Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 218 / 5 / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii
- iii c
  - Solution iii

i

Solution

For

Y \sim N_{n} (Xβ, σ^{2} I)

,

ℓ (β, σ^{2}) = - \frac{n}{2} lo g (2 π) - \frac{n}{2} lo g σ^{2} - \frac{1}{2 σ ^{2}} ∥ Y - Xβ ∥_{2}^{2} .

ii

Solution

At the maximum-likelihood estimates,

σ_{1}^{2} = \frac{1}{n} ∥ Y - X β ∥_{2}^{2},

so

- 2 ℓ (β, σ_{1}^{2}) = n {lo g (2 π σ_{1}^{2}) + 1} .

There are

p

regression coefficients and one variance parameter. Adding twice this parameter count gives

AIC (M_{1}) = n {lo g (2 π σ_{1}^{2}) + 1} + 2 (p + 1) .

iii

Solution

The common terms cancel, and

AIC (M_{2}) - AIC (M_{1}) = n lo g \frac{σ _{2}^{2}}{σ _{1}^{2}} + 2 q .

This is negative exactly when

\frac{σ _{2}^{2}}{σ _{1}^{2}} < e^{- 2 q / n} .

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