Summary Statistics
Study guides: Exam MAS-II, Exam DISC-DA, Exam PCPA
Summary statistics condense a data set, or a fitted model, into a few numbers a reader can take in at once. For data they measure location, spread, shape and association, usually tabulated by group; for a fitted model they measure fit against complexity — the log-likelihood, AIC, BIC and, in a Linear Mixed Model, the estimated variance components.
\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i
s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2}
Describing data — the tabular half of presenting data, beside graphs:
- Location: the mean
\bar{x}and the median. For right-skewed severity the mean sits well above the median, and the gap is itself a finding - Spread: the standard deviation
s(Sample Variance), the interquartile rangeQ_3 - Q_1, the range, and the coefficient of variations/\bar{x}for comparing spread across variables on different scales - Shape and completeness: percentiles (the five-number summary behind a Box Plot), the share of zeros, the largest values, and the count of missing values in each column
- Categorical variables and association: counts, proportions and — in insurance — exposure by level; the correlation matrix for numeric predictors and two-way tables for categorical ones
- Insurance one-way tables show exposure, claim count, frequency, severity and pure premium by level, with pure premium
=frequency\timesseverity. Each is a ratio of sums, not an average of ratios, and each level's figure is confounded with everything correlated with it — the reason final relativities come from a multivariate model - Use a table when the reader needs exact values, a graph when the message is a shape or a comparison
Summarizing a fitted mixed model:
-2 \timeslog-likelihood, by REML or ML, and the likelihood ratio test between nested models. Use REML to compare random effects or covariance structures with identical fixed effects; use ML to compare fixed effects- AIC
= -2\ell + 2kand BIC= -2\ell + k\ln nfor choices that are not nested. BIC's penalty per parameter is heavier once\ln n > 2, i.e.n \geq 8, so it favours smaller models - Variance components and the implied intraclass correlation. A component estimated at zero means the grouping explains nothing. Testing whether a variance is zero puts the null on the boundary of the parameter space, so the naive
\chi^2p-value is too large — for a single variance component it should be halved - Fixed-effect tests — a
torFstatistic per term — guide variable selection, alongside the diagnostic plots that check the assumptions behind them
A One-Way Table by Vehicle Use Personal auto experience by vehicle use:
Use Exposure Claims Losses Pleasure 6,000 300 1,500,000 Commute 3,000 195 1,053,000 Business 1,000 90 540,000 Build the frequency, severity and pure premium table, with pure premium relativities to Pleasure, and the all-uses totals.
Answer
Use Frequency Severity Pure premium Relativity Pleasure 0.0500 5,000 250.00 1.000 Commute 0.0650 5,400 351.00 1.404 Business 0.0900 6,000 540.00 2.160 All 0.0585 5,287 309.30 The Business relativity decomposes as
1.80in frequency times1.20in severity, which is2.16. The overall frequency is585/10{,}000 = 0.0585, not the average of the three rows (0.0683), because Pleasure carries60\%of the exposure. Before quoting2.16as a rating factor, check what else Business use travels with — annual mileage, vehicle type — since a one-way table credits all of it to vehicle use.
Choosing a Random-Effects Structure from Summary Statistics Loss ratios for
500policy-years are fitted by REML with identical fixed effects. Model 1 has a random intercept per policyholder (2covariance parameters); Model 2 adds a random slope on year and its covariance with the intercept (4). The-2REML log-likelihoods are2{,}431.6and2{,}420.2. Compare them by likelihood ratio test, AIC and BIC.Answer
\begin{align*} \text{LRT} &= 2{,}431.6 - 2{,}420.2 \\ &= 11.4 \\ \text{AIC}_1 &= 2{,}431.6 + 2(2) \\ &= 2{,}435.6 \\ \text{AIC}_2 &= 2{,}420.2 + 2(4) \\ &= 2{,}428.2 \\ \text{BIC}_1 &= 2{,}431.6 + 2\ln 500 \\ &= 2{,}444.0 \\ \text{BIC}_2 &= 2{,}420.2 + 4\ln 500 \\ &= 2{,}445.1 \end{align*}The LRT of
11.4on2degrees of freedom exceeds\chi^2_{0.05,\,2} = 5.99; because the slope variance is tested on its boundary, the true p-value is smaller still than the naive one, so the conclusion only strengthens. AIC also prefers Model 2. BIC, charging\ln 500 = 6.21per parameter rather than2, narrowly prefers Model 1. When the criteria split, look at the size of the estimated slope variance and at the residual plots by policyholder before adding structure.
How it appears on the exam
- DISC-DA: Section 8.1 covers traditional data analysis. Questions give a small table of policy or claim data and ask for a mean, median, standard deviation or a one-way frequency, severity or pure premium, or ask what a gap between mean and median says about skewness.
- MAS-II: Questions compare fitted mixed models by log-likelihood, AIC and BIC, as in the second example.
Exam traps
- Average of ratios: An overall frequency is total claims over total exposure, not the average of the levels' frequencies.
- Mean of skewed losses: For right-skewed severity the mean sits well above the median; reporting only the mean overstates the typical claim.
- One-way tables confound: A level's relativity in a one-way table includes the effect of everything correlated with it.