# Compound Poisson Process: Stopping time distribution

Let $X_t$ be a compound Poisson process, i.e.

$$X_t= \sum_{i=0}^{N(t)} Y_i,$$ where $N(t)$ is a Poisson process and $Y_i$ are iid.

I have seen how to calculate the expected value $E(X_t)$ which depends only on the expected value of the $Y$'s and the parameter of the Poisson process.

I am however interested in the time expectation/ time distribution of this process for a given threshold. Let $x>0$ and define the stopping time $T_x= \inf \{t| X_t \geq x\}$.

My question is: How is $T_x$ distributed and what is the expectation of $T_x$. For my application I can assume that I know the distribution function of $Y_i$. Furthermore, we can assume that $Y_i>0$ a.s.

EDIT: I searched a bit further and found that this problem is called the determination of the "first passage time". I found a few papers dealing with this situation, they all focussed on the determination of the distribution of $T_x$. This however seems to be rather complicated (one could relate the question above to the Cramer-Lundberg model).

As I will be working with empirical data I cannot make model assumptions on the distribution of $Y_i$. As the distribution of $T_x$ then seems to much to ask for I would settle with how to compute the expectation.

The basic idea in this setting is to condition on the (exponentially distributed) time $\tau$ of the first event in the Poisson process $N$ and on the value $Y_1$.
• If $Y_1\geqslant x$ then $T_x=\tau$.
• Otherwise, $Y_1=y$ for some $y\lt x$ and the compound Poisson process after time $\tau$ behaves like the original one shifted by level $y$ and time $\tau$, thus, $T_x=\tau+S_{x-y}$ where each $S_{x-y}$ is independent of $\tau$ and distributed like $T_{x-y}$.
In terms of Laplace transforms, this shows that, for every positive $u$, $$E[\mathrm e^{-uT_x}]=E[\mathrm e^{-u\tau}]\,\left(P[Y_1\geqslant x]+\int_0^xE[\mathrm e^{-uT_{x-y}}]\mathrm dP_Y(y)\right).$$ This identity characterizes the distribution of every $T_x$. For example, considering the derivatives at $u=0$ (or going back to the decomposition which led to this identity), one sees that the function $\theta:x\mapsto E[T_x]$ solves the equation $$\theta(x)=E[\tau]+\int_0^x\theta(x-y)\mathrm dP_Y(y).$$ An equivalent formulation is that $$\theta(x)=E[\tau]\,\left(1+P[Y_1\leqslant x]+P[Y_1+Y_2\leqslant x]+P[Y_1+Y_2+Y_3\leqslant x]+\cdots\right).$$