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

B is correct. In the Bayesian credibility (greatest accuracy) framework, the prior $\pi(\theta)$ represents the distribution of the risk parameter $\Theta$ across the heterogeneous population of risks. It captures how different risks are before any data on a specific risk is observed. When we select a risk at random, $\Theta$ is drawn from this prior distribution.

Why each other option is incorrect:
- (D) The distribution of observed losses for a random risk is the marginal (predictive) distribution $f(x) = \int f(x\mid\theta)\pi(\theta)\,d\theta$, not the prior $\pi(\theta)$.
- (A) The conditional distribution of losses given $\theta$ is the likelihood $f(x\mid\theta)$, also called the model or sampling distribution — not the prior.
- (B) The posterior $\pi(\theta\mid x)$ is the updated distribution after observing data; the prior $\pi(\theta)$ is specified before data is collected.
- (C) The marginal distribution of $\sigma^2(\Theta)$ is a functional of the prior, not the prior itself; $\pi(\theta)$ describes the risk parameter, from which $\sigma^2$ and $\mu$ are derived.

C is correct. In the semiparametric empirical Bayes approach, the conditional distribution of losses given the risk parameter, $f(x \mid \theta)$, is assumed to belong to a parametric family (e.g., Poisson, normal). This allows the structural parameters $v = E[\sigma^2(\Theta)]$ and $a = \text{Var}[\mu(\Theta)]$ to be estimated using the known functional form of $\sigma^2(\theta)$ and $\mu(\theta)$. The prior $\pi(\theta)$ remains completely unspecified (nonparametric component).

Why each other option is incorrect:
- (D) Fully specifying the prior and using MLE describes the parametric (fully Bayesian) approach, not the semiparametric approach.
- (A) The semiparametric approach still requires the existence of a risk parameter $\Theta$; it makes fewer assumptions than the fully parametric approach but more than the nonparametric approach.
- (B) When $\sigma^2(\theta)$ is constant, the nonparametric and semiparametric estimators of $v$ coincide, but the methods are not equivalent in general — they differ in how $v$ is estimated when $\sigma^2(\theta)$ varies.
- (C) Kernel density estimation of the prior is a different technique (empirical likelihood / nonparametric Bayes); the semiparametric approach does not estimate $\pi(\theta)$ at all.

E is correct. For a fixed risk $\Theta$, the estimator $\hat{\mu} = Z\bar{X}+(1-Z)\mu$ has MSE
$$E\left[(Z\bar{X}+(1-Z)\mu-\mu(\Theta))^2\right].$$
Conditioning on $\Theta$, $E[\bar{X}|\Theta] = \mu(\Theta)$ and $\text{Var}(\bar{X}|\Theta) = \sigma^2(\Theta)/n$. The overall MSE decomposes as
$$\text{MSE}(Z) = Z^2 \cdot \frac{v}{n} + (1-Z)^2 \cdot a$$
where $v = E[\sigma^2(\Theta)]$ and $a = \text{Var}[\mu(\Theta)]$. Differentiating:
$$\frac{d}{dZ}\text{MSE} = 2Z\frac{v}{n} - 2(1-Z)a = 0 \implies Z\frac{v}{n} = (1-Z)a$$
$$Z\left(\frac{v}{n}+a\right) = a \implies Z^* = \frac{a}{a+v/n} = \frac{na}{na+v} = \frac{n}{n+v/a} = \frac{n}{n+k}.$$

Why each other option is incorrect:
- (B) The MSE-minimizing $Z$ is $n/(n+k)$ in general, not only when $v = na$; the square root form does not arise from this optimization.
- (C) Correctly states $Z^* = na/(na+v) = n/(n+k)$, but the framing as "minimizing variance" rather than MSE is imprecise — both A and C give the same $Z$, but A has the correct MSE decomposition derivation.
- (D) The result is mathematically equivalent to A and C, but the framing as generalized least squares normal equations adds unnecessary complexity.
- (E) The statement that $Z$ is independent of whether $k=v/a$ is incorrect; $k$ is precisely defined as $v/a$ and governs the optimal weight.