Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-30/1/solution
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 30 1 Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-06
For independent identically distributed observations , the likelihood function and log-likelihood areA maximum-likelihood estimator is a measurable choice , when a maximizer exists. It need not be unique. The one-observation score function is , and the Fisher information matrix isUnder the regularity assumptions, differentiation beneath the integral gives . Differentiating again gives the second information identity. Independent observations have total Fisher information .
Assume fixed parameter dimension, an interior true parameter , a positive-definite matrix , the standard differentiability and integrability conditions, and statistical consistency of . Then the asymptotic normality of a maximum likelihood estimator isHere is the information per observation. Thus the leading covariance matrix of the estimator itself is .
To prove this, statistical consistency places the estimator in an interior ball about with probability tending to one, where its score function vanishes. Write and use the integral first-order Taylor formula for a vector map:This integral matrix is needed in a vector problem; one does not have to assert a common scalar mean-value point for every score component. PutThe regular local uniform law of large numbers, continuity of the expected Hessian matrix, and statistical consistency give . The multivariate central limit theorem givesThe limiting information matrix is nonsingular, so . Solving the Taylor identity and applying the Slutsky theorem proves the displayed normal limit, since . Boundary parameters or singular information are outside this regular theorem.
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