 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/28j/a/solution

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

x = (x_{1}, \dots, x_{n})

L (θ; x) = \prod_{i = 1}^{n} f (x_{i}; θ) .

θ_{MLE} (x) \in θ \in Θ argmax L (θ; x) .

S_{θ} (X) = \frac{\partial}{\partial θ} lo g f (X; θ),

I (θ) = E_{θ} [S_{θ} (X)^{2}] = - E_{θ} [\frac{\partial ^{2}}{\partial θ ^{2}} lo g f (X; θ)],

where the second identity requires the usual differentiation-under-the-integral regularity. For

n

I_{n} (θ) = n I (θ)

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