Exponential moments of first passage times and related quantities for random walks

For a zero-delayed random walk on the real line, let $\tau(x)$, $N(x)$ and $\rho(x)$ denote the first passage time into the interval $(x,\infty)$, the number of visits to the interval $(-\infty,x]$ and the last exit time from $(-\infty,x]$, 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, as $x \to \infty$.


Introduction and main results
Let (X n ) n≥1 be a sequence of i.i.d. real-valued random variables and X := X 1 . Further, let (S n ) n≥0 be the zero-delayed random walk with increments S n − S n−1 = X n , n ≥ 1. For x ∈ R, define the first passage time into (x, ∞) τ (x) := inf{n ∈ N 0 : S n > x}, ρ(x) := sup{n ∈ N : S n ≤ x}, if inf n≥1 S n ≤ x, 0, if inf n≥1 S n > x.
For typographical ease, throughout the text we write τ for τ (0), N for N (0) and ρ for ρ(0). Our aim is to find criteria for the finiteness of the exponential moments of τ (x), N (x) and ρ(x), and to determine the asymptotic behaviour of these moments, as x → ∞.
Assuming that 0 < EX < ∞, Heyde [11,Theorem 1] proved that See also [3,Theorem 2] and [6,Theorem 2] for relevant results. When P{X ≥ 0} = 1 and P{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 case when P{X < 0}P{X > 0} > 0, in which is that provided the abscissas of convergence of the moment generating functions of τ (x), N (x) and ρ(x) are positive there exists a unique value R > 0 such that typically In particular, typically Also we prove that whenever the exponential moments are finite they exhibit the following asymptotics: for explicitly given γ > 0 and distinct positive constants C i , 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, The same equivalence also holds for N (x) and ρ(x).
The following theorem provides sharp criteria for the finiteness of exponential moments of τ (x) and N (x) in the case when P{X < 0} > 0. Theorem 1.2. Let a > 0 and P{X < 0} > 0. Then the following conditions are equivalent: n≥1 e an n P{S n ≤ x} < ∞ for some (hence every) x ≥ 0; Ee aτ (x) < ∞ for some (hence every) x ≥ 0; Our next theorem provides the corresponding result for the last exit time ρ(x). Theorem 1.3. Let a > 0 and P{X < 0} > 0. Then the following conditions are equivalent: n≥0 e an P{S n ≤ x} < ∞ for some (hence every) x ≥ 0; Ee aρ(x) < ∞ for some (hence every) x ≥ 0; (8) where γ 0 is the unique positive number such that Ee −γ 0 X = e −R .
Now we turn our attention to the asymptotic behaviour of Ee aτ (x) , Ee aN (x) and Ee aρ(x) and start by recalling 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 for Ee aτ (x) . The phrase 'X is λ-lattice' used in formulations of Proposition 1.4 and Theorem 1.5 is a shorthand for 'The law of X is lattice with span λ > 0'.
where γ is a unique positive number such that Ee −γX = e −a , and in the λ-lattice case the limit is taken over x ∈ λN.
In particular, E γ X = 0. But even if a = R it can occur that E γ X > 0 or, equivalently, ϕ ′ (γ) < 0. Of course, then γ is 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.
(a) Assume that Ee aτ (x) < ∞ for some (hence every) x ≥ 0. Then E γ S τ is positive and finite, and, as x → ∞, Then E γ S τ is positive and finite, and, as x → ∞, (c) Assume that Ee aρ(x) < ∞ for some (hence every) x ≥ 0. Then M := inf n≥1 S n is positive with positive probability, and, as x → ∞, In the λ-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.

Proofs of the main results
Proof of Theorem 1.2. (6) ⇒ (3). Pick any a ∈ (0, R] and let γ be as defined on p. 4. With this γ, the equality where A ⊂ R is a Borel set, defines a measure which is finite on bounded intervals. Furthermore, according to [1, Proposition 1.1 and Theorem  (7) ⇔ (9) of Theorem 1.3).
(3) ⇒ (4). The argument given below will also be used in the proof of Theorem 1.5.
If (3) holds for some x ≥ 0 then, according to the already proved equivalence (3) ⇔ (6), first, a ≤ R and, secondly, (3) holds for every x ≥ 0. For 0 < a ≤ R and x ≥ 0, we have where L j = inf{i ∈ N 0 : S i = M j }, j ∈ N 0 . Since a ≤ R, we can use the exponential measure transformation introduced in (10), which gives Observe that L j = j holds iff j = σ k for some k ∈ N 0 where σ k (σ 0 := 0) denotes the kth strictly ascending ladder epoch of the random walk (S n ) n≥0 . Thus, where U > γ denotes the renewal function of the random walk (S σ k ) k≥0 under P γ , that is, U > γ (·) = k≥0 P γ {S σ k ∈ ·}. Thus, Z > γ (x) is finite for all x ≥ 0 since it is the integral of a directly Riemann integrable function with respect to U > γ . We already know that if the series in (3) converges for x = 0, i.e., if K(a) < ∞, then it converges for every x ≥ 0.
where P{inf n≥1 S n > 0} > 0, since, under the present assumptions, (S n ) n≥0 drifts to +∞ a.s. Hence, Ee aN < ∞. Further, for y ∈ R, is a copy of N (y) that is independent of (τ (x), S τ (x) ). We have Hence, Ee aN (x) < ∞, for every x ≥ 0. The proof is complete.
Proof of Theorem 1.3. The equivalence (7) ⇔ (9) has been proved in [12, Theorem 2.1]. (7) ⇒ (8). According to the just mentioned equivalence, if (7) holds for some x ≥ 0 it holds for every x ≥ 0. It remains to note that for x ≥ 0 . Suppose Ee aρ(x 0 ) < ∞ for some x 0 ≥ 0 and a > 0. Since Ee aρ(x) is increasing in x, we have Ee aρ < ∞. Condition a ≤ R must hold in view of (2) and implication (4) ⇒ (6) of Theorem 1.2. If a < R, we are done. In the case a = R it remains to show that Define the measure V by for Borel sets A ⊂ R. Then from (20) we infer that Under the present assumptions, the random walk (S n ) n≥0 drifts to +∞ a.s. Thus, P{inf n≥1 S n > ε} > 0 for some ε > 0. With such an ε, Therefore, Hence (S n ) n≥0 must be transient under P γ 0 , which yields the validity of (9) in view of (11) and E γ 0 S 1 = e R EXe −γ 0 X . The proof is complete.
Proof of Theorem 1.5. (a) In view of (15), (16) and (17), in order to find the asymptotics of Ee aτ (x) , it suffices to determine the asymptotic behaviour of Z > γ (x) defined in (17). By the key renewal theorem on the positive half-line, where the limit x → ∞ is taken over x ∈ λN when X is lattice with span λ > 0. It remains to check that E γ S τ is finite. As pointed out in (11), either E γ X ∈ (0, ∞) or E γ X = 0. In the first case, S n → ∞ a.s. under P γ and, therefore, E γ τ < ∞, see, for instance, [4, Theorem 2, p. 151], which yields E γ S τ < ∞ by virtue of Wald's identity. If, on the other hand, E γ X = 0, then E γ τ = ∞ and we cannot argue as above. But in this case, by [5,Formula (4a) implies the finiteness of E γ (S + 1 ) 2 , and the proof of part (a) is complete. (b) We only consider the case when X is non-lattice since the lattice case can be treated similarly. Denote by R x := S τ (x) − x the overshoot. Since Ee aτ (x) = E γ e γS τ (x) , we have in view of the already proved part (a) By Theorem 1.2, if Ee aN (x) < ∞, then Ee aτ (x) < ∞. Therefore, according to part (a), we have 0 < E γ S τ < ∞. This implies (see, for instance, [10, Theorem 10.3 on p. 103]) that, as x → ∞, R x converges in distribution to a random variable R ∞ satisfying In particular, under P γ , e γRx converges in distribution to e γR∞ . Further, Therefore, (25) can be rewritten as follows: 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. that R x → R ∞ P γ -a.s. From (19) we infer that for f (y) := Ee aN (y) , y ∈ R we have f is an increasing function and, therefore, has only countably many discontinuities. Hence e γRx f (−R x ) converges P γ -a.s. to e γR∞ f (−R ∞ ). Further, and e γRx f (0) converges P γ -a.s. to e γR∞ f (0). Finally, Therefore the assumptions of Pratt's lemma are fulfilled and an application of the lemma yields (c) From (20) and (22) (with R replaced by a and M = inf k≥1 S k ), we infer Assume that X is non-lattice and set D 1 := e −a γEXe −γX . It follows from (9) that D 1 ∈ (0, ∞) and from [12,Theorem 2.2 The latter implies that for any ε > 0 there exists an x 0 > 0 such that Letting first x → ∞ and then ε → 0 we conclude that Together with (27) the latter yields Under the present assumptions, the random walk (S n ) n≥0 drifts to +∞ a.s. Therefore, P{M > 0} > 0 which implies that 1−Ee −γM + > 0 and completes the proof in the non-lattice case. The proof in the lattice case is based on the lattice version of [12, Theorem 2.2] and follows the same path.

Examples
In this section, retaining the notation of Section 1, we illustrate the results of Theorem 1.