 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/29j/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Write

f_{θ} (x)

for the joint density and assume regularity permits differentiation under the integral. Since

\int f_{θ} (x) d x = 1,

one differentiation gives

E_{θ} S_{n} (θ) = \int \frac{\partial _{θ} f _{θ}}{f _{θ}} f_{θ} d x = \partial_{θ} \int f_{θ} d x = 0.

Differentiating the score itself,

\partial_{θ} S_{n} = \frac{\partial _{θ}^{2} f _{θ}}{f _{θ}} - (\frac{\partial _{θ} f _{θ}}{f _{θ}})^{2} .

Taking expectation gives

E_{θ} [\partial_{θ} S_{n} (θ)] = \int \partial_{θ}^{2} f_{θ} (x) d x - E_{θ} [S_{n} (θ)^{2}] = - I_{n} (θ) .

I_{n} (θ) = - E_{θ} [\frac{\partial}{\partial θ} S_{n} (θ)] = - E_{θ} [ℓ_{n}^{''} (θ)] .

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