Non trivial limit distributions for transient renewal chains

0.12 MB·

by  Dalia Terhesiu

Non trivial limit distributions for transient renewal chains
DaliaTerhesiu∗
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1 December, 2016
0
2

n
a
J Abstract

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Inthisworkwe studytheasymptoticofrenewalsequencesassociated withcertaintran-

sientrenewalMarkovchainsandenquireabouttheexistenceoflimitlawsinthissetup.
]
R
P
. 1 Introduction
h
t
a
m In the first part of this work we are interested in the asymptotic behaviour of renewal sequences
[ associated with transient Markov renewal chains with regularly varying tails of the return time

1 to the state [0]. The precise meaning of the transient renewal chains considered here is given in
v
9 Subsection 2.2 (in particular, see equation (2.3)). In this set up, we show that up to a constant,
9
independentoftheindexofregularvariation,therenewalsequencesareasymptoticallyequivalent
7
0 to the tails of the return to state [0]: see Proposition 3.2 (and its weaker version Proposition 3.1)
0
. in Section 3. The result in Proposition 3.2 is implicit in the work [5], which focuses on transient
1
0 random walks on Zd, d ≥ 1. In short, Proposition 3.2 is a result of similar flavour to that in [5,
7
1 Theorem 4]. The analytical proof of [5, Theorem 4](in its full generality) in [5, Section 2] relies
:
v on [2,Theorem 1], of which proof is based on deep Banach algebra techniques. Theproof of the
i
X presentProposition 3.2isentirelyelementary.
r In the second part, restricting to indices of regular variation that, provided that the renewal
a
chain is recurrent, would imply it is null recurrent, we enquire about the existence of limit laws.
The main result of this paper, Theorem 4.1 in Section 4, shows the existence of an arcsine law
for the transient chain; the proof of this result exploits the asymptotic behaviour of the renewal
sequence obtained in Proposition 3.2. In Section 5, weprovide an asymptotic characterization of
the random variable S describing the number of visits to the state [0] in the interval [0,n] when
n
appropriately scaled: seeProposition 5.1.
We believe that the techniques in this work can be extended to dynamical systems, in which
anyformofindependencefails.Typicalsystemsthat,apartfromindependence,resemblearenewal
chain are the so called interval maps with indifferent fixed points such as the one studied in [8].
The task of extending the present results to such systems is beyond the scope of this work, but
onceaccomplished itcouldofferanalternativetotheresultsin[4].
∗UniversityofExeter,NorthParkRoad,Exeter,UK,EX44QF.Email:[email protected].
1
2
2 Set up. Notation
2.1 Renewal chain, induced renewal chain
Let(X ) ,X ∈ N = N∪{0}beaMarkovrenewalchainwithtransition probabilities
n n≥0 n 0
f ℓ = 0,
k
p := P(X = k|X = ℓ)= 1 k =ℓ−1,
ℓ,k n+1 n 

0 otherwise.


We assume f = 1 and recall that depending on the asymptotics of f , (X ) is a
k k k>n k n n≥0
P P
positiverecurrent oranull-recurrent renewalchain(see,forinstance, [7]).
LetX = NN0 andletT : X → X betheshiftmap.Thenanycylinder [e e ...e ]hasmea-
0 0 1 k−1
sureµ([e e ...e ])= µ([e ])p ···p .Thiscanbecomputediftheinitialdistribution
0 1 k−1 0 e0e1 ek−2ek−1
µ([j]),j ∈ N ,isgiven.TheMarkovmeasureµisT invariant.
0
LetY = [0] = {x ∈ X : x = 0},anddecompose
0
Y = ∪ C , whereC = [0,k,k−1,k−2,...,0].
k≥0 k k
Thecylinders C arepairwisedisjoint, andtheirmeasuresaregivenby
k
µ(C )= µ(Y)p p ···p = µ(Y)f .
k 0,k k,k−1 1,0 k
We recall the definition of the induced shift on Y and associated ’induced renewal chain’.
For y ∈ Y, let τ(y) = min{n ≥ 1 : Tn(y) ∈ Y} and T = Tτ. The probability measure
Y
ν = µ(Y)−1µ| isT invariant. WenotethatC = {y ∈ Y : τ(y) = k+1}canberegarded as
Y Y k
N
theshiftonthespace({C } ) 0.
k k≥0
DefinetheinducedMarkovchain(U ) ,U ∈ {C } ,withtransition probabilities:
n n≥0 n k k≥0
P(U = C ∧U = C )
pˆ = P(U = C |U = C ) = n+1 k n ℓ =
ℓ,k n+1 k n ℓ P(U = C )
n+1 k
P(C ∧T−1(C )) p p ···p p
= ℓ Y k = 0,ℓ ℓ,ℓ−1 1,0 0,k = f . (2.1)
P(U = C ) p k
n+1 k 0,ℓ
Notethatpˆ isindependent ofℓ.
ℓ,k
The induced renewal chain (U ) with above transition probabilities pˆ is positive recur-
n n≥0 ℓ,k
rent. Tosee this, fixk ≥ 1and letν∗ = 1 µ|C andϕ : C → N withϕ(y) := min{n ≥ 1 :
µ(Ck) k k
Tn(y) ∈ C } be the first return time of T to C . Since P(U = k) = P(U = k|U =
Y k Y k n ℓ n n−1
ℓ)P(U = ℓ) = pˆ P(U = ℓ) = f P(U = ℓ) = f ,wehPave
n−1 ℓ ℓ,k n−1 ℓ k n−1 k
P P
1
ν∗(ϕ ≥ n) = µ(y ∈ C :ϕ(y) = m)
k
µ(C )
k X
m≥n
1
= P(U = C ∧U 6= C ,0 < j < m∧U = C )
0 k j k m k
µ(C )
k X
m≥n
1
= µ(C )(1−f )m−1f = (1−f )n.
k k k k
µ(C )
k X
m≥n
3
Hence ϕ has an exponential distribution, which shows that (U ) is positive recurrent (since
n n≥0
ν∗(ϕ ≥ n)< ∞).
n
P
2.2 Introducing transience, ’holes’ inthethe originalchain(X )
n n≥0
Recall that T : X → X is the original shift and T = Tτ : Y → Y is the induced shift with
Y
Y = [0] = {x ∈ X : x = 0}.Throughout weassumethat
0
g.c.d.{τ| ,k ≥ 0} = 1, (2.2)
Ck
whichensuresthat(X ) ,X ∈ N isaperiodic.
n n≥0 n 0
We introduce a hole H in X with H ⊂ Y and thus transience1, as follows. Let X˚ = X \H
andY˚ = Y \H.SetX˚n = ∩n T−iX˚and defineT˚= T| s.t.the firstreturn time˚τ ofT˚toY˚
i=0 X˚
satisfies
ν(˚τ = n) =pν(τ = n)= pf := f˚ ,n ≥ 1. (2.3)
n−1 n−1
Here we recall that ν = µ(Y)−1µ| is the T invariant probability measure. In fact, due to the
Y Y
rule above (of introducing a hole in X), ν is also T˚ = T˚˚τ invariant. To see this, let Q be the
Y˚ Y
transition matrix for induced renewal chain (U ) and note that this is an infinite matrix with
n n≥0
(f ,f ,f ,...)ineveryrow.Giventhesetupoftheprevious subsection, ν isthelefteigenvector
0 1 2
of Q (with eigenvalue 1). But, the transition matrix for the modified chain (after introducing a
Y
hole) is simply Q˚ = pQ . While the eigenvalue changes from 1 to p, the left eigenvector ν
Y Y
remainssame.
In what follows we are interested in the asymptotics of the renewal sequence associated with
thetransient renewalchain(X˚ ) ,X˚ ∈N withtransition probabilities
n n≥0 n 0
pf ℓ = 0,
k
˚p := P(X˚ = k|X˚ = ℓ)= 1 k = ℓ−1,
ℓ,k n+1 n 

0 otherwise.


We start by recalling the renewal equation, which can be obtained word by word as in the
recurrent case(see,forinstance, [6]).Forn ∈ N,let˚τ = n−1˚τ ◦T˚j.Recallthatthesequence
n j=0 Y˚
(f˚) isdefinedin(2.3)anddefinetherenewalsequenceP
k k≥1
k n
˚u = 1, ˚u = P ∃k ≤ nsuchthat ˚τ = n = f˚˚u . (2.4)
0 n j j n−j
(cid:16) X (cid:17) X
j=0 j=1
For z ∈ D¯, set f˚(z) = ∞f˚zn and˚u(z) = ∞˚u zn. Since, by assumption f˚ =
1 n 0 n k≥1 k
P P P
p < 1and(2.2)holds,wehavethat
˚u(z) = (1−f˚(z))−1 (2.5)
iswelldefinedonthewholeofD¯.
1Thistypeofruleforintroducingtransience/holesinMarkovchainswassuggestedtomebyRolandZweimu¨ller.In
particular,theresultsinSection3answerhisquestions.Iwishtothankhimforusefuldiscussionsonthistopic.
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3 Non trivial limits for the renewal sequence˚u
n
The first result below gives the asymptotics of the tail renewal sequence, that is ˚u , where
j>n j
(˚u ) is the renewal sequence associated with the chain (X˚ ) , X˚ ∈ NPintroduced in
j j≥1 n n≥0 n 0
Subsection2.2.Throughoutthissection,weassumethesetupofSubsection2.2,inparticular(2.3)
andsuppose that (2.2)holds.
Proposition 3.1. Suppose thatf = O(n−(β+1)),forsomeβ >0.Then
n
˚u = (1−p)−2 f˚(1+o(1)) =p(1−p)−2ν(τ > n)(1+o(1)).
j j
X X
j>n j>n
Proof. Computethat
∞ ∞
˚u(z)−˚u(1) = ˚u (zn −1) = (z−1) ( ˚u )zn.
n j
X X X
n=0 n=0 j>n
Togetherwith(2.5),theaboveequation gives
∞
( ˚u )zn = (z−1)−1(1−f˚(z))−2(f˚(z)−f˚(1))
j
X X
n=0 j>n
= (z−1)−1(1−f˚(1))−2(f˚(z)−f˚(1))
−2
+(1−f˚(1))−3A(z) 1−(1−f˚(1))−1(f˚(z)−f˚(1))
(cid:16) (cid:17)
∞
−2
= (1−p)−2 ( f˚)zn+(1−p)−3A(z) 1−(1−p)−1(f˚(z)−f˚(1)) ,
j
X X (cid:16) (cid:17)
1 j>n
where A(z) = C(z −1)−1(f˚(z)−f˚(1))2, for C > 0. By Lemma A.1, the coefficients of A(z)
are o( f˚). By Wiener’s lemma, the coefficients of (1 − f˚(z))−1, and thus of 1 − (1 −
j>n j
(cid:16)
P −2
p)−1(f˚(z)−f˚(1)) ,areO(n−(β+1)).Convolving, weobtainthatthecoefficientsofA(z)(1−
(cid:17)
(1−f˚(1))−1(f˚(z)−f˚(1))−2 areo( f˚).Theconclusion follows.
j>n j
P
The next result gives the asymptotics of ˚u under a stronger assumption on the asymptotic
n
behaviour off .
n
Proposition3.2. Supposethatnf = C f (1+o(1))andthatf = O(n−(β+1)),forsome
n j>n j n
P
C > 0andβ > 0.Then
˚u = (1−p)−2f˚(1+o(1)) = p(1−p)−2ν(τ = n)(1+o(1)).
n n
Remark 3.3. The above assumption holds under the assumption of regular variation for the se-
quencef = ν(τ = n),thatisiff = ℓ(n)n−(β+1) forℓaslowlyvaryingfunction.
n n
5
Proof. Bydefinition˚u isthecoefficientof(1−f˚(z))−1,soitisn−1C ,whereC isthecoeffi-
n n n
cientofC(z) = d ((1−f˚(z))−1).Computethat
dz
∞ ∞
d
C(z)= (1−f˚(z))−2 (f˚(z)) = (1−f˚(1))−2 nf˚ zn +( f˚ zn
n+1 n+1
dz (cid:16)X X (cid:17)
1 0
−2
+C˜ 1−f˚(1))−3B(z) 1−(1−f˚(1))−1(f˚(z)−f˚(1)) ,
(cid:16) (cid:17)
(cid:0)
whereC˜ > 0(independent ofp)andB(z) = (f˚(1)−f˚(z)) ∞nf˚zn .
0 n
PutD(z) =C˜ 1−f˚(1))−3B(z) 1−(1−f˚(1))−1(f˚(z)P−f˚(1)) −2 = ∞D zn andnote
0 n
(cid:16) (cid:17)
(cid:0) P
that
C = nf˚ +f˚ +O(D ).
n n+1 n+1 n
Byassumption,nf = C f (1+o(1)).WeclaimthatD = o( f˚)andtheconclusion
n j>n j n j>n j
P P
follows.
ToprovetheclaimwenotethatbyWiener’slemma,thecoefficientsof 1−(1−f˚(1))−1(f˚(z)−
(cid:16)
−2
f˚(1)) areO(n−(β+1)).Hence,itsufficestoshowthatthecoefficientsB ofB(z)areo( f˚).
n j>n j
(cid:17)
Since nf = C f (1+o(1)), we have nf˚ = C f˚(1+o(1)). Thus, usPing the
n j>n j n j>n j
P P
definitionofB(z),
(f˚(1)−f˚(z))2 ∞
B(z) = +(f˚(1)−f˚(z)) o f˚ zn.
j
z−1 X (cid:16)X (cid:17)
1 j>n
By Lemma A.1, the coefficients of the first term are o( f˚). By assumption the coeffi-
j>n j
cientsoff˚(1)−f˚(z)areO(n−(β+1))andthus,thecoefficienPtsofthesecondtermareo( f˚),
j>n j
P
asrequired.
4 An arcsine law for β ∈ (0,1)
Recall that (X˚ ) , X˚ ∈ N is the transient renewal chain introduced in Subsection 2.2 with
n n≥0 n 0
associated shiftT˚:X˚→ X˚.Proposition 3.2allowsustoobtainthefollowingarcsinelaw.Let
Z˚ (x) := max{0 ≤ j ≤ n : T˚j(x) ∈ Y˚},
n
be the last visit of the orbit of x under the shift T˚to Y˚ in the interval [0,n]. In what follows,
B(β,1−β)isthestandardBetadistribution withparametersβ,1−β.Also,welet[ ]denotethe
integerpart.
Theorem 4.1. Assume the setting of Proposition 3.2 with f˚ = Cn−(β+1)(1+ o(1)), for some
n
C > 0.Letβ ∈ (0,2)andsetq = 1/(1+2β).Then
Z˚[nq] 1/q →ν0 B(β,1−β),
(cid:16) n (cid:17)
where the convergence is in measure, for any probability measure absolutely continuous w.r.t.
ν = C−2qp−1(1−p)2ν.
0
6
Proof. LetZˆ (x) := max{0 ≤ j ≤ n : T˚[jq](x) ∈ Y˚}andnotethat
n
(Z˚ )1/q = max{[j1/q] :0 ≤ j ≤ [nq] :T˚j(x) ∈ Y˚} = max{j ∈ {0,...,n} :T˚[jq](x) ∈ Y˚}
[nq]
= Zˆ (x). (4.1)
n
Butforanyt > 0,
Zˆ (x)
ν n ≤ t = ν(Zˆ (x) ≤ (nt)1/q)= ν(T˚[jq] ∈ Y˚∩{˚τ > n−[jq]}).
n
(cid:16) n1/q (cid:17) X
0≤j≤(nt)1/q
Duetoindependence,
ν(T˚[jq] ∈ Y˚∩{˚τ > n−[jq]}) = ν({˚τ > n−[jq]})ν(T˚−[jq]Y˚).
Itiseasytoseefromthedefinitionoftherenewalsequencein(2.4)thatν(T˚−[jq]Y˚) =˚u .Propo-
[jq]
sition3.2togetherwithf˚ = Cn−(β+1)(1+o(1))impliesthat˚u = pC(1−p)−2[nq]−(β+1)(1+
n [nq]
o(1)).Puttingtheabovetogether andusingthat2 ν({˚τ > n})= Cpn−β(1+o(1)),
1 1
ν(T˚−[jq] ∈ Y˚∩{˚τ > n−[jq]} = C2p(1−p)−2
[jq]β+1(n−[jq])β
X X
0≤j≤(nt)1/q 0≤j≤(nt)1/q
1 1 1 1
= C2p2(1−p)−2 +O (4.2)
X jq(β+1) (n−[jq])β (cid:16) X j2q(β+1) (n−jq)β(cid:17)
0≤j≤(nt)1/q 0≤j≤(nt)1/q
Forthefirstterm,asn → ∞,
(nt)1/q 1 1
ν(T˚−jq ∈ Y˚∩{˚τ > n−[jq]}) → C2p(1−p)−2 ds.
Z sq(β+1)(n−sq)β
0≤j≤X(nt)1/q 1
Recallq = 1/(1+2β).Withthesubstitution sq → nu
1 (nt)1/q 1 1 n1/q t u1/q−1
q ds = du
nβ Z sq(1+β)(1− sq)β nβnβ+1 Z uβ+1(1−u)β
1 n 1/n
t 1 1 t 1 1
= du = du+O(1/nβ).
Z u1−β (1−u)β Z u1−β (1−u)β
1/n 0
Forthesecond termin(4.2),acalculation similartotheoneaboveshowsthat
1 1
= O(1/nβ).
j2q(β+1) (n−jq)β
X
0≤j≤(nt)1/q
Puttingtheabovetogether, asn → ∞,
Zˆ t 1 1
ν n ≤ t → C2pq−1(1−p)−2 du. (4.3)
(cid:16)n1/q (cid:17) Z0 u1−β (1−u)β
The above displayed equation together with (4.1) ends the proof for the case β ∈ (0,1) of the
claimedconvergencew.r.t.themeasureν = C−2qp−1(1−p)2ν.Theconvergenceinmeasure,for
0
anyprobability ν absolutely continuous w.r.t.ν,followssincethedensity ofν isaconstant.
0
2Here,wealsousetheconventionthatj−γ =0forj =0andγ >0.
7
5 A ratio limit for β ∈ (0,1)
ItisknownthatfornullrecurrentrenewalshiftsT : X → X,X = NN0 withinducedshiftsT =
0 Y
Tτ : Y → Y,Y = [0] = {x ∈ X : x = 0}asrecalled inSubsection 2.1,aDarling Kaclawfor
0
S (1 ) = n−11 ◦Tj holdsunderregularvariationofthetailν(τ > n)(see,forinstance,[7]).
n Y j=0 Y
More preciPsely, simplifying the assumption on the tail, if ν(τ > n) = Cn−β(1 + o(1)) for
some C > 0 and β ∈ (0,1), then as n → ∞, C−1n−βS (1 ) → M , where M is a random
n Y β β
variabledistributedaccordingtotheMittagLefflerdistribution3.Onewayofseeingthisistorecall
that: a) P(τ ≥ n) = P(S (1 ) ≤ m), where τ = m−1τ ◦ Tj; b) under the assumption
m n Y m j=0 Y
ν(τ > n) = Cn−β(1 + o(1)), we have that as m → P∞, m−1/βτ → C Y , where Y is a
m β β β
random variable in the domain of a stable law of index β and C is aconstant that depends only
β
on C and β ; c) M = Y−β. This type of argument for the proof of a Darling Kac law can be
β d β
found,forinstance, in[1],whichgoesbackto[6].
In the case of the transient shift T˚introduced in Subsection 2.2, the duality rule in point b)
above does not hold. Instead, in this section we will exploit Lemma 5.2 below and obtain the
following, moreorlessobvious, limitbehaviour onthesurvivor set:
Proposition5.1. AssumethesetupofSubsection2.2,inparticular(2.3).Assumethat (2.2)holds.
Suppose thatν(τ > n) = Cn−β(1+o(1))withβ ∈ (0,1). LetS˚ (1 ) = n−11 ◦T˚j.Then,
n Y˚ j=0 Y˚
P
foranyt > 0,
lim pn1/βν(n−1/βS˚ ≤ t∩X˚n)
n→∞ n
1 ≤ ≤ 1+p.
P(Y ≤ t)
β
Proof. Write ˚τ = m−1˚τ ◦ T˚j. For notational convenience, from here on we write S ,S˚
m j=0 Y˚ n n
insteadofS (1 ),S˚P(1 ).
n Y n Y˚
ByLemma5.2forwith[nβt] = m,fort > 0,
ν(˚τ ≥ n∩Y˚[nβt]) [nβt]
[nβt] = p[nβt]−kν(S = k). (5.1)
ν(S˚n ≤ [nβt]∩X˚n) Xk=1 n
RewritingtheRHSusingν(S = k) = ν(S ≤ k)−ν(S ≤ k−1)
n n n
[nβt] [nβt] [nβt−1]
p[nβt]−kν(S = k) = p[nβt]−kν(S ≤k)−p p[nβt]−kν(S ≤ k)
n n n
X X X
k=1 k=1 k=0
[nβt−1]
= ν(S ≤ [nβt])+(1−p) p[nβt]−kν(S ≤ k).
n n
X
k=1
Thusfornlargeenough,
[nβt] [nβt−1]
ν(S ≤ [nβt]) ≤ p[nβt]−kν(S = k) ≤ ν(S ≤ [nβt])+(1−p)ν(S ≤ [nβt]) p[nβt]−k
n n n n
X X
k=1 k=1
≤ ν(S ≤ [nβt])(1+p).
n
3WerecallthattheLaplacetransformofthisrandomvariableisgivenbyE(ezMβ)=P∞ Γ(1+β)pzp/Γ(1+
p=0
pβ)forallz ∈C.
8
Equivalently,
[nβt]
ν(τ ≥ n)≤ p[nβt]−kν(S = k) ≤ ν(τ ≥ n)(1+p).
[nβt] n [nβt]
X
k=1
Note that since ν(τ > n) = Cn−β(1+o(1)), for any t > 0, we have ν(τ ≥ [nβt]1/β) →
[nβt]
C P(Y ≥ t1/β).Sinceν(τ ≥ [nβt]1/β)−ν(τ ≥ nt1/β)= o(1),
β β [nβt] [nβt]
ν(τ ≥ n)→ C P(Y ≥ 1).
[nβt] β β
Puttingtogetherthepreviousdisplayedequations, thereexistsaconstantD thatdependsonlyon
β
C andP(Y ≥ 1)suchthat
β β
[nβt]
1 ≤ D−1 p[nβt]−kν(S = k)≤ 1+p. (5.2)
β n
X
k=1
Finally,byLemma5.3,ν(˚τ ≥ n|Y˚[nβt]) → P(Y ≤ t)andthus,
[nβt] β
p−n1/βν(n−1/β˚τ ≥ n∩Y˚[nβt])→ P(Y ≤ t).
[nβt] β
Theconclusion followsbytheaboveequation together with(5.2)and(5.1).
Forn,m ∈ N,theresultbelowrelates S˚ to˚τ = m−1˚τ ◦T˚j anditcanberegarded asan
n m j=0 Y˚
P
analogue ofitemb)mentioned atthebeginning ofthissection.
Lemma5.2. AssumethesetupofSubsection 2.2,inparticular (2.3).Thenforalln,m ∈ N,
m
ν(˚τ ≥n∩Y˚m)= ν(S˚ ≤ m∩X˚n) pm−kν(S (1 ) = k}).
m n n Y
X
k=1
Proof. Usingthatintherecurrent caseP(τ ≥ n) = P(S ≤ m)(foranyprobability measure P
m n
onY),wecomputethat
P(˚τ ≥ n∩Y˚m) = P(τ ≥ n∩Y˚m) = P(S ≤ m∩Y˚m) = P(S ≤ m∩Y˚Sn ∩Y˚m−Sn)
m m n n
= P(S˚ ≤m∩X˚n)P(y ∈ Y : S (y) < m,TSn(y)(y) ∈Y˚m−Sn(y))
n n Y
= P(S˚ ≤m∩X˚n)P(y ∈ Y : TSn(y)(y) ∈ Y˚m−Sn(y))
n Y
m
= P(S˚ ≤m∩X˚n) P(y ∈ Y :Tk(y) ∈ Y˚m−k ∩{y ∈ Y : S (y)= k}).
n Y n
X
k=1
Clearly, the events {y ∈ Y : Tk(y) ∈ Y˚m−k} and {y ∈ Y : S (y) = k} are disjoint. Recalling
Y n
thatν isT invariant, ν({y ∈ Y :Tk(y) ∈ Y˚m−k}) = ν(Y˚m−k)= pm−k.Thus,
Y Y
m m
P(y ∈ Y :Tk(y) ∈ Y˚m−k ∩{y ∈ Y : S (y) = k}) = pm−kν({y ∈ Y : S (y) = k})
Y n n
X X
k=1 k=1
andtheconclusion follows.
9
Lemma 5.3. Assume the set up of Subsection 2.2, in particular (2.3). Assume that (2.2) holds.
Suppose thatν(τ > n)= Cn−β(1+o(1))withβ ∈ (0,1). Then
ν(˚τ ≥ n|Y˚m) → P(Y ≤ t).
m β
Proof. Since we condition on the survivor set Y˚m, the required argument is standard and we
sketch it here only for completeness. It can be regarded as a straightforward modification of, for
instance,theargumentusedintheproofofthecentrallimittheoremforMarkovchainswithquasi
stationary distributions [3,Theorem3.4].
LetR˚bethematrixwithentriesgivenby(2.1).Letr = dν/dLebandnotethatrisconstanton
Y = ∪ C .Also,wenotethatinthesetupofSubsection2.2,R˚r = prandR˚(eiθ˚τr)= peiθ˚τr,
k≥0 k
θ ∈ [−π,π).Next,letR˜ = p−1R˚bethenormalization ofR˚andnotethatform ≥ 0,
E (ei(θ/m1/β)˚τm|Y˚m) = R˜mrei(θ/m1/β)˚τmdLeb = p−m ei(θ/m1/β)˚τmdν.
ν Z Z
Y˚m Y˚m
For m = 1, using the notation in (2.3), E (eiθ˚τ|Y˚) = p−1 ∞ f˚einθ = ∞ f einθ. Since
ν n=0 n n=0 n
byassumption, f = ν(τ > n)= Cn−β(1+o(1))wPithβ ∈ (0,1), asPθ → 0
j>n j
P
1−E (eiθ˚τ|Y˚)= C θβ(1+o(1)),
ν β
whereC isaconstant thatdependsonlyonC andβ (see,forinstance, [7]).Thus,
β
E (ei(θ/m1/β)˚τm|Y˚m)= exp(mlog(E (ei(θ/m1/β)˚τ|Y˚))) = eCβθβ(1+o(1)),
ν ν
asrequired.
A A result used in Proofs of Propositions 3.1 and 3.2
Inthisappendix, weuse“bigO”and≪notation interchangeably, writingA = O(a )orA ≪
n n n
a asn → ∞ifthereisaconstant C > 0suchthatkA k ≤ Ca foralln ≥ 1(forA operators
n n n n
anda ≥ 0scalars).
n
LemmaA.1. LetA(z)andB(z)beoperatorvaluedfunctionsonsomefunctionspacewithnorm
kk,analyticonDsuchthatA(1) =B(1) = 0.SupposethatthecoefficientsA ,B ofA(z),B(z),
n n
z ∈ DaresuchthatkA k ≪ kB k ≪ n−(β+1) forsomeβ > 0.
n n
Define C(z) = (1 − z)−1A(z)B(z), z ∈ D. Then the coefficients of C(z), z ∈ D satisfy
kC k ≪ n−2β ifβ <1,kC k≪ (logn)n−2 ifβ = 1andkC k ≪ n−(β+1) ifβ ≥1.
n n n
Proof. DuringthisproofA′,B′,C′ denotethefirstderivativesofA,B,C andA′ ,B′,C′ denote
n n n
then-thcoefficientofthesefunctions onD.
Clearly, kC k ≪ n−1kC′k.Itremains toestimate thecoefficients ofC′(z). Aneasycalcula-
n n
tionsshowsthat
B(z) A(z) A(z) B(z)
C′(z) = A′(z) + B′(z)+ .
1−z 1−z 1−z1−z
10
Since B(1) = 0, (1 − z)−1B(z) = ( B )zj. Hence, the coefficients (in norm kk) of
n j≥n j
(1−z)−1B(z) are O(n−β). SimilarlyP, thePcoefficients of (1−z)−1A(z) are O(n−β). Also, by
assumption, kA′ k ≪ kB′k ≪ n−β. Putting these together by convolving the coefficients of the
n n
factorscorresponding tothethreetermsintheexpression aboveofC′(z),
n1−2β, 0 < β ≤ 1,
2

n−(2β−1), 1 < β < 1,
kCn′k≪ (logn)n−1, β2 = 1,
n−β, β > 1,





andtheconclusion follows.
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