Solution. In each case \(f_\theta (x) = f_0(x - \theta )\), with respectively \[f_0(z) = \frac {1}{\sqrt {2\pi }}e^{-z^2/2},\qquad f_0(z) = \frac {1}{\pi (1 + z^2)},\qquad f_0(z) = e^{-z}I(z>0) .\] In each, \(X = \theta + Z\) where \(Z\) has the fixed density \(f_0\).

Solution. \(\frac 1n\sum (x_i + a) = \frac 1n\sum x_i + \frac 1n\cdot na = \bar x + a\). Adding \(a\) to every observation does not change their order, so every order statistic, and in particular the median and \(X_{(1)}\), increases by \(a\). By contrast, \(\frac 1n\sum x_i^2\) is not equivariant.

Proof. Write \(X_i = \theta + Z_i\), where \(Z_1, \dots , Z_n\) are independent with the known density \(f_0\). By equivariance with \(a = \theta \), \[\tilde \theta (X_1, \dots , X_n) - \theta = \tilde \theta (Z_1, \dots , Z_n),\] whose distribution involves \(f_0\) only. Hence the distribution of the error \(\tilde \theta - \theta \), and every moment of it, is the same for all \(\theta \). □

Solution. Each such estimator is equivariant. By example 2.5.7, \(X_{(1)} - \theta \) is exponential with mean \(\frac 1n\) and variance \(\frac 1{n^2}\), so \[\MSE \left (X_{(1)} - c\right ) = E\left (X_{(1)} - \theta - c\right )^2 = \frac {1}{n^2} + \left (\frac 1n - c\right )^2 ,\] which, as theorem 3.2.5 says, is free of \(\theta \). It is minimised at \(c = \frac 1n\), giving the unbiased estimator \(X_{(1)} - \frac 1n\) with mean square error \(\frac {1}{n^2}\), half that of the ML estimator \(X_{(1)}\).

Solution.

(a)

\(f_\theta (x) = \frac {1}{\sqrt {2\pi }\,\theta }e^{-\frac 12(x/\theta )^2} = \frac 1\theta f_1\left (\frac x\theta \right )\) with \(f_1(z) = \frac {1}{\sqrt {2\pi }}e^{-z^2/2}\).

(b)

The family is a REF with natural sufficient statistic \(\sum X_i^2\), which is complete. Since \(E(X_i^2) = \theta ^2\), the UMVUE of \(\theta ^2\) is \(\tilde \theta _2 = \frac 1n\sum X_i^2\), and \[\tilde \theta _2(cx_1, \dots , cx_n) = \frac 1n\sum (cx_i)^2 = c^2\,\tilde \theta _2(x_1, \dots , x_n),\] so it is scale equivariant with \(k = 2\).