 Solution

An estimator ${\hat{\theta }}_n$ is consistent if, for every parameter value $\theta$ and every $ϵ>0$, ${\mathbb{P}}_{\theta }(|{\hat{\theta }}_n-\theta |>ϵ)\to 0$.

For the proposed root criterion, fix $ϵ>0$. Convergence in probability at the two points givesOn this event the intermediate value theorem gives a zero between the two points. Since the zero is unique, it is ${\hat{\theta }}_n$. Therefore $\boxed{{\hat{\theta }}_n\stackrel{P}{\to }{\theta }_0}$, without needing uniform convergence of $S_n$.

For independent $N({\theta }_0,1)$ observations, the log-likelihood has derivative $n({\bar{X}}_n-\theta )$ and negative second derivative $-n$. Its unique maximizer is ${\hat{\theta }}_n={\bar{X}}_n$. Apply the criterion to $S_n(\theta )=\theta -{\bar{X}}_n$: the strong law of large numbers gives the limit $S(\theta )=\theta -{\theta }_0$, with the required signs. Hence the maximum-likelihood estimator is consistent.

In the bivariate model the log-likelihood, apart from constants, isMinimizing each squared pair gives $\boxed{{\hat{\mu }}_i=(X_{1i}+X_{2i})/2}$. Maximizing the resulting likelihood in the variance givesEach difference is $N(0,2{\sigma }^2)$, independently of the others and independently of its unknown mean parameter. Thus the strong law of large numbers gives ${\hat{\sigma }}^2\to {\sigma }^2/2$ in probability, not ${\sigma }^2$. This is the Neyman-Scott incidental parameter problem: the number of nuisance means grows with the sample size, but each mean is estimated from only two observations. The usual fixed-dimensional likelihood consistency theory therefore does not apply. The corrected estimator $2{\hat{\sigma }}^2$ is consistent.