Erratum: The remainder in the renewal theorem *

We point out an error in "The remainder in the renewal theorem", and show that the result is essentially correct in two important special cases.

Since Φ ∈ RV (−β) the RHS of (0.1) ∈ RV (1−2β), so this is a substantial improvement on the previously known result that W (x) = o( x 0 Φ(y)dy), particularly for the case β > 1/2. If F is non-lattice, it is natural for φ to be involved, since it is the stationary density for the overshoot process of S, which fact is used in [1] to derive the following relation. First write φ 2 for the convolution φ * φ and define real-valued functions g and G on [0, ∞) by g(y) = 2φ(y) − φ 2 (y), Then the relation which is (2.4) in [1], is key to the results therein. (Note that our W is denoted by m −1 V in [1].) The second crucial fact is that although G is the difference of two functions which are both in RV (−β), it is in RV (−2β), and actually Unfortunately there are mistakes in the proof of (0.1) for the case α ∈ (3/2, 2). Specifically on P5, L9 of [1], it is claimed that having fixed x 0 > 0 such that g * (x) := −g(x) > 0 for x > x 0 , then given ε > 0 we can find where G * (z) := −G(z). But on [0, x 0 ] we have no control over the sign of G * , so this statement cannot be justified. It is also unclear how Lemma 2.1 can be applied, since the condition ∞ 0 Q(y)dy = ∞ fails for α > 3/2. A final error is that it is implicitly assumed in [1] that in the lattice case φ is a stationary density, but of course this is wrong: actually {φ(n), n ∈ Z} is a stationary mass function.
Nevertheless the claimed result (0.1) is essentially correct in the two most important situations. In the lattice case the last mentioned error necessitates a slight change in the definition of W , (see (1.1) below and compare the LHS of (0.1)), but then we are able to give a simple argument to show that (0.1) holds with this new definition. In the absolutely continuous case, under a minor technical assumption we show that (0.1) follows, after some manipulation and use of (0.3), from a stronger result for the density of U which is established in [2].

Lattice case
In this section we assume that F is carried by Z and has period 1, and we specify that the renewal function U and its modification W are given for x ≥ 0 by (1 + Φ(s)), (1.1) where u(r) = ∞ 0 P (S n = r). We start from the observation that the distribution with mass function is stationary for the overshoot process. Φ is the tail function of this discrete distribution, With this definition it is clear that W is piece-wise constant, and it follows that (0.1) will hold in general if it holds as x → ∞ through the integers, and we will now establish this. Again the functions g and G are defined by g(n) = 2φ(n) − φ 2 (n), n = 0, 1, 2, · · · where φ 2 (n) is the discrete convolution    This is the discrete analogue of (0.3). Next we see that the proof in [1] of (0.4) is also valid in this lattice case, with minor changes. Finally when β > 1/2 the condition ∞ 0 G(y)dy = 0 also holds, but because of (1.2) it is equivalent to  It is straightforward to see that the results in Theorem 1.1 of [1] when β ≤ 1/2 hold in this lattice case with x restricted to the integers, and we now show that the same is true when β > 1/2. Recalling that c α < 0 in this case, so that G * (n) = −G(n) is positive for all large n, we assume we know that | n 0 (u n − u n−r )G(r)| = o(nG * (n)) as n → ∞.
which is the required result. Next suppose that with ∆ n := u n − u n−1 we have n∆ n → 0 as n → ∞. For any fixed δ ∈ (0, 1) we can bound the LHS of (1.5) by S 1 + S 2 + S 3 , where Then (1.5) follows by letting n → ∞ and then δ → 0. The fact that n∆ n → 0 can be seen by an application of the Riemann-Lebesgue Lemma: we have the inversion formula (1.6) Integrating by parts and noting that since everything is periodic with period 2π the contribution from the end points cancel, gives (1.8) Known results (see e.g. [3]) give the asymptotic behaviour ofp(t) andp (t) as |t| → 0 and from them we see that |f 1 | is regularly varying as |t| → 0 with index a − 2 > −1.
We deduce that f 1 is integrable over [−π, π] and the result follows.
Remark 1.1. Alternatively, we could appeal to a stronger result on the asymptotic behaviour of ∆ n in [4], but the proof there uses Banach Algebra techniques.

The absolutely continuous case
Assuming that F has a density f and the characteristic functionp(t) = E(e itX ) is such that |p(t)| b is integrable for some b ≥ 1, Isozaki [2] has used an inversion theorem to find an asymptotic estimate of the density u of the renewal measure. This estimate, which is actually valid in the random walk case whenever E|X| γ < ∞ for some γ ∈ (3/2, 2), when specialised to the renewal case becomes Here N is the smallest integer ≥ b + 1, f j is the density of S j , and it is necessary to check, by integration by parts, that the function denoted by r 1 in [2] agrees with our m −1 G. Integrating (1.9) and noting that and this is valid whenever the γ-th moment exists, for some γ ∈ (3/2, 2). In particular under our assumption of asymptotic stability with index α ∈ (3/2, 2), we can choose any γ = α − δ with δ sufficiently small that 1 − γ = δ − β < 1 − 2β and γ > 3/2. Then (0.1) follows, since we know the first term dominates the RHS of (1.11) and we can read off its asymptotic behaviour from (0.4).