 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/28j/c/solution

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

Let

N_{+} = # {i : Y_{i} > 0}

N_{0} = n - N_{+}

, and

a = 1 - e^{- λ}

. Ignoring factors independent of

π

, the likelihood is

L (π) \propto π^{N_{+}} (1 - aπ)^{N_{0}} .

The unconstrained critical point satisfies

\frac{N _{+}}{π} - \frac{a N _{0}}{1 - aπ} = 0,

so, after enforcing

0 \leq π \leq 1

π_{MLE} = min {1, \frac{N _{+}}{n ( 1 - e ^{- λ} )}} .

 New to topics? Read the docs here!