I searched and I didn't find any answer (positive or negative):

Suppose I have a group $G$, and a group family of probability measures . (i.e, there is a probability measure $P$ on $G$, and we define $P_g(A)=P(gA)$ as $g$ in $G$ is a parameter). Also this family is assumed to be complete.

The maximal invariant statistic is always ancillary. Is it always maximal ancillary? Is it always the greatest ancillary?

Are there any reference to a proof or a counter example?


1 Answer 1


the way your question is posed, it suggests that the sample space is $G$. in that case, there is only one orbit and the maximal invariant is a constant [function on $G$], so it is also [trivially] ancillary.

let $G$ be a compact lie group - the orthogonal group on $R^n$, for example. let P be normalized haar measure on $G$. the "group family of probability measures" $G$ generates from P is just P itself.

then any function on $G$ is ancillary - including the identity map of $G$ into itself. the latter is then obviously a maximal ancillary - and is not the same as the maximal invariant.

so in general the maximal invariant is not maximal ancillary.

btw - the above example even works for n = 1, when $G$ contains just 2 elements: the identity map on $R$ and its negative: $x \to -x$.

  • $\begingroup$ Thank you, Indeed I intended the space is a $G^n$ and not G itself. Your example does not fit the conditions since the family formed by shifts of the Haar measure is not complete (It is not even "real" family since all the shifts are identical). I did found a counterexample indicating the maximal invariant statistic is not always the greatest ancillary, but the question about maximality remains. $\endgroup$
    – user18690
    Oct 21, 2011 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.