Solution. The maximum has density \[f_T(t) = \frac {n\,t^{\,n-1}}{\theta ^{n}},\qquad 0<t<\theta .\] Suppose \(h\) is a function with \(E_{\theta }\left [h(T)\right ]=0\) for every \(\theta >0\). Then \[\frac {n}{\theta ^{n}}\int _0^{\theta } h(t)\,t^{\,n-1}\,dt = 0 \qquad \Longrightarrow \qquad \int _0^{\theta } h(t)\,t^{\,n-1}\,dt = 0 \quad \text {for every }\theta >0 ,\] since \(n/\theta ^{n}\neq 0\). The left-hand side is a function of its upper limit which vanishes identically, so differentiating with respect to \(\theta \) by the fundamental theorem of calculus gives \[h(\theta )\,\theta ^{\,n-1} = 0 \qquad \text {for almost every }\theta >0 .\] As \(\theta ^{n-1}\neq 0\) on \((0,\infty )\), this forces \(h(\theta )=0\) almost everywhere, which is exactly completeness.

Solution.

The minimal sufficient statistic

Expanding the exponent, \[-\frac {1}{2a\theta ^{2}}\sum _{i=1}^{n}(x_i-\theta )^{2} = -\frac {1}{2a\theta ^{2}}\sum _{i=1}^{n}x_i^{2} + \frac {1}{a\theta }\sum _{i=1}^{n}x_i - \frac {n}{2a},\] so the density depends on the data only through \(T = \left (\sum _i X_i,\ \sum _i X_i^{2}\right )\), and the ratio criterion shows no further reduction is possible: the two coefficients \(1/(2a\theta ^{2})\) and \(1/(a\theta )\) move independently enough that matching the ratio for all \(\theta \) forces both components to agree. So \(T\) is minimal sufficient.

Why it is not complete

Both components can be scaled to estimate \(\theta ^{2}\) unbiasedly. Since \(E(X_i)=\theta \) and \(\var (X_i)=a\theta ^{2}\), \[E\left (\sum _i X_i^{2}\right ) = n\left (a\theta ^{2}+\theta ^{2}\right ) = n(1+a)\theta ^{2},\] \[E\left [\left (\sum _i X_i\right )^{2}\right ] = \var \left (\sum _i X_i\right ) + \left [E\left (\sum _i X_i\right )\right ]^{2} = na\theta ^{2} + n^{2}\theta ^{2} = n(a+n)\theta ^{2}.\] Hence the non-zero function \[h(T) = \frac {\sum _i X_i^{2}}{n(1+a)} - \frac {\left (\sum _i X_i\right )^{2}}{n(a+n)}\] satisfies \(E_{\theta }\left [h(T)\right ] = \theta ^{2}-\theta ^{2} = 0\) for every \(\theta >0\), while \(h\) is not identically zero. So \(T\) is not complete.

Solution.

Minimal sufficiency

The joint density is one exactly when every observation lies in \((\theta -1,\theta +1)\): \[f_{\theta }(x) = \prod _{i=1}^{n} \frac {1}{2}\,I\left (\theta -1<x_i<\theta +1\right ) = 2^{-n}\,I\left (x_{(n)}-1<\theta <x_{(1)}+1\right ).\] For two samples the ratio \(f_{\theta }(x)/f_{\theta }(y)\) is a constant function of \(\theta \) precisely when the intervals \(\left (x_{(n)}-1,\ x_{(1)}+1\right )\) and \(\left (y_{(n)}-1,\ y_{(1)}+1\right )\) coincide, that is when \(x_{(1)}=y_{(1)}\) and \(x_{(n)}=y_{(n)}\). So \(T=\left (X_{(1)},X_{(n)}\right )\) is minimal sufficient.

It is not complete

Put \(R = X_{(n)}-X_{(1)}\), the sample range. The observations are uniform on an interval of length \(2\), and the range of \(n\) independent uniforms on an interval of length \(L\) has mean \(L(n-1)/(n+1)\). With \(L=2\), \[E_{\theta }(R) = \frac {2(n-1)}{n+1}\qquad \text {for every }\theta ,\] the value being free of \(\theta \) because the range is invariant under the shift \(x\mapsto x+c\) that carries one parameter value to another. Therefore \[E_{\theta }\left [h(T)\right ] = E_{\theta }\left [\frac {X_{(n)}-X_{(1)}}{2}\right ] - \frac {n-1}{n+1} = \frac {n-1}{n+1} - \frac {n-1}{n+1} = 0 \qquad \text {for all }\theta \in \Omega ,\] while \(h\) is plainly not the zero function — the range is genuinely random. So \(T\) is minimal sufficient but not complete, which is the promised counterexample to the converse.

Solution. For the UNIF\((0,\theta )\) model the joint density is \begin {align*} f_{\theta }(x) &= \frac {1}{\theta ^{n}}\prod _{i=1}^{n}I\left (0<x_i<\theta \right )\\ &= \frac {1}{\theta ^{n}}\,I\left (x_{(1)}>0\right )\,I\left (x_{(n)}<\theta \right ). \end {align*}

Sufficiency of \(X_{(n)}\) follows from the factorisation criterion at once, with \(g\left (x_{(n)},\theta \right )=\theta ^{-n}I\left (x_{(n)}<\theta \right )\) and \(h(x)=I\left (x_{(1)}>0\right )\).

For minimality, take two samples in the support and form \[\frac {f_{\theta }(x)}{f_{\theta }(y)} = \frac {I\left (x_{(n)}<\theta \right )}{I\left (y_{(n)}<\theta \right )} .\] Viewed as a function of \(\theta \), this is constant only if the two indicators change value at the same point, that is only if \(x_{(n)} = y_{(n)}\). Hence \(X_{(n)}\) is minimal sufficient.

Combined with the completeness shown above, \(X_{(n)}\) is complete and sufficient, so by Bahadur’s theorem its minimality also follows without this calculation — but the direct argument is shorter than the theorem it instantiates.

Solution. This is the calculation already carried out in the previous section, and it is worth recording that the two are the same problem. The joint density is \[f_{\theta }(x) = e^{n\theta }\,I\left (x_{(1)}>\theta \right )\cdot e^{-\sum _i x_i},\] so \(X_{(1)}\) is sufficient by factorisation, and the ratio \[\frac {f_{\theta }(x)}{f_{\theta }(y)} = e^{\sum _i y_i-\sum _i x_i}\cdot \frac {I\left (x_{(1)}>\theta \right )}{I\left (y_{(1)}>\theta \right )}\] is free of \(\theta \) only when the two indicators switch at the same place, that is only when \(x_{(1)}=y_{(1)}\). Hence \(X_{(1)}\) is minimal sufficient.

Completeness holds as well, by the same upper-limit differentiation used for the UNIF\((0,\theta )\) maximum: \(X_{(1)}\) has density \(n\,e^{-n(t-\theta )}\) on \((\theta ,\infty )\), and \(\int _{\theta }^{\infty }h(t)e^{-nt}\,dt = 0\) for every \(\theta \) forces \(h=0\) almost everywhere. So \(X_{(1)}\) is complete and sufficient, which is what the next problem needs.

Solution. \(T=\sum _{i=1}^{n}X_i\sim \) BIN\((n,\theta )\) is complete and sufficient, so by Lehmann–Scheffé it is enough to find any function of \(T\) that is unbiased for \(\theta (1-\theta )\); it is then automatically the UMVUE.

Try \(T(n-T)\). Using \(E(T)=n\theta \) and \(E\left (T^{2}\right )=n\theta (1-\theta )+n^{2}\theta ^{2}\), \begin {align*} E\left [T(n-T)\right ] &= nE(T) - E\left (T^{2}\right )\\ &= n^{2}\theta - n\theta (1-\theta ) - n^{2}\theta ^{2}\\ &= n^{2}\theta (1-\theta ) - n\theta (1-\theta ) = n(n-1)\,\theta (1-\theta ). \end {align*}

Dividing by the constant \(n(n-1)\), \[\widehat {\tau } = \frac {T(n-T)}{n(n-1)} = \frac {n}{n-1}\,\overline {X}\left (1-\overline {X}\right )\] is unbiased, and therefore is the UMVUE of \(\theta (1-\theta )\).

Solution. \(T=X_{(1)}\) is complete and sufficient for this model, as shown above. Its density is \(n\,e^{-n(t-\theta )}\) on \((\theta ,\infty )\), so \(T-\theta \) is exponential with rate \(n\) and \[E(T) = \theta + \frac {1}{n},\qquad \var (T) = \frac {1}{n^{2}},\qquad E\left (T^{2}\right ) = \frac {1}{n^{2}} + \left (\theta +\frac 1n\right )^{2} = \theta ^{2} + \frac {2\theta }{n} + \frac {2}{n^{2}} .\] Look for an unbiased estimator of the form \(g(T)=T^{2}+\alpha T+\beta \): \[E\left [g(T)\right ] = \theta ^{2} + \left (\frac {2}{n}+\alpha \right )\theta + \left (\frac {2}{n^{2}} + \frac {\alpha }{n} + \beta \right ).\] Matching \(\theta ^{2}\) requires the coefficient of \(\theta \) and the constant to vanish, giving \(\alpha =-2/n\) and then \(\beta =0\). Hence \[\widehat {\theta ^{2}} = X_{(1)}^{2} - \frac {2X_{(1)}}{n},\] and by Lehmann–Scheffé this is the UMVUE of \(\theta ^{2}\).