Statistics and Probability Letters 78 (2008) 3008–3013

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Statistics and Probability Letters

journal homepage: www.elsevier.com/locate/stapro

Higher order moments of renewal counting processes andEulerian polynomialsGeoffrey W. BrownRoyal Military College of Canada, PO Box 17000 Stn Forces, Kingston, ON, Canada, K7K 7B4

a r t i c l e i n f o

Article history:Received 8 May 2007Received in revised form 14 November2007Accepted 1 May 2008Available online 20 May 2008

a b s t r a c t

In this paper we provide explicit formulas for the moments of all orders of Nm, the numberof renewals in [0,m] of discrete-time renewal processes. We decompose the sample spaceof the process, and the formulas are obtained using combinatorial methods. We obtainexplicit formulas for the kth order moments E[Nkm] in this manner, and we also provideasymptotic results in both the transient and recurrent cases.

© 2008 Elsevier B.V. All rights reserved.

1. Introduction

In this paper we study the discrete renewal process Nm for the number of renewals in the interval [0,m]. In particular,we study the higher order moments E[Nkm] of the renewal process. The first and second moments E[Nm] and E[N

2m] may

be obtained by considering the variableWn, the time required for the nth renewal. Calculations involvingWn are simpler,because it is the sum of n identically distributed renewal times T1, T2, . . . , Tn. We can then relate the probabilities of Nm andWn, and wemay obtain the first and secondmoments of Nm by telescoping the variablesWn. See, for example, Chaudhry andGupta (2008). Exact formulas are obtained in this manner, but this telescoping method does not easily extend to momentsof all orders.One way to avoid this difficulty is not to deal with the conventional moments E[Nkm] directly, but instead to compute the

so-called φ-moments of the renewal process. The kth φ-moment for Nm is defined as

φ[k]m = E(Nm + 1)(Nm + 2) · · · (Nm + k) form ≥ 1, and k ≥ 1.

It can be shown (see Hunter (1983)) that the generating function for the kth φ-moments of Nm is given as∞∑m=0

φ[k]m zm=

k!(1− z)(1− f (z))k

, (1)

where f (z) is the probability generating function for the inter-renewal time of the process.Analogous to (1), in this paper we present a method that does give explicit formulas for the moments E[Nkm] of all

orders, and it is based essentially on a decomposition of the sample space of Nm. The resulting formulas are then derivedby combinatorial methods, and they are closely related to the Eulerian polynomials. We obtain asymptotic expressionsfor the higher order moments of transient (and recurrent) renewal processes directly from them. We begin with the basicdefinitions of discrete-time renewal theory, and we closely follow Chaudhry and Gupta (2008) throughout.

2. Basic discrete-time renewal theory

A process Nm,m ≥ 1 whose state space belongs to a denumerable set 0, 1, 2, . . . and for which the inter-renewaltimes Tn = tn − tn−1, n = 1, 2, 3, . . . , t0 = 0 between two successive renewals are positive independent and identically

E-mail address: Geoffrey.brown@rmc.ca.

0167-7152/$ – see front matter© 2008 Elsevier B.V. All rights reserved.doi:10.1016/j.spl.2008.05.017

G.W. Brown / Statistics and Probability Letters 78 (2008) 3008–3013 3009

distributed random variables (distributed as T ) is called a renewal process. For a givenm, we define the random variable Nmto count the number of renewals during a time period [0,m].For each n ≥ 1, the renewal intervals Tn are i.i.d. random variables having common distribution

fk = P(Tn = k), for all n ≥ 1, k ≥ 1, f0 = 0

and common mean

µ = E[Tn] =∞∑k=0

kfk.

The probability generating function for the inter-renewal time T is given as

f (z) = E[zT ] =∞∑k=0

fkzk. (2)

Since fk is a distribution which could possibly be incomplete (or improper) we have

f =∞∑k=0

fk ≤ 1.

In the infinite time horizon, note that the probability that exactly n renewals occur is given by

f n(1− f ), n ≥ 0.

Thus, the long run number of renewals is geometrically distributed, with success probability 1 − f . Success denotes nofurther renewals in this case. For this reason, a renewal process Tn, n ≥ 1 is called recurrent if f = 1 and transient iff < 1. Thus, we observe infinitely many renewals for a recurrent process (with probability 1), whereas we observe onlyfinitelymany renewals for a transient process (with probability 1). With these notions in place, we are ready to discuss themoments E[Nkm] of the renewal counting process Nm.

3. Higher order moments of the process Nm

Let Ω denote the sample space of the renewal process in the infinite time horizon. Let R be the event that a renewaloccurs, and letN be the event that a renewal does not occur. Since successive renewals are independent, wemay decomposethe sample spaceΩ as follows:

Lemma 1. Decomposition of Ω:

Ω =

∞⋃i=0

RiN . (3)

Thus (3) is the decomposition of Ω with respect to the number of renewals, and it is written in terms of disjoint union andCartesian product of sets (or events).We will use (3) to evaluate

∞∑m=0

E[N2m]zm,

the generating series for the second moments of Nm. We formally let R(z) and N(z) be the probability generating functions forthe indicators of the eventsR and N , and we will specify these functions later. If we restrict Ω to the interval [0,m], the eventRiN on [0,m] implies that the random variable N2m will have the value i

2, and so we conclude from (3) that

∞∑m=0

E[N2m]zm=

∞∑k=0

k2R(z)kN(z). (4)

To check (4), we find [zm] on the right-hand side, and we see that

[zm]∞∑k=0

k2R(z)kN(z) =∞∑k=0

k2[zm]R(z)kN(z)

=

∞∑k=0

k2P(Nm = k)

= E[N2m], (5)

3010 G.W. Brown / Statistics and Probability Letters 78 (2008) 3008–3013

exactly as we require. It is important to mention that the decomposition (3) is valid for Ω restricted to any finite interval [0,m],and this is what justifies (4). We now solve for R(z) and N(z) in terms of the probability generating function f (z) of the inter-renewal time T .

Let IR and IN denote the indicator variables of the eventsR andN . Then

R(z) = E[z IR ] =∞∑k=0

fkzk = f (z), (6)

and it is easily shown that

N(z) = E[z IN ] =∞∑k=0

P(Nk = 0)zk =∞∑k=0

P(T > k)zk

=1− f (z)1− z

. (7)

Putting (6) and (7) in (4), we see that∞∑m=0

E[N2m]zm=1− f (z)1− z

·

∞∑k=0

k2f (z)k. (8)

Thus it remains to evaluate∑∞

k=0 k2f (z)k, but this can be done as follows. We begin with

11− f (z)

= 1+ f (z)+ f 2(z)+ · · · . (9)

Note that (9) is well defined as a composition of power series because f0 = 0. Differentiating (9), we obtain

f ′(z)(1− f (z))2

= f ′(z)+ 2f (z)f ′(z)+ 3f 2(z)f ′(z)+ · · · . (10)

Notice that f ′(z) cancels from (10), and we multiply (10) by f (z) to obtain

f (z)(1− f (z))2

= f (z)+ 2f 2(z)+ 3f 3(z)+ · · · =∞∑k=0

kf (z)k. (11)

Continuing this procedure, we again differentiate (11), cancel the f ′(z) factor, and multiply by f (z) to obtain

f (z)+ f 2(z)(1− f (z))3

= f (z)+ 22f 2(z)+ 32f 3(z)+ · · · =∞∑k=0

k2f (z)k. (12)

Hence, from (8) and (12) we conclude that∞∑m=0

E[N2m]zm=1− f (z)1− z

·f (z)+ f 2(z)(1− f (z))3

=11− z

·f (z)+ f 2(z)(1− f (z))2

. (13)

Thus, to extend this procedure to kth order moments, we adapt (4) to equate the following generating series:∞∑m=0

E[Nkm]zm=

∞∑i=0

ikR(z)iN(z)

= N(z) ·∞∑i=0

ikf (z)i. (14)

We may formally regard f (z) as the variable x, and we successively apply to 11−x the operations of differentiating and

multiplying by x, which we will denote as (xD). On the basis of (11) and (12), we now conclude that∞∑i=0

ikf (z)i = (xD)k11− x

∣∣∣∣x=f (z)

. (15)

Note that (15) is justified precisely because the derivative f ′(z) plays no role in the calculations.The expressions (xD)k 1

1−x are well known in combinatorics (see Comtet (1974) or Graham, Knuth and Patashnik (1994))to generate the Eulerian polynomials, and they are the subject of the next section.

G.W. Brown / Statistics and Probability Letters 78 (2008) 3008–3013 3011

4. Eulerian polynomials

In this section we derive the basic facts about the Eulerian polynomials which are relevant for our purposes. We beginby calculating (xD)k 1

1−x for the values k = 1, 2, 3, and 4:

(xD)11− x

=x

(1− x)2. (xD)3

11− x

=x+ 4x2 + x3

(1− x)4.

(xD)211− x

=x+ x2

(1− x)3. (xD)4

11− x

=x+ 11x2 + 11x3 + x4

(1− x)5.

From these calculations we may easily conclude by induction that

(xD)k11− x

=Pk(x)

(1− x)k+1, (16)

where Pk(x) is a monic polynomial of degree kwith no constant term.We now make the following definition:

Definition 1. For k ≥ 1, the polynomial Pk(x) of (16) is the Eulerian polynomial of degree k. For k = 0, we define P0(x) = 1.We now derive the formulas that give the coefficients of Pk(x) explicitly. To this end, let

Pk(x) =k∑r=1

a(k)r xr . (17)

Then from (16) we have

(xD)k11− x

=Pk(x)

(1− x)k+1=

∞∑n=1

nkxn. (18)

We conclude from (18) that

Pk(x) = (1− x)k+1(xD)k11− x

= (1− x)k+1 ·∞∑n=1

nkxn. (19)

Therefore

[xr ]Pk(x) = a(k)r =r∑i=0

(−1)i(k+ 1i

)(r − i)k, for r = 1, 2, . . . . (20)

From (20) we make the important observation that [xr ]Pk(x) = 0 for r > k since Pk(x) is a polynomial of degree k.It is also important for our purposes to know the sum of the coefficients of Pk(x) for each k. Thus, we seek a formula for∑kr=1 a

(k)r . The calculations of P1(x), P2(x), P3(x), and P4(x) suggest that

Sk =k∑r=1

a(k)r = k!. (21)

In general, we may realize the sum of coefficients by multiplying the appropriate generating series by 11−x . Thus from (19)

we calculate:

Pk(x)11− x

=

∞∑n=1

(n∑r=1

a(k)r

)xn = (1− x)k ·

∞∑n=1

nkxn. (22)

Therefore we conclude that

Sk =k∑r=1

a(k)r = [xk](1− x)k ·

∞∑n=1

nkxn =k∑i=0

(−1)i(ki

)(k− i)k. (23)

In order to proceed by induction, we must first notice the crucial fact that

[xk]Pk−1(x) = 0⇒k∑i=0

(−1)i(ki

)(k− i)k−1 = 0 for all k ≥ 1. (24)

Now from the identity i(ki

)= k

(k−1i−1

)and (24), we may show that (23) simplifies as

3012 G.W. Brown / Statistics and Probability Letters 78 (2008) 3008–3013

Sk = k ·k−1∑i=0

(−1)i(k− 1i

)((k− 1)− i)k−1

= k · Sk−1.

Since S1 = 1, we have that Sk =∑kr=1 a

(k)r = k!, and this establishes (21).

Returning to discrete-time renewal processes, let

Mk(z) =∞∑m=0

E[Nkm]zm (25)

be the generating series for the kth order moments of Nm. From (7) and (14)–(17), we may giveMk(z) explicitly as follows:

Mk(z) = N(z) ·∞∑i=0

ikf (z)i

= N(z) · (xD)k11− x

∣∣∣∣x=f (z)

= N(z) ·Pk(x)

(1− x)k+1

∣∣∣∣x=f (z)

=1− f (z)1− z

·

k∑r=1a(k)r f (z)r

(1− f (z))k+1

=

k∑r=1a(k)r f (z)r

(1− z)(1− f (z))k, (26)

where the coefficients a(k)r are given in (20). We also mention that (1), the generating function for the kth φ-moments of Nmmay be computed as N(z) · Dk 1

1−x |x=f (z) in an analogous fashion.In the next section we will use (26) to obtain asymptotic results for the kth order moments of Nm in both the transient

and recurrent cases.

5. Asymptotic results for E[Nkm]

In this section we derive asymptotic expressions for E[Nkm] in both the transient and recurrent cases. The recurrent caseis often established using the asymptotic normality of Nm (see Chaudhry and Gupta (2008)). So (26) provides an alternatederivation. We begin by stating the binomial theorem for negative integers.

Lemma 2 (Negative Binomial Theorem). For |z| < 1, we may write the following power series expansion for all k ≥ 1:

(1− z)−k =∞∑n=0

(n+ k− 1k− 1

)zn.

5.1. Transient case

We assume the inter-renewal time T is transient, so f = f (1) =∑∞

k=0 fk < 1. In this case Nm has a non-degenerateprobability distribution asm→∞, and

limm→∞

P(Nm = k) = f k(1− f ) (27)

from Section 2. Using the long run distribution of Nm, we directly calculate that

limm→∞

E[Nkm] =∞∑i=0

ikf i(1− f )

= (1− f ) · (xD)k11− x

∣∣∣∣x=f

=

k∑r=1a(k)r f r

(1− f )k, (28)

G.W. Brown / Statistics and Probability Letters 78 (2008) 3008–3013 3013

from (16) and (17). Thus, an explicit formula is possible in the transient case, and it is complicated by the fact thatidiosyncrasies do not ‘‘average out’’ in the long run.

5.2. Recurrent case

Wenow consider the casewhere the inter-renewal time T is recurrent, and so f = f (1) =∑∞

k=0 fk = 1. From (26),Mk(z)has a pole of order (k+ 1) at z = 1, so we may write the following Laurent expansion forMk(z) at z = 1:

Mk(z) = Ak(z)+ck

(1− z)k+1, (29)

where limz→1(1− z)k+1Ak(z) = 0. Therefore,

ck = limz→1

(1− z)k+1Mk(z)

= limz→1

(1− z)k

k∑r=1a(k)r f (z)r

(1− f (z))k

=k!µk, (30)

from (21) and repeated applications of L’Hôpital’s rule. From (29), (30) and Lemma 2, we conclude that as n→∞,

[zn]Mk(z) ∼ ck[zn](1− z)−(k+1)

=k!µk

(n+ kk

)∼

(nµ

)k. (31)

Thus for recurrent processes, we expect an average of nµrenewals on [0, n]when n is large, taken to the power k for the kth

order moment.See Hunter (1983) for an excellent discussion of many results of discrete-time renewal theory.

References

Chaudhry, M.L., Gupta, 2008. Models of Discrete-Time Queueing Theory. Working manuscript.Comtet, L., 1974. Advanced Combinatorics. D. Reidel Publishing Company.Graham, Knuth, Patashnik, 1994. Concrete Mathematics, second ed. Addison-Wesley Publishing Company.Hunter, J.J., 1983. Mathematical Techniques of Applied Probability. In: Discrete Time Models: Basic Theory, vol. 1. Academic Press.