For a random variable $X$, the loss function is defined as $$L(r) = \int_{r}^\infty (x-r)f(x)dx,$$ Or, for a discrete distribution, $$L(r) = \sum_{x=r}^\infty (x-r)f(x).$$
Now, we define the compound Poisson distribution where the compounding distribution is a geometric distribution. This is also called the Polya-Aeppli distribution.
The random variable $X$ is distributed according to $GP(\lambda, \theta)$ (where $\lambda$ is the parameter for the Poisson part and $\theta$ is the parameter for the geometric part) with
$$\lambda_k = \lambda(1-\theta)^{k-1}\theta $$ and recall that
$$\mathbb{E}[X] = \frac{\lambda}{\theta}, \mathbb{V}[X] = \frac{\lambda(2-\theta)}{\theta^2}$$.
Also,
$$ \mathbb{P}(X=x) = \sum_{k=1}^x e^{-\lambda}\frac{\lambda^k}{k!}(1-\theta)^{x-k}\theta^k{x-1 \choose k-1}$$, where ${x \choose k}$ is the binomial coefficient and $\mathbb{P}(X=0) = e^{-\lambda}$
For the Poisson distribution, we also know:
$$\mathbb{E}[X] = \mathbb{V}[X] = \lambda$$ and $$\mathbb{P}(X=x) = \frac{\lambda^xe^{-\lambda}}{x!}$$ and we can write the loss function in terms of the CDF ($F(.)$) and the pmf($f(.)$):
$$L(r)_p = -(r-\lambda)[1-F(r)]+\lambda f(r)$$
For the geometric distribution:
$$\mathbb{E}[X] = \frac{1}{\theta}, \mathbb{V}[X] = \frac{(1-\theta)}{\theta^2}$$ and $$\mathbb{P}(X=x) = (1-\theta)^{x-1}\theta$$ and we can write the loss function as:
$$L(r)_g = \left(\frac{1-\theta}{\theta}\right)(1-\theta)^{r-1}$$
Question
How can we use the loss function of the geometric and Poisson distribution to solve the loss function of the geometric Poisson distribution ?
Any help is much appreciated!