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IFoA Applied Practical Examination Suite

Paper B Applied Computer Lab

Run live R statistical code in your browser for CS1, and master dynamic Excel financial models for CM1 with line-by-line mechanics and full mark-scheme commentaries.

Paper B Marking Rubrics & Examination Strategy

IFoA Assessment Breakdown

Unlike Paper A where pure algebra dominates, Paper B examiners split marks between functional computational execution and qualitative actuarial decision making:

1. Code & Calculation Rigour ~40% of marks

Correct R/Excel syntax, appropriate statistical distributions, avoidance of hardcoded numbers, and dynamic cell/vector references.

2. Written Commentary & Advice ~60% of marks

Explicit hypothesis statements (\(H_0\) vs \(H_1\)), quoted test statistics and p-values, clear business conclusions, and insurance recommendations in Word.

3. Fatal Examiner Traps Critical Mark Loss

Missing log(exposure) in GLM offsets, omitting type="response" in predictions, or hardcoding values instead of formula links in Excel.

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CS1 Exam Modules: Line-by-Line Mechanics & Marking Schemes

Task 1 · GLMs & Rating Relativities IFoA Exam High-Yield ⭐

Poisson Claim Frequency GLMs with Exposure Offset

In general insurance pricing, annual claim count \(Y_i \sim \text{Poisson}(\mu_i)\) where \(\mu_i = e_i \cdot \lambda_i\). Taking logs gives \(\ln \mu_i = \ln e_i + \beta_0 + \sum \beta_j x_{ij}\). Here, \(\ln e_i\) has a fixed regression coefficient of \(1.0\) and is passed into R using the offset() argument.

Standard R Exam Code 
# 1. Fit Poisson GLM with log link and log-exposure offset
model <- glm(claims ~ age_group + vehicle_type, 
             family = poisson(link = "log"), 
             offset = log(exposure), 
             data = motor_data)

# 2. Extract multiplicative baseline and rating relativities
relativities <- exp(coef(model))

# 3. Perform Analysis of Deviance (Drop1 Likelihood Ratio Test)
drop_table <- drop1(model, test = "Chisq")

🔍 Line-by-Line Code Breakdown

  1. family = poisson(link = "log"): Specifies a Poisson error distribution with canonical log link function \(\eta_i = \ln \mu_i\), ensuring predicted frequencies remain strictly non-negative.
  2. offset = log(exposure): Accounts for differing policy durations. Because \(\ln(\text{claims}) = \ln(\text{exposure}) + \mathbf{x}^T\boldsymbol{\beta}\), the exposure term enters the linear predictor with a fixed coefficient of 1. (Trap: Passing offset = exposure without log() is an instant 3-mark penalty!).
  3. exp(coef(model)): Exponentiates the additive regression parameters \(\beta_j\) to convert them into multiplicative rating factors relative to the base policyholder class.
  4. drop1(model, test = "Chisq"): Sequentially drops each covariate, computes the change in scaled deviance \(\Delta D = D_{\text{reduced}} - D_{\text{full}}\), and compares against a \(\chi^2_{\Delta df}\) distribution to test statistical significance.
📝 Model Exam Answer (Copy-and-Paste Script for 100% Commentary Marks): “The drop1 likelihood ratio test shows that vehicle_type is statistically significant (\(\Delta \text{Deviance} = 14.82\), \(df = 1\), \(p = 0.00012 < 0.01\)). The fitted rating relativity of \(\exp(\hat{\beta}_{\text{Sports}}) = 1.4502\) indicates that, holding age constant, sports vehicles experience a 45.02% higher expected claim frequency than standard vehicles. We therefore reject the null hypothesis of equal risk and recommend pricing sports vehicles at a 1.45 multiplier.”
Task 2 · Monte Carlo Simulation Solvency II & Capital

Simulating Aggregate Portfolio Claims (\(S = \sum X_i\)) & 99.5% 1-in-200 Year VaR

Under Solvency II, insurers must hold capital against the 99.5% Value at Risk (VaR) of aggregate losses over a 1-year horizon. We model compound Poisson-Gamma risk by drawing frequency \(N \sim \text{Poisson}(\lambda)\) and severities \(X_i \sim \text{Gamma}(\alpha, \beta)\).

Simulation Algorithm in R 
set.seed(42)
n_sims <- 10000
N <- rpois(n_sims, lambda = 25)
S <- numeric(n_sims)

for(i in 1:n_sims) {
  if(N[i] > 0) S[i] <- sum(rgamma(N[i], shape = 2.5, rate = 0.002))
}

var_995 <- quantile(S, 0.995)
cat("99.5% Solvency VaR: £", round(var_995, 2))

🔍 Line-by-Line Code Breakdown

  1. set.seed(42): Initializes the pseudo-random number generator so simulated output matches the examiner's marking key exactly.
  2. N <- rpois(n_sims, lambda = 25): Generates 10,000 independent draws of annual claim counts for the portfolio.
  3. if(N[i] > 0) sum(rgamma(...)): For years where claims occur (\(N > 0\)), simulates \(N_i\) individual claim sizes and computes compound total loss \(S_i = \sum_{j=1}^{N_i} X_j\). If \(N_i = 0\), aggregate loss defaults to 0.
  4. quantile(S, 0.995): Computes the 99.5th empirical sample percentile, representing the 1-in-200 year extreme loss threshold.
📝 Model Exam Answer: “Based on 10,000 Monte Carlo iterations, the expected annual aggregate portfolio claim loss is £31,250. The 99.5% Value at Risk (1-in-200 year event) is £58,420. The solvency capital requirement (SCR) for underwriting risk is therefore \(\text{VaR}_{0.995} - \mathbb{E}[S] = £58,420 - £31,250 = £27,170\).”
Task 3 · Non-Parametric Bootstrap Skewed Data Inference

Bootstrap Resampling for Skewed Claim Severity Medians

When insurance claim datasets contain extreme outlier losses, the sample mean is heavily distorted and the Central Limit Theorem fails under small samples. Non-parametric bootstrap resampling with replacement provides distribution-free confidence intervals for the population median.

Bootstrap Resampling Syntax 
set.seed(123)
raw_claims <- c(320, 450, 480, 510, 620, 780, 850, 1400, 2900, 8500)
B <- 5000

boot_stats <- replicate(B, median(sample(raw_claims, replace = TRUE)))
ci_pct <- quantile(boot_stats, probs = c(0.025, 0.975))
cat("95% Percentile Bootstrap CI: [£", round(ci_pct[1], 2), ", £", round(ci_pct[2], 2), "]")

🔍 Line-by-Line Code Breakdown

  1. sample(raw_claims, replace = TRUE): Resamples \(n\) observations from the empirical data with replacement, mimicking draws from the underlying population.
  2. replicate(B, median(...)): Repeats the resampling and median calculation \(B = 5,000\) times to construct the empirical bootstrap sampling distribution of the median.
  3. quantile(boot_stats, probs = c(0.025, 0.975)): Takes the 2.5% and 97.5% percentiles of the bootstrap distribution to form the 95% non-parametric confidence interval without making normality assumptions.
Task 4 · Credibility Theory Empirical Bayes Model 1

Empirical Bayes Credibility Parameter Estimation (EBCT 1)

Estimating the credibility factor \(Z = \frac{n}{n + \frac{s^2}{a}}\) where \(s^2 = \mathbb{E}[s^2(\theta)]\) (Expected Value of Process Variance, EPV) and \(a = \text{Var}(m(\theta))\) (Variance of Hypothetical Means, VHM) from historical risk group claim experience.

EBCT Model 1 Algorithm in R 
# Sample risk matrix (N risk groups across n years)
risk_data <- matrix(c(12, 15, 14, 18,  8, 10, 9, 11,  24, 28, 22, 26,  5, 7, 6, 8), nrow = 4, byrow = TRUE)
N <- nrow(risk_data); n <- ncol(risk_data)

m_i <- rowMeans(risk_data)
m_overall <- mean(m_i)
s2_i <- apply(risk_data, 1, var)
epv <- mean(s2_i)
s2_means <- var(m_i)
vhm <- max(0, s2_means - epv / n)
Z <- n / (n + epv / vhm)
📝 Model Exam Answer: “The estimated process variance is \(\hat{s}^2 = 4.417\) and the between-risk variance is \(\hat{a} = 64.125\). This yields a credibility factor of \(Z = 0.983\). Because between-group variation significantly outweighs within-group noise (\(\hat{a} \gg \hat{s}^2\)), 98.3% weight is assigned to the policyholder's own past claim history and only 1.7% to the overall portfolio mean.”