# How to construct the likelihood function of compound Poisson process?

Since in the compound Poisson process (CPP), the jumps occur according to the Poisson process with intensity $$\lambda(t)$$. The jumps size is iid random variables and itself independent of the Poisson process.

Two ways are generally found to derive the Poisson process likelihood.

1. By interarrival times:
For fixed $$(0,T]$$, $$m$$ Poisson events at times $$\text{0 = }t_0. With the interarrival times $$t_j-t_{j-1}$$, for $$j = {1, 2, ..., m}$$, representing a random sample from an exponential distribution, then the likelihood function is given as

$$L=\prod _{j=1}^m \left(\lambda e^{-\lambda \left(t_j-t_{j-1}\right)}\right) e^{-\lambda \left(T-t_m\right)}=\lambda ^m e^{-\lambda T}$$

2. By conditional intensity function (referred article @page12):
Using the conditional intensity function (or hazard function),$$\lambda ^* (t)=\frac{f \left(t\left|H_{t_m}\right.\right)}{1-F \left(t\left|H_{t_m}\right.\right)}$$ and conditional density function, $$f \left(t\left|H_{t_m}\right.\right)= \lambda ^* (t) \left(-\int_{t_m}^T \lambda ^* (u) \, du\right)$$ where $$H_{t_m}$$ is history of previous events, in

$$L=f \left(t_1|H_0\right) \left(t_2|H_{t_1}\right)\text{...} \left(t_m|H_{t_{m-1}}\right) \left(1-F \left(T\left|H_m\right.\right)\right)$$

$$L=(\prod _{j=1}^m f \left(t_j|H_{t_{j-1}}\right)) \frac{f \left(T\left|H_{t_m}\right.\right)}{\lambda ^* (T)}$$, then solving furthur we get

$$L=(\prod _{j=1}^m \lambda ^* (t_j)) \exp \left(-\int_0^T \lambda ^* (u) \, du\right)$$
$$L=\lambda ^m \exp(-\lambda T)$$

Such also can be applied for nonhomogeneous Poisson process.

Intuitively, $$(\prod _{j=1}^m \lambda (t_j)) \exp \left(-\int_0^T \lambda(u) \, du\right)$$ could be the part of likelihood of CPP. As the Poisson process which produces such jumps to occur is the primary distribution in the CPP.

But I have no idea how such ways can construct a complete likelihood function for CPP.