which happens with probability Pr[#Fk = n]. Now the lemma follows immediately from Otter-Dwass theorem. Proof of Theorem 2.1. For the random walk Y , T  > n means Y1 = 1 (i.e., Z1 = 0).  Hence for the random walk Y, p0 Pr[T > n,Yn = k] = Pr[T  > n + 1,Yn+1 = k], where p0 = Pr[Z = 0]. Therefore Pr[T > n, Yn = k] =

1 k Pr[Sn+1 = n + 1 − k], p0 (n + 1)

(2.4)

and Pr[T > n] =

1 1 n+1 ∑ k Pr[Sn+1 = n + 1 − k]. p0 (n + 1) k=1

(2.5)

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I. Dumitriu, J. Spencer, and C. Yan

We evaluate the sum in Equation (2.5). 1 n+1 ∑ k Pr[Sn+1 = n + 1 − k] n + 1 k=1 =

1 n ∑ (n + 1 − k) Pr[Sn+1 = k] n + 1 k=0

1 1 n (n + 1) Pr[Sn+1 ≤ n] − ∑ k Pr[Sn+1 = k] n+1 n + 1 k=0

1 = Pr[Sn+1 ≤ n] − E[Sn+1] − ∑ k Pr[Sn+1 = k] n+1 k>n

=

= 1 − Pr[Sn+1 ≥ n + 1] − (1 − ε) + = ε − Pr[Sn+1 ≥ n + 1] +

1 ∑ k Pr[Sn+1 = k] n + 1 k>n

1 ∑ k Pr[Sn+1 = k]. n + 1 k>n

Note that E[Sn+1] = (n + 1)(1 − ε), and var(Sn+1 ) = (n + 1)σ. By Chebyshev’s inequality, Pr[Sn+1 ≥ n + 1] = Pr[Sn+1 − (n + 1)(1 − ε) ≥ (n + 1)ε] ≤

σ = o(1), (n + 1)ε2

and 1 1 k Pr[Sn+1 = k] = Pr[Sn+1 ≥ n + 1] + ∑ ∑ Pr[Sn+1 ≥ k] n + 1 k>n n + 1 k>n+1 ≤

σ (n + 1)σ 1 ∞ + ∑ 2 (n + 1)ε n + 1 t=1 ((n + 1)ε + t)2

∼

σ +σ (n + 1)ε2

 ∞ 1

1 dt ((n + 1)ε + t)2

= O(n−1 ) = o(1). Hence

Pr(T = ∞) = lim Pr(T > n) = ε/p0 . n→∞

For the case that Z is the Poisson distribution with mean 1 − ε, p0 = e−(1−ε) , and hence Pr[T = ∞] = e1−ε ε. Let pi = Pr[Z = i]. In the random walk Y, for 1 ≤ k ≤ n, we have Pr[T > n, Yn = k] =

∑

n

∏ p zi ,

(z1 , z2 ,..., zn ) i=1

Branching Processes with Negative Offspring Distributions

43

where (z1 , z2 , . . . , zn ) are non-negative integers so that Yi > 0 for i = 1, 2, . . . , n − 1, and Yn = k. In terms of zi , these conditions are equivalent to z1 ≤ 1, z1 + z2 ≤ 2, ··· ≤ ···

(2.6)

z1 + z2 + · · · + zn−1 ≤ n − 1, z1 + z2 + · · · + zn = n + 1 − k. When Z has the Poisson distribution with mean µ,

∑

Pr[T > n, Yn = k] =

N

∏ e−µ

(z1 , z2 ,..., zn ) i=1

=e

−µn

µ zi = zi ! (z

µn+1−k (n + 1 − k)! (z

∑

∑

1 , z2 ,..., zn )

1 , z2 ,..., zn )

N

µ zi i=1 zi !

e−µn ∏



 n+1−k , z1 , z2 , . . . , zn

(2.7)

where (z1 , z2 , . . . , zn ) satisfies the condition (2.6). By Equation (2.4), Pr[T > n, Yn = k] = eµ = eµ Combining (2.7) and (2.8), we get 

∑

(z1 , z2 ,..., zn )

k Pr[Sn+1 = n + 1 − k] n+1 k −(n+1)µ ((n + 1)µ)n+1−k e . n+1 (n + 1 − k)!

n+1−k z1 , z2 , . . . , zn

(2.8)

 = k(n + 1)n−k ,

(2.9)

where (z1 , z2 , . . . , zn ) satisfies condition (2.6). The sum in the left-hand side of Equation (2.9) has the following combinatorial interpretation. Put balls B1 , B2 , . . . , Bn+1−k into n different boxes. The sum counts the number of ways one can do it so that the number of balls in the first i boxes does not exceed i, for i = 1, 2, . . . , n. Hence, we have Corollary 2.5. Let 1 ≤ k ≤ n. The number of ways to put n + 1 − k distinct balls into n distinct boxes, so that the number of balls in the first i boxes does not exceed i, for all i = 1, 2, . . . , n, is k(n + 1)n−k . 3. The Asymptotic Formula In this section we consider the asymptotics of the escaping probability of Y when ε → 0+ . Using the Otter-Dwass theorem and Lemma 2.4, we estimate the escaping

44

I. Dumitriu, J. Spencer, and C. Yan

probability after a finite number of steps. More precisely, let Z (ε) be a family of nonnegative integral random variables such that as ε → 0+ , Pr[Z (ε) = 0] → p0 ∈ (0, 1), E[Z (ε) ] = 1 − cε + o(ε) for some positive constant c. Further we assume a uniform bound M such that var(Z (ε) ) ≤ M for all positive ε in some neighborhood of the origin. Under these conditions, Theorem 3.1. Let the random walk Y(ε) be defined by (ε)

(ε)

Y0 = 1, Yi

(ε)

(ε)

= Yi−1 + 1 − Zi ,

(ε)

(ε)

where Zi are independent random variables with identical distribution Z (ε) . Set Sn = (ε) (ε) (ε) (ε) Z1 +Z2 +· · ·+Zn , and T (ε) the least t such that Yt = 0. Then Pr[T (ε) > N] ∼ cε/p0 if N  ε−2 . Proof. From the analytic proof of Theorem 2.1, we have Pr[Z (ε) = 0] Pr[T (ε) > N] = 1 − E[Z (ε)] +

1 (ε) ∑ Pr[SN+1 ≥ k]. N + 1 k>N+1

Let K = Nε2 . If N  ε−2 , then K → ∞ as ε → 0+ . By Chebyshev’s inequality, (N + 1)var(Z (ε) ) 1 1 (ε) Pr[SN+1 ≥ k] ≤ ∑ ∑ N + 1 k>N+1 N + 1 t≥1 ((N + 1)(1 − E[Z (ε)]) + t)2 ≤

M Mε = o(ε). = N(1 − E[Z (ε)]) K(c + o(1))

Hence Pr[T (ε) > N] c + o(1) c = → . ε p0 + o(1) p0 Remark. A typical example is given by Z (ε) = Po(1 − ε), in which case Pr[Z (ε) = 0] = e−(1−ε) → e−1 , E[Z (ε) ] = 1 − ε, and var(Z (ε) ) = 1 − ε ≤ 1 for ε ∈ [0, 1]. Taking, for example, N = ε−2 ln ln ε−1 we have Pr[T (ε) > N] ∼ eε. As an extension to the Otter-Dwass theorem, we have the following formula for the random walk Y. Theorem 3.2. In the random walk Y with any offspring distribution Z, Pr[T = n, Yn = 0] =

1 Pr[Sn = n + 1]. n

Proof. Note that Pr[T = n, Yn = 0] =

∑

n

∏ p zi ,

(z1 , z2 ,..., zn ) i=1

Branching Processes with Negative Offspring Distributions

45

where zi are non-negative integers such that Yi ≥ 1 for i = 1, 2, . . . , n − 1, and Yn = 0. Equivalently, zi satisfies the condition z1 ≤ 1 z1 + z2 ≤ 2, ··· ≤ ···

(3.1)

z1 + z2 + · · · + zn−1 ≤ n − 1, z1 + z2 + · · · + zn = n + 1. Also note Pr[Sn = n + 1] =

∑

n

∏ p zi ,

(z1 , z2 ,..., zn ) i=1

where zi are non-negative integers with z1 + · · · + zn = n + 1. Hence it is sufficient to show that for any sequence of non-negative integers α = (a1 , a2 , . . . , an ) with a1 + a2 + · · · + an = n + 1, there is exactly one cyclic shift αi = (ai , ai+1 , . . . , an , a1 , . . . , ai−1 ) satisfying condition (3.1). The proof is contained in the following two claims. Claim 1. For any sequence α, there is at most one cyclic shift of α satisfying condition (3.1). Prove by contradiction. Assume both α = (a1 , a2 , . . . , an ) and αi = (ai , ai+1 , . . . , an , a1 , . . . , ai−1 ) satisfy condition (3.1), where i is an index between 2 and n. Then a1 +a2 + · · · + ai−1 ≤ i − 1, and ai + · · · + an ≤ n − i. Hence a1 + a2 + · · · + an ≤ n, contradicting the fact that a1 + a2 + · · · + an = n + 1. Note that Claim 1 also implies that all the conjugates of α are distinct. Claim 2. For any sequence α = (a1 , a2 , . . . , an ) of non-negative integers, there exists a cyclic shift αi satisfying condition (3.1). Let xi = i − (a1 + a2 + · · · + ai ) for i = 1, 2, . . . , n − 1. If all xi ≥ 0, then α satisfies condition (3.1). Otherwise, let x = min{xi , i = 1, 2, . . . , n − 1}. Then x < 0. Find the smallest index i such that xi = x. We prove that the cyclic shift αi+1 = (ai+1 , . . . , an , a1 , . . . , ai ) satisfies (3.1). Denote αi+1 by (b1 , b2 , . . . , bn ). 1. For 1 ≤ k ≤ n − 1 − i, b1 + b2 + · · · + bk = ai+1 + ai+2 + · · · + ak+i. Since (k + i) − (a1 + a2 + · · · + ak+i) = xk+i ≥ xi = i − (a1 + a2 + · · · + ai ), b1 + b2 + · · · + bk ≤ k. 2. For k = n − i, we have b1 + b2 + · · · + bk = ai+1 + ai+2 + · · · + an = (n + 1) − (a1 + a2 + · · · + ai ) = n + 1 − (i − x) ≤ n − i, as x < 0.

46

I. Dumitriu, J. Spencer, and C. Yan

3. For n − i < k < n, we have b1 + b2 + · · · + bk = ai+1 + · · · + an + a1 + · · · + ak−(n−i) = n + 1 − (i − x) + a1 + · · · + ak−(n−i). Since k − (n − i) − (a1 + · · · + ak−(n−i)) ≥ x + 1, we have b1 + b2 + · · · + bk ≤ n + 1 − i + x + k − (n − i) − (x + 1) = k. This finishes the proof. Using Theorem 3.2, we have the following result for Z = Po(1 − ε). Theorem 3.3. In the random walk Y with Z = Po(1 − ε), Pr[T > N] ∼ eε iff N  ε−2 . Or equivalently,

Pr[T = ∞|T > N] ∼ 1 iff N  ε−2 .

Proof. The sufficient part is proved in Theorem 3.1. For the necessary part, we need to show ∞

∑ Pr(T = i) = o(ε),

If

then N  ε−2 ,

i=N

where the summation does not contain the term Pr[T = +∞]. We show this by finding a lower bounds for ∑∞ i=N Pr[T = i]. Note that when Z = Po(1 − ε), SN has the distribution Po(N(1 − ε)). Pr[T = N] ≥ Pr[T = N, YN = 0] = =

1 Pr[SN = N + 1] N

t e−tN t N+1 N N (teε )N ∼√ , (N + 1)! 2π (N + 1)3/2

where t = 1 − ε. Note that teε ∼ 1 − ε−2/2. So if N = Cε−2 for a constant C, then ∞

t

∞

(teε )i

∑ Pr[T = i] ≥ √2π ∑ (i + 1)3/2

i=N

i=N

t 2N (teε )i ≥√ ∑ 2π i=N (i + 1)3/2 2N t ≥ √ (teε )2N ∑ (i + 1)−3/2 2π i=N   2t −C 1 ∼√ e 1 − √ ε. 2πC 2

Branching Processes with Negative Offspring Distributions

That is, if N = Cε−2 , then

47

∞

∑ Pr[T = i] > C1 · ε,

i=N

for a constant C1 > 0. This proves that for ∑∞ i=N Pr(T = i) = o(ε), it is necessary that N  ε−2 . References 1. N. Alon and J. Spencer, The Probabilistic Method, 2nd Edition, John Wiley & Sons, Inc., 2000. 2. K.B. Athreya and P. Ney, Branching Processes, Springer-Verlag, 1972. 3. M. Dwass, The total progeny in a branching process, J. Appl. Probab. 6 (1969) 682–686. 4. W. Feller, An Introduction to Probability Theory and Its Applications, John Wiley & Sons, Inc., 3rd Edition, 1970. 5. F. Galton, Problem 4001, Educational Times 1 April (1873) p. 17. 6. T.E. Harris, The Theory of Branching Processes, Springer-Verlag, New York, 1963. 7. P. Jagers, Branching Processes with Biological Applications, Wiley, 1975. 8. D.G. Kendall, Some problems in the theory of queue, J. Roy. Statist. Soc. Ser. B 13 (1951) 151–185. 9. D.G. Kendall, Branching processes since 1873, J. London Math. Soc. 41 (1966) 385–406. 10. C.J. Mode, Multitype Branching Processes, Theory and Applications, American Elsevier, New York, 1971. 11. R. Otter, The multiplicative process, Ann. Math. Statist. 20 (1949) 206–224. 12. M.P. Quine and W. Szczotka, Generalizations of the Bienayme-Galton-Watson branching process via its representation as an embedded random walk, Ann. Appl. Probab. 4 (4) (1994) 1206–1222. 13. M.P. Quine and W. Szczotka, Asymptotic normality of processes with a branching structure, Comm. Statist. Stochastic Models 14 (4) (1998) 833–848. 14. M.P. Quine and W. Szczotka, Existence and positivity of the limit in processes with a branching structure, J. Appl. Probab. 36 (1) (1999) 132–138. 15. J. Spencer and C.H. Yan, An enumeration problem and branching processes, preprint. 16. R. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge University Press, 1999. 17. H.W. Watson, Solution to problem 4001, Educational Times 1 August (1873) 115–116. 18. C.H. Yan, On the enumeration of generalized parking functions, Congr. Numer. 147 (2000) 201–209.