Associate of the Society of Actuaries (ASA/FSA) ASTAM — Advanced Short-Term Actuarial Mathematics Syllabus

Define the loss elimination ratio (LER) for a deductible $d$.

$$\mathrm{LER}(d) = \frac{E[X \wedge d]}{E[X]}$$ It is the proportion of expected losses eliminated by imposing the deductible $d$.

Distinguish proportional reinsurance from excess-of-loss reinsurance.

Proportional (quota share/surplus): reinsurer pays a fixed fraction $\alpha$ of every loss; insurer retains $(1-\alpha)X$. Excess-of-loss (non-proportional): reinsurer pays the amount of each loss above a retention $d$, i.e. $(X-d)_{+}$, up to a limit; insurer retains $X \wedge d$.

Under quota-share reinsurance with retained proportion $\alpha$, what are the insurer's retained mean and variance of an individual loss $X$?

Retained loss is $\alpha X$, so $E[\alpha X] = \alpha E[X]$ and $\mathrm{Var}(\alpha X) = \alpha^{2}\mathrm{Var}(X)$. Variance reduces by the square of the retention proportion.

Under excess-of-loss reinsurance with retention $d$, write the insurer's retained loss and the reinsurer's payment per loss.

Insurer retains $X \wedge d$ with expectation $E[X \wedge d]$. Reinsurer pays $(X-d)_{+}$ with expectation $E[X] - E[X \wedge d]$. The two retained/ceded amounts sum to $X$.

How does an excess-of-loss reinsurance treaty change the frequency of the ceded (reinsurer's) claim process under a compound Poisson model?

Only losses exceeding the retention $d$ generate a reinsurance payment. The reinsurer's claim count is Poisson with thinned rate $\lambda^{*} = \lambda\, S(d) = \lambda\,\Pr(X>d)$, and the ceded severity is the excess $X-d \mid X>d$.

Compare stop-loss reinsurance with excess-of-loss reinsurance.

Excess-of-loss applies a retention to EACH individual loss: reinsurer pays $\sum (X_i - d)_+$. Stop-loss applies a retention to AGGREGATE losses: reinsurer pays $(S - d)_{+}$ where $S$ is total losses. Stop-loss protects against accumulation of many claims; excess-of-loss against large single claims.

Give the formula for the expected cost of stop-loss reinsurance with aggregate retention $d$.

$$E[(S-d)_{+}] = E[S] - E[S \wedge d] = \int_{d}^{\infty} S_{S}(x)\,dx$$ where $S_{S}$ is the survival function of aggregate losses $S$. For discrete $S$, $E[(S-d)_+]=\sum_{s>d}(s-d)f_S(s)$.

State the recursive (one-step) relationship for the stop-loss premium when $S$ is integer-valued, moving the retention from $d$ to $d+1$.

$$E[(S-(d+1))_{+}] = E[(S-d)_{+}] - S_{S}(d) = E[(S-d)_{+}] - \big(1 - F_{S}(d)\big).$$ Each unit increase in retention reduces the stop-loss premium by the survival probability at $d$.

What is the fundamental quantity (credibility premium) in limited fluctuation versus greatest accuracy credibility?

Both compute a credibility-weighted estimate $P_{c} = Z\bar{X} + (1-Z)M$, where $\bar X$ is the observed experience, $M$ the prior/manual mean, and $Z$ the credibility factor. Limited fluctuation sets $Z$ to control fluctuation; greatest accuracy (Bühlmann) derives $Z$ to minimize expected squared error.

Define the structural parameters $\mu$, $v$ (EPV), and $a$ (VHM) in Bühlmann credibility.

Let $\Theta$ be the risk parameter. Hypothetical mean $\mu(\theta)=E[X\mid\theta]$, process variance $v(\theta)=\mathrm{Var}(X\mid\theta)$. Then: collective mean $\mu = E[\mu(\Theta)]$; Expected Process Variance $v = E[v(\Theta)]$; Variance of Hypothetical Means $a = \mathrm{Var}(\mu(\Theta))$.

Give the Bühlmann credibility factor $Z$ and the Bühlmann constant $k$ for $n$ observations.

$$k = \frac{v}{a}, \qquad Z = \frac{n}{n+k} = \frac{n}{n + v/a}.$$ As $n\to\infty$ or $a$ grows relative to $v$, $Z\to 1$ (experience trusted).

Write the Bühlmann credibility premium for the next period given sample mean $\bar{X}$.

$$P_{c} = Z\bar{X} + (1-Z)\mu, \qquad Z = \frac{n}{n+k},\ k=\frac{v}{a}.$$ This is the linear least-squares (best linear) approximation to the Bayesian premium.

How does the Bühlmann-Straub model differ from the basic Bühlmann model?

Bühlmann-Straub allows different exposure/weights $m_{ij}$ per observation (varying volume by period). Process variance scales as $v/m_{ij}$. Credibility uses total exposure $m=\sum m_{j}$: $$Z = \frac{m}{m + k},\quad k=\frac{v}{a},$$ and the experience mean is the exposure-weighted average.

In Bühlmann-Straub, how is the credibility-weighted average $\bar{X}$ computed across periods with exposures $m_{j}$?

$$\bar{X} = \frac{\sum_{j} m_{j} X_{j}}{\sum_{j} m_{j}}$$ the exposure-weighted mean of the per-unit observations $X_j$ (losses divided by exposure).

What does empirical Bayes (nonparametric) credibility estimate, and why is it called 'empirical'?

It estimates the structural parameters $\mu$, $v$, and $a$ from the data itself rather than assuming a parametric prior. 'Empirical' because $\hat\mu$, $\hat v$, $\hat a$ come from observed sample means and variances across risks/policyholders.

Give the nonparametric empirical Bayes (Bühlmann) estimators of $\mu$ and $v$ for $r$ policyholders each with $n$ observations.

$$\hat\mu = \bar{X} = \frac{1}{rn}\sum_{i=1}^{r}\sum_{j=1}^{n} X_{ij},\qquad \hat v = \frac{1}{r}\sum_{i=1}^{r}\frac{1}{n-1}\sum_{j=1}^{n}(X_{ij}-\bar X_{i})^{2}.$$ $\hat v$ averages each risk's sample variance (the within-risk variance).

Give the nonparametric empirical Bayes (Bühlmann) estimator of $a$ (VHM) for $r$ equal-sized risks.

$$\hat a = \frac{1}{r-1}\sum_{i=1}^{r}(\bar X_{i} - \bar X)^{2} - \frac{\hat v}{n}.$$ It is the between-risk variance of the $\bar X_i$ minus the within-risk noise term. If negative, $\hat a$ is set to 0.

What is semiparametric empirical Bayes estimation, and when is it used?

Semiparametric estimation assumes a parametric form for the process (e.g., Poisson frequency, giving $v=\mu$, so $\hat v = \bar X$) but estimates the structure of $a$ empirically from the data. Used when the conditional distribution family is known but the prior is not.

In semiparametric empirical Bayes with Poisson claim counts, how is $\hat{v}$ obtained?

For Poisson, the process variance equals the mean, so $v(\theta)=\mu(\theta)$ and $\hat v = \hat\mu = \bar X$. Then $\hat a = \hat{\mathrm{Var}}(\bar X_i) - \hat v / n$, using the overall sample mean for $\hat v$.

Define the Bayesian premium and contrast it with the Bühlmann premium.

Bayesian premium $= E[X_{n+1}\mid \mathbf{X}]$, the predictive mean using the full posterior $\pi(\theta\mid\mathbf{x})$. The Bühlmann premium is the best LINEAR approximation to this. They coincide exactly when the model/prior pair is a linear-exponential conjugate (e.g., Poisson-gamma, Bernoulli-beta).

For the Poisson-gamma Bayesian model (Poisson$(\lambda)$ likelihood, gamma$(\alpha,\theta)$ prior on $\lambda$), what is the posterior distribution after observing $n$ counts summing to $\sum x_{i}$?

The posterior of $\lambda$ is gamma with shape $\alpha^{*} = \alpha + \sum x_{i}$ and rate updated so the new scale is $\theta^{*} = \frac{\theta}{1 + n\theta}$. The predictive (Bayesian) mean is $\alpha^{*}\theta^{*}$.

Why does exact (Bayesian) credibility equal Bühlmann credibility for conjugate linear-exponential families?

For these families the posterior mean is linear in the observations $\bar X$, so the optimal estimator is already linear. Since Bühlmann is the best linear estimator, it matches the exact Bayesian posterior mean: $E[X_{n+1}\mid\mathbf{X}] = Z\bar X + (1-Z)\mu$.

What is a conjugate prior, and name two common conjugate pairs used in Bayesian credibility.

A conjugate prior yields a posterior in the same family as the prior. Common pairs: Poisson likelihood with gamma prior (gamma posterior); Bernoulli/binomial likelihood with beta prior (beta posterior); normal-normal; exponential/gamma likelihood with inverse-gamma prior.

Explain the chain-ladder method for loss reserving.

Using a run-off triangle of cumulative paid/incurred losses by accident year and development period, compute age-to-age (development) factors $f_{j} = \frac{\sum_i C_{i,j+1}}{\sum_i C_{i,j}}$, then multiply the latest diagonal forward to project ultimate losses. The reserve is ultimate minus paid-to-date.

Define an age-to-age (loss development) factor and the cumulative development factor (CDF).

Age-to-age factor $f_{j}=\frac{C_{\cdot,j+1}}{C_{\cdot,j}}$ measures growth from development period $j$ to $j+1$. The cumulative development factor to ultimate is $$F_{j} = \prod_{k=j}^{n-1} f_{k},$$ which scales a year's current cumulative loss up to its ultimate value.

Describe the Bornhuetter-Ferguson (BF) reserving method and its key advantage over chain-ladder.

BF reserve $= \text{(expected ultimate)} \times (1 - 1/\text{CDF})$, i.e. an a-priori expected ultimate (often premium $\times$ expected loss ratio) times the unreported/undeveloped proportion. Advantage: more stable than chain-ladder for immature/volatile years because it does not rely heavily on the noisy latest diagonal.

Compare the chain-ladder and Bornhuetter-Ferguson reserve estimates as a development year matures.

Chain-ladder relies entirely on reported losses to date (volatile when little is reported). BF blends an a-priori expectation with reported data via the percent-reported $1/F_j$. As a year matures ($1/F_j \to 1$), BF converges to the chain-ladder result; for green years BF leans on the a-priori.

What is the Mack model, and what does it add to the deterministic chain-ladder?

The Mack model is a distribution-free STOCHASTIC chain-ladder that provides the standard error (prediction uncertainty) of reserve estimates. It assumes $E[C_{i,j+1}\mid C_{i,j}]=f_j C_{i,j}$ and $\mathrm{Var}(C_{i,j+1}\mid C_{i,j})=\sigma_j^2 C_{i,j}$, yielding mean-squared-error estimates without specifying a full distribution.

How does the over-dispersed Poisson (ODP) bootstrap produce a distribution of reserves in stochastic reserving?

Fit an ODP GLM (whose mean reserves reproduce the chain-ladder estimate), compute residuals, then resample residuals to create many pseudo-triangles, refit each, and project reserves. The collection of projected reserves forms a predictive distribution, giving percentiles, standard error, and capital estimates.