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.
By interarrival times:
For fixed $(0,T]$, $m$ Poisson events at times $\text{0 = }t_0<t_1<\text{...}<t_m<T$. 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$L=\prod _{j=1}^m \left(\lambda e^{-\lambda \left(t_j-t_{j-1}\right)}\right) e^{-\lambda \left(T-t_m\right)}=\lambda ^m e^{-\lambda T}$
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.
Please help, how do we build the likelihood function for the Compound Poisson process?
