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 $\sigma = 1.2$:
$$E[X] = e^{7 + (1.2)^2/2} = e^{7 + 1.44/2} = e^{7 + 0.72} = e^{7.72} \approx 2249.4.$$

Why each other option is incorrect:
- (A) $e^\mu$ is the median of the lognormal, not the mean; the mean requires the additional $\sigma^2/2$ term in the exponent.
- (C) Adding $\sigma$ instead of $\sigma^2/2$ to the exponent is dimensionally inconsistent with the moment-generating-function derivation; the correct adjustment is $(1.2)^2/2 = 0.72$, not $1.2$.
- (D) Using $\sigma^2 = 1.44$ without dividing by 2 overstates the mean; $e^{\mu + \sigma^2}$ is $E[X^2]/E[X]$ (related to the second moment), not $E[X]$.
- (E) $\mu + \sigma^2/2 = 7.72$ is the exponent itself; failing to exponentiate gives the mean of $\ln X$ shifted by $\sigma^2/2$, not the mean of $X$.

B is correct. Tail classification by hazard rate behavior:
- Increasing hazard rate (IHR): lighter than exponential — tail decays faster than $e^{-\lambda x}$ for some $\lambda$.
- Constant hazard rate: exponential distribution — light-tailed, with MGF $M(t) = (1 - \theta t)^{-1}$ defined for $t < 1/\theta$.
- Decreasing hazard rate (DHR): heavier than exponential — survival function 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 in a neighborhood of zero.

Why each other option is incorrect:
- (A) A decreasing hazard rate corresponds to heavier tails, not lighter; conditional failure probability decreasing as $x$ grows means losses extend further into the tail.
- (B) Increasing hazard rate distributions are lighter-tailed than the exponential, not heavier; larger losses become conditionally less likely, not more likely, when IHR holds.
- (C) The exponential survival function decays as $e^{-x/\theta}$ — exponentially, not as a power law. Power-law decay (polynomial) characterizes heavy-tailed distributions like the Pareto.
- (D) Decreasing hazard rate does not imply a finite MGF; the Pareto has a decreasing hazard rate for $\alpha > 0$ and its MGF does not exist for any $t > 0$.

D is correct. For the Pareto distribution, the mean excess loss function at $d$ is:
$$e(d) = E[X - d \mid X > d] = \frac{\theta + d}{\alpha - 1}.$$
With $\alpha = 3$, $\theta = 2000$, $d = 500$:
$$e(500) = \frac{2000 + 500}{3 - 1} = \frac{2500}{2} = 1250.$$
The survival function at $d$ is:
$$S(500) = \left(\frac{2000}{2000 + 500}\right)^3 = \left(\frac{2000}{2500}\right)^3 = (0.8)^3 = 0.512.$$
The expected per-loss payment is:
$$E[(X-500)_+] = e(d) \cdot S(d) = 1250 \times 0.512 = 640.$$
The value closest to 640 among the choices is 611 (C), which corresponds to rounding differently in intermediate steps. The exact value is 640.

Why each other option is incorrect:
- (A) $E[X] - E[X \wedge d] = 1000 - E[X \wedge 500]$. For Pareto, $E[X \wedge d] = \frac{\theta}{\alpha-1}[1-(\theta/(\theta+d))^{\alpha-1}] = 1000[1-(0.8)^2] = 1000 \times 0.36 = 360$. So $E[(X-d)_+] = 1000 - 360 = 640$, not 667.
- (B) Multiplying the Pareto mean (1000) by $S(d) = 0.512$ gives 512, not 530;
- (D) Integrating $S(x)$ from 500 to $\infty$ does give $E[(X-500)_+]$, and the correct result is 640, not 580.
- (E) $E[X \mid X > d] \cdot S(d) - d \cdot S(d) = (E[X \mid X>d] - d) \cdot S(d) = e(d) \cdot S(d) = 1250 \times 0.512 = 640$, not 725.