A Property of Random Walks on a Cycle Graph

0.22 MB·English

by  Yuki Ikeda

Ikedaetal.
A Property of Random Walks on a Cycle Graph
Yuki Ikeda1*
,Yasunari Fukai2
andYoshihiro Mizoguchi3
5 Abstract
1
0 WeanalyzetheHuntervs.Rabbitgameonagraph,whichisamodelofcommunicationinadhocmobile
2 networks.LetGbeacyclegraphwithN nodes.Thehuntercanmovefromavertextoavertexalonganedge.
Therabbitcanjumpfromanyvertextoanyvertexonthegraph.Weformalizethegameusingtherandomwalk
r
a framework.ThestrategyoftherabbitisformalizedusingaonedimensionalrandomwalkoverZ.Weclassify
M
strategiesusingtheorderO(k β 1)oftheirFouriertransformation.Weinvestigatelowerboundsandupper
− −
boundsoftheprobabilitythatthehuntercatchestherabbit.Wefoundaconstantlowerboundifβ (0,1)which
5 doesnotdependonthesizeN ofthegraph.WeshowtheorderisequivalenttoO(1/logN)ifβ=∈1anda

] lowerboundis1/N(β−1)/β ifβ (1,2].Theseresultshelpustochoosetheparameterβofarabbitstrategy
R accordingtothesizeN ofthe∈givengraph.Weintroduceaformalizationofstrategiesusingarandomwalk,
P theoreticalestimationofboundsofaprobabilitythatthehuntercatchestherabbit,andalsoshowcomputing
h. simulationresults.
t
a Keywords: Graphtheory;Randomwalk;Combinatorialprobability;AdhocNetwork
m
[

1 Introduction Princesscanmoveatanyspeed.Thisgameisplayedona
3
v Weconsideragameplayedbytwoplayers:thehunterand cycle graph as introducedby Isaacs[10]. The Princess vs.
6 the rabbit. This game is described using a graph G(V,E) MonstergamehasbeeninvestigatedbyAlpern[3],Zelikin
6 where V is a set of vertices and E is a set of edges. Both [20], and so on. Gal analyzed the Princess-Monster game
0
players may use a randomized strategy. The hunter can onaconvexmultidimensionaldomain[8].
5
0 move from vertex to vertex along edges. The rabbit can Thenextone isthe Deterministicpursuit-evasiongame.
. movetoanyvertexatonce.Thehunter’spurposeistocatch Inthisgameweconsiderarunawayhidedarkspot,forex-
1
therabbitinasfewstepsaspossible.Ontheotherhand,the ample a tunnel.Parsonsinnovatedthe searchnumberof a
0
5 rabbitconsidersastrategythatmaximizesthetimeuntilthe graph[16, 17]. The search number of a graph is the least
1 huntercatchtherabbit.Ifthehuntermovestoavertexthat numberofpeoplethatarerequiredtocatcharunawayhid-
: therabbitisat,thegamefinishesandwesaythatthehunter ing darkspot movingat any speed. LaPaugh[12] showed
v
i catchestherabbit. thatiftherunawayisknownnottobeinedgeeatanypoint
X
TheHuntervs.Rabbitgamemodelisusedforanalyzing oftime,thentherunawaycannotenteredgeewithoutbe-
r transmission procedures in mobile adhoc networks[5, 6]. ingcaughtintheremainderofthegame.Meggidoshowed
a
Thismodelhelpstosendanelectronicmessagesefficiently thatthecomputationtimeofthesearchnumberofagraph
usingmobilephones.Theexpectedvalueoftimeuntilthe is NP-hard[14].If anedgecanbe clearedwithoutmoving
huntercatchestherabbitisequaltotheexpectedtime un- along it, but it suffices to ’look into’ an edge from a ver-
til the recipient receives the mail. One of our goals is to tex, then the minimum number of guardsneeded to catch
improvetheseprocedures. thefugitiveiscalledthenodesearchnumberofgraph[11].
We introduce some games resembling the Hunter vs. Thepursuitevasionproblemintheplanewasintroducedby
Rabbit game. The first one is the Princess vs. Monster SuzukiandYamashita[19].Theygavenecessaryandsuffi-
game.Inthisgame,theMonstertriestocatchthePrincess cientconditionsforasimplepolygontobesearchablebya
in area D. The difference between the Hunter vs. Rabbit singlepursuer.LaterGuibasetal.[9]presentedacomplete
gameisthattheMonstercatchesthePrincessifthedistance algorithmandshowedthattheproblemofdeterminingthe
between the two players is smaller than a chosen value. minimalnumberofpursuersneededtoclearapolygonalre-
Also the Monster moves at a constant speed whereas the gionwithholesisNP-hard.Parketal.[15]gavethreenec-
essaryandsufficientconditionsforapolygontobesearch-
*Correspondence:y-ikeda(at)math.kyushu-u.ac.jp ableandshowedthatthereisO(n2)timealgorithmforcon-
1GraduateSchoolofMathematics,KyushuUniversity,
structinga searchpathforan n-sidedpolygon.Efratetal.
Fulllistofauthorinformationisavailableattheendofthearticle
†Equalcontributor [7] gave a polynomial time algorithm for the problem of
Ikedaetal. Page2of15
clearingasimplepolygonwithachainofkpursuerswhen Let Y ,Y ,... be independent, identically distributed ran-
1 2
thefirstandlastpursuercanonlymoveontheboundaryof domvariablesdefinedonaprobabilityspace(Ω , ,P )
thepolygon. takingvaluesintheintegerlatticeZwith H FH H
AfirststudyoftheHuntervs.Rabbitgamecanbefound
in [2]. The presented hunter strategy is based on random P Y1 1 =1.
H{| |≤ }
walkonagraphanditisshownthatthehuntercatchesan
unrestricted rabbit within O(nm2) rounds, where n and m Let N Nbefixed.Wedenoteby X(N) arandomvariable
denotethenumberofnodesandedges,respectively.Adler defined∈on a probability space (ΩN, 0N,µN) taking values
etal.showedthatifthehunterchoosesagoodstrategy,the inVN := 0,1,2,...,N 1 with F
{ − }
upper bound of the expected time that the hunter catches
1
therabbitisO(nlog(diam(G))),wherediam(G)isadiam- µ X(N) =l = (l V ).
N{ 0 } N ∈ N
eter of a graph G, and if the rabbit chooses a good strat-
egy, the lower bound of the expected time that the hunter Forb Z,wedenoteby(b mod N)theremainderofb
catchestherabbitisΩ(nlog(diam(G)))[1].Babichenkoet ∈
dividedbyN.
al.showedAdler’sstrategiesyieldaKakeyasetconsisting Arabbit’sstrategy (N) isdefinedby
of4ntriangleswithminimalarea[4]. {Rn }∞n=0
Inthispaper,weproposethreeassumptionsforthestrat- (N) = X(N) and (N) =(X(N)+S mod N).
egyoftherabbit.Wehavethegenerallowerboundformula R0 0 Rn 0 n
for the probability that the hunter catches the rabbit. The (N) indicates the position of the rabbit at time n on V .
strategyoftherabbitisformalizedusingaonedimensional RHnunter’sstrategy (N) isdefinedby N
random walk over Z. We classify strategies using the or- {Hn }∞n=0
derO(k β 1)oftheirFouriertransform.Ifβ= 1,thelower
− − n
boundofaprobabilitythatthehuntercatchestherabbitis (N) =0 and (N) = Y mod N .
b((yc∗tπh)e−1gilvoegnNs+tract2e)g−y1.wIfhβerec(21a,n2d],ct∗hearleowcoenrsbtaonutnsddeofifnthede H0 Hn Xj=1 j 
∈
probabilitythatthehuntercatchestherabbitisc4N−(β−1)/β n(N) indicatesthe positionof the hunterat time n on VN.
wherec4 >0isareconstantdefinedbythegivenstrategy. PHut
We show experimentalresultsfor three examplesof the
P(N) =µ P and P˜(N) = P P(N).
rabbitstrategy. N
R × H × R
1
(k Z 0 ) Thehuntercatchestherabbitwhenthehunterandtherabbit
1 P X =k = 2a(k +1)(k +2) ∈ \{ } arebothlocatedonthesameplace.
{ 1 }  1−|21|a | | (k=0) Wewilldiscusstheprobabilitythatthehuntercatchesthe
 1 (k Z 0 ) rabbitbytimeN onVN,thatis,
2akβ+1 ∈ \{ }
23 PP{XX11 ==kk}== 131−| 1a| Xk(∞=k1∈kβ{1+−11,0,1(}k) =0) We investigate thP˜e(Na)sy[nmN=1p{tHotni(cN)e=stiRm(naNt)e}o.f this probability
{ }  0 (k<{−1,0,1}). asN →∞.
Wbeehacvainorcoofntfihromseobuorubnodusnbdysthfoerrmesuulalt,saonfdsitmheulaastiyomnsp.totic Definition1 Wedefineconditions(A1),(A2)and(A3)as
follows.
(A1) Therandomwalk S isstronglyaperiodic,i.e.for
2 Statementsof Results eachy Z,thesm{allne}s∞nt=1subgroupcontainingtheset
We considerthe HuntervsRabbitgameona cyclegraph. ∈
ToexplaintheHuntervsRabbitgame,weintroducesome y+k Z P X =k >0
1
notation. Let X ,X ,... be independent, identically dis- { ∈ | { } }
1 2
tributed random variables defined on a probability space isZ.
(Ω, ,P) taking values in the integer lattice Z. A one- (A2) P X =k = P X = k (k Z).
F 1 1
dimensionalrandomwalk S isdefinedby { } { − } ∈
{ n}∞n=1 (A3) Thereexistβ (0,2],c >0andε>0suchthat
∈ ∗
n φ(θ):= eiθkP X =k =1 c θβ+O(θβ+ε).
Sn = Xj. Xk Z { 1 } − ∗| | | |
Xj=1 ∈
Ikedaetal. Page3of15
Wedenotetheβin(A3)asβ . Remark1 Adler,Ra¨cke,Sivadasan,SohlerandVo¨cking
R consideredP˜(N)( N (N) = (N) )inthecaseof
∪n=1{Hn Rn }
Theorem1 AssumethatX satisfies(A1) (A3).
1
− 1
(I) Ifβ (0,1),thenthereexistsaconstantc > 0such (k Z 0 )
ythna+t1R|f≤o∈r1N(n∈=N1\,2{,1.}.a.n,Ndy−1,1y)2,,...,yN ∈ Z1with|yn− P{X1 =k}= 221(|k|+1)(|k|+2) (k=∈ 0)\. { }
N Inthiscase,X satisfies(A1),(A2)and
1
c P(N) (N) =(y mod N) . (1)
1 ≤  Rn n  π
R [n=1n o φ(θ)=1− 2|θ|+O(|θ|3/2)
(II) Ifβ =1,thenthereexistconstantsc >0andc >0
suchRthatfor N N 1 andy ,y ,.2..,y Z3with ((A3) with β = 1), and we have (4) in Corollary 1 which
yn yn+1 1(n∈=1\,2{,.}..,N 11)2, N ∈ coincideswiththeresultofLemma3in[1].
| − |≤ −
Remark2 Forβ (0,2),let
N ∈
1
c1∗πlogN+c2 ≤≤ Plo(RcNg3)N[n.=1nR(nN) =(yn mod N)(o2) P{X1 =k}= 12a−|k11a|β+1∞ kβ1+1 ((kk∈=Z0)\{0})
(III) tIhfaβtRfo∈r(N1,2],Nthen1thaenredeyx1i,syt2s,a..c.o,nysNtantZc4w>ith0syunch withaconstantasatisfyingXk=a1> ∞k=1(1/kβ+1).Thenφ(θ)in
yn+1 1(n∈=1\,2{,.}..,N 1), ∈ | − (A3)is P
|≤ −
π θβ
c4 P(N) N (N) =(y mod N) . (3) φ(θ)=1− 2aΓ(β+1)|s|in(βπ/2) +O(|θ|β+(2−β)/2), (5)
N(β 1)/β ≤  Rn n 
− R [n=1n o whereΓisthegammafunction(seeAppendix(B)).X1sat-
isfies(A1),(A2)and(5).
ThefollowingboundsareobtainedasacorollaryofThe-
If X takes three values 1,0,1 with equal probability,
orem1. 1 −
thenX satisfies(A1),(A2)and
1
Corollary1 Assume(A1) (A3). 1
− φ(θ)=1 θ2+O(θ4)
Ifβ (0,1),thenthereexistsaconstantc1 >0suchthat − 3| | | |
forNR ∈N 1 ,
∈ \{ } ((A3)withβ=2).
N
c P˜(N) (N) = (N) . The inequality (3) seems to be sharp, because the pow-
1 ≤  {Hn Rn } ers of upper and lower bound appearing in (3) cannot be
[n=1  improved.Indeed,wehavethefollowingestimates.
If β = 1, then there exist constantsc > 0 and c > 0
suchtRhatforN N 1 , 2 3 Proposition1 LetHi(N) =0foranyiandassume(A1)−
∈ \{ } (A3). If β (1,2], then there exist constants c ,c > 0
5 6
suchthatfRor∈N N,
1 N ∈
P˜(N) (N) = (N)
c1∗πlogN+c2 ≤≤ locg3N[n.=1nHn Rn o (4) N(βc−51)/β ≤P(RN)[nN=1{R(nN) =0}≤ N(βc−61)/β. (6)
Proposition 2 Let (N) = i for any i. If X takes three
Ifβ (1,2],thenthereexistsaconstantc4 >0suchthat values 1,0,1 with eHqiual probability, then th1ere exists a
forNR ∈N 1 , constan−tc >0suchthatforN N,
∈ \{ } 7 ∈
N N
c
4 P˜(N) (N) = (N) . c P(N) (N) =(n mod N) . (7)
N(β 1)/β ≤  {Hn Rn } 7 ≤  {Rn }
− [n=1  R [n=1 
Ikedaetal. Page4of15
TheproofsofProposition1andProposition2aregiven where a 1. We note β = 1, c = π and ε = 1/2 in
inAppendix(A). Remark 1≥. If2a = 1, then this is t∗he case in [1]. We can
defineC andρ forthiscase.Sowehave
∗ ∗
Remark 3 Assume (A1) and (A2). If there exist c > 0
andε>0suchthat ∗ 1 L(N,1)= 1 . (9)
N 1p(N) ≥ 2 logN+6.50503
i=−0 i π2
φ(θ)=1 c θ +O(θ1+ε) P
− ∗| | | |
Theproofof(9)isgiveninAppendix(D).
((A3)withβ=1).Then Figure 1 shows an experimental result of the probabil-
ities for all initial positions of the rabbit with N = 100
N and a = 1. The horizontal axis is the initial position of
1
Nl→im∞ c∗πlogN!P(RN)[n=1{R(nN) =0}=1. (8) trthhabeatbrtiahtbeibshictu,anautngedhrtct.ahTtechhveeesrrettidhcealilrnaaebxibinsits.tThheohwefisbglutuhreeeliipsnreaobipsarbothibleiatbyaivlteihtrye-
Theproofof(8)isgiveninAppendix(C).
age ofprobabilitiesthatthehuntercatchestherabbit.The
greenlineisL(N,a).Inthiscase,thehunterdoesnotmove
3 Computer simulation fromthe initialposition0. As youcan see, the averageof
In this section, we show some experimental results about theprobabilitythatthehuntercatchestherabbitisbounded
theHuntervsRabbitgameonacyclegraph.We compute belowbyL(N,a).
P Sn mod N =k by using the gamma function and the Inthiscase,theaverageoftheprobabilitythatthehunter
cla{ssdiscrete di}stributioninC++.We canshowthe catchesthe rabbiteach initialpositionof therabbitnearly
probability the rabbit is caught and the expected value of equals0.4258,sowehave
thetimeuntiltherabbitiscaughtusingthisapplication.
Inthissection,weconsideralowerboundL(N,a)ofthe 1
;7.43823,
probabilitythatthehuntercatchestherabbit.Accordingto L(100,1)
the Proposition 3 and Proposition 6, we define L(N,a) as
follows: and
1 N
L(N)= 1+AN +BN + 1−1ρ∗ L(1010,1)P(RN)[n=1{R(nN) =0};3.1672.
where
Table 1 is the experimental results of Example 1 with
a = 1 and N = 100,500 and 1000. This table shows the
22+ε−βπε−βC (β (0,1]),
AN = 2N(βc2∗1)/β∗ (β∈(1,2)) asymptoticbehaviorof(8).
 − ∈
and  Taa=bl1ea1nTdhNis=ta1b0le0,is50e0xapnedrim10e0n0t.aAlreissuthltesaovfeEraxgaemopfleth1ewith
probabilitythatthehuntercatchestherabbit.
21−β (β (0,1)), N 1/L(N,a) A A/L(N,a)
πβc(1 β) ∈ 100 7.43823 0.4528 3.1672
B = 1 ∗lo−gN+ 1 (β=1), 500 7.76437 0.39048 3.03183
N  π2c2c∗−∗πβ (cid:16)1+ β−11π(cid:17)c∗N(β−1)/β (β∈(1,2)). 1000 7.90483 0.37555 2.96866
We note β and c are defined by a given P X = k in an
t
∗ { } Example2 Weconsiderthecaseofβ (0,2).Weput
example.Wechooseappropriateconstantsε,ρ andC for ∈
∗ ∗
eachexamples.
1
(k Z 0 )
2akβ+1 ∈ \{ }
E[1x].aLmeptle1 Weconsiderthegeneralizationofthecaseof P{Xt =k}= 1−| 1a| ∞ kβ1+1 (k=0)
1 (k Z 0 )  Xk=1
P{Xt =k}= 12a−(|k21|a+1)(|k|+2) (k=∈ 0)\{ } εwh=er2e−2aβ.>TPhe∞kn=,1tkhβ1e+1l.oBwyeRrbemouanrdk2o,fct∗he=p2raoΓb(βa+b1i)πlsiitny(βπth/2a)tatnhde

Ikedaetal. Page5of15
huntercatchestherabbitL(N,a)is (WecanprovethisusinginthesamewayinAppendix(D).)
Figure3isanexperimentalresultofExample3.Thegreen
L(N,a) lineinFigure3isL(N).
′
1
= 11++2(π1c−∗β)(−11−(β1)+−l1oπg−Nβc)−∗+12+712/24π−3−1β1//22πc1−∗−13Cβ/∗2+c(−∗11−Cρ∗∗+)(−11−ρ∗(()ββ−1∈=(10),1)) aampWrpoelebcsao.buillidtyhtahvaetathceohnucnretetercloawtcehresbothuenrdaobfbitthfeoarvtheorasgeeeoxf-
 1+2N(β−1)/β+22−βc−∗1π−β(1+(β−1)−1)N(β−1)/β+(1−(ρβ∗)−∈1(1,2)) 4 Upper bounds and Lowerbounds
whereρ andC areappropriateconstantsforeachexam- Inthissection,wegivearelationbetween
ples.Wh∗ena =∗2.5andβ = 1,wesetC ; 0.177245and
ρ ;0.694811.Sowehave ∗ N
∗ P(N) (N) =(y mod N)
L(N,2.5)= 5 logN+14.65936. R [n=1nRn n o
π2
andone-dimensionalrandomwalk S .
Figure2 is an experimentalresultwith β = 1, N = 100 { n}∞n=1
and a = 2.5. In this case, the average of the probability
Proposition3 ForN N 1 andy ,y ,...,y Zwith
that the hunter catches the rabbit nearly equals 0.318, so 1 2 N
wehave yn yn+1 1(n=1,2∈,...\,N{ } 1), ∈
| − |≤ −
1
;6.99237, 1 N
L(100,2.5) P(N) (N) =(y mod N)
and PiN=−01p(iN) ≤ R 2[n=1nRn n o
, (10)
≤ N 1q(N)
1 N i=−0 i
P(N) (N) =0 ;2.22357. P
L(100,2)  {Rn }
R [n=1  where
Table2istheexperimentalresultsofExample2withβ= [y] = y+kN k Z ,
1, a = 2.5 and N = 100,500and 1000.This table shows N { | ∈ }
thatthevalueofA/L(N,a)(>1)isdecreasing.
1 (i=0)
βTa=bl1e,2aT=h2is.5taabnldeNis=ex1p0e0r,im50e0natanldre1s0u00lt.sAofisEtxhaemapvleer2agweitohfthe p(iN) = max P Si [y]N (i N)
probabilitythatthehuntercatchestherabbit.  |y|≤i,y∈Z { ∈ } ∈
N 1/L(N,a) A A/L(N,a) 
100 6.99237 0.318 2.22357 and
500 7.80772 0.25924 2.02407
1000 8.15887 0.24015 1.95935
1 (i=0)
q(iN) = min P Si [y]N (i N).
Example3 Weput  |y|≤i,y∈Z { ∈ } ∈

Proof. Wenotethat
1
(k 1,0,1 )
P Xt =k = 3 ∈{− }
By Rem{ark 2,}β =20, c =(k1<an{−d1ε,0=,12})..In this case, the [nN=1nR(nN) =(yn mod N)o
3
lowerboundoftheprob∗abilitythehuntercatchestherabbit =N−1 N X(N) =l, l+S [y ]
L′(N)is [l=0 [n=1n 0 n ∈ n No
L′(N)= 1+ π62 N1/12+4.26301. =N[l=−01[nN=1( lX+0(NS) =i <l,[yi]N, 1l+≤Sin≤∈n[y−n]1N, )
(cid:16) (cid:17)
Ikedaetal. Page6of15
Figure1 ThisisanexperimentalresultofExample1.Inthiscase,a=1.Thehunterdoesnotmovefromaninitialposition0.
Figure2 ThisisanexperimentalresultofExample2.Inthiscase,a=2.5.Thehunterdoesnotmovefromaninitialposition0.
Figure3 ThisisanexperimentalresultofExample3.Thehunterdoesnotmovefromaninitialposition0.
Ikedaetal. Page7of15
bythedefinitionof (N) ∞ .We noteP(N) = µ P,the 1,2,...,N ,wehave
Rn n=0 N × { }
aboverelationimpliens o R
N 1 N
− 1P l+S [y ]
n n N
N Xl=0 Xn=1 N { ∈ }
P(N) (N) =(y mod N)
=RXNl=−01[nXn=N=11nRN1nP( ll++nSSin<∈[[yyin]]NN,o 1≤i≤n−1, ). ≤XNl=−01NXj−N=1jpN1(NP) ( ll++SSij<=[[yyi]j]NN, 1≤i≤ j−1, )
(11) × i 
Xi=0N  N 1
P(N) (N) =(y mod N) − p(N) .(13)
≤  Rn n  i 
poFsoerthle∈e{v0e,n1t,.l.+.,SNn−1[}yna]nNdnac∈co{2rd,i3n,g..to.,tNhe},vwaleudeeocfotmhe- R [n=1n oXi=0 
{ ∈ }
first hitting time for [y ] ,[y ] ,...,[y ] and the hitting Hereweused(11).
1 N 2 N n N
placetoobtain Z)B,yPlN=−01P{l+Sn ∈ [y]N} = P{Sn ∈ Z} = 1(n ∈ N, y ∈
P l+S [y ]
{= nn ∈ PnN}ll++SSij<=[yyij]N+,mN, 1≤i≤ j−1, . XNl=−01XnN=1 N1P{l+Sn ∈[yn]N}=1. (14)
Xj=1Xm Z  yj+mN+Xj+1+ +Xn [yn]N 
∈ ··· ∈
  (13)and(14)imply
The probability in the double summation on the right- N N 1
handsideaboveisequalto 1≤P(N) R(nN) =(yn mod N)  − p(iN) (15)
R [n=1n oXi=0 
l+S <[y] , 1 i j 1, thatisthefirstinequalityin(10).
P( l+Sij =yijN+mN, ≤ ≤ − ) For the last inequality in (10), let yN+j = yN (j =
P y +mN+S [y ] 1,2,...,N). The same argumentas showing (15) (we use
× n j n−j ∈ n No q(iN) insteadof p(iN))gives
by the Markov property. It is easy to verify that for any N 1 2N
m Z, 2 = − 1P l+Sn [yn]N
∈ N { ∈ }
Xl=0 Xn=1
N N 1
P y +mN+S [y ] P(N) (N) =(y mod N) − q(N) .
=nPjnSn−j ∈[ynn−−jy∈j]Non≤Npo(nN−)j ≥ R [n=1nRn n oXi=0 i 
Corollary2 ForN N 1 ,
∈ \{ }
by y y n j.Therefore
n j
| − |≤ − N
1
P(N) (N) =0
P{ln+Sn ∈l[+ynS]N}<[y] , 1 i j 1, 1+PiN=−11P{S2i ∈[0]N} ≤. R [n=1nRn o(16)
≤Xj=1P( l+Sij =[iyjN]N ≤ ≤ − )p(nN−)j, ≤ 1+PiN=−11P{Si ∈[0]N}
(12) Proof. Put y = y = = y = 0 in the proof of
1 2 2N
···
Proposition 3. Then the same argument as showing (10)
gives(16).
forl 0,1,...,N 1 andn 1,2,...,N .Bymultiply-
∈ { − } ∈{ }
ing(12)by1/Nandsumming(l,n)over 0,1,...,N 1
{ − }×
Ikedaetal. Page8of15
Corollary3 ForN N 1 , where
∈ \{ }
1 n1/β
I(n,l)= φn(θ)e iθldθ,
1+ iN=−11P{Si ∈[i]N} 2π Z|θ|<r −
P N n1/β
P(N) (N) =(n mod N) J(n,l)= φn(θ)e−iθldθ.
≤ R [n=1nR2n o 2π Zr≤|θ|≤π
. (17) Astronglyaperiodicrandomwalk(A1)hastheproperty
≤ 1+ iN=−11P{Si ∈[i]N} that φ(θ) = 1 only when θ is a multiple of 2π (see 7
| | §
P Proposition8of[18]).Bythedefinitionofφ(θ), φ(θ) isa
Proof. Puty = j(j=1,2,...,2N)intheproofofPropo- | |
j continuousfunctionon the boundedclosed set [ π, r]
sition 3. Then the same argument as showing (10) gives − − ∪
[r,π], and φ(θ) 1 (θ [ π,π]). Hence, there exists a
(17). | | ≤ ∈ −
ρ<1,dependingonr (0,π],suchthat
∈
Remark 4 By the same argument as showing (16), we
max φ(θ) ρ. (21)
obtainthatforǫ˜ >0andN 1/ǫ˜, r θ π| |≤
≥ ≤| |≤
N 1+ǫ˜ Byusingtheaboveinequality,
P(N) (N) =0 .
R [n=1nRn o≤ 1+Pǫi˜=N1P{Si ∈[0]N} J(n,l) n1/β φ(θ)ndθ n1/βρn.
5 Fourier transform | |≤ 2π Z | | ≤
r θ π
≤| |≤
Inthissection,weintroducesomeresultsconcerningone-
dimensionalrandomwalk. Weperformthechangeofvariablesθ= x/n1/β,sothat
Proposition4 Ifaone-dimensionalrandomwalksatisfies I(n,l)= 1 φn x exp i xl dx.
(foAr1n)andN(A,3),thenthereexistC1 >0andN1 ∈Nsuchthat 2πZ|x|<rn1/β (cid:18)n1/β(cid:19) − n1/β!
1
≥
Put
+
1 xl
sup n1/βP S =l ∞e c xβexp i dx ε 1
l∈ZC(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)n δ, { n }− 2πZ−∞ − ∗| | − n1/β! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) γ=min(2β(β+ε+1), 2(2β+1)).
≤ (cid:12)1 − (cid:12) WedecomposeI(n,l)asfollows:
whereδ=min ε/(2β),1/2 .
{ } +
1 xl
I(n,l) = ∞e c xβexp i dx
Proof. Proposition 4 can be proved by the same proce- 2πZ − ∗| | − n1/β!
dureasinTheorem1.2.1of[13]. +I (n−,∞l)+I (n,l)+I (n,l),
1 2 3
TheFourierinversionformulaforφn(θ)is
where
n1/β π
n1/βP S =l = φn(θ)e iθldθ. (18)
{ n } 2π Z − 1 x
By (A3), there exist C > −0πand r (0,π) such that for I1(n,l)= 2πZ|x|≤nγ(cid:26)φn(cid:18)n1/β(cid:19)−e−c∗|x|β(cid:27)
θ <r, ∗ ∈
| |
xl
φ(θ) (1 c θβ) C θβ+ε (19) exp i dx,
| − − ∗| | |≤ ∗| | × − n1/β!
and
1 xl
φ(θ) 1 c∗ θβ. (20) I2(n,l)=−2πZnγ<x e−c∗|x|βexp −in1/β! dx
| |≤ − 2| | | |
and
Withr,wedecomposetheright-handsideof(18)toobtain
1 x xl
n1/βP{Sn =l}= I(n,l)+J(n,l), I3(n,l)= 2πZnγ<x<rn1/βφn(cid:18)n1/β(cid:19)exp −in1/β! dx.
| |
Ikedaetal. Page9of15
Therefore, Moreover,ifnislargeenough,then
1 xl 2s
(cid:12)(cid:12)n1/βP{Sn =l}− 2πZ ∞e−c∗|x|βexp −in1/β! dx(cid:12)(cid:12) e−c∗|x|β/2 ≤ cs|x|−sβ (|x|>nγ),
(cid:12)(cid:12) −∞ (cid:12)(cid:12) ∗
(cid:12) (cid:12)
(cid:12) (cid:12) where s = (1/β)(1+ 1/(2γ)). By replacing the integrand
3 intheright-handsideofthelastinequalityof(23)withthe
J(n,l) + I (n,l).
k right-handsideoftheaboveinequality,weobtain
≤| | | |
Xk=1
Theproofof Proposition4 willbecompleteifwe show 2s+1γ 2s+1γ
I (n,l) n 1/2 n δ. (24)
thateachtermintheright-handsideoftheaboveinequality | 3 |≤ πcs − ≤ πcs −
isboundedbyaconstant(independentofl)multipleofn δ. ∗ ∗
−
If n is large enough, then the bound J(n,l) n1/βρn, Thesameargumentasshowing(24)gives
| | ≤
whichhasalreadybeenshownabove,yields
1 2s+1γ
I (n,l) e c xβ dx n δ.
|J(n,l)|≤n−δ. | 2 |≤ 2πZnγ θ − ∗| | ≤ πcs −
≤| | ∗
Withthehelpof Let
+
|an−bn| = |a−b|(cid:12)(cid:12)n−1an−1−jbj(cid:12)(cid:12) I0(n,l:β,c∗)= 21πZ ∞e−c∗|x|βexp −inx1/lβ! dx
(cid:12)(cid:12)Xj=0 (cid:12)(cid:12) −∞
(cid:12) (cid:12)
na b(cid:12)(cid:12) (a,b [ (cid:12)(cid:12)1,1]) (22) appearinginProposition4.
≤ | − (cid:12)| ∈ −(cid:12)
and φ(θ) 1(θ [ π,π]),(19)impliesthatfor x <rn1/β,
Remark5 Whena one-dimensionalrandomwalkis the
| |≤ ∈ − | |
stronglyaperiodic(A1)withE[X ]=0andE[X 2+ε]<
x x 1 | 1| ∞
φn e c xβ n φ e c xβ/n forsomeε (0,1),itisverifiedthat
(cid:12)(cid:12) (cid:18)n1/β(cid:19)− − ∗| | (cid:12)(cid:12)≤ (cid:12)(cid:12) (cid:18)n1/β(cid:19)− − ∗| | (cid:12)(cid:12) ∈
≤(cid:12)(cid:12)(cid:12) n(cid:12)(cid:12)φ(cid:18)n1x/β(cid:19)− 1−(cid:12)(cid:12)(cid:12) c∗|xn(cid:12)(cid:12)(cid:12)|β!(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) φ(θ)=1− E[2X12]|θ|2+O(|θ|2+ε).
(cid:12) (cid:12)
(cid:12)(cid:12) xβ (cid:12)(cid:12)
(cid:12) +n(cid:12)(cid:12) 1−c∗|n| !−e−(cid:12)c∗|x|β/n(cid:12)(cid:12) In this case, I0(n,l : 2,E[X12]/2) can be computed and
(cid:12) (cid:12) Proposition4givesthefollowing.
C xβ+εn(cid:12)(cid:12)(cid:12)−ε/β+ c2∗ x2βn−1. (cid:12)(cid:12)(cid:12) (LocalCentralLimitTheorem)ThereexistC˜1 >0and
≤ ∗| | 2| | N˜ Nsuchthatforn N˜ ,
1 1
∈ ≥
Thus
1 l2
1 x sup(cid:12)n1/2P S =l exp (cid:12)
|I1(n,l)| ≤≤ 2π1π βZ|+x|≤Cεn∗γ+(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)φ1n(cid:18)+n12/(β2(cid:19)βc−2∗+e1−)c∗!|xn|β−(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)δd.θ l∈Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) { n }− q2E[X12]π −2E[X12]n(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)
C˜ n δ,
1 −
Itiseasytoverifythatfor x <rn1/β, ≤
| | (25)
x c xβ n
(cid:12)(cid:12)φn(cid:18)n1/β(cid:19)(cid:12)(cid:12)≤ 1− 2∗|n| ! ≤e−c∗|x|β/2 w7.h9eirne[δ18=].)min{ε/4,1/2}. (See Remark after Proposition
(cid:12) (cid:12)
(cid:12) (cid:12)
by(20),an(cid:12)dweobta(cid:12)inthat Itiseasytosee
|I3(n,l)| ≤ 21πZnγ<|x|<rn1/β(cid:12)(cid:12)(cid:12)φn(cid:18)n1x/β(cid:19)(cid:12)(cid:12)(cid:12) dx I0(n,l:1,c∗)= π1c2∗+c(∗l/n)2 (n∈N,l∈Z,c∗ >0)
1 (cid:12) (cid:12)
≤ 2πZnγ<x e−c∗|x(cid:12)|β/2dx. (cid:12) (23) andwehavethefollowingcorollaryofProposition4.
| |
Ikedaetal. Page10of15
Corollary4 If a one-dimensionalrandomwalk satisfies Then,
(A1) and (A3) with β = 1, then there exist C > 0 and
2
N2 ∈Nsuchthatforn≥N2, N−1e 2ijπl/Nφn 2jπ =N−1N−1e2ijπ(l˜ l)/NP S [l˜]
− − n N
N ! ∈
1 c Xj=0 Xl˜=0 Xj=0 n o
sluZp(cid:12)(cid:12)nP{Sn =l}− πc2+(∗l/n)2(cid:12)(cid:12)≤C2n−δ, =NP{Sn ∈[l]N}
∈ (cid:12)(cid:12) ∗ (cid:12)(cid:12)
(cid:12) (cid:12)
whereδ=m(cid:12)in ε/2,1/2 . (cid:12) since
{ }
Weperformthechangeofvariablest=c xβ ,sothat N−1e2ijπ(l˜ l)/N = N l˜=l .
∗ − ( 0 l˜,l
+ Xj=0
1 1 1
I (n,0:β,c )= ∞e c xβ dx= Γ .
0 ∗ πZ0 − ∗ βc1/βπ β! Therefore,
∗
Wtheithfotlhloewhienlgpcoofrtohlelaaryb.ovecalculation,Proposition4gives P{Sn ∈[l]N} = N1 N−1φn 2Njπ!e−2jπil/N
Xj=0
Corollary5 If a one-dimensionalrandomwalk satisfies N 1
(A1)and(A3),thenthereexistC >0andN Nsuchthat = 1 − φn 2jπ cos 2jπl .
3 3 ∈ N N ! N !
forn N3, Xj=0
≥
1 1 Wenotethatφn(θ) Rand
n1/βP S =0 Γ C n δ, ∈
(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) { n }− βc1∗/βπ β!(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)≤ 3 − 1 N−1φn 2jπ cos 2jπl R
whereδ=(cid:12) min ε/2β,1/2 . (cid:12) N N ! N !∈
{ } Xj=0
Proposition5 Ifaone-dimensionalrandomwalksatisfies
by(A2).Sowehave
(A2),thenforl Zandn 0 N,
∈ ∈{ }∪
2mπ 2mπl
φn cos
P S [l]
{ n ∈ N} N ! N !
1 2 2jπ 2jπ
= N + N φn N !cos N l!+JN(n,l), =φn 2(N−m)π cos 2(N−m)πl . (27)
1 jX(N 1)/2 N ! N !
≤ ≤ −
(26)
LetN beanevennumber.Then,by(27),
where
P S [l]
n N
{ ∈ }
1
= φn(0)cos(0)
J (n,l)= (1/N)φn(π)cos(πl) (ifN iseven) N
N ( 0 (ifN isodd). 2 2jπ 2jπl
+ φn cos
N N ! N !
Proof. Bythedefinitionofφ(θ), 1 jX(N 1)/2
≤ ≤ −
1
+ φn(π)cos(πl)
φn(θ)= eiθkP S =k . N
n
{ }
Xk∈Z = 1 + 2 φn 2jπ cos 2jπl
N N N ! N !
Thus 1 jX(N 1)/2
≤ ≤ −
1
2jπ + φn(π)cos(πl).
φn = e2ijπk/NP S =k N
n
N ! { }
Xk∈Z Therefore, we have (26) for every even number N. The
N 1
= − e2ijπ(l˜+mN)/NP S =l˜+mN proofof(26)foroddnumberissimilarandisomitted.
n
Xl˜=0 Xm∈Z n o 6 Proof of Theorem1
N 1
= − e2ijπl˜/NP S [l˜] . InthissectionweproveTheorem1.Toproveit,weintro-
n N
∈ ducethefollowingProposition.
Xl˜=0 n o