getdocd78f. 161KB Jun 04 2011 12:05:02 AM

in PROBABILITY

EXPONENTIAL MOMENTS OF FIRST PASSAGE TIMES AND

RE-LATED QUANTITIES FOR RANDOM WALKS

ALEXANDER IKSANOV

Faculty of Cybernetics, National T. Shevchenko University of Kiev, 01033 Kiev, UA

email: iksan@unicyb.kiev.ua

MATTHIAS MEINERS1

Department of Mathematics, Uppsala University, Box 480, 751 06 Uppsala, SE

email: matthias.meiners@math.uu.se

Submitted28 May 2010, accepted in final form13 September 2010 AMS 2000 Subject classification: 60K05, 60G40

Keywords: first-passage time, last exit time, number of visits, random walk, renewal theory

Abstract

For a zero-delayed random walk on the real line, letτ(x),N(x)andρ(x)denote the first passage time into the interval(x,_∞), the number of visits to the interval(_−∞,x]and the last exit time from(_−∞,x], respectively. In the present paper, we provide ultimate criteria for the finiteness of exponential moments of these quantities. Moreover, whenever these moments are finite, we derive their asymptotic behaviour, asx_{→ ∞}.

1

Introduction and main results

Let(Xn)n≥1be a sequence of i.i.d. real-valued random variables andX:=X1. Further, let(Sn)n≥0

be the zero-delayed random walk with incrementsSn−Sn−1=Xn,n≥1. For x∈R, define the

first passage timeinto(x,_∞)

τ(x) := inf{n∈N_0:Sn>x}, thenumber of visitsto the interval(_−∞,x]

N(x) := #_{n_∈N_{:Sn≤x} =}

X

n≥1 ✶_{Sn≤x},

and thelast exit timefrom(_−∞,x]

ρ(x) :=

sup_{n∈N:Sn≤x}, if infn≥1Sn≤x,

0, if inf_n≥1S_n>x.

Note that, forx≥0,

ρ(x) = sup{n∈N_0:Sn≤x}.

1_{RESEARCH SUPPORTED BY DFG-GRANT ME 3625/1-1}

For typographical ease, throughout the text we writeτforτ(0),NforN(0)andρforρ(0). Our aim is to find criteria for the finiteness of the exponential moments ofτ(x),N(x)andρ(x), and to determine the asymptotic behaviour of these moments, asx→ ∞.

Assuming that 0<E_{X<∞, Heyde[11, Theorem 1]proved that}

E_eaτ(x)_{<∞for somea>0 iff} E_ebX−_{<∞for some b>0.} See also[3, Theorem 2]and[6, Theorem II]for relevant results.

WhenP_{{X≥0}=1 and}P_{{X =0}<1,}

τ(x)₋1 = N(x) = ρ(x), x_≥0. (1) Plainly, in this case, criteria for all the three random variables are the same (Proposition 1.1). An intriguing consequence of our results in the case whenP_{{X<0}>0 and}P_{{X>0}>0, in which}

τ(x)₋1 _≤ N(x) _≤ ρ(x), x_≥0, (2)

is that provided the abscissas of convergence of the moment generating functions ofτ(x),N(x) andρ(x)are positive there exists a unique valueR>0 such that

E_eaτ(x)_<∞, E_eaN(x)_{<∞iffa≤R, and}E_eaρ(x)_<∞ifa<R

whereasE_eRρ(x) is finite in some cases and infinite in others. Also we prove that whenever the exponential moments are finite they exhibit the following asymptotics:

E_eaτ(x) _{∼ C1e}γx_, E_eaN(x) _{∼ C2e}γx_, E_eaρ(x) _{∼ C3e}γx_{, x→ ∞,}

for an explicitly given γ > 0 and distinct positive constants Ci, i = 1, 2, 3 (when the law of X

is lattice with span λ >0 the limit is taken over x ∈λN_{). Our results should be compared (or}

contrasted) to the known facts concerning power moments (see [13, Theorem 2.1 and Section

4.2]and[13, Theorem 2.2], respectively): forp>0

E_(τ(x))p+1_{<∞ ⇔} E_(N(x))p_{<∞ ⇔} E_(ρ(x))p_<∞;

E_(τ(x))p _≍ E_(N(x))p _≍ E_(ρ(x))p _≍

x Emin(X+_,x)

p

, x_{→ ∞}

where f(x)_≍g(x)means that 0<lim inf

x→∞ f(x)

g(x)≤lim sup x→∞

f(x) g(x)<∞.

Proposition 1.1 is due to Beljaev and Maksimov[2, Theorem 1]. A shorter proof can be found in [12, Theorem 2.1].

Proposition 1.1. Assume thatP_{{X ≥0}=1and letβ:=}P_{{X =0} ∈[0, 1). Then for a>0the} following conditions are equivalent:

E_eaτ(x)_{<∞for some (hence every) x≥0;} a<₋logβ

Our first theorem provides sharp criteria for the finiteness of exponential moments ofτ(x)and N(x)in the case whenP_{{X<0}>0. Before we present it, we introduce some notation. Let}

ϕ:[0,_∞)_→(0,_∞], ϕ(t):=E_e−t X ₍₃₎

be the Laplace transform ofX and

R:=₋log inf

t≥0ϕ(t). (4)

Theorem 1.2. Let a>0andP_{{X<0}>0. Then the following conditions are equivalent:}

X

n≥1

ean

n P{Sn≤x} < ∞for some (hence every) x≥0; (5) E_eaτ(x)_{<∞for some (hence every) x≥0; (6)}

E_eaN(x)<_∞for some (hence every) x≥0; (7)

a_≤R. (8)

Our next theorem provides the corresponding result for the last exit timeρ(x).

Theorem 1.3. Let a>0andP_{{X<0}>0. Then the following conditions are equivalent:}

X

n≥0

ean_P{S

n≤x}<∞for some (hence every) x≥0; (9)

E_eaρ(x)_{<∞for some (hence every) x≥0; (10)} a<R or a=R andϕ′(γ₀)<0 (11) whereγ₀is the unique positive number such thatϕ(γ₀) =e−R_.

It is worth pointing out that Theorem 1.3 and Proposition 1.1 could be merged into one result. Indeed, if one setsγ₀:=_∞andϕ′_(∞):=lim

t→∞ϕ′(t)(=0)in the case thatP{X <0}=0, then

(11) includes the criterion given in Proposition 1.1 for the finiteness ofE_eaτ(x)_{which, in this case,} is equivalent to the finiteness ofE_eaρ(x)_{due to Eq. (1).}

Now we turn our attention to the asymptotic behaviour ofE_eaτ(x)_,E_eaN(x)_andE_eaρ(x) _{and start} by quoting a known result which, given in other terms, can be found in[12, Theorem 2.2]. In view of equality (1) we only state it forE_eaτ(x)_{. The phrase ‘The law of X is λ-lattice’ used in} Proposition 1.4 and Theorem 1.5 is a shorthand for ‘The law ofX is lattice with spanλ >0’. Proposition 1.4. Let a>0,P_{{X ≥0}=1and}P_{{X =0}<1. Assume that}E_eaτ(x)_{<∞for some} (hence every) x_≥0. Then, as x_{→ ∞},

E_eaτ(x) _{∼ e}γx _×

( _1−e−a

γE_{X e}−γX, if the law of X is non-lattice,

λ(1−e−a₎

(1−e−λγ₎EX e−γX, if the law of X isλ-lattice

whereγis the unique positive number such thatϕ(γ) =E_e−γX _=e−a_{, and in theλ-lattice case the} limit is taken over x∈λN_.

When 0<a_≤RandP_{{X<0}>0, there exists a minimalγ >0 such thatϕ(γ) =e}−a_{. Thisγcan} be used to define a new probability measureP_γby

for each nonnegative Borel function h onRn+1_{, where} E_{γ denotes expectation with respect to} P_{γ. Since}E_{γX =} E_{γS1 =−e}a_ϕ′_(γ)(whereϕ′ _{denotes the left derivative of ϕ) and since ϕ is} decreasing and convex on[0,γ], there are only two possibilities:

Either E_{γX ∈(0,∞) or} E_{γX=0. (13)}

When a<R, then the first alternative in (13) prevails. Whena= R, then typicallyϕ′(γ) =0 sinceγis then the unique minimizer ofϕ on[0,_∞). In particular,E_{γX =0. But even ifa=Rit} can occur thatE_{γX >0 or, equivalently,ϕ}′_{(γ)<0. Of course, thenγis the right endpoint of the} interval_{t_≥0 :ϕ(t)<_∞}. We provide an example of this situation in Section 3.

Now we are ready to formulate the last result of the paper.

Theorem 1.5. Let a>0andP_{X <0_}>0.

(a) Assume thatE_eaτ(x)_{<∞for some (hence every) x≥0. Then}E_{γSτis positive and finite, and,} as x_{→ ∞},

E_eaτ(x) _{∼ e}γx _×

  

E(eaτ

−1)

γE_γSτ , if the law of X is non-lattice,

λE(eaτ₋₁₎

(1−e−λγ₎E_γSτ, if the law of X isλ-lattice.

(14)

(b) Assume thatE_eaN(x)_{<∞for some (hence every) x≥0. Then}E_{γSτis positive and finite, and,} as x→ ∞,

E_eaN(x) _{∼ e}γx _×

  

e−a_E γ

RSτ

0 e

γy_E[eaN(−y)_]dy

E_γSτ , if the law of X is non-lattice, λe−aE_γPSτ /λ

k=1 eγλkE[eaN(−λk)]

E_γSτ , if the law of X is

λ-lattice.

(15)

(c) Assume thatE_eaρ(x)_{<∞for some (hence every) x≥0. Then M :=inf}

n≥1Snis positive with

positive probability, and, as x_{→ ∞},

E_eaρ(x) _{∼ e}γx _×

  

e−a₍₁₋E_e−γM+)

γEX e−γX , if the law of X is non-lattice,

λe−a₍₁

−Ee−γM+₎

(1−e−λγ₎E_{X e}−γX, if the law of X isλ-lattice.

(16)

In theλ-lattice case the limit is taken over x∈λN_.

The rest of the paper is organized as follows. Section 2 is devoted to the proofs of Theorems 1.2, 1.3 and 1.5. In Section 3 we provide three examples illustrating our main results.

2

Proofs of the main results

Proof of Theorem 1.2. (8)⇒(5). Pick anya∈(0,R]and letγbe as defined on p. 367. With this

γ, the equality

Zγ(A) :=

X

n≥1

whereA_⊂R_{is a Borel set, defines a measure which is finite on bounded intervals. Furthermore,}

as equivalence (9)_⇔(11) of Theorem 1.3).

(5)_⇒(6). The argument given below will also be used in the proof of Theorem 1.5.

If (5) holds for some x _≥0 then, according to the already proved equivalence (5) _⇔(8), first, a_≤Rand, secondly, (5) holds for everyx_≥0. For 0<a_≤Randx_≥0, we have

transformation introduced in (12), which gives

ea jP_{{Lj= j,Sj≤x} =} E_γeγSj✶

γ(x)is finite for all x≥0 since it is the integral of a directly Riemann

integrable function with respect toU_γ>.

By a generalization of Spitzer’s formula[6, Formula (2.6)], the assumptionE_eaτ_{<∞immediately} entails the finiteness ofK(a):

∞ > E_eaτ = 1+ (ea₋₁₎X n≥0

eanP_{Mn=0} = 1+ (ea_−1)eK(a)_.

We already know that if the series in (5) converges for x=0,i.e., ifK(a)<_∞, then it converges for everyx_≥0.

(6)_⇒(7). By the equivalence (5)_⇔(6),E_eaτ(x)_{<∞for everyx≥0. According to[13, Formula} (3.54)],

P_{{N=k} =} P_{inf

n≥1Sn>0}

P_{{τ >k}, k∈}N_0, whereP_{inf

n≥1Sn>0}>0, since, under the present assumptions,(Sn)n≥0drifts to+∞a.s. Hence,

E_eaN_{<∞. Further, for y∈}R_,

b

N(x,y) := X

n>τ(x)

✶_{Sn−Sτ(x)≤y} (20)

is a copy ofN(y)that is independent of(τ(x),Sτ(x)). We have

N(x) = τ(x)₋1+Nb(x,x−Sτ(x)) ≤ τ(x) +Nb(x, 0) (21)

Hence,E_eaN(x)<_∞, for everyx≥0. The proof is complete.

Proof of Theorem 1.3. The equivalence (9)_⇔(11) has been proved in[12, Theorem 2.1]. (9)_⇒(10). According to the just mentioned equivalence, if (9) holds for some x_≥0 it holds for everyx_≥0. It remains to note that for x_≥0

P_{ρ(x) =n} =

Z

(−∞,x]

P_{inf

k≥1Sk>x−y}

P_{{Sn∈dy} ≤} P_{{Sn≤x}. (22)}

(10)⇒ (11). SupposeE_eaρ(x0) <_∞for some x0≥0 anda>0. SinceEeaρ(x)is increasing in

x, we haveE_eaρ<_∞. Conditiona≤Rmust hold in view of (2) and implication (6)_⇒ (8) of Theorem 1.2. Ifa<R, we are done. In the casea=Rit remains to show that

E_{X e}−γ0X>0. (23)

Define the measureV by

V(A) := X

n≥0

eRnP_{{Sn∈A}, (24)}

for Borel setsA_⊂R_{. Then from (22) we infer that}

∞ > E_eRρ =

Z

(−∞,0]

P_{inf

n≥1Sn>−y}V(dy). (25)

Under the present assumptions, the random walk(S_n)_n≥0drifts to+_∞a.s. Therefore,P_{inf

n≥1Sn> ǫ_}>0 for someǫ >0. With such anǫ,

∞ >

Z

(−ǫ,0]

P_{inf

n≥1Sn>−y}V(dy) ≥

P_{inf

Thus,

0, which yields the validity of (11) in view of (13) and

E_γ

0S1=e

R_{EX e}−γ0X_{. The proof is complete.}

Proof of Theorem 1.5. (a) In view of (17), (18) and (19), in order to find the asymptotics of E_eaτ(x), it suffices to determine the asymptotic behaviour of Z_γ>(x)defined in (19). By the key renewal theorem on the positive half-line,

Z_γ>(x) _→ implies the finiteness ofE

γ(S+1)2, and the proof of part (a) is complete.

(b) We only consider the case when the law ofX is non-lattice since the lattice case can be treated similarly. Denote byR_x :=Sτ(x)−xthe overshoot. SinceEeaτ(x)=EγeγSτ(x), we have in view of

the already proved part (a)

lim

converges in distribution to a random variableR_∞satisfying P_γ{R Therefore, (27) can be rewritten as follows:

lim

x→∞

E_γeγRx ₌ E_γeγR∞_{. (28)}

Now we invoke a variant of Fatou’s lemma sometimes called Pratt’s lemma[14, Theorem 1]. To this end, note that, by a standard coupling argument, we can assume w.l.o.g. thatRx→R∞Pγ-a.s.

From (21) we infer that for f(y):=E_eaN(y)_{, y∈}R_{we have}

f is an increasing function and, therefore, has only countably many discontinuities. Hence

Therefore the assumptions of Pratt’s lemma are fulfilled and an application of the lemma yields

lim

and from[12, Theorem 2.2]that

V(x) _∼ D1eγx, x→ ∞. (29)

The latter implies that for anyǫ >0 there exists an x0>0 such that

Together with (29) the latter yields The proof in the lattice case is based on the lattice version of[12, Theorem 2.2]and follows the same path.

3

Examples

In this section, retaining the notation of Section 1, we illustrate the results of Theorem 1.2 and Theorem 1.3 by three examples.

Example 3.1(Simple random walk). Let 1/2<p<1 andP_{X=1}=p=1−P_{X=₋1}=: 1₋q. Then the Laplace transformϕofX is given byϕ(t) =pe−t+qet andR=₋log(2ppq). According to[8, Formula (3.7) on p. 272]and[7, Example 1], respectively,

P_{τ=2n−1} = 1

Example 3.2. LetX=d Y₁₋Y2whereY1andY2are independent r.v.’s with exponential distributions

with parameters α andκ, respectively, 0 < α < κ. Then ϕ(t) =E_e−t X ₌ ακ

According to[9, Formula (5.9) on p. 410],

Note thatSnhas the same law as the difference of two independent random variables with gamma

distribution with parameters (n,α)and (n,κ), respectively, which particularly implies that, for x>0, the density ofS1takes the form ακe Example 3.3. Fix h> 0 and take any probability law µ₁ on R _{such that the Laplace-Stieltjes} transform

positive. For instance, one can take

µ1(dx) := ce−h|x|/(1+|x|r)dx, x∈R

wherer>2 andc:= R_Re−h|x|(1+_|x|r₎−1_dx−1_>0.

Now chooses sufficiently large such that ψ′_{(h)<sψ(h). Then ϕ(t) =e}−st_{ψ(t)is the}

Laplace-Stieltjes transform of the distributionµ:=δ_s∗µ₁. LetX be a random variable with distribution

µ. Plainly,ϕ(t)is finite for 0≤t≤hbut infinite fort>h. Furthermore,

ϕ′(t) = e−st(ψ′(t)₋sψ(t)), _|t_{| ≤}h.

In particular,ϕ′_{(h)<0 which, among other things, implies thatR=−logϕ(h)and thatγ} 0=h.

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