Partial Credibility in Actuarial Science You are given: (i) Claim counts follow a Poisson distribution. (ii) claim size follows a Pareto distribution with parameters $\alpha$ = 3 and $\theta$ = 1. (iii) A full credibility standard is established so that the actual number of claims will be within 5% of the expected number of claims 95% of the time. Determine the number of expected claims needed for 30% partial credibility for the distribution of number of claims. Attempt: E[N]=Var(N)=$\lambda$ (number of claims) E[X]=0.5, Var(X)=0.75 (amount of claims) E[N] _$(1.96/0.05)^2$_ [$\lambda$/0.5*$\lambda$)]^2 = 6146.56 The answers 138. What did I do wrong?

The standard for full credibility in terms of claims is $$n_C = \left(\dfrac{z_{\alpha /2}}{k}\right)^{2}\left(\dfrac{\sigma^2_F}{\mu_F}+\dfrac{\sigma^2_S}{\mu_S^2}\right)\text{,}$$ $F$ denoting frequency, $S$ denoting severity. Since $F \sim \text{Poisson}$, $\sigma^2_F = \mu_F$, so that $\dfrac{\sigma^2_F}{\mu_F} = 1$.

Notice we are NOT looking at _aggregate_ claims, so ignore the $\dfrac{\sigma^2_S}{\mu_S^2}$ (make it $0$). Now using the exam C parametrization, we have $$\mu_{S} = \dfrac{\theta}{\alpha - 1} = \dfrac{1}{2}$$ and $$\mathbb{E}[S^2] = \dfrac{2!\theta}{(\alpha - 1)(\alpha - 2)} = \dfrac{2}{2(1)} = 1\text{.}$$ Thus, $\sigma^2_S = 1 - \left(\dfrac{1}{2}\right)^2 = \dfrac{3}{4}$. With $k = 0.05$ and $\alpha = 1 - 0.95 = 0.05$, we have $$z_{\alpha / 2} = z_{1-0.975} = 1.960$$ so that $n_C = 1536.64$.

Now $n$, the expected number of claims, solves $$\sqrt{\dfrac{n}{n_C}} = 0.3 \Longleftrightarrow n = (0.3)^2n_C = 138.2976\text{.}$$