Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 218 / 6 / b

Table of contents
- Solution b

Solution

With unit error variance, the Gaussian likelihood has log-likelihood

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

Full column rank gives

β = (X^{T} X)^{- 1} X^{T} Y

and the least-squares identity

∥ Y - Xβ ∥_{2}^{2} = ∥ Y - X β ∥_{2}^{2} + (β - β)^{T} X^{T} X (β - β) .

Multiplying the likelihood by

\prod_{j} ϕ (β_{j})

and absorbing all terms independent of

β

into

C

gives

π (β ∣ Y) = C exp [- \frac{1}{2} (β - β)^{T} X^{T} X (β - β) + \sum_{j = 1}^{p} lo g ϕ (β_{j})] .

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