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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/13j/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 13J / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

Fit a Poisson regression to artificial responses

Z_{i} = 1

with means

μ_{i} = Λ (Y_{i}) e^{β^{T} x_{i}} .

With the logarithmic link, this is the generalized linear model

lo g μ_{i} = β^{T} x_{i} + lo g Λ (Y_{i}),

where

lo g Λ (Y_{i})

is a known offset. Its log likelihood, apart from constants, is

ℓ_{P} (β) = \sum_{i} [lo g Λ (Y_{i}) + β^{T} x_{i} - Λ (Y_{i}) e^{β^{T} x_{i}}] .

ℓ (β) - ℓ_{P} (β) = \sum_{i} [lo g λ (Y_{i}) - lo g Λ (Y_{i})]

does not depend on

β

. The Poisson surrogate therefore has exactly the same maximum-likelihood estimate of $β$ . This is the Poisson surrogate for an uncensored proportional-hazards likelihood.

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