The log-likelihood, score, and Fisher information are
$$
\ell_n(\theta)=\sum_{r=1}^n\log f(X_r,\theta),
\qquad
S_n(\theta)=\nabla_\theta\ell_n(\theta),
$$
and
$$
I_n(\theta)=\mathbb E_\theta
\bigl[S_n(\theta)S_n(\theta)^T\bigr]
=-\mathbb E_\theta\bigl[\nabla_\theta^2\ell_n(\theta)\bigr].
$$
For independent identically distributed observations,
$$
I_n(\theta)=nI_1(\theta).
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 29L / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/29l/a/i/solution