ACAS exams & assessments

MAS-II

Modern Actuarial Statistics II

5 discussions

Study strategy

MAS-II: AR(1) stationarity is about a constant long-run distribution

For X_t=c+φX_(t−1)+ε_t with independent zero-mean innovations, the stationary mean is c/(1−φ) when |φ|<1. The process can fluctuate while having a constant unconditional mean and variance. Do not confuse that with a constant realised value.

Worked example or practice scenario

With c=2, φ=.6 and innovation variance 4, stationary mean is 5 and variance is 4/(1−.36)=6.25. Lag-k autocorrelation is .6^k under the model. A conditional forecast from X_t=8 differs from the long-run mean.

Try this next

Compute the one-step forecast as 2+.6×8=6.8. Then compare a multi-step forecast with 5. Check the current content outline’s time-series tasks instead of assuming an older reading list is unchanged.

Reading sources

Official syllabus and resources

ActNet editorial guide · October 1, 2026 · Original illustrative examples.

0 repliesOct 1

Explanation

MAS-II: Clustering and PCA depend on the geometry you create

Unsupervised methods work with relationships among predictors, not a labelled outcome. Scaling and distance choices can dominate the result. Interpret a cluster as a pattern in the selected representation, not as a proven biological or business category.

Worked example or practice scenario

If one feature spans 0–100,000 and another spans 0–1, Euclidean distance on raw units can mostly reflect the first feature. Standardisation changes the geometry. Likewise, PCA on covariance and correlation matrices can produce different directions.

Try this next

Explain the preprocessing choice and compare stability under a sensible alternative. For clustering, examine whether the pattern persists rather than assigning causal meaning to a cluster label. The current MAS-II outline explicitly includes statistical learning; check the sitting’s full tasks.

Reading sources

Official syllabus and resources

ActNet editorial guide · October 1, 2026 · Original illustrative examples.

0 repliesOct 1

Explanation

MAS-II: Tree pruning separates training fit from selection

A deep tree can closely fit a training sample and still predict poorly. Pruning or controlling complexity changes the bias-variance trade-off. Evaluate the whole procedure, including tuning, on data it did not use to select the tree.

Worked example or practice scenario

A maximal tree has training error 1 and validation error 18; a pruned tree has training error 5 and validation error 10. The apparent training deterioration can accompany a useful predictive improvement. The split sample and metric must be the same for a fair comparison.

Try this next

Describe how cross-validation would choose the complexity parameter. Then test sensitivity to a different split. Do not treat one lower validation score as proof that the selected structure is stable.

Reading sources

Official syllabus and resources

ActNet editorial guide · October 1, 2026 · Original illustrative examples.

0 repliesOct 1

Explanation

MAS-II: Random intercepts create dependence within a group

A random-intercept model allows observations in the same group to share a latent effect. That makes them correlated even if the residual errors are independent conditional on the group effect. Repeated observations are not automatically independent data points.

Worked example or practice scenario

In Y_ij=β0+b_i+ε_ij, let Var(b_i)=4 and Var(ε_ij)=12. Each observation has variance 16, and two observations in the same group have covariance 4. The intraclass correlation is 4/16=.25.

Try this next

Compare observations from different independent groups: their covariance from the group effect is zero. State whether a requested prediction concerns an existing group with observed history or a new group; the available information differs.

Reading sources

Official syllabus and resources

ActNet editorial guide · October 1, 2026 · Original illustrative examples.

0 repliesOct 1

Explanation

MAS-II: Bühlmann credibility distinguishes noise from heterogeneity

Credibility is not simply the share of data coming from a policyholder. Under the Bühlmann model, the balance depends on process variance, between-risk variance and the number of observations. Specify which parameter represents each source of variation.

Worked example or practice scenario

For n=5, EPV=120 and VHM=30, K=4 and Z=5/9. If observed mean is 140 and collective mean is 100, the credibility estimate is 122.22. More process noise reduces credibility when other quantities are fixed.

Try this next

Double EPV and recompute the weight. Explain why the estimate moves toward the collective mean. Compare this with increasing VHM: greater real differences between risks make the risk’s own observations more valuable.

Reading sources

Official syllabus and resources

ActNet editorial guide · October 1, 2026 · Original illustrative examples.

0 repliesOct 1