 Solution

A statistic $T$ is sufficient for $\theta$ if the conditional distribution of the full sample given $T$ does not depend on $\theta$. It is minimal sufficient if it is a function of every sufficient statistic.

The likelihood factors asso the Fisher-Neyman factorization theorem shows that $M$, and hence its one-to-one transform $\hat{\theta }=M/2$, is sufficient. Moreover, for two samples $\mathbf{x}$ and $\mathbf{y}$, the ratio $L(\theta ;\mathbf{x})/L(\theta ;\mathbf{y})$ is independent of $\theta$ exactly when their maxima agree. The likelihood-ratio criterion for minimal sufficiency therefore shows that $M$ and $\hat{\theta }$ are minimal sufficient.

For $n\ge 2$, the sample mean $\tilde{\theta }$ is not sufficient: two samples can have the same mean but different maxima, and their likelihood ratio then depends on $\theta$ through the support indicators. It is consequently not minimal sufficient. For the degenerate special case $n=1$, the sample mean and maximum coincide and both conclusions reverse.