 Solution

A transformation model consists of a group $G$ acting on the sample space and on the set of statistical parameters, with the compatibility propertyIn words, transforming a sample from $P_{\theta }$ produces a sample from $P_{g\theta }$. A transitive transformation model is generated from one reference distribution by the group action. For densities, the compatibility property includes the Jacobian determinant of the sample transformation; it is not merely equality of density values at transformed points.

A maximal invariant $A$ is constant on orbits of a group action and separates them: $A(gx)=A(x)$, and $A(x)=A(y)$ implies $y=gx$ for some $g$. An equivariant estimator $T$ of the parameter satisfies $T(gx)=gT(x)$. Equivariance means the estimate follows the transformed parameter, whereas invariance means the reported value stays unchanged.

First suppose the set of statistical parameters can be identified with the group, as happens for a simply transitive action. An equivariant estimator then provides a group-valued fit $\hat{g}(x)$ with $\hat{g}(gx)=g\hat{g}(x)$. Normalize the sample by this fit:Indeed $A(gx)=\hat{g}(x{)}^{-1}{g}^{-1}gx=A(x)$. Conversely, if $A(x)=A(y)$, then $y=\hat{g}(y)\hat{g}(x{)}^{-1}x$, placing $x,y$ in the same orbit of a group action. This establishes maximality, rather than only invariance.

For a transitive action with a nontrivial stabilizer $H$ of a reference parameter ${\theta }_0$, a parameter-valued equivariant estimator does not specify a unique normalizing transformation. Choose a section $h_{\theta }$ with $h_{\theta }{\theta }_0=\theta$ and form $b(x)=h_{T(x)}^{-1}x$. Under $g$,Thus the $H$-orbit of $b(x)$ is invariant. It is maximal: if $b(y)=hb(x)$ with $h\in H$, then $y=h_{T(y)}h{h}_{T(x)}^{-1}x$. Appropriate measurable sections and orbit of a group action coordinates are assumed when treating this as a statistic. This qualification explains why an arbitrary equivariant statistic alone is not automatically a complete set of invariant coordinates.

For the location-scale family, write $X_i=\mu +\sigma Z_i$ with $\sigma >0$ and a fixed joint distribution of $Z$. The group consists of maps $x_i\mapsto a+b{x}_i$ with $b>0$, acting on parameters by $(\mu ,\sigma )\mapsto (a+b\mu ,b\sigma )$. For a nonconstant sample of size $n\ge 2$, take the equivariant estimatorIt gives the maximal invariantIts coordinates satisfy ${\sum }_i{A}_i=0$ and ${\sum }_i{A}_i^2=n$. They stay unchanged under location and positive-scale transformations. If two samples have the same standardized vector, thenso a single group transformation maps the first sample onto the second. The vector retains the observation labels: sorting it would discard information not removed by this location-scale group. The construction is algebraic and does not require the underlying distribution to have finite moments. Constant samples can be assigned a separate invariant value; they form a single exceptional orbit of a group action and have probability zero under an independent continuous base distribution.

Finally, in a transitive transformation model every invariant statistic is ancillary. Choose $g$ taking ${\theta }_0$ to $\theta$ and use $X_{\theta }\stackrel{d}{=}g{X}_{{\theta }_0}$; invariance gives $A(X_{\theta })\stackrel{d}{=}A(X_{{\theta }_0})$. Thus the standardized residual configuration provides a natural conditioning variable, while the equivariant fit contains the location and scale information.