{
  "id": "2108.08267",
  "version": "v1",
  "published": "2021-08-18T17:37:45.000Z",
  "updated": "2021-08-18T17:37:45.000Z",
  "title": "On the existence of moments of the exit time from the positive halfline for a random walk with negative drift",
  "authors": [
    "Sergey Foss",
    "Timofej Prasolov"
  ],
  "comment": "7 pages",
  "categories": [
    "math.PR"
  ],
  "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 $\\varepsilon \\in (0,1)$.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2021-08-18T17:37:45.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "60G50",
      "60G40",
      "60K25"
    ],
    "keywords": [
      "random walk",
      "positive halfline",
      "negative drift",
      "finiteness",
      "first exit time"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 7,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}