Proof. We need to show that \(P\left (U(X) \in B, \, T(X)\in C\right ) = P\left (U(x)\in B\right )\, P\left (T(X)\in C\right )\) for all sets \(B, C\) and \(\theta \in \Omega \).
Let \(g(t) = P\left (U(X) \in B / T(X) = t\right ) - P\left (U(X) \in B\right )\) for all \(t\in A\) where \(P\left (T\in A\right ) = 1\). By sufficiency \(P\left (U(X) \in B / T(X) = t\right )\) does not depend on \(\theta \) and by ancillarity \(P\left (U(X)\in B\right )\) also does not depend on \(\theta \). Therefore \(g(T)\) is a statistic.
Let \[I\left (U(X) \in B\right ) = \begin {cases} 1 & \text {if}\,\,\,\, U(X) \in B\\ 0 & \text {otherwise}\\ \end {cases}\] then \(E\left [I(U(X) \in B)\right ] = P\left (U(X)\in B\right )\), \(\, E\left [I(U(X) = B) / T = t \right ] = P\left (U(X) \in B / T = t\right )\) and \(g(t) = E\left [I\left (U(X) \in B / T = t\right )\right ] - E\left [I\left (U(X) \in B\right )\right ]\). Then \begin {align*} E\left [g(t)\right ] & = E\left \{E\left (I(U(X) \in B / T = t\right )\right \} - E\left [I\left (U(X) \in B\right )\right ]\\ & = E\left [E\left (U(X) \in B\right )\right ] - E\left [I\left (U(X) \in B\right )\right ]\\ & = 0. \end {align*}

Since \(T\) is complete this implies that \(P_{\theta }\left [g(T) = 0\right ] = 1\) for all \(\theta \in \Omega \). Therefore
\(P\left (U(X)\in B / T(X) = t\right ) = P\left (U(X) \in B\right )\) for all \(t \in A\) and \(B\).
Suppose \(T\) has probability density function \(g_{\theta }(t)\), then \begin {align*} P\left (U(X)\in B, T(X) \in C\right ) & = P\left (U(X)\in B / T(X) = t\right )\, P\left (T(X) \in C\right )\\ & = \int _C P\left (U(X) \in B / T = t\right )\, g_{\theta } (t) \, dt\\ & = \int _CP\left (U(X) \in B\right )\, g_{\theta }(t)\, dt\\ & = P\left (U(X)\in B\right )\, P\left (T(X)\in C\right ) \end {align*}

for all sets \(B, C\) and \(\theta \in \Omega \) as required. □

Solution. \(X_i \, \thicksim \, EXP(\theta )\) \[T = \sum ^n_{i = 1} X_i \, \thicksim \, GAM(n,\theta )\] \[Z_i = \frac {X_i}{\theta } \, \thicksim \, EXP(1)\,\,\hspace {0.2cm}\text {is ancillary}.\] \[Z = \frac {T}{\theta } \, \thicksim \, GAM(n ,1)\, \, \text {is ancillary now}\] \[\frac {X_i}{T} = \frac {X_i/\theta }{T/\theta } = \frac {Z_i}{Z}\hspace {0.3cm}\text {is ancillary}\] \[\therefore \, \hspace {0.2cm} U(X) = \left (\frac {X_1}{T}, \, \cdots \, , \, \frac {X_n}{T}\right ) = \left (\frac {Z_1}{Z}, \, \cdots \, , \, \frac {Z_n}{Z}\right )\hspace {0.3cm} \text {is ancillary statistic}\] \(X_i\thicksim EXP(\theta )\) is a REF distribution.
\(\therefore \, T = \sum ^n_{i = 1} X_i\,\) is a complete sufficient statistic. Therefore by Basus’ theorem \(U(X)\) and \(T\) are independent.
Now \(\, E(X_1) = \theta \) \[E\left (\frac {X_1}{T}\, \cdot \, T\right ) = \theta \] \[E\left (\frac {X_1}{T}\right )\, E(T) = \theta \] \[E\left (\frac {X_1}{T}\right )\, n\theta = \theta \] \[\implies \hspace {0.3cm} E\left (\frac {X_1}{T}\right ) = \frac {1}{n}.\] □

Solution. \(X_i \thicksim N(\mu , \sigma ^2)\) assume \(\sigma ^2\) is known, then \(\overline {X}\) is a complete sufficient statistic. \[Z_i = X_i - \mu \, \thicksim \, N(0, \sigma ^2)\,\hspace {0.2cm}\text {is ancillary}\] \(\overline {Z} = \overline {X} - \mu \, \thicksim \, N(0, \sigma ^2/n)\) is ancillary then \(\sum ^n_{i = 1} (Z_i - \overline {Z})^2\) is ancillary \[\sum ^n_{i = 1} (Z_i - \overline {Z})^2 = \sum ^n_{i = 1} (X_i - \mu - (\overline {X} - \mu )^2) = \sum ^n_{i = 1} (X_i - \overline {X})^2\,\hspace {0.2cm} \text {is ancillary}.\] Therefore \(\, S^2 = \frac {1}{n - 1}\sum ^n_{i = 1}(X_i - \overline {X})^2\) is ancillary.
Therefore by Basus’ theorem \(\overline {X}\) and \(S^2\) are independent. Since this argument holds for all values of \(\sigma ^2\).
\(\therefore \, \overline {X}\) and \(S^2\) are independent for all \(\mu \) and \(\sigma ^2\). □