The remainder in the renewal theorem

Ron Doney

doi:10.1214/20-ecp287

Abstract

If the step distribution in a renewal process has finite mean and regularly varying tail with index $-\alpha $, $1<\alpha <2$, the first two terms in the asymptotic expansion of the renewal function have been known for many years. Here we show that, without making any additional assumptions, it is possible to give, in all cases except for $\alpha =3/2$, the exact asymptotic behaviour of the next term. In the case $\alpha =3/2$ the result is exact to within a slowly varying correction. Similar results are shown to hold in the random walk case.

Highlights

Introduction and ResultsWe consider a renewal process (Sn, n ≥ 0), i.e. a random walk with nonnegative, i.i.d increments X1,X2, · · · with a distribution F whose tail F ∈RV (−α) where α ∈ (1, 2] and we assume ∞ 0 y2dF (y) = ∞ if α We write EX1m and define a distribution Φ via its density function φ(y) =
Similar arguments deal with the case β = 1/2
∞ y gr (z )dz and the sequence of functions gr are defined by g2 = g = 2φ − φ ∗ φ and gr+1 = φ + gr − φ ∗ gr, r ≥ 2

Summary

Introduction and Results

We consider a renewal process (Sn, n ≥ 0), i.e. a random walk with nonnegative, i.i.d increments X1,X2, · · · with a distribution F whose tail F ∈. Later Sgibnev showed, in [3], that (3) holds whenever m is finite and EX12 = ∞, so that the assumption of a regularly varying tail is redundant. This in turn suggests that if we do make this assumption we should be able to improve on (3). Since g is bounded in absolute value by the integrable function 2φ + φ2, we can interchange orders of integration to see that (1 − g(λ))/λ = e−λxG(x)dx, and the conclusion follows by letting λ go to 0. For the case β ∈ (1/2, 1) we write g∗, G∗ for −g, −G, and we claim first that.

The Random walk case

Concluding remarks