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

D is correct.

If $\ln X\sim N(\mu ,{\sigma }^2)$, then the mean of $X$ is:
$$E[X]=e^{\mu +{\sigma }^2/2}.$$
With $\mu =7$ and $\sigma =1.2$:
$$E[X]=e^{7+(1.2{)}^2/2}=e^{7+1.44/2}=e^{7+0.72}=e^{7.72}\approx \mathrm{2249.4.}$$

B is correct.

Tail classification by hazard rate behavior:
- Increasing hazard rate (IHR): lighter than exponential, with a tail that decays faster than $e^{-\lambda x}$ for some $\lambda$.
- Constant hazard rate: the exponential distribution, light-tailed, with MGF $M(t)=(1-\theta t{)}^{-1}$ defined for $t<1/\theta$.
- Decreasing hazard rate (DHR): heavier than exponential, with a survival function that decays slower than any pure exponential.

The exponential sits at the boundary between IHR (lighter) and DHR (heavier) distributions. It is classified as light-tailed because its MGF exists on a non-trivial interval containing a neighborhood of zero, namely $t<1/\theta$.

A is correct.

For the lognormal with parameters $\mu$ and $\sigma$, define $z=(\ln d-\mu )/\sigma$. The mean excess loss at $d$ is: $e(d)=[E[X]\cdot \Phi (\sigma -z)]/[1-\Phi (z)]-d$, where $E[X]=e^{\mu +{\sigma }^2/2}$. With $\mu =6$, $\sigma =1.5$, $d=500$: $E[X]=e^{6+1.125}=e^{7.125}\approx 1248.8$. $z=(\ln 500-6)/1.5=(6.2146-6)/1.5=0.2146/1.5\approx 0.143$. $\Phi (0.143)\approx 0.557$, $1-\Phi (0.143)\approx 0.443$. $\Phi (1.5-0.143)=\Phi (1.357)\approx 0.913$. $e(500)=(1248.8\times 0.913)/0.443-500\approx 1139.8/0.443-500\approx 2572-500=2072$.