Yes indeed! I was looking for the name. The paradox from a Bayesian perspective is that the Jeffreys prior on the full model leads to an inadmissible and inconsistent estimator of the variance, while the Jeffreys prior on the MLE model produces a consistent estimator!

]]>Yes, discussing the matter this morning with Jean-Michel, I eventually realised the mistake! Merci!

]]>This is really a good question for an exam despite its simplicity.

Your solution results from direcly expressing the likelihood of the difference Zi=Xi1-Xi2 for i=1 to n that its free of the mu(i)’s; this corresponds exactly to a residual (R Thompson’s acronym) or restricted (D Harville’s one) likelihood.

In classical ML, you have to estimate both the variance sigma2 but also the mu(i)’s jointly from the likellihood of the original data resulting in half the previous estimator.

Incidentally, another quadratic unbiased estimator is

sigma2=(sum over i =1 to n of (Xi1*2-Xi1 x Xi2) )/n but it is not translation invariant. Charles Henderson used to give this example to illustrate the fact that an unbiased estimator is not necessarily translation invariant.

In fact in Ex 1, question 1) d) is the REML (residual maximum likelihood estimator, Patterson & Thompson, 1971) while (e)=(d)/2 is the usual ML (maximum likelihood estimator). The first is unbiased while the second is not. ]]>