Moments of the maximum of the Gaussian random walk

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2 Moments of the maximum of the Gaussian random walk A.J.E.M. Janssen (Philips Research) Johan S.H. van Leeuwaarden (Eurandom)

3 Model definition Consider the partial sums S n = X X n with X 1, X,... i.i.d., X i N( β, 1), β > 0 The Gaussian random walk is then defined as {S n : n 0} ; S 0 = 0 We are interested in the all-time maximum M β = max{s n : n 0} We consider the moments EM k β and the role of the drift β

4 Gaussian random walk with drift =

5 Motivation for studying EMβ k Queues in conventional heavy traffic Scaled version of the queue behaves approximately as M β G/G/1 queue: Kingman (196,1965) Queues in Halfin-Whitt scaling In the limit, the queue behaves exactly as M β G/M/N queue: Halfin-Whitt (1981) G/D/N queue: Jelenković-Mandelbaum-Momčilović (004) Call centers: Borst-Mandelbaum-Reiman (004) Equidistant sampling of Brownian motion Testing for drift: Chernoff (1965) Corrected diffusion approximations: Siegmund (1979,1985) Option pricing: Broadie-Glasserman-Kou (1997)

6 Queues and Halfin-Whitt scaling Example W λ,n+1 = (W λ,n + A λ,n s) + where A λ,n is Poisson distributed with mean λ < s. Let W λ = lim n W λ,n

7 Queues and Halfin-Whitt scaling Example W λ,n+1 = (W λ,n + A λ,n s) + where A λ,n is Poisson distributed with mean λ < s. Let W λ = lim n W λ,n Square-root staffing and gives s = λ + β λ 1 W = lim W λ λ λ W d = (W + N( β, 1)) + d = max{s n : n 0} =: M β

8 Outline 1. Exact expression for EM β. Exact expressions for all moments EM k β 3. Equidistant sampling of Brownian motion

9 "Despite the apparent simplicity of the problem, there does not seem to be an explicit expression even for EM β..., but it is possible to give quite sharp inequalities and asymptotic results for small β." (Sir John Kingman, 1965)

10 "Despite the apparent simplicity of the problem, there does not seem to be an explicit expression even for EM β..., but it is possible to give quite sharp inequalities and asymptotic results for small β." (Sir John Kingman, 1965) Kingman showed that for β 0 EM β = 1 β c + O(β) where c = 1 π n=1 1 n( n + n 1) 0.58

11 "Despite the apparent simplicity of the problem, there does not seem to be an explicit expression even for EM β..., but it is possible to give quite sharp inequalities and asymptotic results for small β." (Sir John Kingman, 1965) Kingman showed that for β 0 EM β = 1 β c + O(β) where c = 1 π n=1 1 n( n + n 1) = ζ(1 ) π

12 Riemann zeta function The Riemann zeta function ζ is defined as ζ(s) = n s, Re s > 1 n=1 This definition is extended by analytic continuation to the entire complex plane except s = 1, where ζ has a simple pole. Calculation of ζ is routine.

13 Theorem For 0 < β < π we have EM β = 1 β + ζ(1 ) π β + β π r=0 ζ( 1 r) r!(r + 1)(r + ) ( β ) r

14 Proof For M β = max{s n : n 0} we have from Spitzer s identity J 1 (β) := EM β = n=1 1 n ES+ n From S n N( βn, n) we get J 1 (β) = n=1 1 n π β n ( nx βn)e x / dx Upon changing variables according to y = x/ n...

15 Proof (cont d)... we can write J 1 (β) as J 1 (β) = n=1 n 1/ π β (y β)e 1 ny dy Differentiating twice w.r.t. β yields J () 1 (β) = n=1 n 1/ π e 1 nβ using d [ ] f(y, β)dy dβ β = f(β, β) + β f (y, β)dy β Idea of Chang-Peres (1997) in their analysis of P(M β = 0)

16 Lerch s transcendent Lerch s transcendent Φ is defined as the analytic continuation of the series Φ(z, s, v) = (v + n) s z n Note that ζ(s) = Φ(1, s, 1). Lemma For ln z < π, s 1,, 3,..., and v 0, 1,,... we have n=0 Φ(z, s, v) = Γ(1 s) (ln 1/z) s 1 + z v z v r=0 (ln z)r ζ(s r, v) r! with ζ(s, v) := Φ(1, s, v) the Hurwitz zeta function. Proof Bateman 1.11(8)

17 Proof (cont d) J () 1 (β) = n=1 n 1/ π e 1 nβ = 1 π e 1 β Φ(z = e 1 β, s = 1, v = 1) Using Bateman s result we get for 0 < β < π J () 1 (β) Γ(3 ) π 3/ 1 β 3 = 1 π r=0 ζ( r 1 )( 1 β ) r The rhs is a well-behaved function of β. Integrating twice yields J 1 (β) 1 β = L 0 + L 1 β + 1 π r=0 where L 1 and L 0 are integration constants r! ζ( r 1 )( 1 )r β r+ r!(r + 1)(r + )

18 Proof (cont d) These can be found using Euler-Maclaurin summation, among other things. L 0 = ζ(1/) π, L 1 = 1 4 This in total gives EM β = 1 β + ζ(1 ) π β + β π r=0 ζ( 1 r) r!(r + 1)(r + ) ( β ) r

19 Cumulants The k-th cumulant of a random variable A is defined as the k-th derivative of log Ee sa evaluated at s = 0.

20 Cumulants The k-th cumulant of a random variable A is defined as the k-th derivative of log Ee sa evaluated at s = 0. Spitzer s identity leads to { E(e sm β ) = exp n=1 1 n E(esS+ n }, 1) Re s 0 Thus log E(e sm β ) = n=1 1 n 0 (sx + 1 s x +...)f S + n (x)dx with f S + n the density function of S n +, and d k (ds) log k E(esM β 1 ) = s=0 n E((S+ n ) k ) =: J k (β), k = 1,,... n=1

21 Recall J 1 (β) = EM β ; J (β) = VarM β ; J 3 (β) = E(M β EM β ) 3 and all moments of M β follow from the cumulants and vice versa. Using S n N( βn, n), it follows that J k (β) = n=1 1 n π β n ( nx βn) k e x / dx It obviously holds that J 0 (β) = n=1 1 n P(S n > 0) From Spitzer s identity we then know that J 0 (β) = ln P(M β = 0)

22 Theorem Assume 0 < β < π. There holds J 0 (β) = ln β ln ζ(1) β 1 π π and for k = 1,,... J k (β) = (k 1)! k β k + + ( 1)k+1 k! π r=0 k j=0 r=1 ( ) k ( 1)j Γ( k j+1 j π ζ( r + 1 )( 1 )r β r+1 ) ζ( k r + 1 )( 1 )r β r+k+1 r!(r + 1) (r + k + 1) r!(r + 1) ζ( 1k 1 k j 1 j + 1) β j Result of Chang-Peres (1997)

23 Proof 1. Use Spitzer s identity and normality of S n to obtain J k (β) = n k 1/ (y β) k e 1 ny dy π n=1 β. Differentiate k + 1 times under the integral sign J (k+1) k (β) = ( 1) k+1 k! n=1 n k 1/ π e 1 nβ 3. Rewrite the expression in terms of Lerch s transcendent 4. Apply Bateman s result 5. Integrate k + 1 times and find the k + 1 integration constants using Euler-Maclaurin summation

24 Equidistant sampling of Brownian motion Let {B t : t 0} be a BM with B 0 = 0, drift β and variance 1, so that B t = βt + W t, where {W t : t 0} is a Wiener process. Let M = max{b t : t 0} It is well known that P( M x) = e βx and so the k-th cumulant of M equals (k 1)! k β k

25 The GRW results from (equidistantly) sampling this BM, and by increasing the sampling frequency the GRW will converge to the BM.

26 4 3 Equidistant sampling of BM (drift= 0.1, variance=1)

27 4 3 Equidistant sampling of BM (drift= 0.1, variance=1). Steps of size:

28 4 3 Equidistant sampling of BM (drift= 0.1, variance=1). Steps of size:

29 4 3 Equidistant sampling of BM (drift= 0.1, variance=1). Steps of size:

30 4 3 Equidistant sampling of BM (drift= 0.1, variance=1). Steps of size:

31 Redefine the GRW as {S n (β, ν) : n = 0, 1,...}, where with Let S n (β, ν) = 0, S n (β, ν) = X ν, X ν,n X ν,1, X ν,,... i.i.d., X i N( β/ν, 1/ν) M ν,β = max{s n (β, ν) : n = 0, 1,...}. Earlier definition corresponds to ν = 1 with M 1,β =: M β. Since M ν,β d = ν 1/ M ν 1/ β, all characteristics of M ν,β can be expressed in those of M β.

32 Say we sample the BM at points 0, 1 ν, ν, 3 ν,... with ν some positive integer. From our results on EM β we find that E M EM ν,β = ζ(1) + O(1/ν) πν Similar result obtained by Asmussen, Glynn and Pitman (1995).

33 Say we sample the BM at points 0, 1 ν, ν, 3 ν,... with ν some positive integer. From our results on EM β we find that E M EM ν,β = ζ(1) + O(1/ν) πν Similar result obtained by Asmussen, Glynn and Pitman (1995). However, we can easily obtain E M EM ν,β = ζ(1 ) πν + β 4ν + O(ν 3/ ) Moreover, our exact analysis of M β leads to asymptotic expressions up to any order, for all cumulants of the maximum. E.g. Var M VarM ν,β = 1 4ν ζ( 1) β π ν + 3/ O(ν )

34 Other, related and future work 1. Exact expressions for P(M = 0), EM β, VarM β (with A.J.E.M. Janssen). Exact expressions for all cumulants J k (β), bounds, analytic continuation for all values of β > 0 (not only for 0 < β < π) (with A.J.E.M. Janssen) 3. Discrete queue and Halfin-Whitt scaling (with A.J.E.M. Janssen and Bert Zwart) lim λ 1 λ W λ d = max{s n : n 0} = M β

CUMULANTS OF THE MAXIMUM OF THE GAUSSIAN RANDOM WALK

CUMULANTS OF THE MAXIMUM OF THE GAUSSIAN RANDOM WALK A.J.E.M. JANSSEN AND J.S.H. VAN LEEUWAARDEN Philips Research Eindhoven University of Technology Abstract. Let X, X 2,... be independent variables, each