Proof. We must show that \(P_\theta (U\in B, T\in C) = P_\theta (U\in B)P_\theta (T\in C)\) for all sets \(B\), \(C\) and all \(\theta \). Fix \(B\) and define \[g(t) = P(U\in B\mid T = t) - P(U\in B) .\] The first term does not depend on \(\theta \) by sufficiency of \(T\), and the second does not by ancillarity of \(U\), so \(g(T)\) is a statistic. By the law of total expectation, \[E_\theta \left [g(T)\right ] = E_\theta \left [P(U\in B\mid T)\right ] - P(U\in B) = P(U\in B) - P(U\in B) = 0\] for every \(\theta \). Completeness gives \(g(T) = 0\) with probability one, that is, \(P(U\in B\mid T = t) = P(U\in B)\) for almost all \(t\). Hence, writing \(f_T(t;\theta )\) for the density of \(T\), \begin {align*} P_\theta (U\in B, T\in C) &= \int _C P(U\in B\mid T = t)f_T(t;\theta )\,dt\\ &= P(U\in B)\int _C f_T(t;\theta )\,dt = P(U\in B)P_\theta (T\in C), \end {align*}

with a sum in place of the integral in the discrete case. □

Solution. The variables \(Z_i = \frac {X_i}{\theta }\) are independent \(\text {EXP}(1)\), whose distribution is free of \(\theta \). Since \[\frac {X_i}{T} = \frac {X_i/\theta }{\sum _j X_j/\theta } = \frac {Z_i}{\sum _j Z_j},\] \(U\) is a function of \((Z_1, \dots , Z_n)\) and is ancillary. By example 1.6.16, \(T\) is complete and sufficient. By theorem 1.7.2, \(T\) and \(U\) are independent.

Now \(X_1 = \frac {X_1}{T}\cdot T\). Taking expectations and using independence, \(\theta = E(X_1) = E\left (\frac {X_1}{T}\right )E(T) = E\left (\frac {X_1}{T}\right )n\theta \), so \(E\left (\frac {X_1}{T}\right ) = \frac 1n\). Conditioning on \(T\), and again using independence, \[E(X_1\mid T) = T\,E\left (\frac {X_1}{T} \mid T\right ) = T\,E\left (\frac {X_1}{T}\right ) = \frac Tn = \bar X .\]

Solution. Fix \(\sigma ^2\) and regard \(\mu \) as the only parameter. The family \(N(\mu , \sigma ^2)\) with \(\sigma ^2\) known is a REF with \(\eta = \frac {\mu }{\sigma ^2}\), so \(\bar X\) is complete and sufficient for \(\mu \). Write \(Z_i = X_i - \mu \), which are \(N(0, \sigma ^2)\) with distribution free of \(\mu \). Then \(X_i - \bar X = Z_i - \bar Z\), so \[S^2 = \frac {1}{n - 1}\sum (X_i - \bar X)^2 = \frac {1}{n - 1}\sum (Z_i - \bar Z)^2\] is a function of the \(Z_i\) and is ancillary for \(\mu \). By theorem 1.7.2, \(\bar X\) and \(S^2\) are independent when the true variance is \(\sigma ^2\). Since \(\sigma ^2\) was arbitrary, they are independent for all \(\mu \) and \(\sigma ^2\).