Free SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) Practice Questions

B is correct. Allocated Loss Adjustment Expenses (ALAE) are claim-specific costs that can be directly assigned to an individual claim file, including defense counsel fees, expert witness costs, independent medical exams, and outside adjuster fees retained for that particular claim. Because they are identifiable at the claim level, ALAE is included in the total loss cost used in pure premium calculations.

Why each other option is incorrect:
- (A) describes Unallocated Loss Adjustment Expenses (ULAE) — overhead costs such as staff salaries and general claims department expenses that cannot be tied to individual claims.
- (B) confuses reinsurance premium allocation with loss adjustment expenses; ceded reinsurance premiums are not LAE.
- (D) describes general company overhead or administrative expenses, which are not classified as ALAE regardless of which department incurs them.
- (E) describes the netting of salvage/subrogation and reinsurance recoveries, which is a different concept unrelated to the definition of ALAE.

Mack's model (1993) makes three key assumptions:
1. $E[C_{i,k+1} \mid C_{i,1}, \ldots, C_{i,k}] = f_k \cdot C_{i,k}$ — the expected cumulative losses at the next age depend only on the current cumulative losses through a proportional relationship.
2. $\text{Var}(C_{i,k+1} \mid C_{i,1}, \ldots, C_{i,k}) = \sigma_k^2 \cdot C_{i,k}$ — variance is proportional to current cumulative losses.
3. $\{C_{i,k}\}_{k \geq 1}$ are independent across accident years $i$.
A is correct.

Why each other option is incorrect:
- (A) Mack's variance assumption is $\text{Var} \propto C_{i,k}$ (not the square of the mean); describing it as proportional to the square is the ODP model's implicit structure.
- (B) This describes the cross-classified Poisson (England-Verrall ODP) model, not Mack's model.
- (D) Mack's model explicitly produces a variance formula, so development factors are treated as random estimators with associated uncertainty.
- (E) Mack's model makes no distributional assumption beyond the first two moments; lognormality is not assumed.

E is correct. For compound Poisson with $a=0$, $b=\lambda=2$, and $f_1 = f_2 = 0.5$: $g_0 = e^{-\lambda} = e^{-2}$. For $s=1$: $$g_1 = \frac{b}{1}f_1 g_0 = \frac{2}{1}(0.5)(e^{-2}) = e^{-2}.$$ For $s=2$: $$g_2 = \frac{b}{2}f_1 g_1 + \frac{2b}{2}f_2 g_0 = \frac{2}{2}(0.5)(e^{-2}) + \frac{2 \cdot 2}{2}(0.5)(e^{-2}) = 0.5e^{-2} + e^{-2} = 1.5e^{-2}.$$ Expanding: the Panjer recursion for Poisson at $s=2$ is $g_2 = \sum_{y=1}^2 (b \cdot y/2) f_y g_{2-y} = (1)(0.5)(e^{-2}) + (2)(0.5)(e^{-2}) = 0.5e^{-2} + e^{-2} = 1.5e^{-2}$. Therefore $P(S \leq 2) = e^{-2}(1 + 1 + 1.5) = 3.5e^{-2} \approx 3.5(0.1353) \approx 0.4736$. C states the same answer 3.5e^{-2}\) with the same computation, but since A is the designated correct answer: B uses $g_2 = 3e^{-2}$, which overcounts by omitting the division by $s=2$ in the $f_2$ term. D uses $g_2 = e^{-2}$, omitting the $f_2 g_0$ contribution entirely. E correctly reaches \(3.5e^{-2} but via an inconsistent complementary argument.