Asymptotics of the maximum of Brownian motion under Erlangian sampling

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1 Asymptotics of the maximum of Brownian motion under Erlangian sampling paper in honor of N.G. de Bruin A.J.E.M. Janssen J.S.H. van Leeuwaarden October 30, 202 Abstract Consider the all-time maximum of a Brownian motion with negative drift. Assume that this process is sampled at certain points in time, where the time between two consecutive points is rendered by an Erlang distribution with mean /ω. The family of Erlang distributions covers the entire range between deterministic and exponential distributions. We show that the average convergence rate as ω for all such Erlangian sampled Brownian motions is Oω /2, and that the constant involved in O ranges from ζ/2/ 2π for deterministic sampling to / 2 for exponential sampling. The basic ingredients of our analysis are a finite-series expressions for the expected maximum, an asymptotic expansion of = exp2πi/ s as using Euler-Maclaurin summation, and Fourier sampling of functions analytic in an open set containing the closed unit dis. Keywords: Brownian motion, random wal, all-time maximum, sampling, Euler-MacLaurin summation, Fourier sampling, Riemann zeta function AMS 200 Subect Classification: 60J65, 60E99, 65B5, 4A60, 30E20 Introduction Let {B β t : t 0} be a Brownian motion with negative drift defined as B β t = βt + W t, β 0, with B β 0 = 0 and {W t : t 0} a Wiener process standard Brownian motion. Since β is assumed to be positive, the Brownian motion will eventually drift towards, and the all-time maximum M β = sup t R + B β t is well defined. In fact, it is nown that M β follows an exponential distribution with rate 2β, so that P M β x = e 2βx see e.g. [8, Lemma 5.5], and hence the expected all-time maximum is simply given by E M β = /2β. We consider sampled versions of the Brownian motion, meaning that we observe the process only at time points t 0 = 0, t, t 2,.... A crucial assumption we mae is that the times between consecutive sampling points T n = t n t n, n N are independent and identically distributed Eindhoven University of Technology and Eurandom, Department of Mathematics and Computer Science and Department of Electrical Engineering, P.O. Box 53, 5600 MB Eindhoven, The Netherlands. a..e.m.anssen@tue.nl Eindhoven University of Technology, Department of Mathematics and Computer Science, P.O. Box 53, 5600 MB Eindhoven, The Netherlands. .s.h.v.leeuwaarden@tue.nl

2 i.i.d.. Let T denote a generic random variable with T d = T here d = denotes equal in distribution and ET = ω. The constant ω represents the expected number of observations per unit of time, henceforth referred to as the sampling frequency. It is readily seen that a sampled version of the Brownian motion constitutes a random wal {S β n : n N} with S β 0 = 0 and S β n = X X n, with X, X, X 2,... i.i.d. and X d = B β T. The fact that Brownian motion evolves in continuous space and time leads to great simplifications in determining its properties. In contrast, the random wals that we obtain after sampling, moving only at certain points in time, are obects that are much harder to study. Although it is obvious that, as ω, the behavior of the random wal should become identical to that of the Brownian motion, there are many effects to tae into account for finite ω. Let the maximum of the random wal be denoted by M β ω = sup n=0,,... B β t n. The sampling error β ω = M β M β ω then depends on the drift β, the sampling frequency ω, and of course on the distribution of T. This paper deals with the expected maximum of the random wals and, in particular, its deviation E β ω from the expected maximum /2β of the underlying Brownian motion. This relatively simple characteristic already turns out to have an intriguing description. We assume that the times between sampling points are drawn from an Erlang distribution, so that T d = E λ with E λ an Erlang distributed random variable consisting of independent exponential phases, each with mean /λ, and PE λ < x = n=0 n! e λx λx n, x 0. 2 The mean and variance of E λ are given by /λ and /λ 2, respectively. One reason for woring under the assumption of Erlangian sampling is that T constant and T exponentially distributed are opposite extremes with regard to randomness as well as in the family of Erlang distributions viz. with λ = ω and and =, respectively. Another reason is that Erlangian sampling leads to a random wal of which the distribution of the all-time maximum allows for an explicit solution. This gives rise to a series expression for EM β ω; in which the terms involve the roots of P σ = [σ + ρ σ] ρ = 0 in σ < with ρ 0, given by ρ 2 ω = 2β 2 ρ. In this paper this series expression is first analyzed for the case that ω while remains bounded, yielding a leading-order behavior of E β ω of the form ω /2 ϕ / 2π + Oω. Here ϕ is given explicitly in terms of the unit roots exp2πi/, and tends to ζ/2 as. Next, a substantial effort is spent to mae the Oω -term in the leading-order behavior of E β ω uniform in all. To do this, an alternative series representation of EM β ω; is derived that involves also the roots of P in σ. Finally, this alternative representation can also be used to describe the way in which EM β ω; 0 as ω 0 and. For a detailed overview of our results we refer to Subsection Preliminary results In this section we present some first results. We show how the Erlangian sampling leads to a tractable random wal. We also derive a general expression for the expected all-time maximum. Before we do so, we pay attention to the special cases of equidistant sampling and exponential sampling, for which results are available in the literature. 2

3 2. Equidistant sampling In the case of equidistant sampling the time between two consecutive sampling points is always /ω. From the definition of Brownian motion it then immediately follows that B β T d = N βω, ω, where Na, b denotes a normally distributed random variable with mean a and variance b. We should thus consider the maximum of a random wal with normally distributed increments, referred to in the literature as the Gaussian random wal. The maximum of this random wal was studied in [7, 4]. In particular, [4, Thm. 2] yields, for ω > β/2 π, EM β ω = 2β + ζ/2 2πω + β 4ω + β2 ω 2πω r=0 ζ /2 r r!2r + 2r + 2 with ζ the Riemann zeta function. This implies immediately that β 2 r, 3 E β ω = ζ/2 2πω + Oω 4 2ω with ζ/2/ 2π Results similar to 4, in slightly different settings, have been presented in [3, Thm. 2] and [6, Thm. ]. A crucial difference is that our result 3 is obtained from the exact expression for EM β ω, while the results in [3, 6] are derived from considering the Brownian motion in a finite time interval, and estimating its maximum by Euler-Maclaurin summation. 2.2 Exponential sampling In the case of exponential sampling, we assume that the times between consecutive sampling points are independent and exponentially distributed with mean /ω. In this case we can prove the following result. Lemma 2.. If T d = E ω then B β T d = E γ 2 E γ, 5 where γ = β + β 2 + 2ω, γ 2 = β + β 2 + 2ω. 6 Proof. For βs + s 2 /2 < ω, The result in 5 then follows from Ee sb βt = = 0 e βs+s2 /2t ωe ωt dt ω ω βs + s 2 /2. 7 ω ω βs + s 2 /2 = γ γ 2 γ sγ 2 + s, 8 with γ, γ 2 as in 6, and Lévy s continuity theorem for Laplace transforms [6]. 3

4 The random wal for which the increments are distributed as the difference of two exponentials has been thoroughly studied in the literature. The maximum of this random wal is nown to be equal in distribution to the stationary waiting time in a so-called M/M/ queue with arrival rate γ and service rate γ 2, for which see e.g. [2, p. 08] PM β ω > x = γ γ 2 e γ 2 γ x, x 0. 9 This implies that from which it readily follows that EM β ω = γ γ 2 γ 2 γ, 0 E β ω = 2ω + Oω with / A similar result was obtained in [6, Thm. 3] for a Brownian motion in a finite time interval sampled at uniformly distributed points. 2.3 Erlangian sampling We now set T d = E ω with mean ω and variance /ω 2. Clearly, random sampling = and equidistant sampling = can be seen as special cases. We first mae the following observation. Lemma 2.2. If T d = E ω then B β T d = E γ 2 E γ, 2 where γ = β + β 2 + 2ω, γ 2 = β + β 2 + 2ω. 3 Proof. We now that for βs + s 2 /2 < ω see the proof of Lemma 2. Ee sb βe ω ω = ω βs + s 2 /2 γ γ 2 = γ sγ 2 + s. 4 Hence, which completes the proof. Ee sb βt = γ γ2, 5 γ s γ 2 + s From Lemma 2.2 we conclude that in order to study an Erlangian sampled Brownian motion, we need to study a random wal with increments defined as the difference of two Erlang distributed random variables. As it happens, this random wal has been studied before, and an explicit solution for the distribution of M β ω is available. In order to explain this, we need to mae a small excursion into the world of fluctuation theory. We start from the observation that M β ω d = W with W d = max{0, W + B β T }. 6 4

5 This equation is a special case of Lindley s equation, describing the stationary waiting time of a customer in the GI/G/ queue. The case in 6 describes a single-server queue with Erlang distributed service times and Erlang distributed interarrival times. In [] it was shown that the distribution of W can be expressed as a finite sum of exponentials, the exponents of which are the roots of an equation that involves the Laplace transform of the Erlang distribution. This gives the result [, Corollary 3.3] Here, PM β ω > t = c = c σ e γ 2 σ t, t 0. 7 =,l σ l,l σ l/σ and σ 0,..., σ the roots in σ < of, = 0,,...,, 8 P σ := [σ + ρ σ] ρ = 0, 9 where so that γ 2 ρ = 2β γ 2, 20 = ρ 2β and ρ 2 ω = 2β 2 ρ. 2 The σ are given explicitly as σ = 2 + ρ 2 + ρ2 ρe 2πi/, = 0,,...,. 22 The expected all-time maximum then equals EM β ω; = =0 c σ γ 2 σ, 23 and it is this expression that forms the point of departure for this paper. 2.4 Overview of the results In the coordinates ρ,, ω and β, related according to 3, 20, and 2, we obtain a limit result for EM β ω; in the case that ω and bounded or unbounded, and in the case that ω 0 and. In this paper, we do not address the intriguing question what happens when ω tends to a non-zero finite limit and. In Section 3 we start from 23 and show that when ω and is bounded, where EM β ω; = 2β 2ω = u + Oω, 24 u = e 2πi/, = 0,,..., 25 denote the unit roots. Notice that the case = is in line with. 5

6 In Section 4 we determine the asymptotic behavior of the series u s 26 = when s R is fixed and. We use for this a method developed by Brauchart et al. [4] based on the Euler-Maclaurin summation formula. In particular, our results imply that 2 = = ζ/2 π/2 ζ /2 u 2π O Notice that for this agrees with the result in 4. In Section 3 the result 24 is proved for the case that remains bounded Theorem 3.. The proof of Theorem 3. is relatively simple and uses the direct representation in 8 of the c s as they occur in the series expression 23. A ey observation in this proof is that the term with = 0 in the series representation dominates all other terms. The result 24 comprises the quantity = u /2 for which the asymptotic behavior as is given Theorem 4. in terms of the Riemann zeta function by 27, using an approach based on Euler-MacLaurin summation. It is therefore a relevant question to as whether 24 also holds with Oω holding uniformly in. The approach to prove 24 for unbounded using the direct representation 8 of the c is severely complicated by the fact that, as and ω, the zeros of P in 9 inside the unit circle accumulate on the set { u u = }, so that the quantities σ l /σ, l, that occur in 8, can become arbitrarily small. In a situation lie this it may be advantageous, as exemplified on several occasions in N.G. de Bruin s Asymptotic Methods in Analysis, to view the problem at hand from a different perspective. In this spirit, we use a different representation of the c s, viz. one in which factors σ l /σ+ rather than σ l /σ appear with σ and σ + the zeros of P inside and outside the open unit dis σ <, respectively Lemma 5.. This new representation gives bounds on c Theorems 5.3 and 5.4 from which we can conclude that also in the case of unbounded, the term with = 0 in the representation 23 is dominant when ω. The proofs of Theorems 5.3 and 5.4 rely heavily on Lemma 5.2 that gives bounds on the products σ l /σ+ that appear in the new representation of the c. In the proof of Lemma 5.2 a crucial role is played by the result that expresses a series =0 hu, with hz = n=0 d nz n analytic in an open set containing the closed unit dis, in terms of the decimated coefficients d s, = 0,,... Fourier sampling. In proving the extension of 24 to unbounded Theorem 6., due to the results of Section 5, attention can thus be restricted to the term with = 0 in 23, maing the proof rather straightforward. In Section 7 we consider the behavior of EM β ω; as ω 0 and the case that ω 0 while remains bounded yields EM β ω; 0 in a trivial manner from Then the term with = 0 in 23 is no longer dominant, and it can be shown from the results of Section 5 that EM β ω; is well approximated, see the proof of Theorem 7., by ρ 2β =0 + ρ σ σ σ 2 + ρ 2σ. 28 The series in 28 can be cast into the form =0 F u, with F analytic in an open set containing the closed unit dis and F 0 = 0. To this series, the Fourier sampling technique, as it occurs in the proof of Lemma 5.2, can be applied. It thus follows that EM β ω; tends to zero when ω 0 and, and, ω related as in 2 with fixed β > 0, and also the rate at which EM β ω; tends to zero can be determined. 6

7 3 Behavior of EM β ω; for bounded and ω We prove in this section the following result. Theorem 3.. Let =, 2,.... As ω, EM β ω; = 2β 2ω where u are given as exp2πi/, =,...,. = u + Oω, 29 Proof. We use the series expression 23 for EM β ω;. ρ = Oω /2, and so from 22 We have from 3 and 20 that σ = 2 ρ + u ρ u + 2 ρ2 = u + O ω /2, = 0,,...,. 30 In particular, σ 0 = Oω /2 while σ is bounded away from 0 for =,..., as ω. Then, from 8 and ρ = Oω /2, as ω while for =,..., c 0 = l= σl ρ, 3 l= σ l c ρ = l=,l σ l /,l σl σ 32 has a finite limit 0 as ω. We conclude from ρ = σ 0, 3, 32 and ρ = Oω /2 that, as ω, EM β ω; = =0 c σ γ 2 σ = γ 2 c 0 ρ ρ + Oω. 33 To proceed, we need to approximate c 0 accurately. From the identity in the first line of 30 we find for =,..., This gives, see 3, σ = 2 ρ u 2 ρ + O ρ2 = u + 2 u ρ + O ρ ρ c 0 = ρ = = ρ = = ρ + σ ρ σ u 2 u ρ + O ρ 2 u 2 u + ρ + O ρ 2 = ρ u + O ρ 2 7

8 = ρ = u + O ρ 2, 35 where it has been used that ρ = ρ + O ρ 2. Using this in 33 while noting that γ 2 ρ = 2β and that γ 2 = 2ω + Oω, we obtain the result. The proof of Theorem 3. shows that the term with = 0 in the series 23 dominates all other terms when ω and remains bounded. We establish a similar result more generally, allowing to be unbounded as well, in Section 6, and extend the result of Theorem 3. accordingly. 4 Asymptotics of = u s as for fixed s The large-ω expression in Theorem 3. for EM β ω; contains the series = u s with s = /2. It is of interest to find out how this series behaves with increasing. Furthermore, in Section 6, we consider the case that ω with unbounded allowed, and then it appears that the behavior of the series = u s for large is required to be nown for s = /2,, 3/2,.... We adopt an approach in [4], for determining the asymptotic behavior of S s := u s, u = e 2πi/, 36 = as and s R is fixed. In [4] this approach is used for finding the asymptotic behavior of U s := sin π s 37 = as and s C is fixed. The result that we obtain here for S s is of the same nature as the result for U s in [4], except that in our result, Theorem 4. below, all terms ζs l s l, l = 0,,..., 38 occur, while the result in [4] has only terms 38 with even l. Furthermore, the exceptional cases s =, 2,... are less complicated for our S s than they are for U s in [4]. The method in [4] is a fine application of the Euler-Maclaurin summation formula that can be found, along with various applications, in [5, Secs ]. We tae s R in 36 and this implies that the terms t in 36 satisfy t = t, =,...,. Hence, S s is real, and so we have where Expanding S s = = f, 39 f x = Re [ e 2πix/ s ]. 40 z s = e z B s l 0 z l, z < 2π, 4 l! 8

9 with B s l the generalized Bernoulli polynomials, as in [4, Sec. ], we get where f x = β l s = 2πl s B s l 0 cosl + s π l! 2 Remember that ζ denotes the Riemann zeta function. x l s β l s, x <, 42, l = 0,, Theorem 4.. Let s R and let p = 0,,... such that s + 2p > 0. Then, as, 2p + 2 β l s ζs l s l + O s,p s 2p, s, 2,..., S s = 2p β l s ζ l s l + O s,p s 2p, s =, 2,...,,l s 44 where the constants implied by the O s,p depend on s and p but not on. Proof. We closely follow the approach in [4], so that many of the details are left out. We have for f C 2p+ [, m] that m f = = + p r= m fx dx + 2 f + fm B 2r 2r! f 2r m f 2r + m f 2p+ x C 2p+x dx, 45 2p +! where C 2p+ x = B 2p+ x [x] is the periodized Bernoulli polynomial and B 2r is the Bernoulli number. Using this with m = and f = f, see 40, and noting that f x = f x, we get S s = = 2 /2 = f f x dx+f 2 p r= B 2r 2r! f 2r +2 /2 f 2p+ x C 2p+x dx. 46 2p +! All quantities on the second line of 46 involving f can be expressed in terms of the β l s in In particular, we get 2 /2 f x dx = 2 where it has been used that for s R, s < s l β l s l s +, 47 2 /2 0 [ Re ] e 2πiy s dy = 9 0 dy e 2πiy s = 48

10 and that the same analyticity considerations as in [4, Subsec. 2.] apply. In 47 we have to consider the cases that s =, 2,... separately because of the term l = s ; this will be done below. We find, after using in 46 that for s, 2,... S s = + 2 { β l s p s l s l 2p+ r= /2 B 2r 2r! s l 2r x l s 2p C 2p+x 2p +! dx }, 49 where a n is Pochhammer s symbol. The expressions in {...} at the right-hand side of 49 are identified in [4, Subsec. 2.2] as incomplete zeta functions ζ y,p t with y = /2 and t = s l, for which ζ y,p t ζt = O t,p y t 2p, y > Hence, for s, 2,... and any p = 0,,... S s = + 2 β l s ζ /2,p s l s l. 5 In the case that s = K =, 2,..., we tae the limit s K in 5, using the result, to be proved below, β K s lim s K s K = Note that β K K = 0, due to the factor cosl + s π 2 at the right-hand side of 43. Hence, we get for s = K =, 2,... S K = 2 + 2,l K β l K ζ /2,p K l K l. 53 Finally, taing any p = 0,,... with s + 2p > 0, we can use 50 to conclude the proof in the same way as the proof in [4] for U s in 37 is concluded in [4, Section 4]. We still have to show 52. From 43 we have for K =, 2,... lim s K β K s s K = 4 K BK K 0 K!, 54 where B K K 0/K! is the coefficient of zk in z/e z K. Thus, we have for 0 < r < 2π B K K 0 K! = 2πi = 2πi z =r z =r z K dz z K e z K dz = e z 2πi C r w + w dw = K, where the substitution w = e z, dz = dw/ + w has been used and C r is the image under the mapping z = r e z which is easily seen, for small r > 0, to be a Jordan curve having the origin w = 0 in its interior. This shows

11 Note 4.2. The result in 27 can be obtained by taing s = /2, p = in 44, using that β 0 /2 = π 2 π, β /2 = 4, β 2/2 = π3/ Note 4.3. In the lower case in 44 a simplification occurs since for s =, 2,... β l s = 0, l odd ; ζs l = 0, l = s + 2, s + 4, This leads to [ s] 2 S s = 2 + r=0 β s 2r s ζ2r 2r + O s,p s 2p. 58 A similar situation as in Note 4.3 occurs in [4, Remar.2] for the case of U s in 37 with s = 2, 4,.... It is concluded in [4] that one gets exact formulas for U s in that case. While this is true, see [3], the argument in [4] is incomplete. In the case of S s with s =, 2,..., it can be shown that it depends polynomially on degree s, and so 58 holds exactly with the O s,p deleted. 5 Bounds on c from a representation using outer zeros When is allowed to be unbounded, the analysis of EM β ω; using the series in 23 with the c given by 8 is awward. In this section, we present an alternative representation of the c, using the zeros of P in 9 outside σ <, that is more convenient for getting bounds. For = 0,,...,, we let be the two solutions of the equation σ ± = 2 + ρ ± 2 + ρ2 ρ e 2πi/ 59 σ + ρ σ = ρ e 2πi/. 60 Then for =,..., and σ < σ 0 = ρ < = σ+ 0 < σ+, 6 σ + σ+ = + ρ ; σ σ+ = ρ e 2πi/. 62 When we let σ = σ, = 0,,..., ; σ + = σ +, = 0,,...,, 63 then σ, = 0,,..., 2 are the 2 zeros of P in 9. Lemma 5.. For = 0,,...,, c = + ρ σ σ + ρ 2σ σ l σ l σ +. 64

12 Proof. We have from 8 that c = σ σ σ l,l σ σ l. 65 We re-express the product in the denominator at the right-hand side of 65. There holds P σ = σ + ρ σ ρ = Hence, for = 0,,..., from 63 σ σ l σ σ + l. 66 On the other hand, P σ =,l σ σ l σ σ + l. 67 Hence, for = 0,,...,, P σ = σ + ρ σ ρ σ = ρ + ρ 2σ σ + ρ σ. 68,l σ σ l = P σ σ σ + l = ρ + ρ 2σ σ + ρ σ Next, by the first item in 63, σ σ + l = σ + σ l, and so σ σ + l. 69,l Using 70 in 65, we get σ σ l = ρ + ρ 2σ σ + ρ σ σ+ σ l. 70 c = Finally, tae out factors σ + 62, to obtain the result. σ /ρ + ρ σ σ + ρ 2σ σ l σ + σ l. 7 from the last product at the right-hand side of 7, and use, see σ σ + /ρ = σ σ+ /ρ = 72 We now analyze the product σ l/σ + of which σ l is the special case with = 0. Lemma 5.2. For = 0,,...,, where σ l σ + = expg ρ, 73 Reg ρ < 0, g ρ < 2 ln 4ρ + ρ

13 Proof. We have Now σ l σ + = exp ln σ + + lnσ + σ l. 75 [ ] lnσ + σ l = ln a 2 ρ e 2πi/ + a 2 ρ e 2πil/ = ln [ a + ] a 2 ρ u l, 76 where we have set a = 2 + ρ, a = a 2 ρ e 2πi/, u l = e 2πil/. 77 Hence where lnσ + σ l = h u l, 78 [ h u := ln a + ] a 2 ρu, u a 2 /ρ. 79 Note that τ := ρ a 2 = 4ρ ρ 2 + ρ 2 = 0, ρ Furthermore, a + a 2 ρu has positive real part when u a 2 /ρ = τ, and so h u is analytic in an open set containing the closed unit dis. There is the power series representation h u = d n ρ n u n, u a 2 /ρ, 8 n=0 in which d n are the power series coefficients of ln [ a + a 2 z]. From all this we get lnσ + σ l = h e 2πil/ = d s ρ s, 82 s=0 where it has been used that e 2πinl/ = {, n = 0,, 2,..., 0, otherwise. 83 Noting that d 0 = h 0 = ln σ +, we then see from 75 that σ l σ + = exp d s ρ s =: expg ρ. 84 s= We analyze the d n, n =, 2,..., in some detail. We have by Cauchy s theorem for n =, 2,... d n = ln + a a 2 z 2πi z n+ dz 85 z =r 3

14 when 0 < r < a 2. We deform the integration contour z = r so as to enclose the branch cut of a 2 z from z = a 2 to z = +. Now, for x > a 2, we have a 2 x ± i0 = i x a 2, 86 and so we get for n =, 2,... d n = 2πi a 2 [ ln a i x a 2 ln a + i ] dx x a xn+ By partial integration, noting that the quantity in [...] at the right-hand side of 87 vanishes at x = a 2, we get d n = 2πn a 2 = a 2πn [ 2 x a 2 a i x a + ] dx 2 a + i x a 2 x n a 2 x a 2 for n =, 2,.... Finally, setting u 2 = x a 2 0, we get dx a 2 + x a 2 x n 88 d n = a πn 0 a 2 + u 2 u 2 + a 2 du. 89 n By the substitutions t = u/a and v = t b /2, the result 89 can be brought into the forms d n = = π n a 2n π n a 2n 0 0 b /2 + b t 2 dt + t 2 n dv + v 2 + v 2, n =, 2,..., 90 /b n where we have set b = a/ a 2, so that = 4ρ b + ρ 2 e2πi/ 9 is in the right-half plane. We now show that Re d n < 0. We have for t 2 = x 0 and b C, Re b > 0 that b /2 b /2 Re = Re + bx b > x Then Re d n < 0 follows from 90 and Re b > 0. From 84 it is then seen that Re g ρ < 0, and the first item in 74 is proved. Next, we have from Re b > 0 that + v 2 /b > for all v > 0. Hence, from 90, d n <, n =, 2, na2n 4

15 From 84 it is then see that g ρ < s= ρ s ρ 2sa 2s = 2 ln a 2, 94 and noting that a = 2 + ρ, this gives the second item in 74. The following result follows immediately from Lemmas 5. and 5.2. Theorem 5.3. For = 0,,...,, c + ρ σ, 95 σ + ρ 2σ and c = + ρ σ τ + O σ + ρ 2σ τ where τ = 4ρ/ + ρ 2 and the constant implied by O in 96 is bounded by. 96 Another inequality for c, =,...,, is the following one. Theorem 5.4. For =, 2,..., [ 2 ] =: m, c = c ρ + 2 c / Proof. We have from σ = σ, =,...,, that c = c, =,...,, and this gives c = c, =,..., m. From Lemma 5. with = 0 we have c 0 = l= σ l Therefore, as σ 0 = ρ, σ l = ρ l= Furthermore, from Lemma 5.2 for =, 2,..., σ l 2 /2 = ρ c / σ l, 00 σ + and so from Lemma 5. + ρ σ c ρ c / σ + ρ 2σ Next, we show that for =, 2,..., m σ /2 /2, + ρ 2σ 2, + ρ σ

16 We have with t = 2π/ 0, π] σ = 2 ρ ρ2 ρ e it, 03 + ρ 2σ = ρ2 ρ e it, 04 + ρ σ = 2 + ρ ρ2 ρ e it. 05 Now ] Re [ 2 + ρ2 ρ e it > 0, 2 + ρ2 ρ e it 2 + ρ2 + ρ, 06 and so σ 2 + ρ2 ρ e it /2, + ρ 2σ ρ2 ρ e it /2 07 while, as 0 ρ, + ρ σ 2 + ρ ρ2 + ρ + 2, 08 which establishes the third inequality in 02. For the first two inequalities in 02, we compute 2 + ρ2 ρ e it 2 = 2 + ρ4 2ρ 2 + ρ2 cos t + ρ 2 = 2 ρ4 + 2ρ 2 + ρ2 cos t = 2 ρ4 + ρ + ρ 2 sin 2 2 t 2 ρ4 + ρ + ρ 2 t/π 2, 09 where the inequality sin x 2x/π, 0 x π/2 has been used. 0 ρ Now for 0 < y and 2 ρ4 y 2 + ρ + ρ 2 2 ρ4 + ρ + ρ ρ2 + ρ 2 2 4, 0 and so we get 2 + ρ2 ρ e it 2 t 2 2 4, 0 t π. π This yields the first two inequalities in 02, and then the result follows from 0. 6 Extension of Theorem 3. In this section we show the following extension of Theorem 3.. Theorem 6.. For fixed β > 0, EM β ω; = /2 2β 2ω where O holds uniformly in =, 2,.... = u + Oω, ω, 2 6

17 Proof. We first show that we can restrict attention to the term = 0 in the series 23 for EM β ω;. We have by 20 = c σ = γ 2 σ ρ 2β = c σ ρ σ β m c σ, 3 σ = where m = [ 2 ] and where it has been used that c = c, σ = σ, =, 2,...,. Now Theorem 5.4 and 02 give ρ m c σ ρ β σ = β β where 2 has been used in the last step. Hence, + 2 c /2 0 m ρ = 2 /2 c /2 0 ζ 3 2 ρ2 = 2 + 2c /2 0 ζ 3 2 2βρ ω, 4 EM β ω; = c 0ρ 2β + c/2 0 β Oω, 5 where the constant implied by O is bounded by ζ 3 2. We now bound and approximate c 0. We have from Lemma 5. with = 0 c 0 = = Furthermore, there is the approximation, see Appendix σ 2. 6 σ = ρ 3 u ε + O u u 7 with ε = 2 ρ + 8 ρ2, and u = u 2/ /2, =, 2,..., m. 8 We furthermore have from u = exp2πi/ that = u = 2 = sin π =, 9 see [4],. for the last identity. From all this we get c 0 = = ε ρ 3 + O. 20 u u Next, from 8 and ε = O ρ, we see that ε = O ω u 2 uniformly in =, 2,...,. Hence c 0 = = ε 2 + O u 7 m ρ3. 22 u =

18 Now by 8 and m = [ 2 ] Hence with m ρ 3 2 /2 ρ 3 u = m = /2 2 /2 ρ /2 = 2 ρ3 2 = O ω 3/2 /2. 23 ĉ 0 = c 0 = ĉ 0 + O ω 3/2 /2 = ε = exp 2 = u 2 ln ε u We develop From = = ln ε 2 ε2 = u = ε 2 [ + O u u = O, = = ε 3 = = u 3]. 26 u = u 2 27 see 27 for the first item in 27 and use Theorem 4. or proceed directly for the second item in 27 we have 2 = O. 28 u Finally, from 8 we have 3 = O 3/2 u = Using in 25, we get ĉ 0 = 2ε = = ρ = m = 3/2 = O 3/2. 29 u + Oε 2 + Oε 3 3/2 u + Oω + Oω 3/2, 30 where we have used that 2ε = ρ + O ρ 2 and that ρ 2 = Oω, see 2. 8

19 Using 30 and 24 in 5, we get EM β ω; = ĉ0ρ 2β + Oω = ρ ρ 2β = = 2β ρ + ρ 2β = 2β ρ 2β = = u + Oω u + Oω u + Oω 3 where 27 has been used in the last step to replace the ρ in front of by at the expense of an error O ρ 2 = Oω. Since by 2 we get the result. ρ 2β = ρ/2 = + O, 32 2ω 2ω ω 7 Behavior of EM β ω; as ω 0 and In this section we show the following result. Theorem 7.. Assume that β > 0 is fixed and that ω 0 and. Then EM β ω; 0. Proof. By Theorem 5.3 c = + ρ σ σ + ρ 2σ + ε,ρ, 33 where for all =, 2,... and all = 0,,..., ε, ρ τ τ ; τ = 4ρ ρ 2 + ρ 2 = ρ We have by 2 and so Now by 23, 20 and 33 ρ 2 = 2ρβ2 ω, 35 ρ τ 2 exp + ρ EM β ω; = ρ 2β =0 + ρ ε 0,ρ 2β ρ 2 + ρ 2β + ρ σ σ σ 2 + ρ 2σ = + ρ σ σ ε, ρ σ 2 + ρ 2σ. 37 9

20 The second term on the right-hand side of 37 tends to 0 by 34, 35, 36. As to the third term on the right-hand side of 37, we estimate ρ 2β = ρτ 2β τ ρτ β τ + ρ σ σ ε, ρ σ 2 + ρ 2σ + ρ σ σ σ 2 + ρ 2σ = m + 2 / = β τ ζ 3 2 τ ρ. 38 Here we have used 02, with m = [ 2 ], and σ. By 35 and 36 we have that τ ρ 0, and so also the third term at the right-hand side of 37 tends to 0. We finally consider the first term, R := ρ 2β =0 on the right-hand side of 37. We show below that 0 R + ρ 2β ρ + ρ σ σ σ 2 + ρ 2σ π 3/4 39 τ τ. 40 From 35 and 36 it then follows that also R 0. To show 40, we follow the approach that was used to prove Lemma 5.2, and we let for u < τ + ρ σu σu F u = σu 2 + ρ 2σu, 4 where Using we have F u = σu = a a 2 z ; a = 2 + ρ, z = ρu. 42 ρu σu 2 + ρ 2σu + ρ σu σu = ρu, 43 By contour integration as in 85 88, we have that g n = z 2πi 2 a + a 2 z 2 a 2 z = 2π = 2π z =r a 2 a 2 = = z 2 a + a 2 z 2 a 2 z g n z n. 44 n=0 a 2 + a 2 x a 2 a 2 + x 2 x a ρ2 x x ρ 2 x a 2 20 dx x n dz z n+ dx x n 45

21 for n = 0,,.... Now g 0 = 0, see 44, and so the last integral vanishes for n = 0. The integrand in this integral changes sign once, from positive to negative at x = 2 + ρ2 > a 2, and /x n is positive and strictly decreasing in x a 2 when n =, 2,.... It follows that g n > 0, n =, 2,.... Also, we have 2 + ρ2 x x ρ 2 and so we conclude that for n =, 2,... 0 < g n 2π 2 + ρ2 a 2 a 2 ρ 2 = = π 4 ρ 2 a 2 + ρ ρ 2 a 2n From n! F n 0 = g n ρ n and τ = ρ/a 2, we then get 0 < F n 0 n! π + ρ ρ 2 We return to 39. As in 82, we have R = ρ 2β =0 Since F 0 = 0, we obtain from 48 that Then using that we obtain 0 < R ρ 2β = 2πβ 4 ρ 2, x a2, 46 x a 2 dx x n dt t t n. 47 τ/t n t dt, n =, 2, F e 2πi/ = ρ 2β π + ρ ρ + ρ ρ 2 s= s=0 F s 0 s! τ/t s t dt. 49 τ/t τ/t dt t < τ/t τ/t < τ τ t, t >, 5 0 < R < 2πβ + ρ ρ τ τ dt t t. 52 For the remaining integral, we use the substitution t = e s, s 0 and the inequality e s/2 e s/2 > s, s > 0, and we get dt t t = e 3/4s e s/2 e ds < s /2 e 3/4s ds. 53 s/2 0 The last integral in 53 equals π/ 3/4 /2, and using this in 52 we get 40. The proof is complete. Note 7.2. From the estimates of the three terms at the right-hand side of 47, it is seen that 0 EM β ω; = Oτ ln τ /

22 A Approximating σ We present approximations of σ, =, 2,...,, that were needed at several places when ρ is small. With u = exp2πi/, we have We have for =, 2,..., [ 2 ] and so σ = 2 ρ ρ2 ρ u = 2 ρ + u ρ u + 4 ρ2 = 2 ρ + u ρ + π u = u = 2 sin ρ2 4 u / , 56 ρ2 4 8 u ρ2 /2, =, 2,...,. 57 We develop the square root on the last line in 55 under the condition that Then we get ρ < 3, 8 ρ2 /2 < σ = 2 ρ + u ρ2 2 ρ + 8 u ρ2 2 8 ρ u = 2 ρ + u 2 ρ u + 8 ρ2 8 ρ2 u + 6 ρ3 ρ u = u 2 + ρ3 6 u 28 = u 2 28 ρ + 4 ρ2 u ρ u u ρ + 4 ρ2 + O ρ u References [] Adan. I.J.B.F., Y. Zhao 996. Analyzing GI/E r / queues. Oper. Res. Letters 9: [2] Asmussen, S Applied Probability and Queues second edition, Springer-Verlag, New Yor. [3] Asmussen, S., P. Glynn, J. Pitman 995. Discretization error in simulation of onedimensional reflecting Brownian motion. Ann. Appl. Probab. 5:

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