{
  "id": "math/0507034",
  "version": "v1",
  "published": "2005-07-02T06:24:36.000Z",
  "updated": "2005-07-02T06:24:36.000Z",
  "title": "Levy Processes: Hitting time, overshoot and undershoot - part I: Functional equations",
  "authors": [
    "Bernard Roynette",
    "Pierre Vallois",
    "Agnes Volpi"
  ],
  "comment": "Manuscript P660 submitted to SPA, October 2004. 30 pages, 5 figures, 25 references",
  "categories": [
    "math.PR"
  ],
  "abstract": "Let (X_t, t >=0) be a Levy process started at 0, with Levy measure nu, and T_x the first hitting time of level x>0: T_x := inf{t>=0; X_t>x}. Let F(theta,mu,rho,.) be the joint Laplace transform of (T_x, K_x, L_x): F(theta,mu,rho,x) := E (e^{-theta T_x - mu K_x - rho L_x} 1_{T_x<+infinity}), where theta>=0, mu>=0, rho>=0, x>0, K_x := X_{T_x} - x and L_x := x - X_{T_{x^-}}. If nu(R) < + \\infinity and integral_1^{+\\infty} e^{sy} nu (dy) < +infinity for some s>0, then we prove that F(theta,mu,rho,.) is the unique solution of an integral equation and has a subexponential decay at infinity when theta>0 or theta=0 and E(X_1)<0. If nu is not necessarily a finite measure but verifies integral_{-infinity}^{-1} e^{-sy} nu (dy) < +infinity for any s>0, then the x-Laplace transform of F(theta,mu,rho,.) satisfies some kind of integral equation. This allows us to prove that F(theta,mu,rho,.) is a solution to a second integral equation.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2005-07-02T06:24:36.000Z"
    }
  ],
  "analyses": {
    "keywords": [
      "levy processes",
      "functional equations",
      "undershoot",
      "levy measure nu",
      "joint laplace transform"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 30,
      "language": "en",
      "license": "arXiv",
      "status": "editable",
      "adsabs": "2005math......7034R"
    }
  }
}