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 $f$.

doestell 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 Aug 21 '17 at 8:57