Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 1 / 13K (Statistical Modelling)

For one observation,Thus this is an exponential family with natural parameter $\theta =\log \lambda$, statistic $T(y)=y$, base measure $1/y!$, and cumulant $A(\theta )=e^{\theta }$.

For nonnegative $y_i$ summing to $m$, divide the joint Poisson mass by $P(M=m)=e^{-{\lambda }_{+}}{\lambda }_{+}^m/m!$:This is independent Poisson conditioning.

In the log-linear model fit1 has only three main effects, so its fitted mean factors into a species factor, a diameter factor, and a height factor: mutual independence. Fit2 adds the species-diameter interaction, allowing those variables to associate, while asserting $\text{Height}\perp (\text{Species},\text{Diameter})$.

The likelihood-ratio deviance drop $12.606$ on one degree of freedom has $p=0.0003845$, strong evidence against species-diameter independence. Test fit2's remaining independence claim by comparing it with a model adding both Species:Height and Diameter:Height interactions (equivalently compare fit2 with the saturated model); its residual deviance $12.431$ is assessed against ${\chi }_3^2$.