We derive an explicit expression for the probability density of the first passage time to state 0 for the Ehrenfest diffusion model in continuous time

Flegg, Mark B.

Pollett, Philip K.

English

We derive an explicit expression for the probability density of the first passage time to\ud
state 0 for the Ehrenfest diffusion model in continuous time

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First passage time density for the Ehrenfest model

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   COVER SHEET   This is the author version of article published as:  M. B., Flegg and P. K., Pollett (2007) First passage time density for the ehrenfest model. Copyright 2007 (The authors)   Accessed from   http://eprints.qut.edu.au           Applied Probability Trust (February 26, 2007)FIRST PASSAGE TIME DENSITY FOR THE EHRENFEST MODELM.B. FLEGG, ∗ Queensland University of TechnologyP.K. POLLETT, ∗∗ University of QueenslandAbstractWe derive an explicit expression for the probability density of the first passage time tostate 0 for the Ehrenfest diffusion model in continuous time.Keywords: Urn models; hitting times.AMS 2000 SUBJECT CLASSIFICATION: PRIMARY 60J27SECONDARY 60K401. IntroductionThe Ehrenfest model was introduced by Paul and Tatyana Ehrenfest [7] as a model for gasdiffusion, to help explain why the entropy of a closed system must increase. Mathematicaltreatments were later given by Kac [12] and Feller [8], and since then the model hasappeared in variety of contexts: where a closed system comprises units of two types andtransmutation occurs from one to the other. The particular application that motivatedthe present work comes from the study of thermal fragmentation of aerosols [9, 10, 11].Particles are suspended in a gas and weakly interacting (dispersion forces) molecules bondpairs of particles. For any given pair, there are K bonding sites. It is assumed, as part ofthe model, that fragmentation/evaporation of existing bonds occurs at rate μ for each bondand rebonding/condensation occurs at rate λ for each vacant site. We assume that μ andλ are both strictly positive. If X(t) is the number of bonds at time t, then (X(t), t ≥ 0)is assumed to be a continuous-time Markov chain taking values in S = {0, 1, . . . , K} withtransition rates qn,n+1 = λ(K − n) and qn,n−1 = μn for n = 1, 2, . . . , K. However, in ourmodel q0n = δ0n, because once there are no bonds present, the two particles dissociate andrebonding does not occur. Thus, our model differs from the usual Ehrenfest model in that 0is an absorbing state. We are interested in the time T it takes for the particles to dissociatestarting with X(0) = N bonds. T is therefore the first passage time to state 0 in thestandard Ehrenfest model. We derive an explicit analytical expression for the probabilitydensity function of T .If K were large, as it would be in the classical context, T would have an approximateexponential distribution (see for example [14]), but in the aerosols application K is usuallysmall (K  10), and thus it is useful to have an explicit expression. This would also be truein other modelling contexts, for example in animal population networks (metapopulations),∗ Postal address: School of Physical and Chemical Sciences, Queensland University of Technology, GPOBox 2434, Brisbane, Queensland 4001, Australia. E-mail address: m.flegg@qut.edu.au∗∗ Postal address: Department of Mathematics, The University of Queensland, Queensland 4072,Australia. E-mail address: pkp@maths.uq.edu.au12 M.B. Flegg and P.K. Pollettwhere the state is the number of occupied habitat patches and the “per-capita” upward anddownward rates are the rates of colonisation and local extinction; here the total number ofpatches would be moderate.The theory of hitting times is well developed (see the books of Syski [17] and Kemper-man [15]), but there are few explicit formulae available for specific models. Exceptions tothis are in the recent work of Di Crescenzo and colleagues [3, 4, 5, 16, 6], where specialstructure in a wide range of continuous-time models has been exploited. The property of“central symmetry” exploited in [5, 6] would, in the present context, require λ = μ, and thusis far too restrictive for our purposes. Di Crescenzo studied the Ehrenfest model with λ = μand K even in Section 4.1 of [5], and obtained an explicit expression for the probabilitydensity of the first passage time to the “symmetry state” K/2 starting from any state N .Comprehensive early treatments of the Ehrenfest model in continuous-time were given byKarlin and McGregor [13] and Bellman and Harris [2]. In Section 4 of [13] an expressionis given for the transition probabilities Pij(t) = Pr(X(t) = j|X(0) = i), both in termsof generating functions and explicitly in terms Krawtchouk (orthogonal) polynomials. Itis therefore not surprising that the first passage time density can be evaluated explicitly,because its Laplace transform is the ratio of the Laplace transforms of Pi0(t) and P00(t).However, we will find it convenient to work directly from the Kolmogorov forward equations.2. First passage time denistyFor the Ehrenfest model with absorption at 0 the forward equations aredP0dt= μP1dP1dt= −(λ(K − 1) + μ)P1 + 2μP2dPidt= λ(K − (i− 1))Pi−1 − (λ(K − i) + μi)Pi + μ(i + 1)Pi+1 (i = 2, . . . , K − 1)dPKdt= λPK−1 − μKPK .with P0(0) = δN0, where Pi(t) = PNi(t). In terms of the probability generating functionH(z, t) =∑Ki=0 Pi(t)zi, they are summarised by the partial differential equation∂H∂t+ (λz + μ) (z − 1)∂H∂z− λK(z − 1)H = −λK(z − 1)P0(t),with the boundary conditions H(0, t) = P0(t), H(1, t) = 1 and H(z, 0) = zN . In orderto make the boundary conditions homogeneous, it will be convenient to work in terms ofG = H + 1:∂G∂t+ (λz + μ) (z − 1)∂G∂z− λK(z − 1)G = −λK(z − 1)Q(t), (1)where Q(t) = P0(t) − 1, with G(0, t) = Q(t), G(1, t) = 0 and G(z, 0) = zN − 1. Ourimmediate aim is to evaluate Q(t) and thus determine P0(t) = Pr(T ≤ t|X(0) = N).First passage time density for the Ehrenfest model 3By considering the homogeneous form of (1), a separation of variables argument suggeststhat we should look for a solution of the formG(z, t) =K∑i=1Ai(t) (λz + μ)K−i (z − 1)i, (2)for suitable functions (Ai(t)) that do not depend on z. Notice that G given by (2) satisfiesthe boundary condition G(1, t) = 0. On substituting (2) into (1) we obtainK∑i=1(A ′i(t) + (λ + μ)iAi(t)) (λz + μ)K−i (z − 1)i = −λK (z − 1)Q(t).Thus, if we can find constants (Ci) such that∑Ki=1 Ci (λz + μ)K−i (z − 1)i = −λK (z − 1),then it is clear that (Ai) will satisfyA ′i(t) + i (λ + μ)Ai(t) = CiQ(t) (i = 1, 2, . . . , K). (3)The following lemma shows that this is possible, and at the same time establishes theexistence of functions (Ai(t)) satisfying (2).Lemma 1. Let K ≥ 1, and let f be a polynomial with real coefficients that satisfies f(1) = 0and has degree no greater than K. Then, ∀ a, b > 0, ∃ uniquely, constants (Bi) such thatf(z) =K∑i=1Bi (az + b)K−i (z − 1)i . (4)They are given by Bi = gi(1), wheregi(z) =1i!di−1dzi−1(az + b)i−K(ddz− aKaz + b)f(z). (5)Proof. First observe that(az + b)i−K(ddz− aKaz + b)f(z) =K∑j=1Bj(a + b)j(az + b)i−j−1(z − 1)j−1.Then, by Leibniz Theorem,gi(z) =1i!di−1dzi−1(az + b)i−K(ddz− aKaz + b)f(z)=i−1∑k=0K∑j=1Bjji!(a + b)(i− 1k)di−k−1dzi−k−1(az + b)i−j−1dkdzk(z − 1)j−1.Now, since for j ≥ 1, (dk/dzk)(z − 1)j−1∣∣z=1= (j − 1)!δk,j−1, we getgi(1) =i∑j=1Bjj!i!(a + b)(i− 1j − 1)di−jdzi−j(az + b)i−j−1∣∣∣∣∣z=1= Bi,because (di−j/dzi−j) (az + b)i−j−1∣∣z=1= (a + b)−1δij. 4 M.B. Flegg and P.K. PollettIndeed we can evaluate (Ci) explicitly. Putting a = λ and b = μ, setting f(z) =−λK (z − 1), and evaluating the derivatives in (5), we find thatCi = −λK(K − 1i− 1)(−λ)i−1(λ + μ)K−1(i = 1, . . . , K). (6)(This can be established more simply by direct substitution in the right-hand side of (4).)Next, we take Laplace transforms in (3), writing F˜ (s) for the Laplace transform of F (t).We find that sA˜i(s)− Ai(0) + i (λ + μ) A˜i(s) = CiQ˜(s), and hence thatA˜i(s) =CiQ˜(s) + Ai(0)s + i (λ + μ).Since we require G(0, t) = Q(t), (2) yields Q˜(s) =∑Ki=1 A˜i(s)μK−i(−1)i. Therefore,Q˜(s) =(1−K∑i=1Ci(−1)iμK−is + i (λ + μ))−1 K∑i=1Ai(0)(−1)iμK−is + i (λ + μ). (7)The condition G(z, 0) = zN − 1 entails zN − 1 =∑Ki=1 Ai(0) (λz + μ)K−i (z− 1)i, and so theconstants (Aj(0)) can be determined from Lemma 1. Putting a = λ and b = μ and settingf(z) = zN − 1, we find thatAi(0) =min{i,N}∑j=1(Nj)(K − ji− j)(−λ)i−j(λ + μ)K−j(i = 1, . . . , K). (8)We therefore have an explicit expression for Q˜(s), and it remains for us to invert the Laplacetransform. The change of variable s → s/(λ+ μ) makes the calculations more manageable.If R˜(s) = (λ+μ)Q˜((λ+μ)s), then R˜(s) will be the Laplace transform of R(t) = Q(t/(λ+μ)).Set ρ = μ/(λ + μ) and α = λ/μ. Then, on substituting (6) and (8) into (7) we find thatR˜(s) =∑Ki=1 aiUi(s), whereai = ρKαimin{i,N}∑j=1(Nj)(K − ji− j)(−1)j(1− ρ)−jandUi(s) =1(s + i)(1− ρK ∑Kj=1 (Kj )αj ( js+j)) .Notice that there is no singularity in Ui(s) at s = −i. But, we shall prove that Ui(s)has precisely K (first-order) singularities, r1, . . . , rK , which satisfy ri ∈ (−i,−i + 1) (i =1, 2, . . . , K). Observe that1− ρKK∑j=1(Kj)αj(js + j)= ρKK∑j=0(Kj)αj(ss + j)=φ(s)∏Kk=1(s + k),First passage time density for the Ehrenfest model 5whereφ(s) = ρKK∑j=0(Kj)αjK∏k=0k =j(s + k)is a degree K polynomial. Its K zeros, r1, . . . , rK , are distinct and satisfy ri ∈ (−i,−i + 1)because φ(−j) = K!ρK(−α)j (j = 0, 1, . . . , K) are K + 1 distinct values of φ that alternatein sign, and, since the leading term of φ(s) is sK , we may write φ(s) =∏Km=1(s − rm).Therefore, Ui(s) = Vi(s)/∏Km=1(s − rm), where Vi(s) =∏Kk=1, k =i(s + k) is a degree K − 1polynomial that does not vanish at any of r1, . . . , rK (and hence r1, . . . , rK are all thesingularities of Ui(s)). Using partial fractions we getUi(s) =K∑m=1(1s− rm)Vi(rm)∏k =m(rm − rk),and so the inversion of R˜(s) is straightforward:R(t) =K∑m=1ermtK∑i=1aiVi(rm)∏k =m(rm − rk)=K∑m=1ermt1∏k =m(rm − rk)K∑i=1ai∏k =i(rm + k).(Both products are over k = 1, 2, . . . , K.) Finally, since R(t) = Q((λ + μ)t), we obtainP0(t) = 1 + Q(t) = 1 +K∑m=1erm(λ+μ)t1∏k =m(rm − rk)K∑i=1ai∏k =i(rm + k).On substituting for ai, putting si = −ri to have si > 0 ∀ i, and differentiating with respectto t, we arrive at our main result.Theorem 1. The probability density function f of the first passage time to 0 starting in Nis given byf(t) =K∑m=1sm(λ + μ)e−sm(λ+μ)t∏k =m(sk − sm)K∑i=1ρK−i∏k =i(k − sm)min{i,N}∑j=1(Nj)(K − ji− j)(−1)j−1(1− ρ)i−j,where ρ = μ/(λ + μ) and s1, . . . , sK are the roots ofK∑j=0(Kj)(λμ)j K∏k=0k =j(s− k) = 0,arranged so that sm ∈ (m− 1,m) (m = 1, 2, . . . , K).Remarks (1) Notice that in the limit as λ → 0, rm → m and P0(t) → (1 − e−μt)N . Byreworking our arguments, this can be obtained as an exact result, P0(t) = (1−e−μt)N , when6 M.B. Flegg and P.K. Pollettλ = 0. It is obviously true because when λ = 0 the N bonds fragment independently at thesame rate μ, each bond lasting for an exponentially distributed amount of time, and thusT is the maximum of these times.(2) From more elementary considerations (see for example Section 8.1 of [1]), the expectedfirst-passage time from N to 0 isE(T ) =N∑n=11μnπnK∑m=nπm,where μn = μn and πn =(Kn)(1 − ρ)nρK−n (n = 0, 1, . . . , K), which leads to the explicitexpressionE(T ) =1μN∑j=1K∑i=j(K − ji− j)(ij)−1(λμ)i−j.This can be shown to be consistent with Theorem 1 by evaluating E(T ) either as∫∞0Q(t)dtor as −Q˜(0).AcknowledgementsThe work of Phil Pollett is supported by the Australian Research Council Centre ofExcellence for Mathematics and Statistics of Complex Systems.References[1] W.J. Anderson. Continuous-Time Markov Chains: An Applications-Oriented Ap-proach. Springer-Verlag, New York, 1991.[2] R. Bellman and T. Harris. Recurrence times for the Ehrenfest model. Pacific J. Math.,1:179–193, 1951.[3] A. Di Crescenzo. On certain transformation properties of birth-and-death processes.In R. Trappl, editor, Cybernetics and Systems ’94, pages 839–846. World-Scientific,Singapore, 1994.[4] A. Di Crescenzo. On some transformations of bilateral birth-and-death processes withapplications to first passage time evaluations. In Proc. 17th Symposium on InformationTheory and Its Applications (SITA ’94), pages 739–742. Hiroshima, Japan, December6–9, 1994.[5] A. Di Crescenzo. First-passage-time densities and avoiding probabilities for birth-and-death processes with symmetric sample paths. J. Appl. Probab., 35(2):383–394, 1998.[6] A. Di Crescenzo and A. Nastro. On first-passage-time densities for certain symmetricMarkov chains. Sci. Math. Jpn., 60(2):381–390, 2004.First passage time density for the Ehrenfest model 7[7] P. Ehrenfest and T. Ehrenfest. U¨ber zwei bekannte Einwa¨nde gegen das BoltzmannscheH-Theorem. Phys. Z., 8:311–314, 1907.[8] W. Feller. An Introduction to Probability Theory and Its Applications. Vol. I. JohnWiley & Sons Inc., New York, N.Y., 1950.[9] D.K. Gramotnev and G. Gramotnev. Modeling of aerosol dispersion from a busy roadin the presence of nanoparticle fragmentation. Journal of Applied Meteorology, 44:888–899, 2005.[10] D.K. Gramotnev and G. Gramotnev. A new mechanism of aerosol evolution near abusy road: fragmentation of nanoparticles. Journal of Aerosol Science, 36:323–340,2005.[11] D.K. Gramotnev and G. Gramotnev. Kinetics of stochastic degradation/evaporationprocesses in polymer-like systems with multiple bonds. Journal of Applied Physics, toappear.[12] M. Kac. Random walk and the theory of Brownian motion. Amer. Math. Monthly,54:369–391, 1947.[13] S. Karlin and J. McGregor. Ehrenfest urn models. J. Appl. Probab., 2:352–376, 1965.[14] J. Keilson. Markov Chain Models–Rarity and Exponentiality. Springer-Verlag, NewYork, 1979.[15] J. H. B. Kemperman. The passage problem for a stationary Markov chain. 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First passage time density for the Ehrenfest model

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