Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 30 / 4 / Solution

Let ${\Delta }_n={\sup }_{\theta \in \Theta }|Q_n(\theta )-Q(\theta )|$. For any $\epsilon >0$, the setis compact. If it is empty there is nothing to prove. Otherwise continuity and the unique minimum give a strictly positive separation gapTake any measurable attained minimizer of $Q_n$. Its defining inequality givesConsequentlyThis proves argmin consistency under uniform convergence in probability. No continuity of $Q_n$ is needed once the minimizer exists; compactness and continuity concern the deterministic separation gap. The uniform convergence in probability assumption controls all candidate parameters, including the random minimizer.

For the estimating equation, fix $\epsilon >0$ and put $a={\theta }_0-\epsilon$, $b={\theta }_0+\epsilon$. The prescribed signs make $\eta =\frac{1}{2}\min \{-S(a),S(b)\}>0$. Pointwise convergence in probability at just these two points impliesand similarly $\mathrm{Pr}(S_n(b)\le 0)\to 0$. With probability tending to one, $S_n(a)<0<S_n(b)$. The intermediate value theorem then gives a zero inside $(a,b)$, and uniqueness identifies it with ${\hat{\theta }}_n$. HenceThis is consistency of a uniquely bracketed zero. It needs no monotonicity, no continuity of the limit $S$, and no uniform convergence of the $S_n$. The bracket interval must lie in the domain of $S_n$: read literally, the printed sign condition for every positive $\epsilon$ puts every real point into $\Theta$, so this requirement is satisfied. More generally it is enough to have an interval about ${\theta }_0$ and sign brackets arbitrarily close to it.