# Maximum likelihood estimators of the parameter of an aggregate loss (Poisson frequency, exponential loss)

Question:

$$N\sim Poisson(\theta), X\sim exp(\theta)$$

$$S = X_1 + X_2 + ... + X_N$$

With $$4$$ observed aggregate loss $$s_1, s_2, s_3, s_4$$.

What's the maximum likelihood estimator of $$\theta$$?

My attempts:

$$(S|N)\sim Gamma(N,\theta)$$

$$f(s)=\prod_{i=1}^{4}\mathrm{f}_{S}(s_i)$$

Use law of total probability to write : \begin{align*} f(s) & = \prod_{i=1}^4 \sum_{i=1}^{\infty}f_{S | N}(s_i | n) f_{N}(n)\\ & = \prod_{i=1}^4 \sum_{i=1}^{\infty} \frac{s_{i}^{n-1}e^{-s_{i}/\theta}e^{-\theta}\theta^n}{\Gamma(n)n!\theta^n} \\ & = \prod_{i=1}^4 \sum_{i=1}^{\infty} \frac{s_{i}^{n-1}e^{-s_{i}/\theta}e^{-\theta}\theta^n}{(n-1)!n!\theta^n} \\ & = \prod_{i=1}^4 \frac{e^{-s_{i}/\theta}e^{-\theta}}{s_{i}}\sum_{i=1}^{\infty} \frac{s_{i}^{n}}{(n-1)!n!} \end{align*} since $$\Gamma(n) = (n-1)!$$ for $$n \geq 1$$. Then check this answer :