Free SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) Construction and Selection of Parametric Models Practice Questions

D is correct. The Akaike Information Criterion is defined as:
$$\text{AIC} = -2\hat{\ell} + 2k$$
where $\hat{\ell}$ is the maximized log-likelihood and $k$ is the number of estimated parameters. The criterion balances fit (via $-2\hat{\ell}$) against complexity (via $2k$). Lower AIC is preferred.

Why each other option is incorrect:
- (B) This is the Bayesian Information Criterion (BIC/SBC), not AIC; BIC uses $k\ln n$ instead of $2k$ as the penalty.
- (C) The parameter penalty in AIC is linear ($2k$), not quadratic ($2k^2$); a quadratic penalty is not used in standard information criteria.
- (D) The sign convention is reversed; AIC uses $-2\hat{\ell}$ (positive for typical negative log-likelihoods), and lower values are preferred, not higher.
- (E) Normalizing the log-likelihood by $n$ before adding the penalty is not part of the AIC formula; AIC uses the raw maximized log-likelihood.

E is correct. AIC $= -2\hat{\ell} + 2p$ penalizes each parameter by $2$. (__T(B)MP__)IC $= -2\hat{\ell} + p\ln n$ penalizes each by $\ln n$. The (__T(B)MP__)IC penalty exceeds AIC's when $\ln n > 2$, i.e., $n > e^2 \approx 7.4$. For any dataset with $n \ge 8$ — which includes virtually every actuarial application — (__T(B)MP__)IC penalizes additional parameters more heavily than AIC and therefore selects simpler (fewer-parameter) models.

Why each other option is incorrect:
- (A) Claims (__T(B)MP__)IC always penalizes more for any $n > 1$; this is false for $n \le 7$, where $\ln n < 2$ and AIC has the larger per-parameter penalty.
- (C) Reverses the formulas: AIC uses $2p$ and (__T(B)MP__)IC uses $p\ln n$, not the other way around.
- (D) Reverses the consistency properties of AIC and (__T(B)MP__)IC; (__T(B)MP__)IC is consistent (converges to the true model as $n\to\infty$) while AIC tends to overfit asymptotically.
- ((B)) States both criteria minimize the same objective but with different coefficients; while true as a narrow arithmetic statement, the more important error is calling them equivalent in purpose — AIC minimizes expected Kullback-Leibler divergence (prediction) while (__T(B)MP__)IC approximates the log marginal likelihood (model probability).

D is correct. The survival function is $S(t) = e^{-t/\theta}$. By the delta method:
$$\text{Var}(\hat{S}(t)) \approx \left(\frac{dS}{d\theta}\right)^2 \text{Var}(\hat{\theta})$$
$$\frac{dS}{d\theta} = \frac{t}{\theta^2}e^{-t/\theta}$$
$$\text{Var}(\hat{\theta}) = \frac{\theta^2}{n}$$
$$\text{Var}(\hat{S}(t)) = \left(\frac{t}{\theta^2}e^{-t/\theta}\right)^2 \cdot \frac{\theta^2}{n} = \frac{t^2 e^{-2t/\theta}}{n\theta^2}$$
At $t = 500$, $\theta = 1000$: $\text{Var} = \frac{500^2 e^{-1}}{n \times 10^6} = \frac{250000 e^{-1}}{10^6 n} = \frac{e^{-1}}{4n}$.

Why each other option is incorrect:
- (B) The binomial variance $S(1-S)/n$ applies to a Bernoulli proportion, not to a parametric MLE of a continuous survival function; the delta method is the correct approach.
- (C) Omitting the $e^{-2t/\theta}$ factor drops the squared exponential term that arises when squaring the derivative $dS/d\theta$.
- (D) A linear (first-order) Taylor expansion retains $(dS/d\theta)^2$, which still includes two powers of the exponential; there is no single-power simplification.
- (E) While algebraically similar, option E uses $\theta^2$ in the numerator implicitly and then factors differently; the exact simplified form is $t^2 e^{-2t/\theta}/(n\theta^2)$, which is choice A.