3
$\begingroup$

Let us define the compound Poisson process $$Y_{N_t}=a+\sum_{n=1}^{N_t}X_n$$ where $X_n \sim f(x)$ such that $\langle X_n \rangle > 0$ $\forall n$ and $N_t$ is an independent Poisson random variable of parameter $\lambda t$. I want to study the survival probability above $0$ at infinite time $\mathbb{P}(Y_{N_t}\geq0, \; \forall t\;|\;a)=S(a)$. This turns out to be very similar to the Cramér-Lundberg model for ruin / collective risk theory: $$\tilde{Y}_{N_t}=a+ct-\sum_{n=1}^{N_t}\tilde{X}_n \quad \text{where } \langle \tilde{X}_n \rangle > 0$$ for which we know some asymptotic estimations (cf. intro here) of the ruin probability (complementary to the survival). I wonder if we could define $$X_n=cT_n-\tilde{X}_n \quad \text{where } T_n \sim \lambda e^{-\lambda t}\; t>0$$ to use the result of the Cramér-Lundberg model for the random walk (which looks different because $T_n$ is certainly not independent of $N_t$).

This question is a reformulation of some aspects of these related questions: 1 and 2 where we have set a renewal equation for $S(a)$. I find in this paper a quite different renewal equation for the risk model (p. 67) which suggests that the solution is different, of course. But I would expect the same asymptotic. Any comment about this would be appreciated.

EDIT :

  • The function $f$ can be any distribution, but I am first interested in pdf of the form $f=\sum_{i=0}^n a_i \delta_{+c_i}+b_i \delta_{-c_i}$.

  • The motivation for this would be to estimate the first moment of the cdf in $a$ from $f$.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ This is the basic question in fluctuation theory of Lévy processes, which provides you with a lot of tools to study it. For compound Poisson processes, the "discrete" variant for random walks is typically sufficient, and a standard reference here is Feller. For processes in continuous time, see either Bertoin's book on Lévy processes, or any book that combines "fluctuation" and "Lévy" in its title. In principle, if $\langle X_n\rangle>0$, then the survival probability has a positive limit at infinity, so I suppose you meant $\langle X_n\rangle<0$. (to be continued) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 20, 2017 at 19:29
  • $\begingroup$ (second part of the above comment) In this case the asymptotic behaviour of the survival probability depends on the properties of the distribution of $X_n$, and in order to tell something more, it would be convenient to know what do we assume about (the tail of) $X_1$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 20, 2017 at 19:32
  • $\begingroup$ Thank you for these references. Yes I am interested in the case of a positive bias, since the survival probability is zero otherwise, and I expect a distribution which is asymptotically exponential. I will add some details. $\endgroup$
    – Alexandre
    Commented Aug 20, 2017 at 19:54
  • $\begingroup$ Ok, reading about fluctuation of Lévy processes, I found a theorem: Cramér's estimate (p.207), which gives the conditions for the exponential behaviour of the tail of the survival probability. This is sufficient to say that its moments are finite, but we cannot say much more. For the moment, I can not find a generalization of the Lundberg inequality which states the conditions such that the whole survival probability is bounded by above by the exponential distribution (with the same exponent as the tail). $\endgroup$
    – Alexandre
    Commented Aug 20, 2017 at 23:52
  • 2
    $\begingroup$ Well, it does tell you more. If you look into the proof of the result that you linked to, you will realise that $e^{\nu x}\mathbf{P}(\overline{X}_\infty<x)=\mathbf{P}(\overline{X}_\infty^{(\nu)}<\infty)$ for a Lévy process $X_t^{(\nu)}$ obtained by an exponential change of measure. This implies that if $X_t$ is as in your question, then $e^{\nu a}(1-S(a))$ is the tail of survival probability of a different compound Poisson process (with jumps distributed as $ce^{-\nu x}f(x)$ for appropriate $\nu$ and $c$). In particular, $1-S(a)\leqslant e^{-\nu a}$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 21, 2017 at 8:57

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .