ON THE RECURRENCE OF RANDOM WALKS

ON THE RECURRENCE OF RANDOM WALKS SIMON LAZARUS Abstact. We define andom walks on and ecuent points and demonstate that a andom walk s ecuence to implies its ecuence to each of its possible points. We then pove two diffeent necessay and sufficient conditions fo the ecuence of andom walks. Finally, we employ these esults to povide conditions unde which andom walks on R and on R 2 ae ecuent as well as pove that when d 3, no tuly d-dimensional andom walk on is ecuent. Contents. Peliminaies 2. Random Walks and Recuence 4 3. Sufficient Conditions fo Recuence and Tansience 6 Acknowledgments 23 Refeences 23. Peliminaies In this section, we povide some basic definitions fom pobability theoy and intoduce some notation so that ou use of vaious tems late in this pape shall be clea. We also povide without poof seveal basic but impotant esults fom pobability theoy that we shall employ in ou discussion of andom walks. Fo poofs of these esults, see [2]. Lastly, we pesent two examples which we shall late find useful. Below, A denotes the collection of all subsets of A. Definition.. A pobability space is a tiple Ω, F, P, whee Ω is a set, F Ω is a σ-algeba, and P : F R is a measue a nonnegative countably additive set function such that P Ω =. We efe to Ω as the sample space; its elements ae outcomes. F is the set of events, and P is the pobability measue of the space. Given A F we call P A the pobability of the event A. Definition.2. Let d N. i A function X : Ω is said to be a andom vaiable on if fo all Boel sets B we have X B F. ii Given a Boel set B and a andom vaiable X on, we define the event that X B denoted {X B} as X B. Date: Septembe 5, 23.

2 SIMON LAZARUS If d =, we simply call X a andom vaiable. One impotant elementay esult is that sums, poducts, supemums, and limit supeios of andom vaiables ae themselves andom vaiables; and in geneal any Boel measuable function of andom vaiables on R n that maps to R m is itself a andom vaiable on R m. Given a Boel set B and a andom vaiable X on, we can natually define the pobability that X is in B, denoted P X B, as P {X B}. This is well defined since Definition.2 guaantees that {X B} F. Letting B d denote the collection of Boel sets in, it follows fom the above definitions that if we also define µ X : B d R by µ X B = P X B, then µ X is a measue on B d such that µ X =. We call µ X the pobability measue of X. In geneal, any measue µ on B d such that µ = is called a pobability measue. Definition.3. Let X be a andom vaiable. The expectation o expected value o mean of X is defined as E[X] = XdP wheneve this integal exists this Ω includes the possibilities E[X] = and E[X] =. We next pesent a theoem that will help us compute expected values. Theoem.4. Let X be a andom vaiable on and let g : R be Boel measuable. If g o if E gx <, then E[gX] = g dµ X. We now discuss density functions and chaacteistic functions of pobability measues. Note that since evey andom vaiable uniquely detemines a pobability measue, ou coming definitions of density functions and chaacteistic functions of pobability measues immediately define density functions and chaacteistic functions of andom vaiables. In the definition that follows and thoughout this pape, we shall use B fxdx to denote the integal of f ove B with espect to Lebesgue measue. Definition.5. Let µ be a pobability measue on B d. If f : R is measuable such that fo all B B d we have µb = fxdx, we say f is a density function B fo µ. It is clea that any measuable f : R can only be a density function fo some pobability measue on B d if fxdx = and f almost eveywhee. It is also appaent that evey f satisfying these conditions is in fact the density fo some pobability measue on B d. Fo given such an f, we can define a pobability measue µ f on B d by µ f B = B fxdx. In this way, we see that thee may be multiple density functions coesponding to a single pobability measue as such densities can diffe on a set of measue, but thee is exactly one pobability measue coesponding to a given density. Definition.6. Let µ be a pobability measue on B d. Then the function ϕ µ : R given by ϕ µ t = e it y dµy is called the chaacteistic function of µ. Hee, e it y dµy simply means cost ydµy + i sint ydµy. We note that given any t, these integals always exist and ae always finite, as follows. Fo all x, let fx = cost x and gx = sint x. Then f, g

ON THE RECURRENCE OF RANDOM WALKS 3 on, so since µ is a pobability measue, fx dµx dµx = and similaly gx dµx. We also note that given a andom vaiable X on, we can define the chaacteistic function of X, denoted ϕ X, as ϕ µx. In this case, by simila easoning to the above paagaph we can apply Theoem.4 to find that fo all t, ϕ X t = E[cost X] + ie[sint X]. We wite this as ϕ X t = E[e it X ]. We now pesent some faily simple facts about chaacteistic functions and an example of a chaacteistic function that will be elevant late. Theoem.7. Let X, X 2 be independent andom vaiables on and let ϕ, ϕ 2 be thei chaacteistic functions. Then ϕ is unifomly continuous, ϕ =, and fo all t we have ϕ t and ϕ t = ϕ t. Additionally, ϕ ϕ 2 is the chaacteistic function of X + X 2. Example.8. Let >. Conside the tiangula andom vaiable T given by the density function g, whee fo all x R we have gx = x if x 2 and gx = othewise. It can be shown that T is equivalent to the sum of two independent andom vaiables call them X and X 2 which ae unifomly distibuted on [ /2, /2]. By Theoem.7, then, ϕ T = ϕ X ϕ X2 = ϕ 2 X. Now X has density f, whee fo all x R, fx = / if x [ /2, /2] and fx = othewise. Hence, fo all t R \ {} we have ϕ X t = e itx fxdx = /2 e itx dx = R e 2 it e 2 it = 2 sin t 2. it t /2 So since ϕ T = ϕ 2 X, this means that fo all t R \ {}, 2 sin t 2 ϕ T t = 2 = 4 sin2 t 2 t t 2 We also have that ϕ X = R e ix fxdx = = 4 cos2 t 2 2 t 2 = 2 t 2 cost. /2 /2 dx =, so ϕ T = 2 =. Next, we pesent two impotant theoems about chaacteistic functions the Invesion Fomula and Pólya s Citeion and employ them in anothe example that will be elevant late. Theoem.9. Let µ be a pobability measue on B and let ϕ be its chaacteistic function. If ϕx dx <, then µ has a density function f given by: y R, R fy = e ity ϕtdt. 2π In this case, f is continuous and bounded on R. R Theoem.. Let ϕ : R R be even, continuous on R, and convex on, and suppose ϕ = and lim x ϕx =. Then ϕ is the chaacteistic function fo some pobability measue µ on B. Example.. Let >. Fo all t R, let ϕt = t if t and ϕt = othewise. Then ϕ clealy satisfies the hypotheses of Theoem., so ϕ is the chaacteistic function of some pobability measue µ on B. Since R ϕx dx = <, by Theoem.9 we theefoe know that µ has density f,

4 SIMON LAZARUS whee fo all y R we have fy = 2π / / e ity t dt. By consideing the two cases y = and y, evaluating this integal is staightfowad albeit somewhat tedious, and it can be shown that f = 2π and that fo all y R \ {}, fy = cos y πy 2. Having now discussed chaacteistic functions and density functions at length, we close this section by defining thee types of convegence which we shall late employ and discussing the notion of infinitely often. Below, is the Euclidean nom on, and A is said to be a continuity set of X if A is a Boel set and P X A =. Definition.2. Let X, X, X 2,... be andom vaiables on. i If we have P lim n X n = X =, then we say {X n } conveges almost suely to X and a.s. we wite X n X. ii If fo all ɛ > we have lim n P X n X ɛ =, p then we say {X n } conveges in pobability to X and we wite X n X. iii If lim n P X n A = P X A fo all A that ae continuity sets of X, then we say {X n } conveges in distibution to X and we wite X n d X. It is a fact that convegence almost suely implies convegence in pobability, which in tun implies convegence in distibution. Definition.3. Let A, A 2,... F. Then the event A n i.o. is defined as k= n=k A n. A n i.o. is ead A n infinitely often. Note that A n i.o. F since F is a σ-algeba. We see that A n i.o. is the collection of outcomes ω which ae elements of infinitely many of the events A, A 2,. Intuitively, A n i.o. occus if and only if infinitely many of the events A, A 2, occu. As a final note in this section, we pove the Boel-Cantelli Lemma. Theoem.4. Let A, A 2,... F. If P A n <, then P A n i.o. =. Poof. If P A n <, then cetainly inf k N n=k P A n =. Now note that fo any events B, B 2,... F we have P =B inf m N P B m since fo all m N we have =B B m. Thus, P A n i.o. = P A n inf P A n inf P A n =, k N k N k= n=k whee the second inequality follows fom the countable subadditivity of P. n=k 2. Random Walks and Recuence Definition 2.. Let X, X 2,... be i.i.d. andom vaiables on. Then the sequence {S n } given by S = and S n = X +... + X n is said to be a andom walk on. n=k

ON THE RECURRENCE OF RANDOM WALKS 5 In this section, we shall take up the question: what ae necessay and sufficient conditions such that with pobability, a andom walk will visit abitaily close to a given point in infinitely many times? In ode to pecisely state and answe this question, we intoduce some useful definitions. Below, denotes the Euclidean nom on. Definition 2.2. Let {S n } be a andom walk on and let x. i x is said to be a possible point fo {S n } if ɛ >, N N such that P S N x < ɛ >. ii x is said to be a ecuent point fo {S n } if ɛ >, P S n x < ɛ i.o. =. That is, x is a possible point fo a andom walk if given any desied level of closeness ɛ, thee is some time at which thee is a nonzeo pobability of the andom walk being within distance ɛ of x. Similaly, x is a ecuent point fo a andom walk if with pobability one the andom walk comes abitaily close to x infinitely many times. Although the notions of ecuent points and possible points may at fist seem quite dispaate, we now pove that they ae elated. We fist note that fom Definition 2.2 it is obvious that all ecuent points of a andom walk ae possible points of a andom walk. The following theoem establishes the convese in the case whee a andom walk has any ecuent points. Theoem 2.3. Let {S n } be a andom walk on and let R and P be espectively the sets of its ecuent points and its possible points. If R, then R = P. Poof. Suppose R. Since ecuent points ae possible points, R P. We now show that P R by fist establishing if R and p P then p R by contadiction. So let R and p P and suppose p R. Then by definition thee exists ɛ > such that P S n p < 2ɛ i.o. <, i.e. thee exists M N such that P n M, S n p 2ɛ >. Since p is a possible point fo {S n }, by definition thee exists N N such that P S N p < ɛ >. Now note that fo all n > N, S n S N = X N+ +... + X n is independent of the fist N steps of the andom walk and has the same distibution as S n N as it is simply n N steps stating fom the oigin. Thus, cetainly P n N + M, S n S N p 2ɛ = P n M, S n p 2ɛ >. Note also that the combination of the event that S N p < ɛ and the event that n N + M, S n S N p 2ɛ cetainly by the Tiangle Inequality implies the event that n N + M, S n ɛ. Since these fome two events ae independent as established above, since M, we thus have P n N + M, S n ɛ P n N + M, S n S N p 2ɛ and S N p < ɛ = P n N + M, S n S N p 2ɛ P S N p < ɛ >, a contadiction to the fact that is a ecuent point fo {S n }. Thus, holds. Using, we now show that R and then show that if x R then x R. Since R, let x R. Then cetainly x P, so applying with = p = x gives R. Similaly, applying with = and p = x gives x R. To pove the theoem itself, we now let y P and show that y R. Applying with = and p = y gives y R, which by the above means y R.

6 SIMON LAZARUS By Theoem 2.3, we see that ou oiginal question is a given possible point fo {S n } a ecuent point fo {S n }? educes to the question does {S n } have a ecuent point? It thus makes sense to simply call a andom walk ecuent if it has at least one ecuent point. We hencefoth adopt this definition, and we call a andom walk tansient if it has no ecuent points. Additionally, within the poof of Theoem 2.3, we found that if {S n } has any ecuent points, then is a ecuent point of {S n }. Thus, ou oiginal question educes to is a ecuent point fo {S n }? In summay, we have: Coollay 2.4. Let {S n } be a andom walk on. Then i if R then R = P, and ii if R then R =. Befoe continuing ou discussion of geneal andom walks on, we give a moe concete example of such a andom walk. Specifically, we conside a simple andom walk on Z d. In this case, the distibution of ou i.i.d. andom vaiables X i is given by j {,..., d}, P X i = e j = 2d = P X i = e j. That is, each step of the andom walk has an equal chance of moving unit in any of the 2d cadinal diections and neve moves anywhee else. In this case, given any x Z d, thee is at least one possible path fom the oigin to x in a finite numbe of steps; say this numbe of steps is N. Thee is a 2d N chance of ou simple andom walk following this pecise path fom the fist step to the Nth step. Thus, cetainly P S N = x 2d N >, meaning x is a possible point fo {S n }. In this way, we see that if a simple andom walk on Z d has as a ecuent point, then its set of ecuent points is all of Z d ; and if is not a ecuent point, then it has no ecuent points. With this example in mind, we now move on to the conditions unde which is a ecuent point fo any geneal andom walk on. In ode to simplify some of the poofs below, we intoduce a new nom on and pesent a useful lemma. Definition 2.5. Let x = x,..., x d. We define x = max{ x,..., x d }. Lemma 2.6. Let ɛ > and m N such that m >. Then 2m d P S n < mɛ P S n < ɛ Poof. Let A = {x,..., x d : i {,..., d} such that x i = mɛ} and B = {x : x < mɛ}. We patition A B into 2m d half-open cubes as follows. Let I = { m, m +,..., m } d. Fo all v = v,..., v d I, let C v = [v ɛ, v + ɛ [v d ɛ, v d + ɛ. Then clealy v I C v = A B and the C v ae paiwise disjoint. So since B A B, this means P S n B P S n A B = P S n C v. v I Since each of these pobabilities is cetainly nonnegative, by Fubini s Theoem we can exchange the ode of summation to get 2 P S n C v. P S n B v I Now fix v I and let T v = inf{k {,,...} : S k C v }. Note that this infimum may be. Note that given any n {,,...}, the event that S n C v occus if and

ON THE RECURRENCE OF RANDOM WALKS 7 only if pecisely one of the events S n C v and T v =, S n C v and T v =,..., S n C v and T v = n occus. Fo if the fome event does not occu, cetainly none of the latte events occued; and if the fome event does occu, then the andom walk landed in C v by time n and hence thee was some ealiest time k {,,..., n} at which it did so. We theefoe have n P S n C v = P S n C v and T v = k k= Since each of these tems is nonnegative, we can again exchange the ode of summation to get 3 P S n C v = P S n C v and T v = k. k= n=k Now fix k {,,...}. Note that if T v = k, then the andom walk has eached C v at time k, meaning in ode fo the walk to each C v again at time n we cetainly must have that the next n k steps of the walk do not in sum displace the walk by a distance geate than ɛ in any cadinal diection fo C v is a cube with side length ɛ. This means P S n C v and T v = k P S n S k < ɛ and T v = k n=k n=k But those next n k steps ae independent of the fist k steps and have the same distibution as the fist n k steps of a sepaate andom walk stating at. Hence, P S n C v and T v = k P S n S k < ɛ P T v = k n=k n=k n=k = P T v = k P S n k < ɛ = P T v = k P S i < ɛ whee the last equality is simply e-indexing. Since this holds fo any fixed k {,,...}, by letting k vay and summing we obtain P S n C v and T v = k P T v = k P S i < ɛ. k= n=k Similaly, since this holds fo any fixed v I, by letting v vay we get P T v = k P S i < ɛ. v I k= k= n=k P S n C v and T v = k v I Applying 3 and then 2, this implies P T v = k P S i < ɛ. P S n B v I k= We now note that by definition S n B if and only if S n < mɛ. We also note that the innemost sum on the ight-hand side of the above does not depend on k o v and hence can be pulled outside of the othe two sums. Thus, we have P S n < mɛ P S i < ɛ P T v = k. i= k= i= i= i= v I k= i=

8 SIMON LAZARUS Lastly, we note that k= P T v = k since given v I, T v takes exactly one value in {,,...} { } and hence has pobability of taking a value in {,,...} and we note that thee ae 2m d elements in I. Thus, we have P S n < mɛ P S i < ɛ 2m d as desied. i= With this lemma in hand, we ae eady to pove an impotant necessay and sufficient condition fo ecuence which we shall employ in the poofs of Theoems 2.8, 3., and 3.3 below. Theoem 2.7. Let {S n } be a andom walk on and let >. Then {S n } is ecuent if and only if P S n < =. Poof. We must show that i P S n < < implies {S n } is tansient and ii P S n < = implies {S n } is ecuent. To do so, we show that in case i is not a ecuent point fo {S n } but in case ii it is a ecuent point fo {S n }. i By Theoem.4, if P S n < conveges, then P S n < i.o. =. Now note that the event that S n < i.o. cetainly implies the event that S n < i.o. since the Euclidean nom of S n being less than fo some n cetainly implies that the lagest coodinate of S n is less than. Thus, P S n < i.o. P S n < i.o.. That is, P S n < i.o. =, so by definition is not a ecuent point fo {S n }. ii Suppose P S n < =. Let ɛ >. We wish to show P S n < ɛ i.o. =. We do so by showing that fo evey k N, P n k, S n ɛ =. To do this, we fist establish that P S n < ɛ/2 =. Let M N such that M > and M < ɛ 2. Let s = d M 2M d which by substituting s = M 2M d. Then since s >, Lemma 2.6 gives P S n < Ms gives P S n < P P S n < s, S n < M P S n < ɛ 2 d since M < ɛ 2. Now note that given any n {,,...}, the event that S d n < ɛ 2 d implies the event that S n < ɛ/2. Fo if S n = x,..., x d and S n < ɛ 2, then d by definition x,..., x d < S n = ɛ 2 d so 2 ɛ x 2 +... + x2 d d < 2 = ɛ d 2. Thus, we cetainly have that P S n < ɛ 2 P S n < ɛ/2, d

ON THE RECURRENCE OF RANDOM WALKS 9 which combined with the above gives 2M d P S n < P S n < ɛ/2, meaning necessaily P S n < ɛ/2 =. Now let k N. Fo all n N, let A n be the event that both S n < ɛ/2 and m k + n, S m ɛ/2. Then clealy at most k of the events A n can occu. Fo if j N is the smallest n fo which A n occus, then by definition m k + j, S m ɛ/2, meaning all of the events that can possibly occu ae A j, A j+,..., A j+k. Thus, P A n k since at most k of the events have any chance of occuing. Now note that given any n {,,...}, the event that both S n < ɛ/2 and m k + n, S m S n ɛ cetainly implies the event A n. Thus, P A n P S n < ɛ/2 and m k + n, S m S n ɛ. But since the event { m k + n, S m S n ɛ} is independent of the event that S n < ɛ/2 and occus with the same pobability that the event { i k, S i ɛ} occus, this means k P A n P S n < ɛ/2 P m k + n, S m S n ɛ = P S n < ɛ/2 P i k, S i ɛ. Now suppose c > such that P i k, S i ɛ = c. Then the above gives k c P S n < ɛ/2 =, a contadiction. Thus, P i k, S i ɛ =. So since k N was abitay, we have shown that fo all k N, P n k, S n ɛ =, as desied. Having pesented in Theoem 2.7 a necessay and sufficient condition fo the ecuence of a andom walk {S n }, we now establish anothe such condition oiginally given in [] that is sometimes moe useful. Theoem 2.8. Let {S n } = {X +...+X n } be a andom walk on and let >. Let ϕ be the chaacteistic function of X. Then {S n } is ecuent if and only if dx =. cϕx sup c, [,] d Re Remak 2.9. We note that given c,, the above integal always exists and has value in [, ]. Fo since < c < and by Theoem.7 ϕ, cetainly fo all x we have creϕx <, so Lemma 3.7 below gives that Re fo all x. cϕx This new condition is sometimes moe tactable than the one in Theoem 2.7, fo instead of equiing one to find o estimate pobabilities elated to {S n }, it equies finding ϕ and computing o estimating the above integal. Fo this eason,

SIMON LAZARUS we employ it in poving Theoem 3.9 below. Befoe poceeding with the poof of Theoem 2.8, we povide fou lemmas whose uses will soon be appaent. Lemma 2.. Fo all x [ π/3, π/3] we have cosx x2 4. Poof. Since cos is an even function, it suffices to show this fo x [, π/3]. Fist, note that if t [, π/3] then clealy cost /2. Hence, if y [, π/3] then Thus, if x [, π/3] then siny = cosx = y x costdt sinydy y x 2 dt = y 2. y x2 dy = 2 4. Lemma 2.. Let Y,..., Y d be andom vaiables on R with chaacteistic functions ϕ,..., ϕ d. Let Z = Y,..., Y d. Then ϕ Z, the chaacteistic function of Z, is given by: x = x,..., x d, ϕ Z x = d i= ϕ ix i. Poof. Let x = x,..., x d. By definition, ϕ Z x = E[e ix Z ] = e ix t,...,td dµ Z t,..., t d. Since fo all t we have e ix t = and since µ Z is a pobability measue, necessaily e ix t,...,td dµ Z t,..., t d =. So by Fubini s Theoem, we have ϕ Z x = e ix t,...,td dµ Y t dµ Yd t d R R = e ixt e ix dt d dµ Y t dµ Yd t d R R = e ixt dµ Y t e ix dt d dµ Yd t d R R = E[e ixy ] E[e ix dy d ] = ϕ x ϕ d x d. Lemma 2.2. Let µ and ν be pobability measues on B d the Boel sets in with chaacteistic functions ϕ µ and ϕ ν. Then ϕ µ dν = ϕ ν dµ. Poof. By definition ϕ µ tdνt = expit xdµxdνt. Since fo all R d t, x we have expit x = and since µ and ν ae pobability measues, necessaily expit x dµxdνt =, so Fubini s Theoem gives R d expit xdµxdνt = expit xdνtdµx = ϕ ν xdµx, as desied. Lemma 2.3. Fo all n {,,...}, let a n. Then sup c, c n a n = a n.

ON THE RECURRENCE OF RANDOM WALKS Poof. Since fo all c, and fo all n {,,...} we have c n a n a n, cetainly sup c, cn a n a n, so it suffices to show the evese inequality. We do so by consideing two cases. Fist, suppose a n =. Let R >. Choose N N such that N a n > 2R. Then choose c, such that fo all n {,,..., N} we have c n 2. Then cn a n N cn a n N 2 a n > 2 2R = R. Thus, fo all R >, thee exists c, such that cn a n > R, meaning sup c, cn a n =. Now suppose instead that thee exists S [, such that a n = S. Let ɛ >. Choose M N such that M a n > S ɛ 2. Then choose c, such that fo all n {,,..., M} we have c n ɛ 2S+. Then M M c n a n c n ɛ ɛ a n a n > S ɛ 2S + 2S + 2 = S ɛs 2S + ɛ 2 + ɛ 2 4S + > S ɛ 2 ɛ 2 + = S ɛ. Thus, fo all ɛ >, thee exists c, such that cn a n > a n ɛ, meaning necessaily sup c, cn a n a n, as desied. With these tools in hand, we ae eady to pove Theoem 2.8. Poof of 2.8. We fist show that i if sup c, Re [,] d P S n < / < and then that ii if sup c, cϕx [,] d Re dx < then dx = cϕx then P S n < / =. By 2.7, showing i and ii will complete the poof. i Suppose thee exists C [, such that sup c, [,] d Re cϕx dx = C. Let T,..., T d be i.i.d. andom vaiables with density g, whee fo all t R, gt = t if t and gt = othewise. Then by Example.8, each T 2 i has chaacteistic function ϕ T, whee ϕ T = and t R \ {} we have ϕ T t = 2 cost t. 2 Note that if t [, ], then t < π/3. Thus, by Lemma 2., fo all t [, t2 ] we have cost 4, meaning fo all t [, ] \ {} we have 2 2 cost t. Thus, fo all t [ 2, ] we have 2ϕ T t since ϕ T =. So cetainly fo all x,..., x d [, ]d we have 4 2 d ϕ T x i = i= 2ϕ T x i since each tem in the ightmost poduct of the above is. Now let n {,,...}. Then we know P S n < = P S n, d =, dµ d Sn x, whee µ Sn is the pobability measue of S n. Then by 4, P S n < 2 d ϕ T x i dµ Sn x, i=, d i=

2 SIMON LAZARUS whee dµ Sn x is shothand fo dµ Sn x,..., x d. Now fo all y R \ {}, we know cosy y since cosy. Thus, fo all t R we have ϕ 2 T t since ϕ T t = 2 cost t 2 if t and ϕ 2 T t = if t =. Theefoe, fo all x,..., x d we cetainly have d i= ϕ T x i. So since, d, we see P S n < 2 d ϕ T x i dµ Sn x. It should be noted that in the steps that follow up to and including 8 below, evey manipulation we pefom only substitutes fo the ight-hand side of the above something that is equivalent to it. In paticula, it is clea since ϕ T is a eal function on R that the ight-hand side of the above inequality is eal; this does not and cannot change in any of the steps below. Let Z = T,..., T d. Then by Lemma 2., the above inequality gives P S n < 2 d ϕ Z xdµ Sn x, which by Lemma 2.2 means P S n < 2 d ϕ Sn xdµ Z x. Since S n = X +... + X n and each of the X j has chaacteistic function ϕ, by Theoem.7 we know ϕ Sn = ϕ n. If n =, we must use othe easoning to show that ϕ Sn = ϕ n on. Fist, note that since S =, we know ϕ S = on, meaning if x such that ϕx, then ϕ x = = ϕ S x. Theefoe, we do have that ϕ S x = ϕ x fo all x such that ϕx. Howeve, we may un into touble if x such that ϕx =, as in this case ϕ x = is not conventionally defined. Theefoe, we now define ϕ x to be wheneve x such that ϕx =. In this way, ϕ S = ϕ holds on all of. Thus, P S n < 2 d ϕ n xdµ Z x. Now note that if I,..., I d R ae intevals, then since T,..., T d ae independent, we know P Z I I d = P T I,..., T d I d = d i= P T i I i. So since each of the T i has the density g given above, we see that Z has the density h, whee fo all x = x,..., x d, hx = d i= gx i. Theefoe, we have P S n < 2 d ϕ n xhxdx = 2 d ϕ n x gx i dx. Since g is given by t R, gt = t if t and gt = othewise, the above 2 clealy implies P S n < 2 d ϕ n xi x [,] 2 dx. d Now let c,. Then we have c n P S n < 2 d c n ϕ n xi x [,] 2 dx. d i= i= i= i=

ON THE RECURRENCE OF RANDOM WALKS 3 This inequality holds fo any given n {,,...}. So letting n vay and summing, we see c n P S n < 2 d c n ϕ [,] n xi x 2 dx d i= 5 d 2 = c n ϕ n xi x dx. [,] d We now seek to swap the ode of summation and integation on the ight-hand side of 5. By Fubini s Theoem, we can do so as long as we can show that both 6 cn ϕ n xi x dx < and 7 [,] d [,] d i= cn ϕ n xi x i= i= dx <. To pove that 6 and 7 hold, ecall that ϕ on and that fo all t [, ], t. Thus, fo all x = x,..., x d [, ] d we have cn ϕ n xi x cn n d = c n. This means [,] d i= i= cn ϕ n xi x dx c n dx = [,] d 2 d c n < since the ight-hand side of the above is a geometic seies and c <, thus poving 6; and 7 follows similaly. This means we can switch the ode of summation and integation in 5 and then move the poduct outside of the sum to obtain c n P S n < 2 d [,] d xi c n ϕ n xdx. Now note that given x, the sum on the ight-hand side of the above is a convegent complex geometic seies with value cϕx. If ϕx, this is obvious since < cϕx < ; and if ϕx = then cϕx = = + + +, which we see equals c ϕ x+c ϕ x+ when we ecall that we defined ϕ x = when ϕx =. Thus, c n P S n < d 2 xi [,] cϕx dx. d i= i= As noted above, the ight-hand side of this inequality is and has always been eal, so we can take its eal pat without affecting its value. Hence, 8 c n P S n < d 2 xi Re dx. cϕx [,] d i=

4 SIMON LAZARUS By Remak 2.9, we know that fo all x, Re know that d xi i= wheneve t [, ], 8 implies that c n P S n < cϕx. So since we also wheneve x,..., x d [, ] d as t 2 d [,] d Re cϕx dx d 2 C. The above inequality holds fo any c,. Theefoe, taking the supemum of both sides ove all such c and using the fact given by Lemma 2.3 that sup c, cn P S n < = P S n <, we have P as desied. ii Now suppose instead that sup c, S n < [,] d Re d 2 C <, dx =. cϕx Let Y,..., Y d be i.i.d. andom vaiables with density g, whee g = 2π and fo all t R\{}, gt = cost/ πt. Then by Example., each Y 2 i has chaacteistic function ϕ Y, whee fo all t R we have ϕ Y t = t if t and ϕ Y t = othewise. Let n {,,...}. Note that if t [, ], then t. Thus, cetainly fo all x,..., x d [, ]d we have d i= x i. Theefoe, P S n < = dµ Sn x x i dµ Sn x., d, d i= As befoe, it should be noted that the manipulations that follow up to and including below meely substitute the ight-hand side of the above inequality with something equivalent and hence do not change the fact that that side of the inequality is eal. Since ϕ Y t = t if t, and ϕ Y t = if t,, the above clealy gives P S n < ϕ Y x i dµ Sn x = ϕ Y x i dµ Sn x., d i= i= So letting Z = Y,..., Y d, Lemma 2. and then Lemma 2.2 imply P S n < ϕ Z xdµ Sn x = ϕ Sn xdµ Z x. Since defining ϕ as befoe ϕ Sn = ϕ n on and since as befoe Z has the density h given by x = x,..., x d, hx = d i= gx i, the above gives P S n < ϕ n xdµ Z x = ϕ n xhxdx = ϕ n x gx i dx. i=

ON THE RECURRENCE OF RANDOM WALKS 5 Let c,. Then the above gives c n P S n < c n ϕ n x gx i dx. This inequality holds fo any n {,,...}. So letting n vay and summing, we see 9 c n P S n < c n ϕ n x gx i dx. As befoe, we wish to apply Fubini s Theoem to switch the ode of summation and integation on the ight-hand side of the above, and to do so we must show cn ϕ n x gx i dx < and cn ϕ n x gx i dx <. i= i= We show the fist of these inequalities; the second follows similaly. Let n {,,..., }. Fist, note that since g = 2π and fo all t R \ {}, gt = cost/ πt, we see that g on R since cos on R. That is g = g 2 on R. So since ϕ on, we see cn ϕ n x gx i dx cn gx i dx = cn gx gx d dx dx d. i= i= Since g on R, Fubini s Theoem thus gives cn ϕ n x gx i dx cn i= i= R i= gtdt = c n i= = c n since g is a pobability density function. The above inequality holds fo any n {,,..., }. So letting n vay and summing, we see cn ϕ n x gx i dx c n < i= since c,. Thus, the fist half of holds; the second half follows similaly. Theefoe, we can exchange the ode of summation and integation in 9 and move the poduct outside of the sum to obtain c n P S n < gx i c n ϕ n xdx. As befoe, we know that given x, cn ϕ n x = cϕx. Thus, we have c n P S n < i= i= gx i i= cϕx dx. As noted above, the integal on the ight-hand side of must be eal, so we can take the eal pat of both sides to find c n P S n < gx i Re dx. cϕx i=

6 SIMON LAZARUS By Remak 2.9, fo all x R we have Re since [, ] d, we see c n P S n < [,] d cϕx gx i i=. So since g on R and Re cϕx dx. Now note that since fo all t [, ] we have t < π/3, by Lemma 2., fo all t [, ] we have cost/ t/2 4. Thus, fo all t [, ] \ {}, we have gt = cost/ πt 2 4π all t [, ], gt d i= gx i. Since Re 4π d c n P S n <. So since g = 2π > 4π, we see that fo 4π. Thus, fo all x,..., x d [, ] d, we cetainly have cϕ on, this means 4π d [,] d Re cϕx dx. The above inequality holds tue fo any c,. Theefoe, taking the supemum of both sides ove all such c and employing Lemma 2.3 gives P S n < 4π d dx =, cϕx as desied. sup c, [,] d Re 3. Sufficient Conditions fo Recuence and Tansience Having established in the pevious section two diffeent necessay and sufficient conditions fo ecuence, we now employ these conditions to discove some simple conditions that imply ecuence o tansience. Theoem 3.. Let {S n } be a andom walk on R. If S n /n p, then {S n } is ecuent. Poof. Let c N \ {}. Since S n /n p, by definition P S n n < c, i.e. P S n < n. c Thus, we can choose N N\{} such that fo all n > N we have P S n < n c > 2. By doing so, we find that 2 cn n=n+ P S n < n c cn n=n+ Now since N >, by Lemma 2.6, P S n < P S n < N 2N 2N = c N 2 2. cn n=n+ P S n < N whee the latte inequality follows since each of the tems of the sum is nonnegative. Now note that fo all n {N +, N + 2,..., cn} we have n c N. Thus, fo all n {N +,..., cn} we have P S n < n c P Sn < N. Theefoe, P S n < 2N cn n=n+ P S n < n, c

ON THE RECURRENCE OF RANDOM WALKS 7 which by 2 above means P S n < c N 2N 2 = c 4. So since c N\{} was abitay, necessaily P S n < =. By Theoem 2.7, then, {S n } is ecuent fo in R the noms and ae equivalent. Coollay 3.2. Let {S n } = {X +... + X n } be a andom walk on R. Suppose E X <. Then {S n } is ecuent if and only if E[X ] =. Poof. Note that E X < implies that E[X ] exists. By the Stong Law of Lage Numbes, S n /n a.s. E[X ]. So since convegence almost suely implies convegence in pobability, we see that if E[X ] =, then by Theoem 3., {S n } is ecuent. If instead E[X ] >, then since n fom above and S n/n a.s. E[X ], necessaily a.s. S n, which pecludes {S n } fom having any ecuent points. Similaly, if a.s. E[X ] <, then S n, meaning {S n } has no ecuent points. Theoem 3.3. Let {S n } be a andom walk on R 2. If thee exists a andom vaiable Y on R 2 such that S n / n d Y and if Y has a density function f such that f > and f is continuous at, then {S n } is ecuent. Poof. We wish to pove that P S n < = ; by Theoem 2.7, showing this will establish the theoem. By Lemma 2.6, we know that fo all m {2, 3,...}, P S n < 4m 2 P S n < m. Thus, it suffices to show that fo all R >, thee exists m {2, 3,...} such that 4m 2 P S n < m > R. Theefoe, it cetainly suffices to show that S lim inf m m 2 P S n < m =, as in this case we can always choose an m such that m 2 P S n < m > 4R. Now fo all y >, let gy = fxdx = P Y < y. Note that g y,y 2 on,. Now since f is continuous at and f >, by definition thee exists > such that fo all x [, ] 2 we have fx f < f 2. Thus, cetainly fo all x [, ] 2 we have fx > f. Theefoe, fo all y, ], gy = fxdx y,y 2 Now note that since g on,, gt /2 dt / 2 gt /2 dt 2 y,y 2 f f dx = 2y 2 = 2fy 2. 2 2 / 2 2ft /2 2 dt = 2f dt =, / t 2 whee the second inequality follows since t implies t /2, ]. Thus, it 2 suffices to pove S2 lim inf m m 2 P S n < m gt /2 dt

8 SIMON LAZARUS since the integal on the ight-hand side of the above is, meaning S2 implies S. Now note that fo all m N we have P S n < m = P S y < mdy, whee is the floo function. Thus, given m N, we can substitute y = tm 2 in the above to find P S n < m = P S tm 2 < m m 2 dt; that is, 3 m 2 P S n < m = P S tm 2 < mdt. Now suppose fo a moment that fo all t > we have S3 lim P S tm m 2 < m = gt /2. Then by taking the limit infeio of both sides of 3 and applying Fatou s Lemma which we can do since pobabilities ae nonnegative, we find lim inf m m 2 P S n < m lim inf P S tm m 2 < mdt = gt /2 dt, which is pecisely S2. Theefoe, it suffices to show that S3 holds. So let t >. Note that since S n / n d Y as n, cetainly S tm2 / tm 2 d Y as m since the sequence { tm 2 } tends to as m and is a sequence of natual m numbes. So since tm2 t /2 as m, by the definition of convegence in distibution we see that S tm P S tm2 < m = P 2 tm2 < m tm2 thus establishing S3. m P Y < t /2 = gt /2, Remak 3.4. If {S n } = {X +... + X n } is a andom walk on R 2 and if E[X ] = and the coodinates of X have finite covaiances, then the Cental Limit Theoem guaantees that S n / n conveges in distibution to a multivaiate nomal andom vaiable on R 2 with mean. If this nomal andom vaiable on R 2 is non-degeneate, then its density function is positive and continuous at, so by Theoem 3.3 {S n } is ecuent. If instead the nomal andom vaiable on R 2 is degeneate, then necessaily {S n } is not tuly 2-dimensional in the sense of Definition 3.6 below and hence we have educed {S n } to a andom walk on R. In this case as well, since E[X ] = and finite covaiances imply E X <, by Coollay 3.2 {S n } is ecuent. Example 3.5. Let {S n } = {X +... + X n } be a simple andom walk on Z and {T n } = {Y +... + Y n } be a simple andom walk on Z 2. Then both {S n } and {T n } ae ecuent, as follows. Fist, {S n } is ecuent by Coollay 3.2 since E[X ] = and E X = <. Second, {T n } is ecuent by Remak 3.4 since E[Y ] = and, witing Y = Y,, Y,2, we have CovY,, Y,2 = < since Y, = when Y,2 and vice-vesa, V ay, = 2 <, and V ay,2 = 2 <.

ON THE RECURRENCE OF RANDOM WALKS 9 We have now shown that unde the appopiate conditions, andom walks on R and on R 2 ae ecuent. One may now ask: unde what conditions ae andom walks on R 3 ecuent? What about andom walks on in geneal? As we shall pove below, if d 3, then no andom walks on ae ecuent. Howeve, since, fo example, one could easily embed a simple andom walk on Z which is ecuent into R 3 and call it a andom walk on R 3, we see that we must be at least somewhat caeful with ou definition of a andom walk on. Hence, we shall intoduce the notion of tuly d-dimensional andom walks, as defined below. Definition 3.6. Let {S n } = {X +... + X n } be a andom walk on. If thee exists θ such that θ and P X θ = =, then we say {S n } is not tuly d-dimensional. Essentially, if with pobability the steps of a andom walk eside in some space with dimension stictly less than d, then that andom walk is not tuly d- dimensional. Fo example, any simple andom walk on Z d is tuly d-dimensional; but embedding a simple andom walk {S n } = {X +... + X n } on Z k into R m when k < m esults in a andom walk on R m which is not tuly m-dimensional, as necessaily and hence with pobability X e k+ =. We shall now pove that esticting ouselves to tuly d-dimensional andom walks on is a sufficient condition fo tansience when d 3. This will esolve the emainde of the question posed above, fo if a andom walk on is not tuly d-dimensional, then it educes to the case of a andom walk on. We begin by poving two lemmas, the latte of which will be necessay late. Lemma 3.7. Let a, b R such that a. Then taking = Re a + bi a. Poof. Let z = a + bi. Then Re = Re z z z z that is, Re a + bi = z = Re z 2 = Rez z 2 ; a a 2 + b 2. Since a, the ight-hand side of the above equation has both a nonnegative numeato and a nonnegative denominato and hence is nonnegative even if a =. Additionally, if a <, since b 2 we cetainly have Re a a + bi a 2 = a As stated above, when a = we still have Re a+bi we tivially have Re a+bi < a =. ; and also when a = Lemma 3.8. Let a, b R such that a and let c,. Then Re ca + bi a.

2 SIMON LAZARUS Poof. Since a and c <, cetainly ac a, meaning ac a. Thus, ac a. Also, since a and c <, Reca + bi = ca <. So by Lemma 3.7, Re ca + bi ca a. Theoem 3.9. Let d N such that d 3. Let {S n } = {X +... + X n } be a tuly d-dimensional andom walk on. Then {S n } is tansient. Poof. Let ϕ be the chaacteistic function of X, let µ be the pobability measue associated with X, and let A = {x : x = }. We seek to bound the quantity Reϕ fom below. We fist pove that thee exist p > and s > such that fo all s, s, 4 inf θ y 2 dµy p. θ A θ y <π/3s Hee, θ y <π/3s means. The above integal is clealy nonnegative fo all s > and θ A since fo all θ, y we have θ y 2. So {y : θ y <π/3s} to pove 4, assume fo a contadiction that * fo all p > and s >, thee exists s, s such that the infimum in 4 is < p. Let n N. We claim that 5 inf inf θ y 2 dµy =. s, n θ A θ y <π/3s Fo if this wee not the case, thee would exist c > such that inf inf θ y 2 dµy = c, s, n θ A θ y <π/3s which would in tun imply that fo all s, n, the infimum in 4 is c, thus contadicting ou assumption * above when we take s = n. Thus, 5 holds. Now note that if s, n then π 3s > nπ 3. Thus, given any θ A and s, n we cetainly have θ y 2 dµy θ y 2 dµy θ y <nπ/3 θ y <π/3s since the integand is nonnegative and the integal on the left is ove a smalle set. Taking the infimum of both sides ove all such θ and s and ecalling 5, this means that inf inf θ y 2 dµy inf inf θ y 2 dµy =. s, n θ A s, n θ A θ y <nπ/3 θ y <π/3s The integal on the left-hand side of the above is independent of s, so we can dop the infimum ove s, n on the left-hand side. Futhemoe, the above inequality holds fo evey n N since ou fixed n was abitay. That is, fo all n N we have inf θ A θ y <nπ/3 θ y2 dµy =. By definition, this implies that fo all n N, thee exists θ n A such that 6 θ n y 2 dµy < n. θ n y <nπ/3

ON THE RECURRENCE OF RANDOM WALKS 2 Since A is compact, this means thee exists θ A and a subsequence {θ nk } of {θ n } such that θ nk θ. Now fo all k N and fo all y, let { θ nk y 2 if θ nk y < n kπ f k y = 3 else. Since n kπ 3 and θ nk θ as k, clealy fo all y we have that f k y θ y 2 as k. Since each f k is nonnegative on, by Fatou s Lemma this means 7 θ y 2 dµy lim inf k By 6, we have that lim inf k so 7 eads 8 f k ydµy = lim inf k θ nk y <n k π/3 θ nk y 2 dµy lim inf θ y 2 dµy =. θ nk y <n k π/3 k n k =, θ nk y 2 dµy. But this contadicts the fact that {S n } is tuly d-dimensional, as follows. Since θ, so by definition we have P θ X >, i.e. P θ X 2 > >. That is, < P X {y : θ y 2 > } = µ{y : θ y 2 > } meaning < {y :θ y 2 >} θ y 2 dµy since θ y 2 is stictly positive on {y : θ y 2 > } and that set has positive measue. So since fo all y, θ y 2 and since {y : θ y 2 > }, we cetainly have < θ y 2 dµy θ y 2 dµy, {y :θ y 2 >} a contadiction to 8. Thus, 4 holds. Now fo all >, let B = {x : x d}. Since by Theoem.7 ϕ = and ϕ is unifomly continuous on, we can choose, s d such that Reϕx > wheneve x B. Then cetainly Reϕx > fo all x [, ] d, as we know [, ] d B. Now let θ A and s, d]. Then by ou choice of we know s d < s. Since µ is a pobability measue, we also have dµ =. So since fo all z R, cosz, we have Reϕsθ = cossθ ydµy cossθ ydµy. sθ y <π/3 By Lemma 2., this means Reϕsθ sθ y <π/3 sθ y 2 4 dµy = s2 4 θ y <π/3s θ y 2 dµy.

22 SIMON LAZARUS Since s, s, applying 4 to the ight-hand side of the above thus gives Reϕsθ s2 4 p >. Since θ A and s, d] wee abitay, we see that fo all θ A and fo all s [, d] we have 9 < Reϕsθ 4 ps 2 fo if s = then this follows tivially fom. Now let c, and ecall that Reϕx > fo all x [, ] d. So since ϕ on, we see that fo all x [, ] d, the hypotheses of Lemma 3.8 ae satisfied with a + bi = ϕx. Thus, fo all x [, ] d we have This means that [,] d Re Re cϕx Reϕx. dx cϕx [,] Reϕx dx. d Now since ϕ on, cetainly fo all x B we have Reϕx. So since [, ] d B, the above inequality implies [,] d Re cϕx dx B Reϕx dx. By changing to spheical coodinates and noting that sint if t [, π], meaning we can ignoe all of the volume element of this change except fo the s d below and ecalling that A = {x : x = }, we theefoe have d dx s d [,] d Re cϕx [,] d Re A Reϕsθ dθds, which by 9 means d Re dx s d 4 [,] cϕx d A ps 2 dθds. Now let C = 4 p A dθ. Then C, since p > and since dθ is simply the A suface aea of the d -dimensional unit sphee. Thus, d dx C s d 3 ds. cϕx Since d 3, the integal on the ight-hand side of the above equals some D,. Thus, we have that fo any given c,, Re [,] d cϕx dx C D. So cetainly sup Re dx C D <. c, [,] cϕx d By Theoem 2.8, this means {S n } is tansient.

ON THE RECURRENCE OF RANDOM WALKS 23 Acknowledgments. I would like to thank Gegoy Lawle fo giving the lectues that spaked my inteest in andom walks. I would like to thank Bowei Zheng fo ecommending mateials on pobability theoy and andom walks fo me to study, fo patiently answeing my questions egading said mateial and othe topics, and fo poviding valuable input on dafts of this pape. Finally, I would like to thank all of the instuctos and oganizes of the 23 Univesity of Chicago Mathematics REU fo making the pogam as valuable and engaging as it was. Refeences [] K. L. Chung and W. H. J. Fuchs. On the distibution of values of sums of andom vaiables. Mem. Ame. Math. Soc. 95 : 6 95. [2] Rick Duett. Pobability: Theoy and Examples, 4th ed. Cambidge Univesity Pess, 2.