Solution. (a). Under \(H_0\) the density is \(f_0(x)=1\) — the uniform on \((0,1)\) — and under \(H_1\) it is \(f_1(x)=2x\). By the Neyman–Pearson lemma the most powerful test rejects when the ratio is small: \[\frac {f_0(x)}{f_1(x)}=\frac {1}{2x}\leq k.\] Since \(\frac {1}{2x}\) decreases in \(x\), this is equivalent to \(x\geq c\) for some \(c\). Fixing the size, \[\alpha =P(X\geq c\mid \theta =1)=\int _c^1 1\,dx=1-c=0.05 \quad \implies \quad c=0.95.\] \[\therefore \quad \text {reject } H_0 \text { when } X\geq 0.95.\]

(b). The power is the same probability computed under the alternative: \[\text {power}=P(X\geq 0.95\mid \theta =2)=\int _{0.95}^1 2x\,dx =\left [x^2\right ]_{0.95}^1=1-0.9025=0.0975.\]

(c). For a sample of size \(n\), \[\frac {L(1)}{L(2)}=\frac {1}{2^n\prod x_i}\leq k \quad \Longleftrightarrow \quad \prod _{i=1}^n x_i\geq c.\] Taking logarithms and reversing the inequality (the logarithms are negative), \[-\sum _{i=1}^n\ln x_i\leq c'.\] Under \(H_0\), \(X_i\sim \text {UNIF}(0,1)\), so \(-\ln X_i\sim \text {EXP}(1)\) and \[-\sum _{i=1}^n\ln X_i\sim \text {GAM}(n,1),\qquad \text {equivalently}\qquad -2\sum _{i=1}^n\ln X_i\sim \chi ^2(2n).\] \[\therefore \quad \text {reject } H_0 \text { when } -2\sum _{i=1}^n\ln X_i\leq \chi ^2_{1-\alpha }(2n),\] the lower tail, because large values of \(\prod x_i\) favour the alternative.

Solution. (a). The Neyman–Pearson ratio is \[\frac {f_0(x)}{f_1(x)} =\frac {e^{-\theta _0}\theta _0^x/x!}{e^{-\theta _1}\theta _1^x/x!} =e^{\theta _1-\theta _0}\left (\frac {\theta _0}{\theta _1}\right )^x.\] Because \(\theta _1>\theta _0\) the base \(\frac {\theta _0}{\theta _1}\) is less than \(1\), so the ratio decreases in \(x\) and “ratio small” means “\(x\) large”. The test is \[\text {reject } H_0 \text { when } X\geq c,\qquad \text {where } P\left (X\geq c\mid \theta _0\right )=\alpha .\]

(b). For a sample, \(\sum X_i\) is sufficient and \(\sum _{i=1}^nX_i\sim \text {POI}(n\theta )\). The ratio becomes \[\frac {L(\theta _0)}{L(\theta _1)}=e^{n(\theta _1-\theta _0)} \left (\frac {\theta _0}{\theta _1}\right )^{\sum x_i},\] decreasing in \(\sum x_i\) by the same argument, so \[\text {reject } H_0 \text { when } \sum _{i=1}^nX_i\geq c,\qquad P\left (\sum X_i\geq c\ \Big |\ n\theta _0\right )=\alpha .\]

Solution. Note first that \(0.5^k0.5^{x-k}=0.5^x\) whatever \(k\) is, so \[f(x)=\binom {x-1}{k-1}(0.5)^x.\] Under \(H_0\) \((k=1)\): \(f_0(x)=(0.5)^x\) for \(x=1,2,\ldots \) Under \(H_1\) \((k=2)\): \(f_1(x)=(x-1)(0.5)^x\) for \(x=2,3,\ldots \)

(a). The ratio is \[\frac {f_0(x)}{f_1(x)}=\frac {1}{x-1},\] which decreases in \(x\), so the most powerful test rejects for large \(X\). For the size, the \(H_0\) tail sums as a geometric series: \[P(X\geq c\mid k=1)=\sum _{x=c}^{\infty }(0.5)^x=(0.5)^{c-1}.\] Setting this equal to \(0.125=(0.5)^3\) gives \(c-1=3\), so \[\boxed {\ \text {reject } H_0 \text { when } X\geq 4\ }\] and the size is exactly \(0.125\) — one of the rare discrete cases where \(\alpha \) is attained on the nose.

(b). The power is \[P(X\geq 4\mid k=2)=1-P(X=2)-P(X=3) =1-(1)(0.5)^2-(2)(0.5)^3=1-0.25-0.25=0.5.\]

Solution. For any \(\theta _1>\theta _0\), \[\frac {L(\theta _1)}{L(\theta _0)} =\left (\frac {\theta _0}{\theta _1}\right )^n \exp \!\left [-\sum x_i^3\left (\frac {1}{\theta _1}-\frac {1}{\theta _0}\right )\right ] =\left (\frac {\theta _0}{\theta _1}\right )^n \exp \!\left [\sum x_i^3\left (\frac {1}{\theta _0}-\frac {1}{\theta _1}\right )\right ].\] Since \(\theta _1>\theta _0\) we have \(\frac {1}{\theta _0}-\frac {1}{\theta _1}>0\), so the ratio is an increasing function of \[T=\sum _{i=1}^nX_i^3,\] and it depends on the sample only through \(T\). The family therefore has a monotone likelihood ratio in \(T\), and the critical region is free of \(\theta _1\) — which is what makes one test uniformly most powerful for the whole composite alternative: \[\text {reject } H_0 \text { when } T=\sum _{i=1}^nX_i^3\geq c.\]

To find \(c\), substitute \(Y=X^3\). Then \(Y\sim \text {EXP}(\text {mean }\theta )\), so under \(H_0\) \[T\sim \text {GAM}(n,\theta _0)\qquad \text {and}\qquad \frac {2T}{\theta _0}\sim \chi ^2(2n).\] \[\therefore \quad \text {reject } H_0 \text { when } \frac {2}{\theta _0}\sum _{i=1}^nX_i^3\geq \chi ^2_{\alpha }(2n).\]

Solution. This is the geometric distribution with success probability \(p=\frac {1}{\theta +1}\), counted as the number of failures, so \(E(X)=\theta \). The likelihood is \[L(\theta )=\frac {1}{(\theta +1)^n}\left (\frac {\theta }{\theta +1}\right )^{\sum x_i}.\] For \(\theta _1>\theta _0\), \[\frac {L(\theta _1)}{L(\theta _0)} =\left (\frac {\theta _0+1}{\theta _1+1}\right )^{n} \left [\frac {\theta _1(\theta _0+1)}{\theta _0(\theta _1+1)}\right ]^{\sum x_i}.\] Compare the two products inside the bracket: \[\theta _1(\theta _0+1)=\theta _0\theta _1+\theta _1,\qquad \theta _0(\theta _1+1)=\theta _0\theta _1+\theta _0,\] and since \(\theta _1>\theta _0\) the first exceeds the second, so the bracket is greater than \(1\). The ratio is therefore increasing in \(\sum x_i\): a monotone likelihood ratio in \(T=\sum X_i\). Hence \[\text {reject } H_0 \text { when } T=\sum _{i=1}^nX_i\geq c,\qquad P\left (T\geq c\mid \theta _0\right )=\alpha ,\] and this same region is most powerful against every \(\theta _1>\theta _0\), so the test is UMP. Under \(H_0\), \(T\) is negative binomial — the sum of \(n\) independent geometrics — which fixes \(c\).

Solution. (a)(i). With \(\beta \) known, \[L(\theta )=\frac {1}{\theta ^{n\beta }\Gamma (\beta )^n} \left (\prod x_i\right )^{\beta -1}e^{-\frac {1}{\theta }\sum x_i},\] so for \(\theta _1>\theta _0\) \[\frac {L(\theta _1)}{L(\theta _0)} =\left (\frac {\theta _0}{\theta _1}\right )^{n\beta } \exp \!\left [\sum x_i\left (\frac {1}{\theta _0}-\frac {1}{\theta _1}\right )\right ].\] The exponent’s coefficient is positive, so the ratio increases in \(T=\sum X_i\): the family has a monotone likelihood ratio in \(T\).

(ii). By the Karlin–Rubin theorem, a family with MLR in \(T\) gives a UMP test of \(H_0:\theta \leq \theta _0\) against \(H_1:\theta >\theta _0\) that rejects for large \(T\). Under \(\theta =\theta _0\), \[T\sim \text {GAM}(n\beta ,\theta _0)\qquad \text {so}\qquad \frac {2T}{\theta _0}\sim \chi ^2(2n\beta ),\] \[\therefore \quad \text {reject } H_0 \text { when } \frac {2}{\theta _0}\sum _{i=1}^nX_i\geq \chi ^2_\alpha (2n\beta ).\]

(iii). The power function \[\pi (\theta )=P\!\left (\frac {2T}{\theta _0}\geq \chi ^2_\alpha (2n\beta )\ \Big |\ \theta \right )\] increases in \(\theta \): it is below \(\alpha \) throughout \(\theta <\theta _0\), equals \(\alpha \) exactly at \(\theta =\theta _0\), and rises towards \(1\) as \(\theta \) grows. That the maximum over \(H_0\) is attained at the boundary \(\theta =\theta _0\) is what makes the size of the composite null equal to \(\alpha \).

(b). With \(\theta \) known and \(\beta \) unknown, collect the terms in \(\beta \): \[L(\beta )=\frac {\left (\prod x_i\right )^{\beta -1}} {\theta ^{n\beta }\Gamma (\beta )^n}e^{-\frac {1}{\theta }\sum x_i},\] \[\frac {L(\beta _1)}{L(\beta _0)} =\left [\frac {\Gamma (\beta _0)}{\Gamma (\beta _1)}\right ]^n \theta ^{-n(\beta _1-\beta _0)} \left (\prod x_i\right )^{\beta _1-\beta _0}.\] For \(\beta _1>\beta _0\) the exponent \(\beta _1-\beta _0\) is positive, so the ratio increases in \(\prod x_i\), equivalently in \[T=\sum _{i=1}^n\ln X_i.\] By Karlin–Rubin again, \[\text {reject } H_0 \text { when } \sum _{i=1}^n\ln X_i\geq c,\qquad P\left (T\geq c\mid \beta _0\right )=\alpha ,\] and the power function increases in \(\beta \) with the same shape as in (iii), crossing \(\alpha \) at \(\beta =\beta _0\).

Solution. The likelihood is \(L(\theta )=\theta ^{-n}e^{-\frac {1}{\theta }\sum x_i}\), and the unrestricted maximum is at \(\hat \theta =\bar x\). The GLR statistic is \[\lambda =\frac {L(\theta _0)}{L\left (\hat \theta \right )} =\frac {\theta _0^{-n}e^{-\frac {\sum x_i}{\theta _0}}} {\bar x^{-n}e^{-\frac {\sum x_i}{\bar x}}}.\] Since \(\frac {\sum x_i}{\bar x}=n\), this simplifies to \[\lambda =\left (\frac {\bar x}{\theta _0}\right )^n e^{\,n-\frac {n\bar x}{\theta _0}}.\] Writing \(u=\frac {\bar x}{\theta _0}\), \[\lambda =u^ne^{n(1-u)}=\left (ue^{1-u}\right )^n.\]

Now examine \(h(u)=ue^{1-u}\). Its derivative is \[h'(u)=(1-u)e^{1-u},\] which is positive for \(u<1\) and negative for \(u>1\), so \(h\) rises to a single maximum \(h(1)=1\) and falls away on both sides. Consequently \[\lambda \leq c\quad \Longleftrightarrow \quad u\leq c_1\ \text { or }\ u\geq c_2,\] that is, the critical region is two-sided in \(\bar X\) — as it must be against a two-sided alternative. Equivalently, in terms of the sum, \[\text {reject } H_0 \text { when } \frac {2}{\theta _0}\sum _{i=1}^nX_i\leq \chi ^2_{1-\frac {\alpha }{2}}(2n) \quad \text {or}\quad \frac {2}{\theta _0}\sum _{i=1}^nX_i\geq \chi ^2_{\frac {\alpha }{2}}(2n),\] using \(\frac {2}{\theta _0}\sum X_i\sim \chi ^2(2n)\) under \(H_0\).