Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 13K (Statistical Modelling)

Conditionally on $X$, the log likelihood differs fromonly by terms independent of $\beta$. Differentiation gives the weighted normal equationsand therefore

Dividing row $i$ of both $Y$ and $X$ by $\sqrt{v(X_i)}$ transforms the model to one with covariance ${\sigma }^{2}I$. Ordinary least squares without an intercept on these transformed data is exactly the estimator in part (a). In R, the unweighted estimator is returned by

R
fit2 <- lm(Y ~ X - 1)
fit2$coefficients

Write $Y_i=X_i^T\beta +{\epsilon }_i$, where conditionally $\mathbb{E}({\epsilon }_i|X_i)=0$ and $\mathrm{Var}({\epsilon }_i|X_i)={\sigma }^{2}v(X_i)$. ThenThe law of large numbers sends the matrix towhile the vector tends to zero, proving consistency. The central limit theorem and Slutsky's theorem give

Similarly, with $B=\mathbb{E}(X{X}^T)$ and $C=\mathbb{E}(v(X)X{X}^T)$,These conclusions require the displayed matrices to be finite and nonsingular.

For $p=1$, part (c) givesConsequently

Weighted least squares uses the known conditional variances and should be at least as efficient, so $\rho \le 1$. Indeed, apply Cauchy--Schwarz toIt giveswhich is exactly $\rho \le 1$. Equality holds precisely when $v(X_1)$ is constant on the part of the support where $X_1\ne 0$, up to null sets.