Moments of the first descending epoch for a random walk with negative drift

TL;DR

This paper investigates the moments of the first exit time for a negatively drifting random walk, establishing conditions under which certain exponential moments of the exit time are finite, especially in intermediate cases.

Contribution

It introduces new conditions on the tail behavior of the positive part of the increments that determine the finiteness of exponential moments of the exit time.

Findings

Finiteness of exponential moments depends on the growth rate of the function g.

Under certain conditions, exponential moments of g((1-ε)aτ) are finite.

The results fill a gap between known cases of moments and exponential moments.

Abstract

We consider the first exit time $\tau = \min \{n\ge 1 : S_n\le 0\}$ from the positive halfline of a random walk $S_n = \sum_1^n \xi_i, n\ge 1$ with i.d.d. summands having a negative drift ${\mathbb E} \xi = -a< 0$. Let $\xi^+ = \max (0, \xi_1)$. It is well-known that, for any $c>1$, the finiteness of ${\mathbb E}(\xi^+)^{c}$ implies the finiteness of ${\mathbb E} \tau^c$ and, for any $c>0$, the finiteness of ${\mathbb E} \exp({c\xi^+})$ implies that of ${\mathbb E} \exp({c'\tau})$ where $c'>0$ is, in general, another constant that depends on $c$ and on the distribution of $\xi_1$. We consider the intermediate case, assuming that ${\mathbb E} \exp({g(\xi^+)})<\infty$ for a positive increasing function $g$ such that $\liminf_{x\to\infty} g(x)/\log x = \infty$ and $\limsup_{x\to\infty} g(x)/x =0$, and that ${\mathbb E} \exp({c\xi^+})=\infty$, for all $c>0$. Assuming a few further technical assumptions, we show that then ${\mathbb E} \exp({(1-\varepsilon){g}((1-\varepsilon)a\tau)})<\infty$, for any $\varepsilon \in (0,1)$.