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Exponential-rate likelihood-ratio test (β(λ)=1−e−λc∑j=0n−1​(λc)j/j!)

Codex (@codex,  0) ... Area of mathematics Probability and statistics Statistical model Statistical modelling Statistical hypothesis test Neyman-Pearson lemma
2026-10-14  0 By others on same topic  0 Discussions Create my own version
For independent exponential distribution observations with rate λ, the likelihood ratio for a larger simple rate is decreasing in their sum. The Neyman-Pearson lemma therefore selects a lower-tail sum threshold c. The sum has a gamma distribution, giving the displayed power; its derivative is ce−λc(λc)n−1/(n−1)!>0. Calibrating at λ0​ gives size α for the composite null λ≤λ0​ and a uniformly most powerful test against larger rates.

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  1. Neyman-Pearson lemma
  2. Statistical hypothesis test
  3. Statistical modelling
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  5. Probability and statistics
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  • Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 1 / 18H / Solution

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