A Property of Random Walks on a Cycle Graph
aa r X i v : . [ m a t h . P R ] M a r Ikeda etal.
A Property of Random Walks on a Cycle Graph
Yuki Ikeda , Yasunari Fukai and Yoshihiro Mizoguchi Abstract
We analyze the Hunter vs. Rabbit game on a graph, which is a model of communication in adhoc mobilenetworks. Let G be a cycle graph with N nodes. The hunter can move from a vertex to a vertex along an edge.The rabbit can jump from any vertex to any vertex on the graph. We formalize the game using the random walkframework. The strategy of the rabbit is formalized using a one dimensional random walk over Z . We classifystrategies using the order O ( k − β − ) of their Fourier transformation. We investigate lower bounds and upperbounds of the probability that the hunter catches the rabbit. We found a constant lower bound if β ∈ (0 , whichdoes not depend on the size N of the graph. We show the order is equivalent to O (1 / log N ) if β = and alower bound is / N ( β − /β if β ∈ (1 , . These results help us to choose the parameter β of a rabbit strategyaccording to the size N of the given graph. We introduce a formalization of strategies using a random walk,theoretical estimation of bounds of a probability that the hunter catches the rabbit, and also show computingsimulation results. Keywords:
Graph theory; Random walk; Combinatorial probability; Adhoc Network
We consider a game played by two players: the hunter andthe rabbit. This game is described using a graph G ( V , E )where V is a set of vertices and E is a set of edges. Bothplayers may use a randomized strategy. The hunter canmove from vertex to vertex along edges. The rabbit canmove to any vertex at once. The hunter’s purpose is to catchthe rabbit in as few steps as possible. On the other hand, therabbit considers a strategy that maximizes the time until thehunter catch the rabbit. If the hunter moves to a vertex thatthe rabbit is at, the game finishes and we say that the huntercatches the rabbit.The Hunter vs. Rabbit game model is used for analyzingtransmission procedures in mobile adhoc networks[5, 6].This model helps to send an electronic messages e ffi cientlyusing mobile phones. The expected value of time until thehunter catches the rabbit is equal to the expected time un-til the recipient receives the mail. One of our goals is toimprove these procedures.We introduce some games resembling the Hunter vs.Rabbit game. The first one is the Princess vs. Monstergame. In this game, the Monster tries to catch the Princessin area D . The di ff erence between the Hunter vs. Rabbitgame is that the Monster catches the Princess if the distancebetween the two players is smaller than a chosen value.Also the Monster moves at a constant speed whereas the * Correspondence: y-ikeda(at)math.kyushu-u.ac.jp Graduate School of Mathematics, Kyushu University,Full list of author information is available at the end of the article † Equal contributor
Princess can move at any speed. This game is played on acycle graph as introduced by Isaacs[10]. The Princess vs.Monster game has been investigated by Alpern [3], Zelikin[20], and so on. Gal analyzed the Princess-Monster gameon a convex multidimensional domain [8].The next one is the Deterministic pursuit-evasion game.In this game we consider a runaway hide dark spot, for ex-ample a tunnel. Parsons innovated the search number of agraph[16, 17]. The search number of a graph is the leastnumber of people that are required to catch a runaway hid-ing dark spot moving at any speed. LaPaugh [12] showedthat if the runaway is known not to be in edge e at any pointof time, then the runaway can not enter edge e without be-ing caught in the remainder of the game. Meggido showedthat the computation time of the search number of a graphis NP-hard[14]. If an edge can be cleared without movingalong it, but it su ffi ces to ’look into’ an edge from a ver-tex, then the minimum number of guards needed to catchthe fugitive is called the node search number of graph [11].The pursuit evasion problem in the plane was introduced bySuzuki and Yamashita [19]. They gave necessary and su ffi -cient conditions for a simple polygon to be searchable by asingle pursuer. Later Guibas et al. [9] presented a completealgorithm and showed that the problem of determining theminimal number of pursuers needed to clear a polygonal re-gion with holes is NP-hard. Park et al. [15] gave three nec-essary and su ffi cient conditions for a polygon to be search-able and showed that there is O ( n ) time algorithm for con-structing a search path for an n -sided polygon. Efrat et al.[7] gave a polynomial time algorithm for the problem of keda etal. Page 2 of 15 clearing a simple polygon with a chain of k pursuers whenthe first and last pursuer can only move on the boundary ofthe polygon.A first study of the Hunter vs. Rabbit game can be foundin [2]. The presented hunter strategy is based on randomwalk on a graph and it is shown that the hunter catches anunrestricted rabbit within O ( nm ) rounds, where n and m denote the number of nodes and edges, respectively. Adleret al. showed that if the hunter chooses a good strategy, theupper bound of the expected time that the hunter catchesthe rabbit is O ( n log( diam ( G ))), where diam ( G ) is a diam-eter of a graph G , and if the rabbit chooses a good strat-egy, the lower bound of the expected time that the huntercatches the rabbit is Ω ( n log( diam ( G ))) [1]. Babichenko etal. showed Adler’s strategies yield a Kakeya set consistingof 4 n triangles with minimal area [4].In this paper, we propose three assumptions for the strat-egy of the rabbit. We have the general lower bound formulafor the probability that the hunter catches the rabbit. Thestrategy of the rabbit is formalized using a one dimensionalrandom walk over Z . We classify strategies using the or-der O ( k − β − ) of their Fourier transform. If β =
1, the lowerbound of a probability that the hunter catches the rabbit is(( c ∗ π ) − log N + c ) − where c and c ∗ are constants definedby the given strategy. If β ∈ (1 , c N − ( β − /β where c > P { X = k } = a ( | k | + | k | +
2) ( k ∈ Z \ { } )1 − a ( k = P { X = k } = a | k | β + ( k ∈ Z \ { } )1 − a ∞ X k = k β + ( k = P { X = k } =
13 ( k ∈ {− , , } )0 ( k < {− , , } ) . We can confirm our bounds formula, and the asymptoticbehavior of those bounds by the results of simulations.
We consider the Hunter vs Rabbit game on a cycle graph.To explain the Hunter vs Rabbit game, we introduce somenotation. Let X , X , . . . be independent, identically dis-tributed random variables defined on a probability space( Ω , F , P ) taking values in the integer lattice Z . A one-dimensional random walk { S n } ∞ n = is defined by S n = n X j = X j . Let Y , Y , . . . be independent, identically distributed ran-dom variables defined on a probability space ( Ω H , F H , P H )taking values in the integer lattice Z with P H {| Y | ≤ } = . Let N ∈ N be fixed. We denote by X ( N )0 a random variabledefined on a probability space ( Ω N , F N , µ N ) taking valuesin V N : = { , , , . . . , N − } with µ N { X ( N )0 = l } = N ( l ∈ V N ) . For b ∈ Z , we denote by ( b mod N ) the remainder of b divided by N .A rabbit’s strategy {R ( N ) n } ∞ n = is defined by R ( N )0 = X ( N )0 and R ( N ) n = ( X ( N )0 + S n mod N ) . R ( N ) n indicates the position of the rabbit at time n on V N .Hunter’s strategy {H ( N ) n } ∞ n = is defined by H ( N )0 = H ( N ) n = n X j = Y j mod N . H ( N ) n indicates the position of the hunter at time n on V N .Put P ( N ) R = µ N × P and ˜ P ( N ) = P H × P ( N ) R . The hunter catches the rabbit whenthe hunter and the rabbitare both located on the same place.We will discuss the probability that the hunter catches therabbit by time N on V N , that is,˜ P ( N ) N [ n = {H ( N ) n = R ( N ) n } . We investigate the asymptotic estimate of this probabilityas N → ∞ . Definition 1
We define conditions (A1), (A2) and (A3) asfollows.( A
1) The random walk { S n } ∞ n = is strongly aperiodic, i.e. foreach y ∈ Z , the smallest subgroup containing the set { y + k ∈ Z | P { X = k } > } is Z .( A P { X = k } = P { X = − k } ( k ∈ Z ).( A
3) There exist β ∈ (0 , c ∗ > ε > φ ( θ ) : = X k ∈ Z e i θ k P { X = k } = − c ∗ | θ | β + O ( | θ | β + ε ) . keda etal. Page 3 of 15 We denote the β in ( A
3) as β R . Theorem 1
Assume that X satisfies ( A − ( A β R ∈ (0 , c > N ∈ N \ { } and y , y , . . . , y N ∈ Z with | y n − y n + | ≤ n = , , . . . , N − c ≤ P ( N ) R N [ n = n R ( N ) n = ( y n mod N ) o . (1)(II) If β R =
1, then there exist constants c > c > N ∈ N \ { } and y , y , . . . , y N ∈ Z with | y n − y n + | ≤ n = , , . . . , N − c ∗ π log N + c ≤ P ( N ) R N [ n = n R ( N ) n = ( y n mod N ) o ≤ c log N . (2)(III) If β R ∈ (1 , c > N ∈ N \ { } and y , y , . . . , y N ∈ Z with | y n − y n + | ≤ n = , , . . . , N − c N ( β − /β ≤ P ( N ) R N [ n = n R ( N ) n = ( y n mod N ) o . (3)The following bounds are obtained as a corollary of The-orem 1. Corollary 1
Assume ( A − ( A β R ∈ (0 , c > N ∈ N \ { } , c ≤ ˜ P ( N ) N [ n = {H ( N ) n = R ( N ) n } . If β R =
1, then there exist constants c > c > N ∈ N \ { } ,1 c ∗ π log N + c ≤ ˜ P ( N ) N [ n = n H ( N ) n = R ( N ) n o ≤ c log N . (4)If β R ∈ (1 , c > N ∈ N \ { } , c N ( β − /β ≤ ˜ P ( N ) N [ n = {H ( N ) n = R ( N ) n } . Remark 1
Adler, R¨acke, Sivadasan, Sohler and V¨ockingconsidered ˜ P ( N ) ( ∪ Nn = {H ( N ) n = R ( N ) n } ) in the case of P { X = k } = | k | + | k | +
2) ( k ∈ Z \ { } )12 ( k = . In this case, X satisfies ( A A
2) and φ ( θ ) = − π | θ | + O ( | θ | / )(( A
3) with β = Remark 2
For β ∈ (0 , P { X = k } = a | k | β + ( k ∈ Z \ { } )1 − a ∞ X k = k β + ( k = a satisfying a > P ∞ k = (1 / k β + ) . Then φ ( θ ) in( A
3) is φ ( θ ) = − π a | θ | β Γ ( β +
1) sin( βπ/ + O ( | θ | β + (2 − β ) / ) , (5)where Γ is the gamma function (see Appendix (B)). X sat-isfies ( A A
2) and (5).If X takes three values − , , X satisfies ( A A
2) and φ ( θ ) = − | θ | + O ( | θ | )(( A
3) with β = Proposition 1
Let H ( N ) i = i and assume ( A − ( A β R ∈ (1 , c , c > N ∈ N , c N ( β − /β ≤ P ( N ) R N [ n = {R ( N ) n = } ≤ c N ( β − /β . (6) Proposition 2
Let H ( N ) i = i for any i . If X takes threevalues − , , c > N ∈ N , c ≤ P ( N ) R N [ n = {R ( N ) n = ( n mod N ) } . (7) keda etal. Page 4 of 15 The proofs of Proposition 1 and Proposition 2 are givenin Appendix (A).
Remark 3
Assume ( A
1) and ( A c ∗ > ε > φ ( θ ) = − c ∗ | θ | + O ( | θ | + ε )(( A
3) with β = N →∞ c ∗ π log N ! P ( N ) R N [ n = {R ( N ) n = } = . (8)The proof of (8) is given in Appendix (C). In this section, we show some experimental results aboutthe Hunter vs Rabbit game on a cycle graph. We compute P { S n mod N = k } by using the gamma function and theclass discrete distribution in C++ . We can show theprobability the rabbit is caught and the expected value ofthe time until the rabbit is caught using this application.In this section, we consider a lower bound L ( N , a ) of theprobability that the hunter catches the rabbit. According tothe Proposition 3 and Proposition 6, we define L ( N , a ) asfollows: L ( N ) = + A N + B N + − ρ ∗ where A N = + ε − β π ε − β C ∗ c ∗ ( β ∈ (0 , , N ( β − /β ( β ∈ (1 , B N = − β π β c ∗ (1 − β ) ( β ∈ (0 , , π c ∗ log N + π c ∗ ( β = , − β c ∗ π (cid:16) + β − (cid:17) N ( β − /β ( β ∈ (1 , . We note β and c ∗ are defined by a given P { X t = k } in anexample. We choose appropriate constants ε , ρ ∗ and C ∗ foreach examples. Example 1
We consider the generalization of the case of[1]. Let P { X t = k } = a ( | k | + | k | +
2) ( k ∈ Z \ { } )1 − a ( k =
0) where a ≥ . We note β = c ∗ = π and ε = / a =
1, then this is the case in [1]. We candefine C ∗ and ρ ∗ for this case. So we have1 P N − i = p ( N ) i ≥ L ( N , = π log N + . . (9)The proof of (9) is given in Appendix (D).Figure 1 shows an experimental result of the probabil-ities for all initial positions of the rabbit with N = a =
1. The horizontal axis is the initial position ofthe rabbit, and the vertical axis shows the probability therabbit is caught. The red line in the figure is a probabilitythat the hunter catches the rabbit.The blue line is the aver-age of probabilities that the hunter catches the rabbit. Thegreen line is L ( N , a ). In this case, the hunter does not movefrom the initial position 0. As you can see, the average ofthe probability that the hunter catches the rabbit is boundedbelow by L ( N , a ).In this case, the average of the probability that the huntercatches the rabbit each initial position of the rabbit nearlyequals 0 . L (100 , ; . , and 1 L (100 , P ( N ) R N [ n = {R ( N ) n = } ; . . Table 1 is the experimental results of Example 1 with a = N = ,
500 and 1000. This table shows theasymptotic behavior of (8).
Table 1 This table is experimental results of Example 1 with a = and N = , and . A is the average of theprobability that the hunter catches the rabbit. N / L ( N , a ) A A / L ( N , a )100 Example 2
We consider the case of β ∈ (0 , P { X t = k } = a | k | β + ( k ∈ Z \ { } )1 − a ∞ X k = k β + ( k = a > P ∞ k = k β + . By Remark 2, c ∗ = π a Γ ( β +
1) sin( βπ/ and ε = − β . Then, the lower bound of the probability that the keda etal. Page 5 of 15 hunter catches the rabbit L ( N , a ) is L ( N , a ) = + − β (1 − β ) − π − β c − ∗ + − β/ π − β/ c − ∗ C ∗ + (1 − ρ ∗ ) − ( β ∈ (0 , + ( π c ∗ ) − (1 + log N ) + / π − / c − ∗ C ∗ + (1 − ρ ∗ ) − ( β = + N ( β − /β + − β c − ∗ π − β ( + ( β − − ) N ( β − /β + (1 − ρ ∗ ) − ( β ∈ (1 , ρ ∗ and C ∗ are appropriate constants for each exam-ples. When a = . β =
1, we set C ∗ ; . ρ ∗ ; . L ( N , . = π log N + . . Figure 2 is an experimental result with β = N = a = .
5. In this case, the average of the probabilitythat the hunter catches the rabbit nearly equals 0 . L (100 , . ; . , and 1 L (100 , P ( N ) R N [ n = {R ( N ) n = } ; . . Table 2 is the experimental results of Example 2 with β = a = . N = ,
500 and 1000. This table showsthat the value of A / L ( N , a )( >
1) is decreasing.
Table 2 This table is experimental results of Example 2 with β = , a = . and N = , and . A is the average of theprobability that the hunter catches the rabbit. N / L ( N , a ) A A / L ( N , a )100 Example 3
We put P { X t = k } =
13 ( k ∈ {− , , } )0 ( k < {− , , } ) . By Remark 2, β = c ∗ = and ε =
2. In this case, thelower bound of the probability the hunter catches the rabbit L ′ ( N ) is L ′ ( N ) = (cid:16) + π (cid:17) N / + . . (We can prove this using in the same way in Appendix (D).)Figure 3 is an experimental result of Example 3. The greenline in Figure 3 is L ′ ( N ).We could have a concrete lower bound of the average ofa probability that the hunter catches the rabbit for those ex-amples. In this section, we give a relation between P ( N ) R N [ n = n R ( N ) n = ( y n mod N ) o and one-dimensional random walk { S n } ∞ n = . Proposition 3
For N ∈ N \ { } and y , y , . . . , y N ∈ Z with | y n − y n + | ≤ n = , , . . . , N − P N − i = p ( N ) i ≤ P ( N ) R N [ n = n R ( N ) n = ( y n mod N ) o ≤ P N − i = q ( N ) i , (10)where [ y ] N = { y + kN | k ∈ Z } , p ( N ) i = i = | y |≤ i , y ∈ Z P { S i ∈ [ y ] N } ( i ∈ N )and q ( N ) i = i = | y |≤ i , y ∈ Z P { S i ∈ [ y ] N } ( i ∈ N ) . Proof.
We note that N [ n = n R ( N ) n = ( y n mod N ) o = N − [ l = N [ n = n X ( N )0 = l , l + S n ∈ [ y n ] N o = N − [ l = N [ n = ( X ( N )0 = l , l + S n ∈ [ y n ] N , l + S i < [ y i ] N , ≤ i ≤ n − ) keda etal. Page 6 of 15 Figure 1
This is an experimental result of Example 1. In this case, a = . The hunter does not move from an initial position . Figure 2
This is an experimental result of Example 2. In this case, a = . . The hunter does not move from an initial position . Figure 3
This is an experimental result of Example 3. The hunter does not move from an initial position .keda etal. Page 7 of 15 by the definition of n R ( N ) n o ∞ n = . We note P ( N ) R = µ N × P , theabove relation implies P ( N ) R N [ n = n R ( N ) n = ( y n mod N ) o = N − X l = N X n = N P ( l + S i < [ y i ] N , ≤ i ≤ n − , l + S n ∈ [ y n ] N ) . (11)For l ∈ { , , . . . , N − } and n ∈ { , , . . . , N } , we decom-pose the event { l + S n ∈ [ y n ] N } according to the value of thefirst hitting time for [ y ] N , [ y ] N , . . . , [ y n ] N and the hittingplace to obtain P { l + S n ∈ [ y n ] N } = n X j = X m ∈ Z P l + S i < [ y i ] N , ≤ i ≤ j − , l + S j = y j + mN , y j + mN + X j + + · · · + X n ∈ [ y n ] N . The probability in the double summation on the right-hand side above is equal to P ( l + S i < [ y i ] N , ≤ i ≤ j − , l + S j = y j + mN , ) × P n y j + mN + S n − j ∈ [ y n ] N o by the Markov property. It is easy to verify that for any m ∈ Z , P n y j + mN + S n − j ∈ [ y n ] N o = P n S n − j ∈ [ y n − y j ] N o ≤ p ( N ) n − j by | y n − y j | ≤ n − j . Therefore P { l + S n ∈ [ y n ] N }≤ n X j = P ( l + S i < [ y i ] N , ≤ i ≤ j − , l + S j = [ y j ] N ) p ( N ) n − j , (12)for l ∈ { , , . . . , N − } and n ∈ { , , . . . , N } . By multiply-ing (12) by 1 / N and summing ( l , n ) over { , , . . . , N − } × { , , . . . , N } , we have N − X l = N X n = N P { l + S n ∈ [ y n ] N }≤ N − X l = N X j = N P ( l + S i < [ y i ] N , ≤ i ≤ j − , l + S j = [ y j ] N ) × N − j X i = p ( N ) i ≤ P ( N ) R N [ n = n R ( N ) n = ( y n mod N ) o N − X i = p ( N ) i . (13)Here we used (11).By P N − l = P { l + S n ∈ [ y ] N } = P { S n ∈ Z } = n ∈ N , y ∈ Z ), N − X l = N X n = N P { l + S n ∈ [ y n ] N } = . (14)(13) and (14) imply1 ≤ P ( N ) R N [ n = n R ( N ) n = ( y n mod N ) o N − X i = p ( N ) i (15)that is the first inequality in (10).For the last inequality in (10), let y N + j = y N ( j = , , . . . , N ). The same argument as showing (15) (we use q ( N ) i instead of p ( N ) i ) gives2 = N − X l = N X n = N P { l + S n ∈ [ y n ] N }≥ P ( N ) R N [ n = n R ( N ) n = ( y n mod N ) o N − X i = q ( N ) i . Corollary 2
For N ∈ N \ { } ,11 + P N − i = P { S i ∈ [0] N } ≤ P ( N ) R N [ n = n R ( N ) n = o ≤ + P N − i = P { S i ∈ [0] N } . (16) Proof.
Put y = y = · · · = y N = keda etal. Page 8 of 15 Corollary 3
For N ∈ N \ { } ,11 + P N − i = P { S i ∈ [ i ] N }≤ P ( N ) R N [ n = n R ( N ) n = ( n mod N ) o ≤ + P N − i = P { S i ∈ [ i ] N } . (17) Proof.
Put y j = j ( j = , , . . . , N ) in the proof of Propo-sition 3. Then the same argument as showing (10) gives(17). Remark 4
By the same argument as showing (16), weobtain that for ˜ ǫ > N ≥ / ˜ ǫ , P ( N ) R N [ n = n R ( N ) n = o ≤ + ˜ ǫ + P ˜ ǫ Ni = P { S i ∈ [0] N } . In this section, we introduce some results concerning one-dimensional random walk.
Proposition 4
If a one-dimensional random walk satisfies( A
1) and ( A C > N ∈ N such thatfor n ≥ N ,sup l ∈ Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n /β P { S n = l } − π Z + ∞−∞ e − c ∗ | x | β exp − i xln /β ! dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C n − δ , where δ = min { ε/ (2 β ) , / } . Proof.
Proposition 4 can be proved by the same proce-dure as in Theorem 1.2.1 of [13].The Fourier inversion formula for φ n ( θ ) is n /β P { S n = l } = n /β π Z π − π φ n ( θ ) e − i θ l d θ. (18)By ( A C ∗ > r ∈ (0 , π ) such that for | θ | < r , | φ ( θ ) − (1 − c ∗ | θ | β ) | ≤ C ∗ | θ | β + ε (19)and | φ ( θ ) | ≤ − c ∗ | θ | β . (20)With r , we decompose the right-hand side of (18) to obtain n /β P { S n = l } = I ( n , l ) + J ( n , l ) , where I ( n , l ) = n /β π Z | θ | < r φ n ( θ ) e − i θ l d θ, J ( n , l ) = n /β π Z r ≤| θ |≤ π φ n ( θ ) e − i θ l d θ. A strongly aperiodic random walk ( A
1) has the propertythat | φ ( θ ) | = θ is a multiple of 2 π (see § φ ( θ ), | φ ( θ ) | is acontinuous function on the bounded closed set [ − π, − r ] ∪ [ r , π ], and | φ ( θ ) | ≤ θ ∈ [ − π, π ]). Hence, there exists a ρ <
1, depending on r ∈ (0 , π ], such thatmax r ≤| θ |≤ π | φ ( θ ) | ≤ ρ. (21)By using the above inequality, | J ( n , l ) | ≤ n /β π Z r ≤| θ |≤ π | φ ( θ ) | n d θ ≤ n /β ρ n . We perform the change of variables θ = x / n /β , so that I ( n , l ) = π Z | x | < rn /β φ n (cid:18) xn /β (cid:19) exp − i xln /β ! dx . Put γ = min ( ε β ( β + ε + , β + ) . We decompose I ( n , l ) as follows: I ( n , l ) = π Z + ∞−∞ e − c ∗ | x | β exp − i xln /β ! dx + I ( n , l ) + I ( n , l ) + I ( n , l ) , where I ( n , l ) = π Z | x |≤ n γ (cid:26) φ n (cid:18) xn /β (cid:19) − e − c ∗ | x | β (cid:27) × exp − i xln /β ! dx , I ( n , l ) = − π Z n γ < | x | e − c ∗ | x | β exp − i xln /β ! dx and I ( n , l ) = π Z n γ < | x | < rn /β φ n (cid:18) xn /β (cid:19) exp − i xln /β ! dx . keda etal. Page 9 of 15 Therefore, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n /β P { S n = l } − π Z ∞−∞ e − c ∗ | x | β exp − i xln /β ! dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | J ( n , l ) | + X k = | I k ( n , l ) | . The proof of Proposition 4 will be complete if we showthat each term in the right-hand side of the above inequalityis bounded by a constant (independent of l ) multiple of n − δ .If n is large enough, then the bound | J ( n , l ) | ≤ n /β ρ n ,which has already been shown above, yields | J ( n , l ) | ≤ n − δ . With the help of | a n − b n | = | a − b | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − X j = a n − − j b j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n | a − b | ( a , b ∈ [ − , | φ ( θ ) | ≤ θ ∈ [ − π, π ]), (19) implies that for | x | < rn /β , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ n (cid:18) xn /β (cid:19) − e − c ∗ | x | β (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ (cid:18) xn /β (cid:19) − e − c ∗ | x | β / n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ (cid:18) xn /β (cid:19) − − c ∗ | x | β n !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − c ∗ | x | β n ! − e − c ∗ | x | β / n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C ∗ | x | β + ε n − ε/β + c ∗ | x | β n − . Thus | I ( n , l ) | ≤ π Z | x |≤ n γ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ n (cid:18) xn /β (cid:19) − e − c ∗ | x | β (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d θ ≤ π C ∗ β + ε + + c ∗ β + ! n − δ . It is easy to verify that for | x | < rn /β , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ n (cid:18) xn /β (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ − c ∗ | x | β n ! n ≤ e − c ∗ | x | β / by (20), and we obtain that | I ( n , l ) | ≤ π Z n γ < | x | < rn /β (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ n (cid:18) xn /β (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dx ≤ π Z n γ < | x | e − c ∗ | x | β / dx . (23) Moreover, if n is large enough, then e − c ∗ | x | β / ≤ s c s ∗ | x | − s β ( | x | > n γ ) , where s = (1 /β )(1 + / (2 γ )) . By replacing the integrandin the right-hand side of the last inequality of (23) with theright-hand side of the above inequality, we obtain | I ( n , l ) | ≤ s + γπ c s ∗ n − / ≤ s + γπ c s ∗ n − δ . (24)The same argument as showing (24) gives | I ( n , l ) | ≤ π Z n γ ≤| θ | e − c ∗ | x | β dx ≤ s + γπ c s ∗ n − δ . Let I ( n , l : β, c ∗ ) = π Z + ∞−∞ e − c ∗ | x | β exp − i xln /β ! dx appearing in Proposition 4. Remark 5
When a one-dimensional random walk is thestrongly aperiodic ( A
1) with E [ X ] = E [ | X | + ε ] < ∞ for some ε ∈ (0 , φ ( θ ) = − E [ X ]2 | θ | + O ( | θ | + ε ) . In this case, I ( n , l : 2 , E [ X ] /
2) can be computed andProposition 4 gives the following. (Local Central Limit Theorem)
There exist ˜ C > N ∈ N such that for n ≥ ˜ N ,sup l ∈ Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n / P { S n = l } − q E [ X ] π exp − l E [ X ] n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ˜ C n − δ , (25)where δ = min { ε/ , / } . (See Remark after Proposition7.9 in [18].)It is easy to see I ( n , l : 1 , c ∗ ) = π c ∗ c ∗ + ( l / n ) ( n ∈ N , l ∈ Z , c ∗ > keda etal. Page 10 of 15 Corollary 4
If a one-dimensional random walk satisfies( A
1) and ( A
3) with β =
1, then there exist C > N ∈ N such that for n ≥ N ,sup l ∈ Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) nP { S n = l } − π c ∗ c ∗ + ( l / n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C n − δ , where δ = min { ε/ , / } . We perform the change of variables t = c ∗ x β , so that I ( n , β, c ∗ ) = π Z + ∞ e − c ∗ x β dx = β c /β ∗ π Γ β ! . With the help of the above calculation, Proposition 4 givesthe following corollary.
Corollary 5
If a one-dimensional random walk satisfies( A
1) and ( A C > N ∈ N such thatfor n ≥ N , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n /β P { S n = } − β c /β ∗ π Γ β !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C n − δ , where δ = min { ε/ β, / } . Proposition 5
If a one-dimensional random walk satisfies( A l ∈ Z and n ∈ { } ∪ N , P { S n ∈ [ l ] N } = N + N X ≤ j ≤ ( N − / φ n j π N ! cos j π N l ! + J N ( n , l ) , (26)where J N ( n , l ) = ( (1 / N ) φ n ( π ) cos( π l ) ( if N is even )0 ( if N is odd ) . Proof.
By the definition of φ ( θ ), φ n ( θ ) = X k ∈ Z e i θ k P { S n = k } . Thus φ n j π N ! = X k ∈ Z e i j π k / N P { S n = k } = N − X ˜ l = X m ∈ Z e i j π (˜ l + mN ) / N P n S n = ˜ l + mN o = N − X ˜ l = e i j π ˜ l / N P n S n ∈ [˜ l ] N o . Then, N − X j = e − i j π l / N φ n j π N ! = N − X ˜ l = N − X j = e i j π (˜ l − l ) / N P n S n ∈ [˜ l ] N o = NP { S n ∈ [ l ] N } since N − X j = e i j π (˜ l − l ) / N = ( N ˜ l = l l , l . Therefore, P { S n ∈ [ l ] N } = N N − X j = φ n j π N ! e − j π il / N = N N − X j = φ n j π N ! cos j π lN ! . We note that φ n ( θ ) ∈ R and1 N N − X j = φ n j π N ! cos j π lN ! ∈ R by ( A φ n m π N ! cos m π lN ! = φ n N − m ) π N ! cos N − m ) π lN ! . (27)Let N be an even number. Then, by (27), P { S n ∈ [ l ] N } = N φ n (0) cos (0) + N X ≤ j ≤ ( N − / φ n j π N ! cos j π lN ! + N φ n ( π ) cos ( π l ) = N + N X ≤ j ≤ ( N − / φ n j π N ! cos j π lN ! + N φ n ( π ) cos ( π l ) . Therefore, we have (26) for every even number N . Theproof of (26) for odd number is similar and is omitted. In this section we prove Theorem 1. To prove it, we intro-duce the following Proposition. keda etal. Page 11 of 15
Proposition 6
Assume ( A A
2) and ( A β ∈ (0 , c > N − X i = p ( N ) i ≤ c . (28)If β =
1, then there exists a constant c > N − X i = p ( N ) i ≤ c ∗ π log N + c . (29)If β ∈ (1 , c > N − X i = p ( N ) i ≤ c N ( β − /β . (30) Proof.
There exist C ∗ and r ∈ (0 , π ) such that for | θ | < r , | φ ( θ ) − (1 − c ∗ | θ | β ) | ≤ C ∗ | θ | β + ε (31)by ( A r ∗ ∈ (0 , r ] small enough so that C ∗ r ε ∗ ≤ c ∗ and c ∗ r β ∗ ≤ . (32)Then for | θ | < r ∗ ,12 c ∗ | θ | β ≤ | − φ ( θ ) | (33)and | − φ ( θ ) | ≤ c ∗ | θ | β ≤ . (34)There exists a ρ ∗ ∈ [0 , r ∗ , such thatmax r ∗ ≤| θ |≤ π | φ ( θ ) | ≤ ρ ∗ (35)by the same reason as (21). (Here we used the condition( A i ∈{ , , . . . , N − } , p ( N ) i = max | l |≤ i P { S i ∈ [ l ] N }≤ N + X ≤ j ≤ ( N − / N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ j π N !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) i + | J N ( i , |≤ N + X ≤ j < ( r ∗ / (2 π )) N N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ j π N !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) i + ρ i ∗ . Therefore N − X i = p ( N ) i ≤ + Φ N + − ρ ∗ , (36) where Φ N = X ≤ j < ( r ∗ / (2 π )) N N − (cid:12)(cid:12)(cid:12)(cid:12) φ (cid:16) j π N (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) N − (cid:12)(cid:12)(cid:12)(cid:12) φ (cid:16) j π N (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) . Because of ( A φ ( θ ) takes a real number. Then (33), (34)and ( A
1) mean that12 < φ ( θ ) = | φ ( θ ) | < θ ∈ ( − r ∗ , ∪ (0 , r ∗ )) (37)and Φ N ≤ X ≤ j < ( r ∗ / (2 π )) N N − φ (cid:16) j π N (cid:17) . (38)We will calculate Φ N in the case β ∈ (0 , X ≤ j < ( r ∗ / (2 π )) N N − φ (cid:16) j π N (cid:17) = ˜ Φ N + E N , (39)where˜ Φ N = − β π β c ∗ N β − X ≤ j < ( r ∗ / (2 π )) N j − β , E N = X ≤ j < ( r ∗ / (2 π )) N N − φ (cid:16) j π N (cid:17) − c ∗ (cid:16) j π N (cid:17) β . To estimate E N , we use (31) and (33) which imply thatfor j ∈ [1 , ( r ∗ / (2 π )) N ) ∩ Z ,2 N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − φ (cid:16) j π N (cid:17) − c ∗ (cid:16) j π N (cid:17) β (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = N (cid:12)(cid:12)(cid:12)(cid:12) − φ (cid:16) j π N (cid:17) − c ∗ (cid:16) j π N (cid:17) β (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − φ (cid:16) j π N (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) · (cid:12)(cid:12)(cid:12)(cid:12) c ∗ (cid:16) j π N (cid:17) β (cid:12)(cid:12)(cid:12)(cid:12) ≤ c N β − ε − j ε − β , where c = + ε − β π ε − β C ∗ / c ∗ . By noticing that 1 + ε − β > X ≤ j < ( r ∗ / (2 π )) N j ε − β ≤ Z N x ε − β dx = N + ε − β + ε − β . Thus | E N | ≤ c / (1 + ε − β ) . (40) keda etal. Page 12 of 15 It is easy to see that˜ Φ N ≤ − β π β c ∗ N β − + Z N x − β dx ! ≤ − β π β c ∗ (1 − β ) ( β ∈ (0 , π c ∗ log N + π c ∗ ( β = . (41)Put the pieces ((36), (38)-(41)) together, we have (28) and(29).In the case β ∈ (1 , Φ N ≤ Φ (1) N + Φ (2) N , (42)where N ( β ) = min { N ( β − /β , ( r ∗ / (2 π )) N } and Φ (1) N = X ≤ j < N ( β ) N (cid:12)(cid:12)(cid:12)(cid:12) − φ (cid:16) j π N (cid:17) N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − φ (cid:16) j π N (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) , Φ (2) N = X N ( β ) ≤ j < ( r ∗ / (2 π )) N N (cid:12)(cid:12)(cid:12)(cid:12) − φ (cid:16) j π N (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) . We use (22)(set n = N and a = , b = φ (cid:16) j π N (cid:17) ), then Φ (1) N ≤ N ( β ) ≤ N ( β − /β . (43)We notice that β − >
0, (33) gives Φ (2) N ≤ − β c ∗ π β N β − X N ( β ) ≤ j < ( r ∗ / (2 π )) N j − β ≤ − β c ∗ π β N β − N − β + + Z + ∞ N ( β − /β x − β dx ! ≤ − β c ∗ π β + β − ! N ( β − /β . (44)Put the pieces ((36), (42)-(44)) together, we have (30).It remains to show the last inequality in (2). To achievethis, we will use Proposition 3 and Corollary 4.There exist C > N ∈ N such that for i ≥ N and l ∈ Z , P { S i = l } ≥ π c ∗ c ∗ + ( l / i ) i − C i − − δ by Corollary 4. Let c : = π c ∗ c ∗ + N + C ∞ X i = N i − − δ . We can choose N ∗ ∈ N large enough so that12 1 π c ∗ c ∗ + N ∗ ≥ c . Then for N ≥ N ∗ + N − X i = q ( N ) i ≥ N − X i = N min | l |≤ i P { S i = l }≥ π c ∗ c ∗ + N − X i = N i − C ∞ X i = N i − − δ ≥ π c ∗ c ∗ + N − c ≥
12 1 π c ∗ c ∗ + N . (45)It follows from Proposition 3 and (45) that for N ∈ [ N ∗ + , + ∞ ) ∩ N and y , y , . . . , y N ∈ Z with | y n − y n + | ≤ n = , , . . . , N − P ( N ) R N [ n = {R ( N ) n = ( y n mod N ) } ≤ π ( c ∗ + c ∗ log N . It is clear that P ( N ) R (cid:16)S Nn = {R ( N ) n = ( y n mod N ) } (cid:17) is boundedby 1. Put c = max { π ( c ∗ + / c ∗ , log N ∗ } . The last inequal-ity in (2) holds.The proof of Theorem 1 is complete. We formalized the Hunter vs Rabbit game using the ran-dom walk framework. We generalize a probability distri-bution of the rabbit’s strategy using four assumptions. Wehave the general lower bound formula of a probability thatthe rabbit is caught. Let P { X = k } = O ( k − β − ). If β ∈ (0 , c where c > β =
1, the lowerbound of a probability that the rabbit is caught is c ∗ π log N + c where c and c ∗ are constants defined by the given strategy.If β ∈ (1 , c N ( β − /β where c > N →∞ c ∗ π log N ! P ( N ) R N [ n = {R ( N ) n = } = . In this paper, we consider the lower bound of a proba-bility that the rabbit is caught to show the worst expectedvalue of time until the rabbit caught. Our motivation is to keda etal. Page 13 of 15 find the best strategy of the rabbit. Our results help to findthe best strategy of the rabbit. On the other hands, what isthe best strategy of the hunter? And what is the worst strat-egy of the hunter? Future works include to show the beststrategy of the hunter is Y j + = Y j +
1, and the worst strat-egy of the hunter is Y j = H ( N )0 for any j . I would like to express my deepest gratitude to ProfessorHiroyuki Ochiai for his valuable advice and guidance. Iwould like to thank Mr. Norikazu Ishii for his help.
Author details Graduate School of Mathematics, Kyushu University,. Faculty of Mathematics,Kyushu University,. Institute of Mathematics for Industry, Kyushu University,.
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Appendix (A)
Proof of Proposition 1.
The first inequality in (6)comes from (3) in Theorem 1. To prove the last inequalityin (6), we will use Corollary 2 and 5 instead of Proposi-tion 3 and Corollary 4. The same argument as showing thelast inequality in (3) gives the last inequality in (6). (cid:3)
Proof of Proposition 2.
We consider the case when X takes three values − , , X satisfies ( A A
2) and φ ( θ ) = − | θ | + O ( | θ | ) . We can show that there exist ˜ C > N ∈ N such thatfor i ≥ ˜ N and l ∈ Z , P { S i = l } ≤ √ √ π i / exp − l i ! + ˜ C i − (46)by (25). We notice that P {| X | ≤ } =
1, then we obtain thatfor N ∈ N \ { } ,1 + N − X i = P { S i ∈ [ i ] N } = + N − X i = P { S i = i } + X N / ≤ i ≤ N − P { S i = i − N } and N − X i = P { S i = i } = N − X i = ! i ≤ . With the help of e − x ≤ / x ( x > N ≥ N , X N / ≤ k ≤ N − P { S k = k − N }≤ X N / ≤ k ≤ N − √ √ π k / exp − k − N ) k ! + ˜ C k − ≤ r π N / X ≤ k ≤ N / exp − k N ! + ˜ C X ≤ k ≤ N / N ≤ r π N / X ≤ k ≤ N / + X N / < k N k + C ≤ r π + √ √ π N / N + Z + ∞ N / x dx ! + C ≤ c , keda etal. Page 14 of 15 where c = √ / (2 π ) + √ / √ π + C . Thus for N ∈ N \ { } ,1 + N − X i = P { S i ∈ [ i ] N } ≤ max { N , (3 / + c } . Combining the above inequality with Corollary 3, we have(7). (cid:3) (B) To obtain (5), we use the formula Z + ∞ sin bxx α dx = π b α − Γ ( α ) sin( απ/
2) (47)for α ∈ (0 ,
2) and b >
0. By the definition of X ,1 − φ ( θ ) = a ∞ X k = (1 − cos | θ | k ) 1 k β + . A simple calculation shows that the absolute value of thedi ff erence between the right-hand side of the above and1 a Z + ∞ − cos | θ | xx β + dx is bounded by a constant multiple of | θ | β + (2 − β ) / . It remainsto show that1 a Z + ∞ − cos | θ | xx β + dx = π a | θ | β Γ ( β +
1) sin( βπ/ . (48)We perform integration by part for the left-hand side of (48)and use (47). Then we have (48) and (5).(C) Proof of (8).
Let ǫ > C > N ∈ N such that for i ≥ N , P { S i = } ≥ c ∗ π i − C i − − δ . (49)(49) implies that for N ≥ (4 /ǫ )( N + + X ≤ i ≤ ( ǫ/ N P { S i ∈ [0] N } ≥ X N ≤ i ≤ ( ǫ/ N P { S i = }≥ X N ≤ i ≤ ( ǫ/ N c ∗ π i − C i − − δ ! ≥ c ∗ π Z ( ǫ/ NN x dx − C N + δ + Z + ∞ N x − − δ dx = c ∗ π log N + c ∗ π log ǫ − c , (50)where c = (1 / ( c ∗ π )) log 4 + (1 / ( c ∗ π )) log N + C { / N + δ + / ( δ N δ ) } . We can choose N ∈ N which satisfiesmin ( , ǫ ) c ∗ π log N ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − c ∗ π log ǫ + c (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (51)and ǫ c ∗ π log N ≥ c , (52)where c is the same constant in (2).Combining Remark 5 with (50) and using the left-handside of (2), we obtain that for N ≥ max { N , (4 /ǫ )( N + } ,1 c ∗ π log N + c ≤ P ( N ) R N [ n = {R ( N ) n = } ≤ + ( ǫ/ c ∗ π log N + c ∗ π log ǫ − c . Hence for N ≥ max { N , (4 /ǫ )( N + } , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c ∗ π log N ! P ( N ) R N [ n = {R ( N ) n = } − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ E (1) N + E (2) N , where E (1) N = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c ∗ π log N c ∗ π log N + c − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) and E (2) N = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1 + ( ǫ/ c ∗ π log N c ∗ π log N + c ∗ π log ǫ − c − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . The proof is complete if we show that for N ≥ max { N , (4 /ǫ )( N + } , E (1) N + E (2) N ≤ ǫ. (53)We use (52), then E (1) N ≤ c c ∗ π log N ≤ ǫ N ≥ max { N , (4 /ǫ )( N + } . We can show that E (2) N ≤ ( ǫ/ c ∗ π log N + (cid:12)(cid:12)(cid:12)(cid:12) − c ∗ π log ǫ + c (cid:12)(cid:12)(cid:12)(cid:12) c ∗ π log N − (cid:12)(cid:12)(cid:12)(cid:12) − c ∗ π log ǫ + c (cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ + (cid:12)(cid:12)(cid:12)(cid:12) − c ∗ π log ǫ + c (cid:12)(cid:12)(cid:12)(cid:12) (1 / c ∗ π log N ≤ ǫ keda etal. Page 15 of 15 for N ≥ max { N , (4 /ǫ )( N + } by (51). The above twoinequalities yield (53). (cid:3) (D) Proof of (9).
We show the lower bound of Example1. In this case, a = β = c ∗ = π a and ε = . We have | E N | = c by (40). We note c = + ε − β π ε − β C ∗ c ∗ = / π − / C ∗ . We can choose C ∗ = .