I am solving an actuarial math problem.
Suppose that the number of claims, $N$, in a year follows a Poisson distribution $\mathrm{Po}(\Theta)$, where $\Theta$ is a gamma distribution $\Gamma(5, 1/2)$, and its probability density function (pdf) is given by $\frac{(1/2)^5}{\Gamma(5)}x^4\mathrm{e}^{-x/2}$.
In the next year, there is an expected 10% decrease in contracts. The problem is to find the distribution of $N$ for the next year.
Here is what I did. Initially, without considering the 10% decrease,$N$ follows a negative binomial distribution $\mathrm{NB}(5, 1/3)$. However, I got stuck at this point.
The solution suggests that due to the 10 % decrease in contracts, $N$ will follow $\mathrm{NB}(5*0.9, 1/3)$. I have not encountered this computation before and based on my understanding, the interpretation of $\mathrm{NB}(a, p)$ is the number of failure $a$ before a successful trial with probability $p$. Consequently, I assumed the parameter $a$ cannot take non-integer values.
My question is: Is this solution correct? If so, what is the interpretation of it? Additionally, what is the probability density function (pdf) associated with this distribution?
Your calculation of the marginal distribution of $N$ is correct. The hierarchical model is $$N \mid \Theta \sim \operatorname{Poisson}(\Theta), \\ \Theta \sim \operatorname{Gamma}(5, 1/2),$$ hence $$N \sim \operatorname{NegativeBinomial}(5, 1/3)$$ with parametrization $$\Pr[N = n] = \binom{n+4}{4} (1/3)^5 (2/3)^n, \quad n \in \{0, 1, 2, \ldots \}.$$
It is worth noting that in the negative binomial PMF with general parameters $r$ and $p$ $$\Pr[N = n] = \binom{n+r-1}{r-1} p^r (1-p)^n, \quad n \in \{0, 1, 2, \ldots \},$$ we can write the binomial coefficient as $$\binom{n+r-1}{r-1} = \frac{(n+r-1)!}{(r-1)! \, n!} = \frac{(r)(r+1)(r+2)\cdots(r+n-1)}{n!} = \prod_{j=1}^n \frac{j+r-1}{j}.$$ And this product does not require $r$ to be an integer. So for instance, the choice $n = 3, r = 7/3$ yields $$\binom{13/3}{4/3} = \prod_{j=1}^3 \frac{j+4/3}{j} = \frac{455}{81}.$$ Consequently, the parameter $r$ can be any positive real number.
That said, I take issue with the claim that $10\%$ decrease in contracts implies that $$N_{\text{new}} \sim \operatorname{NegativeBinomial}(4.5, 1/3).$$ There is no reason to believe that the reduced frequency manifests through a change in the parameter $r$ only. The expectation of a negative binomial distribution is $$\operatorname{E}[N] = \frac{r(1-p)}{p},$$ and we could certainly obtain a $10\%$ decrease in the expected number of contracts by adjusting $p$ instead of $r$, or even $r$ and $p$ simultaneously. For instance, $\operatorname{E}[N] = 10$, so next year, we expect $\operatorname{E}[N_{\text{new}}] = 0.9(10) = 9$. But this could be achieved by choosing $r = 5$, $p = \frac{5}{14}$. The question doesn't adequately specify how the probability distribution of $N_{\text{new}}$ is modified. One possible way that the question could have uniquely identified the parameters is if it said that there is an expected $10\%$ decrease in contracts but the variance of the number of contracts next year will remain the same, in which case the solution requires $r = \frac{27}{7}$ and $p = \frac{3}{10}$; the proof of which I leave to you as an exercise.