Local Behaviour of First Passage Probabilities - MIMS - The ...

Local Behaviour of First Passage ProbabilitiesR. A. DoneyFirst version: 12 June 2010Research Report No. 1, 2010, Probability and Statistics GroupSchool of Mathematics, The University of Manchester

2 ResultsNotation In what follows the phrase "S is an a.s.r.w.", (asymptotically stablerandom walk) will have the following meaning. S = (S n ; n 0) is a 1-dimensional random walk with S 0 = 0 and S n =P n1 X r for n 1 where X 1 ; X 2 ; are i.i.d. with F (x) = P (X 1 x) : S is either non-lattice, or it takes values on the integers and is aperiodic: there is a monotone increasing continuous function c(t) such that theprocess de…ned by X (n)t = S [nt] =c n converges weakly as n ! 1 to astable process Y = (Y t ; t 0): the process Y has index 2 (0; 2]; and := P (Y 1 > 0) 2 (0; 1):Remark 1 The case = 1; 2 (1; 2] is the spectrally negative case, and wewill sometimes need to treat this case separately. If < 1 then < 2 and theLévy measure of Y has a density equal to c + x 1 on (0; 1) with c + > 0;and then we can also assume that the norming sequence satis…esBut if = 1 we will havenF (c n ) ! 1 as n ! 1: (7)nF (c n ) ! 0 as n ! 1: (8)Here are our main results, where we recall that h y () denotes the densityfunction of the passage time over level y > 0 of the process Y: We will alsoadopt the convention that both x and are restricted to the integers in thelattice case.Theorem 2 Assume that S is an asrw. Then(A) uniformly for x such that x=c n ! 0;P (T x = n) v U(x)P (T = n) v U(x)n 1 L(n) as n ! 1 : (9)(B) uniformly in x n := x=c(n) 2 [D 1 ; D]; for any D > 1;If, in addition, < 1; andP (T x = n) v n 1 h xn (1) as n ! 1: (10)f x := P (S 1 2 [x; x + )) is regularly varying as x ! 1; (11)then(C) uniformly for x such that x=c n ! 1;P (T x = n) v F (x) as n ! 1: (12)3

From this we get immediately a strengthening of (2).Corollary 3 If S is an a.s.r.w. the estimateholds uniformly as x=c n ! 0:P (T x > n) v U(x)n L(n) as n ! 1Remark 4 In view of (3) the result (10) might seem obvious. However (3)could also be written as P (max rn S r x) v R x nm(y)dy; where m denotes the0density function of sup t1 Y s ; and in the recent paper [22], Wachtel has shownthat the obvious local version of this is only valid under an additional hypothesis.Remark 5 The asymptotic behaviour of h x (1) has been determined in [16], andis given byh x (1) v k 1 x as x # 0; h x (1) v k 2 x as x ! 1:(We mention here that k 1; k 2 ; will denote particular …xed positive constantswhereas C will denote a generic positive constant whose value can change fromline to line.) It is therefore possible to compare the exact results in (9) and(12) with the behaviour of n 1 h xn (1) when x=c n ! 0 or x=c n ! 1: It turnsout that the ratio of the two can tend to 0; or a …nite constant, or oscillate,depending on the s.v. functions involved and the exact behaviour of x; except inthe special case that c n v Cn : (Here, and throughout, we write 1= = :) Infact, if F (x) v C=(x L 0 (x)); one can check that, when x=c n ! 1;nF (x)h xn (1) v L 0(c n )L 0 (x) ;and of course L 0 is asymptotically constant only in the aforementioned specialcase. Similarly, the RHS of (9) only has the same asymptotic behaviour asn 1 h xn (1) in this same special case.Remark 6 In the spectrally negative case = 1; without further assumptionswe know little about the asymptotic behaviour of F ; so it is clear that (C) doesn’tgenerally hold in this case, and indeed it is somewhat surprising that parts (A)and (B) do hold.3 PreliminariesThroughout this section it will be assumed that S is an a.s.r.w.. With ( 0 ; H 0 ) :=(0; 0) we write ( ; H) = (( n ; H n ); n 0) for the bivariate renewal process ofstrict ladder times and heights, so that 1 = T and H 1 = S T is the …rst ladderheight; we also write and H for 1 and H 1 . It is known that there aresequences a n and b n such that ( n =a n ; H n =b n ) converges in distribution to abivariate law whose marginals are positive stable laws with parameters and4

(ii) Since the probability of the limiting conditioning event is positive, thisfollows from the weak convergence of X (n) to Y:Until further notice we assume we are in the lattice case, and x; y; z willbe assumed to take non-negative integer values only.We will write1X1Xg(m; y) = P (T n = m; H n = y) and g(y) = P (H n = y)n=0for the bivariate renewal mass function of ( ; H) and the renewal mass functionof H respectively: Our proofs are based on the following obvious representation:P (T x = n + 1) = X y0P (T x > n; S n = x y)F (y); x 0; n > 0; (16)n=0To exploit this we need good estimates of P (T x > n; S n > x y); and we derivethese from the formulaXy^xnXP (S n = x y; T x > n) = g(r; x z)g (n r; y z); (17)z=0 r=0where g denotes the bivariate mass function in the weak downgoing ladderprocess of S: Formula (17), which extends a result originally due to Spitzer,follows by decomposing the event on the LHS according to the time and positionof the maximum, and using the well-known duality result:g (m; u) = P (S m = u; > m): (18)(See Lemma 2.1 in [4]). Of course we also haveg(m; u) = P (S m = u; > m); (19)and our main tool in estimating P (S n = x y; T x > n) will be the following estimatesfor g and g : The results for g are established in [21], where they are statedas estimates for the conditional probability P (S m = uj > m): (See Theorems5 and 6 in [21].) The results for g can be derived by applying those resultsto S; and then using the calculation given on page 100 of [3] to deduce theresult for the weak ladder process. Recall that V (x) = P 1 P xm=0 u=0g (m; u),and write f for the density of Y 1 ; where Y is the limiting stable process. Alsop and ~p stand for the densities of Z (1)1 and Z ~ (1)1 ; the stable meanders of length1 at time 1 corresponding to Y and Y:Lemma 10 Uniformly in x 1 and x 0; respectively,c n g(n; x)P ( > n) = p(x=c n)+o(1) and c ng (n; x)P ( > n) = ~p(x=c n)+o(1) as n ! 1: (20)Also, uniformly as x=c n ! 0;g(n; x) vf(0)U(x 1)nc nfor x 1 and g (n; x) vf(0)V (x)nc nfor x 0: (21)6

From this we can deduce the following result, which is a minor extension ofLemma 20 in [21].Lemma 11 Given any constant C 1 there exists a constant C 2 such that for alln 1 and 0 x C 1 c ng(n; x) C 2U(x); and g (n; x) C 2V (x):nc nnc n(22).Proof. Observe that on any interval [c n ; C 1 c n ]; p(x=c n ) is bounded, andU(x) U(c n ); so by Lemma 7 we see that the ratioP ( > n)= U(x) nP ( > n)U(c n )c n nc nis also bounded above. A similar proof works for g :Just as these local estimates for the distribution of S n on the event > nplayed a crucial rôle in the proof of (5) in [21], we need similar information onthe event T x > n: This is given in the following result, where for x > 0 we writeq x () for the density of P (Y 1 2 x : sup t1 Y t < x): We also write x=c n = x nand y=c n = y n :Proposition 12 (i) Uniformly as x n _ y n ! 0P (S n = xy; T x > n) vU(x)f(0)V (y)nc n: (23)(ii) For any D > 1; uniformly for y n 2 [D 1 ; D];P (S n = x y; T x > n) v U(x)P ( > n)~p(y n)c nas n ! 1 and x n ! 0; (24)and uniformly for x n 2 [D 1 ; D];P (S n = x y; T x > n) v V (y)P ( > n)p(x n)c nas n ! 1 and y n ! 0: (25)(iii) For any D > 1; uniformly for x n 2 [D 1 ; D] and y n 2 [D 1 ; D];P (S n = x y; T x > n) v q x n(y n )c nas n ! 1: (26)The proof of this result is given in the next section. We can repeat theargument used in Lemma 11 to get the following corollary.Corollary 13 Given any constant C 1 there exists a constant C 2 such that forall n 1 and 0 x C 1 c n ; 0 y C 1 c n ;P (S n = xy; T x > n) C 2U(x)f(0)V (y)nc n:7

It is apparent that we will also need information about the behaviour ofg(n; x); or equivalently of P (S n = x; > n); in the case x=c n ! 1. Fortunatelythis has been obtained recently in [19], and we quote Propositions 11and 12 therein as (28) and (30). The related unconditional results (27) and (29)have been proved in special cases in [12], [13], and [19], and the general resultscan be deduced from Theorem 2.1 of [9].Proposition 14 If S is an asrw with < 1; then, uniformly for x such thatx=c n ! 1;P (S n > x) v nP (S 1 > x) as n ! 1; and (27)P (S n > x; > n) v 1 P (S n > x)P ( > n) as n ! 1: (28)If, additionally, (11) holds, thenP (S n 2 [x; x + )) v nf x as n ! 1 (29)andP (S n 2 [x; x + ); > n) v 1 nf x P ( > n) as n ! 1: (30)3.1 Some identities for stable processes.Proposition 15 (i) For any stable process Y which has < 1 there are positiveconstants k 7 and k 8 such that the following identities hold:Z 1h x (t) = k 7 q x (w)w dw and (31)Z1q u (v) = k 800u^v Z0u(t; u z)u (1 t; v z)dzdt; (32)where u and u denote the bivariate renewal densities for the increasing ladderprocesses of Y and Y:(ii) If = 1 there is a positive constant k 9 such thatp(x) = k 9 h x (1): (33)Proof. All three results are special cases of results for Lévy processes. Thegeneral version of (31) is given in [15], and (32) follows from the followingobservation, which is a minor extension of Theorem 20, p176 of [5].Assume X is a Lévy process which is not compound Poisson. Then there isa constant k 7 > 0 such that for x > 0 and w < x;P (X t 2 dw; t < T x )dt = k 7Z xy=w + Z ts=0U(ds; dy)U (dt s; y dw); (34)where U and U are the bivariate renewal measures in the increasing ladderprocesses of X and X : Clearly it su¢ ces to prove that the Laplace transform,8

in t; of the LHS of (34) is the same as that of the RHS, which isk 7Z x= k 7Z xZ 1y=w + t=0Z 1y=w + t=0e qt Z ts=0e qt U(dt; dy)U(ds; dy)U (dt s; y dw)Z 1s=0e qs U (ds; y dw): (35)Note that if we introduce an independent Exp(q) random variable e q the Wiener-Hopf factorisation allows us to write X eq = S eqe Seq ; where S denotes thesupremum process of X and S ~ denotes an independent copy of the supremumprocess S of X: Let and denote the bivariate Laplace exponents of theladder processes of X and X :Then, using the identity (q; 0) (q; 0) = q=k 7which follows from the Wiener-Hopf factorisation, (see e.g. (3), p166 of [5]),Z 1t=0But (1), p 163 of [5] givessoe qt P (X t 2 dw; t < T x )dtE(e Seq )(q; 0)== q 1 P (X eq 2 dw; e q < T x )= q 1 P (S eq x; S eqe Seq 2 dw)Z x= q 1 P (S eq 2 dy)P (S eq2 y dw)wZ + xP (S eq 2 dy) P (S eq2 y dw)= k 7 :w (q; 0) (q; 0)+Z11(q; ) =P (S eq 2 dy)=(q; 0)0Z 10Z 10e (qt+y) U(dt; dy);e qt U(dt; dy):Using the analogous expression for P (S eq2 y dw); (35) is immediate, and then(34) follows. Specializing this to the stable case then gives (32). Finally, if wewrite n for the characteristic measure of the excursions away from zero of X I;with I denoting the in…mum process of X, then p t (dx) := n(" t 2 dx)=n( > t)is a probabilty measure which coincides with that of the meander of length t attime t in the stable case. (Here denotes the life length of the generic excursion":) In the special case of spectrally negative Lévy processes, we havep t (dx) = Cxt 1 P (X t 2 dx)=n( > t)= Ch x (t)dx=n( > t):The …rst equality here comes from Cor 4 in [2], and the second is the Lévyversion of the ballot theorem (see Corollary 3, p 190 of [5]). Specialising to thestable case and t = 1 gives (33). (I owe this observation to Loic Chaumont.)9

4 Proof of Proposition 12Proof. We will be applying the results in Lemma 10 to formula (17) and wewrite the RHS of (17) as P 1 + P 2 + P 3 ; where with 2 (0; 1=2);andP i = X Xg(r; x z)g (n r; y z); 1 i 3;y^xr2A i z=0A 1 = f0 r ng; A 2 = fn < r (1 )ng; and A 3 = f(1 )n < r ng:(i) We introduce d(); an increasing and continuous function which satis…esd(n) = nc n ; and with bc standing for the integer part function, use Lemma 10to writeXy^xP 1 f(0)=bncXz=0 r=0V (y z)g(r; x z)d(n r)y^xbncf(0) X XV (y z) g(r; x z)d(n(1 ))z=0r=0f(0) Xy^xV (y z)[u(x z)d(n(1 ))z=01Xr=bnc+1g(r; xz)]:We can apply Lemma 10 again to get the estimate, uniform for w=c n ! 0;1Xr=bnc+1Since we know thatg(r; w) 1Xr=bnc+1nu(n)lim infn!1 U(n) > 0;f(0)U(w) Cf(0) U(w):rc r(see e.g. Theorem 8.7.4 in [6]) we see that this is o(u(w)). Also we haveXy^xbncXz=0 r=0V (y z)g(r; x z)d(n r) 1d(n)Xy^xbncXz=0 r=0c ng(r; x z)V (y z);so we see thatd(n)P 1limn; f(0) P y^x= 1; (36)V (y z)u(x z)z=010

where lim n; () = 1 is shorthand for lim !0 lim sup n!1 () = lim !0 lim inf n!1 () =1: Similarly, once we observe thatP 3 ==Xy^xnXz=0 r=n bncXy^xbncXz=0 r=0g(r; x z)g (n r; y z)g(n r; x z)g (r; y z);an entirely analogous argument givesd(n)P 3limn; f(0) P y^x= 1; (37)U(x z 1)v(y z)z=0where v is the renewal mass function in the down-going ladder height process.Noting thatU(x z 1)v(y z)+V (y z)u(x z) = U(x z)V (y z) U(x z 1)V (y z 1);we get the formulaXy^xfU(x z 1)v(y z) + V (y z)u(x z)g = V (y)U(x);z=0and the result will follow by letting n ! 1 and then # 0 provided d(n)P 2 =o(V (y)U(x)) for each …xed > 0: In fact, using Lemma 10 again,d(n)P 2 = d(n) X Xy^xg(r; x z)g (n r; y z)A 2z=0Xy^xXf(0) 2 V (y z)U(x z)d(n)d(r)d(n r)z=0 A 2(y ^ x)f(0)2 d(n)nV (y)U(x)d(bnc) 2= O((y ^ x)V (y)U(x)) = o(V (y)U(x));c nand the result follows.(ii) In this case we can assume WLOG that y ^ x = x = o(y); so thatLemma 10 gives g (n r; y z) v P ( > n r)~p(y=c n r )=c n r uniformly forr 2 A 1 and 0 z x: With e denoting a continuous and monotone interpolantof c m =P ( > m), and noting that ~p() is uniformly continuous and boundedaway from zero and in…nity on [D 1 ; D]; see [16], we can use a similar argument11

to that in (i) to show thatP 1 =bncX0xXz=0~p (y)e(n(1 ))~p(y=c n r )g(r; x z)(1 + o(1))e(n r)!xXu(xz=0= U(x)~p (y)(1 + o(1));e(n(1 ))z)(1 + o(1))where ~p (y) = sup 0rn ~p(y=c n r ) = ~p(y n ) + "(n; ) and lim n; "(n; ) =0: In the same way we get P 1 U(x)~p (y)=e(n)(1 + o(1)), where ~p (y) =inf 0rn ~p(y=c n r ); and we deduce thate(n)P 1lim= 1: (38)n; ~p(y n )U(x)Since inff~p(y) : y 2 [D 1 ; D] > 0; the result will follow if we can show that forany …xed > 0e(n)(P 2 + P 3 )lim sup= 0: (39)n!1 U(x)However (37) still holds, but note now thatXy^xU(x 1 z)v(y z) =z=0xXU(x 1 z)v(y z)z=0 U(x)(V (y) V (y x 1)) = o(U(x)V (y)):and since the analogue of (14) holds, viz V (c n ) v k 10 =P ( > n); we see that e(n)P 3 U(x)V (y) e(n)= oU(x)d(n)U(x)1= o(nP ( > n)P ( > n) ) = o(1):Finallye(n)P 2 = e(n) X xXg(r; x z)g (n r; y z)A 2z=0 f(0)e(n) X A 2xXz=0U(x z)~p((y z)=c n r )d(r)e(n r)Ce(n)nxU(x)d(bnc)e(bnc) = O(xU(x) ) = o(U(x)):Thus (39) is established, and the result (24) follows. Since (25) is (24) for Swith x and y interchanged, modi…ed to take account of the di¤erence betweenstrict and weak ladder epochs, we omit it’s proof.c n;12

(iii) In this case it is P 2 that dominates. To see this, note that if we denoteby b() a continuous interpolant of c n =P ( > n); we have b(n)e(n) v nc 2 n:Then using Lemma 10 twice gives, uniformly for x n ; y n 2 (D 1 ; D) and for any…xed 2 (0; 1=2);X Xx^yc n P 2 = c nA 2z=0X Xx^y= c nA 2z=0p((x z)=c r )~p((y z)=c n r )b(r)e(n r)p((x z)=c r )~p((y z)=c n r )b(r)e(n r)+ o(c n (x ^ y) X A 21b(r)e(n r) )+ o(1):Making the change of variables r = nt; y = c n z; recalling that p and ~p areuniformly continuous on compacts, and thatb(r)e(n r)b(n)e(n)! t 1+ (1 t) 1 uniformly on A 2 ; we see that for each …xed > 0 we get the uniform estimatec n P 2 = I (x n ; y n ) + o(1); whereI (u; v) =1Zu^v Z0p( u zt )~p( v z(1 t) )t 1+ (1 t) dtdz:If we introduce p t (z) = t p(zt ); which is the density function of Z t , themeander of length t at time t; according to Lemma 8 of [16] we have that therenewal measure of the increasing ladder process of Y has a joint density which isgiven by u(t; z) = Ct 1 p t (z) = Ct + 1 p(zt ): Similarly for the decreasingladder process we have u (t; z) = C t + ~p(zt ); and (32) in Proposition 15givesZ 1I 0 (u; v) = Cso we conclude that0u^v ZTurning to P 1 ; if K = sup y0 ~p(y); we havebncXc n P 1 v c n00Kc ne(n(1Xx^yz=0u(t; u z)u (1 t; v z)dzdt = Cq u (v);lim lim c n P 2= C: (40)!0 n!1 q xn (y n )))~p((y z)=c n r )g(x z; r)e(n r)bncX0xXg(xz=0z; r) CP ( > n) (bnc);where is the renewal function in the increasing ladder time process. SinceT 2 D(; 1); we know that (bnc) v (n) and P ( > n) (n) ! C; (see e.g.13

p 361 of [6]), so we conclude thatlim lim sup c n P 1 = 0:!0 n!1Exactly the same argument applies to P 3 , and since q u (v) is clearly boundedbelow by a positive constant for u; v 2 [D 1 ; D] we have shown that (26) holds,except that the RHS is multiplied by some constant C: However if C 6= 1, bysumming over y we easily get a contradiction, and this …nishes the proof.5 Proof of Theorem 25.1 Proof when x=c n ! 0:Proof. As already indicated, the proof involves applying the estimates in Proposition12 to the representation (16), which we recall isP (T x = n + 1) = X y0P (S n = x y; T x > n)F (y); x 0; n > 0: (41)In the case < 1, given " > 0 we can …nd K " and n " such that nF (K " c n ) "for n n " and, using (23) from Proposition 12,P (S n = xy; T x > n) 2U(x)f(0)V (y)nc nfor all x_y "c n and n n " : (42)We can then use (24) of Proposition 12 to show that we can also assume, increasingthe value of n " if necessary, that for x "c n and y 2 ("c n ; K " c n );1 " c nP (S n = x y; T x > n)U(x)P ( > n)~p(y n ) 1 + " for all n n " : (43)For …xed " it is clear that as n ! 1XZ K"Z K"~p(y n )F (y)=c n v ~p(z)F (c n z)dz v n 1 ~p(z)z dz: (44)"c""n n)F (y) = 0;"#0 n!1 U(x)P ( = n)y2[0;"c n][[K "c n;1)(45)it will follow from (43) that P (T x = n) v U(x)k 11 1 P ( = n): Since this holdsin particular for x = 0; we see that k 11 = ; so it remains only to verify (45).Note …rst thatXP (S n = x y; T x > n)F (y) F (K " c n )P (T x > n)y2[K "c n;1)"n 1 P (T x > n);14

so one part of (45) will follow if we can show thatObviously, for any B > 0P (T x > n) CU(x)P ( > n): (46)P (T x > n) = X y0P (S n = x y; T x > n)X0yBc nP (S n = x y; T x > n) + P (S n < x Bc n ; T x > n):Since x=c n ! 0; we can use the invariance principle in Proposition 9 to …x Blarge enough to ensure that P (S n < x Bc n jT x > n) 1=2 for all su¢ cientlylarge n: We can also use Corollary 13 to getXP (S n = x y; T x > n) CU(x) XV (y)nc n0yBc n 0yBc nand thus (46) is established. = 1:) Now (42) giveslim lim sup"#0 n!1 CU(x)Bc nV (Bc n ) CU(x)nc n nP ( > n) CU(x)P ( > n);1U(x)P ( = n)(Note that this proof is also valid for the caseXy2[0;"c n]2f(0)lim lim sup"#0 n!1 nc n P ( = n)P (S n > xXy2[0;"c n]y; T x > n)F (y)F (y)V (y);and since < 1; (105) of [21] shows that this is zero.In the case = 1; a di¤erent proof is required. We make use of theobservation in [21] that there is a sequence n # 0 with n c n ! 1 and suchthat nF ( n c n ) ! 0: This givesXP (S n = x y; T x > n)F (y) P (T x > n)F ( n c n )y> nc n= o(U(x)P ( > n)=n) = o(U(x)P ( = n));where we have used (46). Using (23) of Proposition 12 givesXy nc nP (S n = xy; T x > n)F (y) v f(0)U(x)!(n)nc nas n ! 1;uniformly in x, where !(n) = P 0y nc nV (y)F (y): But in [21], it is shownthat nc n P ( = n) v f(0)!(n), so we deduce from (41) that P (T x = n + 1) vU(x)P ( = n); as required.15

5.2 Proof when x=c n = O(1)Proof. Again we treat the case < 1 …rst; and start by noting that for anyB > 0;XP (S n = x y; T x > n)F (y) F (Bc n ) v 1 as n ! 1;nB y>Bc nuniformly in x 0: Also by Corollary 13, for b > 0,Xybc nP (S n = xy; T x > n)F (y) CU(x) P ybc nV (y)F (y)nc n: (47)Since V F 2 RV ( ); P yzV (y)F (y) v zV (z)F (z)=(1 ); and we see thatwhen x Dc n ;U(x) P ybc nV (y)F (y)c nv b 1 U(x)V (c n )F (c n ) Cb 1 D U(c n )V (c n )=n Cb 1 D ;where we have used (15) from Corollary 8. We conclude the proof by showingthatlim lim n XP (S n = x y; T x > n)F (y) = C; (48)b#0; B"1 n!1 h xn (1)bc n n)F (y) nF ( n c n ) ! 0;y> nc nandnX0y nc nP (S n = xv nP ( > n)p(x n)c nvXy; T x > n)F (y)0y nc nV (y)F (y)np(x n)!(n)v p(x n)P ( = n)v p(x n );P ( > n)nc n P ( > n)and the result follows from the identity (33) in Proposition 15, since again therewould be a contradiction if k 9 6= 1.16

5.3 Proof when x=c n ! 1Proof. This time we write P (T x = n + 1) = P 41 P (i) ; where P (i) =P fA (i) \ (T x = n + 1)g andA (1) = (S n x); A (2) = (x < S n x Kc n );A (3) = (x Kc n < S n x c n ); and A (4) = (S n > x c n ):We note …rst that for 2 (0; 1);P (1)lim supn!1 F (x)P (1)lim infn!1 F (x)P (S n x)F ((1lim supn!1 F (x))x)= (1 ) ;P (max rn S r x; jS n j x)F ((1 + )x)lim infn!1F (x)= (1 + ) : (49)Next, using (27),P (2)lim supn!1 F (x)P (S n > x)F (Kc n )lim supn!1 F (x)nF (x)F (c n )lim sup= Cn!1 F (x)K (K) : (50)To deal with the next term, we use (27) again to see that for any …xed K;P (x Kc n < S n x) is uniformly o(nF (x)): Since P (3) F (c n )P (x Kc n n) vnP ( > n)x 1 F (x); so we can assume thatxg(n; x)P ( > n)sup< 1: (52)n>0; x>Kc n F (x)17

Then for large enough n and x > Kc nP (4) ===nX0bcX ncy=0bc ncCF (x)xP ( > n)CF (x)xP ( > n)CF (x)xP ( > n)CF (x)xP ( > n)Xz=0nX0ybcX ncy=0bcX ncz=0bcX ncz=0g(n m; x y)g (m; z)F (y + z)bcX ncy=0bc ncXz=0bc ncXy=0bcX ncw=zbc ncXz=0yzyg (m; z)F (y + z)v(z)F (y + z)v(z)F (y + z)v(z)F (w):Now a summation by parts and the fact that V F 2 RV (y ! 1) shows that asyX yXyXv(z)F (w) v V (z)F (z) v yV (y)F (y)=(1z=0 w=zz=0):So for all large enough n we have the boundP (4) CF (x)1 c n V (c n )nxP ( > n)CF (x) 1 c nnxP ( > n)P ( > n) v C1 c n F (x) (53)xThe result follows from (49)-(53) and appropriate choice of ; K; and :Remark 16 The assumption (11) is not strictly necessary for (52) to hold,and this is the only point where we use this assumption. In fact, if the followingslightly weaker version of (52),c n g(n; x)P ( > n)sup< 1; (54)n>0; x>Kc n F (x)were to hold, then (53) would hold with c n replacing x in the denominator, andthe proof would still be valid, by choosing small.18

6 The non-lattice caseWe indicate here the main di¤erences between the proof in the lattice and nonlatticecases. First, we haveG(n; dy) : =G (n; dy) : =1XP (T r = n; H r 2 dy) = P (S n 2 dy; r=0> n);1XP (T r = n; H r 2 dy) = P ( S n 2 dy; > n);r=0and the following analogue of Lemma 10 is given in Theorems 3 and 4 of [21].Lemma 17 For any 0 > 0; uniformly in x 0 and 0 < 0 ;c n G(n; [x; x + ])P ( > n)Also, uniformly as x=c n ! 0;G(n; [x; x+]) v f(0) R x+U(w)dwxnc n= p(x=c n )+o(1) and c nG (n; [x; x + ])P ( > n)= ~p(x=c n )+o(1) as n ! 1:(55)and G (n; [x; x+]) v f(0) R x+V (w)dwx:nc n(56)Remark 18 Again, only the results for G are given in [21], but it is easy to getthe reults for G : Actually the result in [21] has U(w ) rather than U(w) in(56), but clearly the two integrals coincide. Finally the uniformity in is notmentioned in [21], but a perusal of the proof shows that this is true, essentiallybecause it holds in Stone’s local limit theorem. See e.g. Theorem 8.4.2 in [6].In writing down the analogues of (16) and (17) care is required with the thelimits of integration, since the distribution of S n and the renewal measures arenot necessarily di¤use. These analogues areZP (T x = n + 1) = P (S n 2 x dy; T x > n)F (y); (57)and for w 0P (S n 2 x dw; T x > n) =nXZr=0[0;1)[0;x)\[0;w]The key result, the analogue of Proposition 12, isG(r; x dz)G (n r; dw z): (58)Proposition 19 Fix > 0: Then (i) uniformly as x n _ y n ! 0;P (S n 2 (x y ; x y]; T x > n) v U(x)f(0) R y+V (w)dwy: (59)nc n19

(ii) For any D > 1; uniformly for y n 2 [D 1 ; D];P (S n 2 (x y ; x y]; T x > n) v U(x)P ( > n)~p(y n)c nas n ! 1 and x n ! 0;and uniformly for x n 2 [D 1 ; D];P (S n 2 (x y ; x y]; T x > n) v V (y)P ( > n)p(x n)c nas n ! 1 and y n ! 0:(iii) For any D > 1; uniformly for x n 2 [D 1 ; D] and y n 2 [D 1 ; D];(60)(61)P (S n 2 (x y ; x y]; T x > n) v q x n(y n )as n ! 1: (62)c nOnce we have these results, we deduce Theorem 2 by applying them to amodi…ed version of (57), viz1XP (S n 2 (x (n + 1); x n]; T x > n)F (n) P (T x = n + 1)01XP (S n 2 (x (n + 1); x n]; T x > n)F ((n + 1));0and letting ! 0: So the key step is establishing Proposition 19, and weillustrate how this can be done by proving (59).Proof. We want to apply Lemma 17 to (58), but technically the problem isthat we can’t do this directly, as we did in the lattice case. The …rst step is toget an integrated form of (58), and it is useful to separate o¤ the term r = 0 ;so that for x; y 0;P (S n 2 (x y ; x y]; T x > n) = G (n; [fy xg + ; y + x))1 fxy+g + P ~ ;(63)wherenXZ Z~P =G(r; x dz)G (n r; dw z);==r=1nXZr=1nXZr=1yw

An asymptotic lower bound is given byr>nf(0)d(n)bncXr=1Z0z

After reading o¤ the asymptotic behaviour of the …rst term in (64) from(56), the proof is now completed by using (65), (66), (67), and the followingresult.Lemma 20 For x; y 0 and > 0 the following identity holdsZZU(x z)dzV (dw z) + U(x dz)V (w z)dw0z

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