 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-34/3/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

u_{i} = μ_{i 1}

S_{i} = \sum_{k} e^{η_{ik}}

. The correct Poisson mass function has

e^{- μ}

; the positive sign in the PDF's reminder is erroneous. The group likelihood is

p (y_{i} ∣ u_{i}, β) = \frac{u _{i}^{n_{i}} e x p { \sum _{k} y _{ik} η _{ik} } e ^{- u_{i} S_{i}}}{\prod _{k} y _{ik} !} .

Multiply by the shape–rate Gamma distribution prior

b^{a} u_{i}^{a - 1} e^{- b u_{i}} /Γ (a)

and integrate. The gamma integral gives

p (y_{i} ∣ β, a, b) = \frac{b ^{a} Γ ( n _{i} + a )}{Γ ( a ) \prod _{k} y _{ik} !} \frac{exp { \sum _{k} y _{ik} η _{ik} }}{( S _{i} + b ) ^{n_{i} + a}} .

Independent baseline priors give the product over groups. For fixed

a, b

, the leading factors do not depend on

β

 New to topics? Read the docs here!