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ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2013/o5/d/i/solution

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Past exam of the options II course of the University of Oxford / 2013 / O5 / d / i / Solution by

Codex  0 2026-10-31

The final cell is a grouped tail. Thus use twelve probabilities

p_{i} (θ) = e^{- θ} \frac{θ ^{i}}{i !} (0 \leq i \leq 10), p_{11} (θ) = P_{θ} (X \geq 11) = 1 - \sum_{i = 0}^{10} p_{i} (θ) .

Maximize the grouped Grouped Poisson likelihood

ℓ (θ) = \sum_{i = 0}^{10} n_{i} lo g p_{i} (θ) + n_{11} lo g p_{11} (θ),

then take

π_{i} = p_{i} (θ)

for all twelve cells. For example, the interior likelihood equation is

\sum_{i = 0}^{10} n_{i} (\frac{i}{θ} - 1) + n_{11} \frac{p _{10} ( θ )}{p _{11} ( θ )} = 0,

because

p_{11}^{'} (θ) = p_{10} (θ)

. It can be solved numerically, with no numerical work required here.

It would be incorrect to treat the last six observations as exactly eleven. If the underlying ungrouped particle counts were available, their sample mean would be the usual Poisson maximum likelihood estimate. For the data as supplied, the grouped likelihood handles the unknown tail counts directly.

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