Random Intercept and Slope

Study guide: Exam MAS-II

A random intercept lets each group have its own baseline; a random slope lets each group have its own response to a predictor. Together they are the two building blocks of a Linear Mixed Model's random part.

$y_{ij} = (\beta_0 + b_{0i}) + (\beta_1 + b_{1i})x_{ij} + \varepsilon_{ij}$

$\begin{pmatrix} b_{0i} \\ b_{1i}\end{pmatrix} \sim N\!\left(\mathbf{0},\; \mathbf{D} = \begin{pmatrix}\sigma_0^{2} & \sigma_{01} \\ \sigma_{01} & \sigma_1^{2}\end{pmatrix}\right)$

Random intercept only: parallel lines, one per group, at different heights — implies a compound-symmetric covariance structure
Random intercept and slope: fanning lines, so within-group variance grows with $x$ and the correlation between two observations depends on their $x$ values
$\sigma_{01}$ is the intercept-slope covariance: negative means groups that start high grow more slowly (a common convergence pattern)
A random slope costs two extra parameters ( $\sigma_1^2$ , $\sigma_{01}$ ) and often fails to converge when groups are small — test it before keeping it
The predictor carrying a random slope must also appear as a fixed effect; the random part is a mean-zero deviation from it

Does the Development Slope Vary by Accident Year? Log paid losses are modelled against development age with a random effect by accident year. A random-intercept fit has $-2\ell = 812.4$ ; adding a random slope gives $-2\ell = 803.1$ . At the 5% level, keep the random slope?

Answer Adding a random slope adds two parameters ( $\sigma_1^2$ and $\sigma_{01}$ ), so the Likelihood Ratio Test statistic is $\chi^{2} = 812.4 - 803.1 = 9.3 \quad \text{on } 2 \text{ df}$ $\chi^2_{0.95, 2} = 5.99 < 9.3$ , so reject — accident years genuinely develop at different rates, not just from different levels. (The boundary problem makes the true $p$ -value smaller still, so the conclusion is safe.)