Exam MAS-I — Frequency, Severity & Aggregate Models Practice Flashcards

**Answer: (C).** Since $0 < a < 1$, the distribution is **negative binomial**, where $a = \dfrac{\beta}{1+\beta}$ and $b = (r-1)\dfrac{\beta}{1+\beta} = (r-1)a$. From $a = \dfrac{\beta}{1+\beta} = 0.25$: $\beta = 0.25(1+\beta) \Rightarrow 0.75\beta = 0.25 \Rightarrow \beta = \tfrac{1}{3}$. From $b = (r-1)a = 0.50$: $r - 1 = \dfrac{0.50}{0.25} = 2 \Rightarrow r = 3$. So negative binomial with $r = 3$, $\beta = \tfrac{1}{3}$. A geometric (E) would require $b = 0$; a Poisson (A) requires $a = 0$; $a < 0$ would be binomial (B).

**Answer: (A).** Since $a < 0$, the distribution is **binomial**, with $a = -\dfrac{q}{1-q}$ and $b = (m+1)\dfrac{q}{1-q}$. From $a = -\dfrac{q}{1-q} = -0.25$: $\dfrac{q}{1-q} = 0.25 \Rightarrow q = 0.25(1-q) \Rightarrow 1.25q = 0.25 \Rightarrow q = 0.20$. From $b = (m+1)\dfrac{q}{1-q} = (m+1)(0.25) = 1.25 \Rightarrow m + 1 = 5 \Rightarrow m = 4$. So binomial with $m = 4$, $q = 0.20$. Note $\dfrac{q}{1-q} = 0.25$ does **not** mean $q = 0.25$ (distractor E uses that error).

**Answer: (B).** The variance-to-mean ratio screens the family: $\dfrac{\text{Var}(N)}{E[N]} = \dfrac{0.50}{0.50} = 1$. A ratio of exactly $1$ is the signature of the **Poisson**, for which $E[N] = \text{Var}(N) = \lambda$. Moment matching gives $\hat\lambda = \bar N = 0.50$. A ratio below $1$ would indicate a binomial (underdispersed); above $1$, a negative binomial/geometric (overdispersed). Here it equals $1$, so Poisson with $\lambda = 0.50$.

**Answer: (C).** The $(a,b,0)$ recursion $\dfrac{p_k}{p_{k-1}} = a + \dfrac{b}{k}$ at $k=1,2$ gives $a + b = 0.45$ and $a + \dfrac{b}{2} = 0.30$. Subtract the second from the first: $\dfrac{b}{2} = 0.15 \Rightarrow b = 0.30$, then $a = 0.45 - 0.30 = 0.15$. Since $0 < a < 1$ with $b \neq 0$, this is a **negative binomial** with $a = 0.15$, $b = 0.30$. (Reversing the algebra to $a = 0.30$, $b = 0.15$ is distractor D; a geometric would need $b = 0$.)

**Answer: (A).** Thinning a frequency distribution by an independent per-loss survival probability $v$ preserves the family. For a **negative binomial$(r,\beta)$**, only $\beta$ is scaled by $v$: $\text{payments} \sim \text{negative binomial}(r,\,\beta v).$ Here $r = 2$ stays fixed and $\beta v = 1.5(0.4) = 0.6$, giving negative binomial with $r = 2$, $\beta = 0.6$. Scaling $r$ instead of $\beta$ (B) is the common error; for a Poisson it would be the rate $\lambda$ that gets scaled, which is not the family here.

**Answer: (B).** For an exponential$(\theta)$, $E[X] = \theta$ and $\text{Var}(X) = \theta^{2}$. $\text{Var}(X) = 500^{2} = 250{,}000$, and $\text{SD}(X) = \sqrt{250{,}000} = 500$. For the exponential the standard deviation equals the mean ($\theta = 500$). Distractor (D) wrongly uses $E[X^2] = 2\theta^2 = 500{,}000$ as a variance; (C) takes the root of $E[X^2]$ but reports the variance as $250{,}000$, an inconsistent mix.

**Answer: (C).** Pareto$(\alpha,\theta)$: $E[X] = \dfrac{\theta}{\alpha - 1} = \dfrac{3000}{3} = 1000$. $E[X^2] = \dfrac{2\theta^{2}}{(\alpha-1)(\alpha-2)} = \dfrac{2(3000)^{2}}{(3)(2)} = \dfrac{18{,}000{,}000}{6} = 3{,}000{,}000$. $\text{Var}(X) = E[X^2] - E[X]^2 = 3{,}000{,}000 - 1{,}000{,}000 = 2{,}000{,}000$. (Forgetting to subtract $E[X]^2$ leaves $E[X^2] = 3{,}000{,}000$, distractor D; using $\dfrac{\theta}{\alpha} = 750$ for the mean is distractor A.)

**Answer: (B).** For a lognormal, $E[X] = e^{\mu + \sigma^{2}/2}$. $E[X] = e^{8 + 0.5} = e^{8.5}$. Since $e^{8} \approx 2980.958$ and $e^{0.5} \approx 1.648721$: $E[X] \approx 2980.958(1.648721) \approx 4{,}914.77$. (Using $e^{\mu} = e^{8} \approx 2980.96$ forgets the $\tfrac{1}{2}\sigma^2$ term, distractor A; $e^{\mu+\sigma^2} = e^{9} \approx 8103.08$ over-counts the variance term, distractor C.)