Random Effects

Study guide: Exam MAS-II

Random effects are the group-level departures from the population-average fit in a Linear Mixed Model. Unlike fixed effects they are treated as draws from a distribution, so a group with thin data is pulled toward the overall mean rather than fitted on its own.

\mathbf{y} = \mathbf{X}\boldsymbol{\beta} + \mathbf{Z}\mathbf{b} + \boldsymbol{\varepsilon}

\mathbf{b} \sim N(\mathbf{0}, \mathbf{D}), \quad \boldsymbol{\varepsilon} \sim N(\mathbf{0}, \mathbf{R}), \quad \mathbf{b} \perp \boldsymbol{\varepsilon}

• A random effect costs one variance parameter, not one coefficient per level — 68 territories add a single \sigma_b^2 to the model
• \mathbf{Z} is the design matrix mapping each observation to its group; \mathbf{D} holds the variance components
• Random effects induce correlation between observations in the same group — this is what lets a mixed model handle repeated measures on one policyholder
• Their predicted values are shrunk toward zero by exactly a credibility factor — see Best Linear Unbiased Predictor
• The effects are assumed mean zero: the population average lives entirely in the fixed part, so a random effect adds no intercept of its own

Why the Random Effect Shrinks a Thin Territory A random-intercept model of loss ratio by territory has \hat\sigma_b^2 = 0.010 (between territories) and \hat\sigma^2 = 0.090 (within). Territory A has 9 observations averaging 0.15 above the grand mean; territory B has 1 observation 0.15 above. What are their predicted effects?

Answer The shrinkage factor is Z_i = \dfrac{n_i \sigma_b^2}{n_i \sigma_b^2 + \sigma^2}: Z_A = \frac{9(0.010)}{9(0.010) + 0.090} = \frac{0.09}{0.18} = 0.50 \Rightarrow \hat b_A = 0.50(0.15) = 0.075 Z_B = \frac{0.010}{0.010 + 0.090} = 0.10 \Rightarrow \hat b_B = 0.10(0.15) = 0.015 Same raw deviation, very different predictions — the model trusts nine observations five times as much as one. This is Bühlmann Credibility in another notation.