Solution. (a)(i). The order statistics of a random sample \(X_1,\ldots ,X_n\) are the sample values rearranged in increasing order, \[X_{(1)}\leq X_{(2)}\leq \cdots \leq X_{(n)},\] so \(X_{(1)}=\min (X_1,\ldots ,X_n)\) and \(X_{(n)}=\max (X_1,\ldots ,X_n)\).

(ii). If \(U\sim \chi ^2(m)\) and \(V\sim \chi ^2(n)\) are independent, then \[F=\frac {U/m}{V/n}\] has the \(F\)-distribution with \(m\) and \(n\) degrees of freedom.

(b)(i). \[M_X(t)=\int _0^\infty e^{tx} \frac {x^{\frac {n}{2}-1}e^{-\frac {x}{2}}} {2^{\frac {n}{2}}\Gamma \left (\frac {n}{2}\right )}dx =\frac {1}{2^{\frac {n}{2}}\Gamma \left (\frac {n}{2}\right )} \int _0^\infty x^{\frac {n}{2}-1}e^{-x\left (\frac {1-2t}{2}\right )}dx.\] The integral is a gamma integral with shape \(\frac {n}{2}\) and scale \(\frac {2}{1-2t}\), so it equals \(\Gamma \left (\frac {n}{2}\right )\left (\frac {2}{1-2t}\right )^{\frac {n}{2}}\) and \[M_X(t)=\frac {1}{2^{\frac {n}{2}}}\left (\frac {2}{1-2t}\right )^{\frac {n}{2}} =\left (\frac {1}{1-2t}\right )^{\frac {n}{2}},\qquad t<\tfrac 12.\qquad \blacksquare \]

(ii). It is easier to differentiate the cumulant generating function \[K(t)=\ln M_X(t)=-\frac {n}{2}\ln (1-2t),\] \[K'(t)=\frac {n}{1-2t},\qquad K''(t)=\frac {2n}{(1-2t)^2},\] \[\therefore \quad E(X)=K'(0)=n,\qquad \text {Var}(X)=K''(0)=2n.\]

(iii). The chi-square distribution is the gamma distribution with shape \(\frac {n}{2}\) and scale \(2\): \[\chi ^2(n)\equiv \text {GAM}\left (\alpha =\tfrac {n}{2},\ \beta =2\right ).\] Equivalently, if \(X\sim \text {GAM}(\alpha ,\beta )\) then \(\frac {2X}{\beta }\sim \chi ^2(2\alpha )\) — the fact used throughout Chapter 3 to turn a sum of exponentials into a chi-square critical value.

(c)(i). With \(T=\sum x_i\) the likelihood is \[L(\theta )=\theta ^n(1-\theta )^{T-n},\] so the Neyman–Pearson ratio is \[\frac {L(\theta _0)}{L(\theta _1)} =\left (\frac {\theta _0}{\theta _1}\right )^n \left (\frac {1-\theta _0}{1-\theta _1}\right )^{T-n}.\] Since \(\theta _1>\theta _0\) we have \(1-\theta _0>1-\theta _1\), so the base of the second factor exceeds \(1\) and the ratio increases in \(T\). The lemma rejects where the ratio is small, hence \[\text {reject } H_0 \text { when } T=\sum _{i=1}^nX_i\leq c,\qquad P\left (T\leq c\mid \theta _0\right )=\alpha .\] This is the right direction on inspection: \(E(X)=\frac {1}{\theta }\), so a larger \(\theta \) produces smaller observations.

(ii). For any \(\theta _1>\theta _0\), \[\frac {L(\theta _1)}{L(\theta _0)} =\left (\frac {\theta _1}{\theta _0}\right )^n \left (\frac {1-\theta _1}{1-\theta _0}\right )^{T-n},\] and \(\frac {1-\theta _1}{1-\theta _0}<1\), so the ratio is a monotone (decreasing) function of \(T=\sum X_i\) and depends on the sample only through \(T\). The family therefore has a monotone likelihood ratio in \(T\) — equivalently, an increasing one in \(-T\).

(iii). For the alternative \(H_1:\theta <\theta _0\) take \(\theta _1<\theta _0\). Then \(\frac {1-\theta _1}{1-\theta _0}>1\) and the ratio \(\frac {L(\theta _1)}{L(\theta _0)}\) increases in \(T\). By the Karlin–Rubin theorem the test rejecting for large \(T\) is uniformly most powerful: \[\text {reject } H_0 \text { when } T=\sum _{i=1}^nX_i\geq c,\qquad P\left (T\geq c\mid \theta _0\right )=\alpha ,\] where under \(H_0\) the statistic \(T\) is the sum of \(n\) independent geometric variables, so \(T\sim NB(n,\theta _0)\) on \(\{n,n+1,n+2,\ldots \}\), which fixes \(c\).