Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 219 / 4 / d

Solution

Every move can be drawn from a full conditional distribution, producing a Gibbs sampler with acceptance probability one. Write

s_{x, i}^{2} = σ_{x, i}^{2}

and

s_{y, i}^{2} = σ_{y, i}^{2}

. First update independently

ξ_{i} ∣ - \sim N (V_{ξ i} [\frac{μ}{τ ^{2}} + \frac{β ( η _{i} - α )}{σ ^{2}} + \frac{x _{i}}{s _{x, i}^{2}}], V_{ξ i}), V_{ξ i}^{- 1} = \frac{1}{τ ^{2}} + \frac{β ^{2}}{σ ^{2}} + \frac{1}{s _{x, i}^{2}},

and then

η_{i} ∣ - \sim N (V_{η i} [\frac{α + β ξ _{i}}{σ ^{2}} + \frac{y _{i}}{s _{y, i}^{2}}], V_{η i}), V_{η i}^{- 1} = \frac{1}{σ ^{2}} + \frac{1}{s _{y, i}^{2}} .

Let

X

have rows

(1, ξ_{i})

and

η = (η_{i})

(α β) - \sim N_{2} ((X^{T} X)^{- 1} X^{T} η, σ^{2} (X^{T} X)^{- 1}),

and update

μ ∣ - \sim N (\overset{ˉ}{ξ}, \frac{τ ^{2}}{N}) .

Finally, the flat positive variance priors give the following full conditionals, each an inverse-gamma distribution:

σ^{2} ∣ - \sim InvGamma (\frac{N}{2} - 1, \frac{1}{2} \sum_{i} (η_{i} - α - β ξ_{i})^{2}),

τ^{2} ∣ - \sim InvGamma (\frac{N}{2} - 1, \frac{1}{2} \sum_{i} (ξ_{i} - μ)^{2}) .

A systematic sweep in the displayed order, using the newly sampled values immediately, defines the chain. The shapes are positive for

N > 2

X

is also required.

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