Proof. Let \(R_1\) be any critical region with \(P_{\theta _0}(X\in R_1)\le \alpha \). We write \(f_0\) and \(f_1\) for \(f(x;\theta _0)\) and \(f(x;\theta _1)\), and use integrals; for a discrete distribution replace them by sums. Split each region into its part shared with the other and its own part: \[R = (R\cap R_1)\cup (R\cap R_1^c),\qquad R_1 = (R\cap R_1)\cup (R_1\cap R^c).\] The shared part cancels in differences, so \begin {equation} \label {eq:npdiff1} P_{\theta _1}(X\in R) - P_{\theta _1}(X\in R_1) = \int _{R\cap R_1^c} f_1\,dx - \int _{R_1\cap R^c} f_1\,dx , \end {equation} and in the same way \begin {equation} \label {eq:npdiff0} \int _{R\cap R_1^c} f_0\,dx - \int _{R_1\cap R^c} f_0\,dx = P_{\theta _0}(X\in R) - P_{\theta _0}(X\in R_1) \ge \alpha - \alpha = 0 . \end {equation} On \(R\cap R_1^c\) we are inside \(R\), so \(f_1>k f_0\); on \(R_1\cap R^c\) we are outside \(R\), so \(f_1\le k f_0\). Hence \begin {equation} \label {eq:npbound} \int _{R\cap R_1^c} f_1\,dx \ge k\int _{R\cap R_1^c} f_0\,dx,\qquad \int _{R_1\cap R^c} f_1\,dx \le k\int _{R_1\cap R^c} f_0\,dx . \end {equation} Putting (3.3) into (3.1) and then using (3.2), \[P_{\theta _1}(X\in R) - P_{\theta _1}(X\in R_1) \ge k\left [\int _{R\cap R_1^c} f_0\,dx - \int _{R_1\cap R^c} f_0\,dx\right ]\ge 0 .\] So no region of size at most \(\alpha \) has more power than \(R\). □

Solution. The likelihood ratio is \[\frac {L(12;x)}{L(10;x)} = \frac {(32\pi )^{-n/2}\exp \left (-\frac {1}{32}\sum (x_i - 12)^2\right )} {(32\pi )^{-n/2}\exp \left (-\frac {1}{32}\sum (x_i - 10)^2\right )} = \exp \left (\frac {1}{32}\left [\sum (x_i - 10)^2 - \sum (x_i - 12)^2\right ]\right ).\] Expanding the squares, \((x_i - 10)^2 - (x_i - 12)^2 = 4x_i - 44\), so the ratio is \[\exp \left (\frac {4\sum x_i - 44n}{32}\right ),\] an increasing function of \(\sum x_i\), and hence of \(\bar x\). The ratio exceeds \(k\) exactly when \(\bar x>c\) for some \(c\), so \(R = \{x : \bar x>c\}\). Under \(H_0\), \(\bar X\sim N\left (10,\frac {16}{n}\right )\) and \[0.05 = P_{10}(\bar X>c) = P\left (Z>\frac {c - 10}{4/\sqrt n}\right ) \quad \Longrightarrow \quad c = 10 + 1.645\,\frac {4}{\sqrt n} = 10 + \frac {6.58}{\sqrt n} .\] The MP test rejects \(H_0\) if \(\bar x>10 + \frac {6.58}{\sqrt n}\).

The power is \[\pi (12) = P_{12}(\bar X>c) = P\left (Z>\frac {c - 12}{4/\sqrt n}\right ) = P\left (Z>1.645 - \frac {\sqrt n}{2}\right ).\] For power \(0.90\) we need \(1.645 - \frac {\sqrt n}{2} = -1.2816\), so \(\sqrt n = 2(2.9266) = 5.853\) and \(n = 34.3\). A sample of \(n = 35\) is needed. For \(n = 16\), \(c = 11.645\) and the power is \(0.639\), as in example 3.1.8; figure 3.2 shows the whole power function for \(n = 16\).

Solution. As in example 3.2.4, \[\frac {L(\mu _1;x)}{L(\mu _0;x)} = \exp \left (\frac {1}{2\sigma ^2}\sum \left [(x_i - \mu _0)^2 - (x_i - \mu _1)^2\right ]\right ) = \exp \left (\frac {(\mu _1 - \mu _0)}{\sigma ^2}\sum x_i - \frac {n(\mu _1^2 - \mu _0^2)}{2\sigma ^2}\right ).\] Since \(\mu _1 - \mu _0>0\) this increases with \(\bar x\), so \(R = \{x : \bar x>c\}\). Writing \(z_\alpha \) for the point with \(P(Z>z_\alpha ) = \alpha \), the size condition gives \[c = \mu _0 + z_\alpha \,\frac {\sigma }{\sqrt n} ,\] and the power is \[\pi (\mu _1) = P\left (Z>\frac {c - \mu _1}{\sigma /\sqrt n}\right ) = 1 - \Phi \left (z_\alpha - \frac {(\mu _1 - \mu _0)\sqrt n}{\sigma }\right ).\] The power grows with the sample size \(n\) and with the distance \(\frac {\mu _1 - \mu _0}{\sigma }\) measured in standard deviations.

Solution. Write \(q = \sum _{i=1}^n (x_i - \mu )^2\). The likelihood ratio is \[\frac {L(\sigma _1;x)}{L(\sigma _0;x)} = \frac {(2\pi \sigma _1^2)^{-n/2} e^{-q/(2\sigma _1^2)}}{(2\pi \sigma _0^2)^{-n/2} e^{-q/(2\sigma _0^2)}} = \left (\frac {\sigma _0}{\sigma _1}\right )^n \exp \left (-\frac {q}{2}\left [\frac {1}{\sigma _1^2} - \frac {1}{\sigma _0^2}\right ]\right ).\] Because \(\sigma _1<\sigma _0\), the bracket is positive and the ratio decreases in \(q\). So the ratio exceeds \(k\) exactly when \(q<c\), and \(R = \{x : q<c\}\): small spread around \(\mu \) is evidence for the smaller variance.

Under \(H_0\), \(\frac {Q}{\sigma _0^2} = \sum \left (\frac {X_i - \mu }{\sigma _0}\right )^2 \sim \chi ^2(n)\) by corollary 1.3.40. Writing \(\chi ^2_{n,\gamma }\) for the point with upper-tail probability \(\gamma \), \[\alpha = P_{\sigma _0}(Q<c) = P\left (\chi ^2(n)<\frac {c}{\sigma _0^2}\right ) \quad \Longrightarrow \quad c = \sigma _0^2\,\chi ^2_{n,1-\alpha } .\] The MP test rejects \(H_0\) if \(\sum (x_i - \mu )^2<\sigma _0^2\,\chi ^2_{n,1-\alpha }\).

Solution. Write \(y = \sum x_i\). The likelihood ratio is \[\frac {L(\theta _1;x)}{L(\theta _0;x)} = \frac {\theta _1^{-n}e^{-y/\theta _1}}{\theta _0^{-n}e^{-y/\theta _0}} = \left (\frac {\theta _0}{\theta _1}\right )^n \exp \left (-\left [\frac {1}{\theta _1} - \frac {1}{\theta _0}\right ] y\right ).\] Since \(\theta _1<\theta _0\) the bracket is positive and the ratio decreases in \(y\), so \(R = \{x : \sum x_i<c\}\): short lifetimes point to the smaller mean.

To find \(c\) we need the distribution of \(Y = \sum X_i\). The sum of \(n\) independent \(\text {EXP}(\theta )\) variables is \(\text {GAM}(n,\theta )\), and then \(\frac {2Y}{\theta }\sim \text {GAM}(n,2) = \chi ^2(2n)\). Hence \[\alpha = P_{\theta _0}(Y<c) = P\left (\chi ^2(2n)<\frac {2c}{\theta _0}\right ) \quad \Longrightarrow \quad c = \frac {\theta _0}{2}\,\chi ^2_{2n,1-\alpha } .\] The MP test rejects \(H_0\) if \(\sum x_i<\frac {\theta _0}{2}\chi ^2_{2n,1-\alpha }\).

For \(n = 10\), \(\theta _0 = 2\) and \(\alpha = 0.05\), \(\chi ^2_{20,0.95} = 10.851\), so \(c = 10.851\): reject \(H_0\) if the ten lifetimes total less than \(10.85\). Under \(\theta _1 = 1\), \(2Y\sim \chi ^2(20)\), so the power is \[P_1(Y<10.851) = P\left (\chi ^2(20)<21.70\right ) = 0.643 .\]

Solution. With \(y = \sum x_i\), \[\frac {L(\frac 14;x)}{L(\frac 34;x)} = \frac {(\frac 14)^y(\frac 34)^{3 - y}}{(\frac 34)^y(\frac 14)^{3 - y}} = 3^{3 - 2y},\] which decreases in \(y\). So the MP test rejects for small \(Y = \sum X_i\), and \(Y\sim \text {BIN}(3,\theta )\). Under \(H_0\) the probabilities are

Rejecting only when \(y = 0\) gives size \(0.0156\); rejecting when \(y\le 1\) gives size \(0.1562\). Neither is \(0.05\). So reject when \(y = 0\), and when \(y = 1\) reject with probability \(\gamma \), chosen so that \[\frac {1}{64} + \gamma \,\frac {9}{64} = 0.05 \quad \Longrightarrow \quad \gamma = \frac {0.05 - 0.015625}{0.140625} = 0.2444 .\] The test is \(\phi (x) = 1\) if \(y = 0\), \(\phi (x) = 0.2444\) if \(y = 1\), and \(\phi (x) = 0\) otherwise. Under \(\theta = \frac 14\), \(P(Y = 0) = P(Y = 1) = \frac {27}{64}\), so the power is \[\pi \left (\tfrac 14\right ) = \frac {27}{64} + 0.2444\cdot \frac {27}{64} = 0.525 .\] In practice one usually prefers a non-randomised test of level \(\alpha \), here the test that rejects only when \(y = 0\), and reports its actual size \(0.0156\).