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

D is correct. The Negative Binomial distribution with parameters $r > 0$ and $\beta > 0$ belongs to the $(a,b,0)$ class with $a = \beta/(1+\beta) > 0$ and $b = (r-1)\beta/(1+\beta)$. When $r > 1$, both $a > 0$ and $b > 0$; when $r = 1$ (geometric case), $b = 0$. So the Negative Binomial is the $(a,b,0)$ member with $a > 0$, $b \geq 0$.

Why each other option is incorrect:
- (B) The Binomial has $a < 0$, so it does not satisfy $a > 0$ as required.
- (C) The Poisson has $a = 0$, not $a > 0$, so it does not meet the strict inequality.
- (D) The Logarithmic distribution is a member of the $(a,b,1)$ class (zero-truncated/modified), not the $(a,b,0)$ class.
- (E) The uniform discrete distribution does not satisfy the $(a,b,0)$ linear recursion for any constants; it is not a member of the class.

E is correct.

A key property of compound Poisson distributions is reproductive closure under independent summation: if $S_1 \sim \text{CompPois}(\lambda_1, F_X)$ and $S_2 \sim \text{CompPois}(\lambda_2, F_X)$ are independent, then $S_1 + S_2 \sim \text{CompPois}(\lambda_1 + \lambda_2, F_X)$. This follows from the additive property of Poisson distributions and the independence of the sums.

This closure property does NOT hold for compound Negative Binomial distributions in the same clean way; the sum of two independent compound Negative Binomials is not generally compound Negative Binomial with the same severity distribution.

Why each other option is incorrect:
- (A) The formula $E[S] = E[N] \cdot E[X]$ holds for ALL compound distributions by Wald's identity, not just compound Poisson.
- (C) MGF existence in a neighborhood of zero depends on the severity MGF and holds broadly, not exclusively for compound Poisson.
- (D) Panjer recursion applies to any (a,b,0) or (a,b,1) frequency class, including Poisson, Binomial, and Negative Binomial.
- (E) The law of total variance formula holds for all compound distributions regardless of the frequency family.

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.