Proof. For simplicity assume \(f_\theta (x)>0\) for all \(x\) in the sample space and all \(\theta \).

\(T\) is sufficient

For each value \(t\) of \(T\) choose one sample point \(x_t\) with \(T(x_t) = t\). For any \(x\), the points \(x\) and \(x_{T(x)}\) have the same value of \(T\), so by hypothesis \(h(x) = \frac {f_\theta (x)}{f_\theta (x_{T(x)})}\) does not depend on \(\theta \). Then \[f_\theta (x) = f_\theta \left (x_{T(x)}\right )h(x) = g(T(x);\theta )h(x)\] with \(g(t;\theta ) = f_\theta (x_t)\), and \(T\) is sufficient by theorem 1.3.4.

\(T\) is minimal

Let \(U\) be any sufficient statistic, so \(f_\theta (x) = g'(U(x);\theta )h'(x)\). If \(U(x) = U(y)\), then \[\frac {f_\theta (x)}{f_\theta (y)} = \frac {g'(U(x);\theta )h'(x)}{g'(U(y);\theta )h'(y)} = \frac {h'(x)}{h'(y)},\] which does not depend on \(\theta \), and so \(T(x) = T(y)\) by hypothesis. Thus \(T\) takes the same value wherever \(U\) does, that is, \(T\) is a function of \(U\). □

Solution. \(f_\theta (x) = \theta ^n\left (\prod x_i\right )^{\theta - 1}\), so for two sample points \(x\) and \(y\), \[\frac {f_\theta (x)}{f_\theta (y)} = \left (\frac {\prod x_i}{\prod y_i}\right )^{\theta - 1} .\] This is constant in \(\theta \) if and only if \(\prod x_i = \prod y_i\). By theorem 1.4.2, \(T(X) = \prod X_i\) is minimal sufficient, and so is \(\sum \ln X_i\).

Solution. \[f_\theta (x) = (2\pi \theta ^2)^{-n/2}\exp \left (-\frac {1}{2\theta ^2}\left [\sum x_i^2 - 2\theta \sum x_i + n\theta ^2\right ]\right ),\] so \[\frac {f_\theta (x)}{f_\theta (y)} = \exp \left (-\frac {1}{2\theta ^2}\left [\sum x_i^2 - \sum y_i^2\right ] + \frac 1\theta \left [\sum x_i - \sum y_i\right ]\right ).\] The functions \(\frac {1}{\theta ^2}\) and \(\frac 1\theta \) are linearly independent, so the exponent is constant in \(\theta \) only if both brackets vanish. The ratio is constant if and only if \(\sum x_i^2 = \sum y_i^2\) and \(\sum x_i = \sum y_i\), and \(T(X) = \left (\sum X_i, \sum X_i^2\right )\) is minimal sufficient. Although there is only one parameter, the minimal sufficient statistic is two-dimensional.

Solution. Here the density can be zero, so we apply the idea of theorem 1.4.2 directly. For sample points \(x\) and \(y\) with positive entries, \[\frac {f_\theta (x)}{f_\theta (y)} = \frac {I\left (x_{(n)}<\theta \right )}{I\left (y_{(n)}<\theta \right )} .\] If \(x_{(n)} = y_{(n)}\), numerator and denominator are equal for every \(\theta \), and the ratio is \(1\) wherever it is defined. If \(x_{(n)}>y_{(n)}\), the ratio is \(1\) for \(\theta >x_{(n)}\) but \(0\) for \(y_{(n)}<\theta \le x_{(n)}\), so it depends on \(\theta \); the case \(x_{(n)}<y_{(n)}\) is symmetric. Hence the ratio is free of \(\theta \) exactly when \(x_{(n)} = y_{(n)}\), and \(X_{(n)}\) is minimal sufficient. Problem 1.5.4 gives a direct proof from the definition.

Solution. \[\frac {f_\theta (x)}{f_\theta (y)} = \exp \left (\sum y_i - \sum x_i\right ) \prod _{i=1}^n\left (\frac {1 + e^{-(y_i - \theta )}}{1 + e^{-(x_i - \theta )}}\right )^2 .\] If the \(x_i\) are a rearrangement of the \(y_i\) the ratio is \(1\). Conversely, put \(u = e^\theta \); the ratio is constant in \(\theta \) only if \(\prod (1 + u e^{-y_i})^2\) and \(\prod (1 + u e^{-x_i})^2\) are proportional as polynomials in \(u\). Two polynomials that are proportional have the same roots, and the roots here are \(-e^{y_i}\) and \(-e^{x_i}\), so the sets \(\{x_i\}\) and \(\{y_i\}\) agree. Hence the ratio is constant if and only if \(x\) and \(y\) have the same order statistics, and \((X_{(1)}, \dots , X_{(n)})\) is minimal sufficient. Even for a one-parameter location family, no reduction beyond the order statistics may be possible.