Chapter 9 Exercises

9.6

Let $X_1,X_2,\dots ,X_{10}$ be a random sample from a distribution that is $N({\theta }_1,{\theta }_2)$¹. Find a best test of the simple hypothesis $H_0:{\theta }_1={\theta }_1^{\prime }=0,{\theta }_2={\theta }_2^{\prime }=1$ against the alternative simple hypothesis $H_1:{\theta }_1={\theta }_1^{\prime \prime }=1,{\theta }_2={\theta }_2^{\prime \prime }=4$.

Answer

Untuk $H_0:{\theta }_1=0,{\theta }_2=1$, fungsi likelihood adalah:

$$\begin{array}{rl}{L}_0(x_1,\dots ,x_{10})& =\prod _{i=1}^{10}\frac{1}{\sqrt{2\pi \cdot 1}}\exp \left(-\frac{(x_i-0{)}^2}{2\cdot 1}\right)\\ & =(2\pi {)}^{-5}\exp \left(-\frac{1}{2}\sum _{i=1}^{10}{x}_i^2\right)\end{array}$$

Untuk $H_1:{\theta }_1=1,{\theta }_2=4$, fungsi likelihood adalah:

$$\begin{array}{rl}{L}_1(x_1,\dots ,x_{10})& =\prod _{i=1}^{10}\frac{1}{\sqrt{2\pi \cdot 4}}\exp \left(-\frac{(x_i-1{)}^2}{2\cdot 4}\right)\\ & =(8\pi {)}^{-5}\exp \left(-\frac{1}{8}\sum _{i=1}^{10}(x_i-1{)}^2\right)\end{array}$$

Berdasarkan Neyman-Pearson Theorem, best critical region didefinisikan oleh:

$$\begin{array}{rl}\frac{L_1}{L_0}& \ge k\\ \frac{(8\pi {)}^{-5}\exp \left(-\frac{1}{8}{\sum }_{i=1}^{10}(x_i-1{)}^2\right)}{(2\pi {)}^{-5}\exp \left(-\frac{1}{2}{\sum }_{i=1}^{10}{x}_i^2\right)}& \ge k\\ \frac{(2\pi {)}^{-5}}{(8\pi {)}^{-5}}\exp \left(-\frac{1}{8}\sum _{i=1}^{10}(x_i-1{)}^2+\frac{1}{2}\sum _{i=1}^{10}{x}_i^2\right)& \ge k\\ \frac{1}{4^5}\exp \left(-\frac{1}{8}\sum _{i=1}^{10}(x_i-1{)}^2+\frac{1}{2}\sum _{i=1}^{10}{x}_i^2\right)& \ge k\end{array}$$

Perhatikan bahwa:

$$\begin{array}{rl}-\frac{1}{8}\sum _{i=1}^{10}(x_i-1{)}^2+\frac{1}{2}\sum _{i=1}^{10}{x}_i^2& =-\frac{1}{8}\sum _{i=1}^{10}(x_i^2-2x_i+1)+\frac{1}{2}\sum _{i=1}^{10}{x}_i^2\\ & =-\frac{1}{8}\sum _{i=1}^{10}{x}_i^2+\frac{1}{4}\sum _{i=1}^{10}{x}_i-\frac{10}{8}+\frac{1}{2}\sum _{i=1}^{10}{x}_i^2\\ & =\frac{3}{8}\sum _{i=1}^{10}{x}_i^2+\frac{1}{4}\sum _{i=1}^{10}{x}_i-\frac{5}{4}\end{array}$$

Ingat bahwa $k$ adalah suatu konstanta positif. Misalkan $k^{\prime }=\ln (k)+5\ln (4)+\frac{5}{4}$. Dapat diperoleh:

$$\begin{array}{rl}\frac{L_1}{L_0}=\frac{1}{4^5}\exp \left(\frac{3}{8}\sum _{i=1}^{10}{x}_i^2+\frac{1}{4}\sum _{i=1}^{10}{x}_i-\frac{5}{4}\right)& \ge k\\ \frac{3}{8}\sum _{i=1}^{10}{x}_i^2+\frac{1}{4}\sum _{i=1}^{10}{x}_i-\frac{5}{4}& \ge \ln (k\cdot 4^5)\\ 3\sum _{i=1}^{10}{x}_i^2+2\sum _{i=1}^{10}{x}_i& \ge k^{\prime }\end{array}$$

Karena koefisien dari $\sum x_i^2$ dan $\sum x_i$ keduanya positif, critical region dapat ditulis sebagai:

$\boxed{C:3\sum _{i=1}^{10}{X}_i^2+2\sum _{i=1}^{10}{X}_i\ge k^{\prime }}$

untuk suatu konstanta $k$ positif dengan $k^{\prime }=\ln (k)+5\ln (4)+\frac{5}{4}$.

$\therefore$ Best test untuk hipotesis yang diberikan adalah tolak $H_0$ jika $3{\sum }_{i=1}^{10}{X}_i^2+2{\sum }_{i=1}^{10}{X}_i\ge k^{\prime }$.

9.10

Let $X_1,X_2,\dots ,X_{10}$ denote a random sample of size $10$ from a Poisson distribution with mean $\theta$. Show that the critical region $C$ defined by ${\sum }_{i=1}^{10}{x}_i\ge 3$ is a best critical region for testing $H_0:\theta =0.1$ against $H_1:\theta =0.5$. Determine, for this test, the significance level $\alpha$ and power at $\theta =0.5$.

Jawab

Diketahui pmf dari distribusi Poisson adalah $f(x;\theta )=\frac{{\theta }^x{e}^{-\theta }}{x!},x=0,1,2,\dots$

Fungsi likelihood untuk sampel $X_1,X_2,\dots ,X_{10}$ adalah:

$$\begin{array}{rl}L(\theta )& =\prod _{i=1}^{10}\frac{{\theta }^{x_i}{e}^{-\theta }}{x_i!}\\ & =\frac{{\theta }^{{\sum }_{i=1}^{10}{x}_i}{e}^{-10\theta }}{{\prod }_{i=1}^{10}{x}_i!}\end{array}$$

Untuk $H_0:\theta =0.1$: $L_0=\frac{(0.1{)}^{\sum x_i}{e}^{-1}}{\prod x_i!}$

Untuk $H_1:\theta =0.5$: $L_1=\frac{(0.5{)}^{\sum x_i}{e}^{-5}}{\prod x_i!}$

Sehingga diperoleh

$$\begin{array}{rl}\frac{L_1}{L_0}& =\frac{(0.5{)}^{\sum x_i}{e}^{-5}}{(0.1{)}^{\sum x_i}{e}^{-1}}\\ & ={\left(\frac{0.5}{0.1}\right)}^{\sum x_i}{e}^{-4}\\ & =5^{\sum x_i}{e}^{-4}\end{array}$$

Berdasarkan Neyman-Pearson Theorem, best critical region didefinisikan oleh:

$$\begin{array}{rl}\frac{L_1}{L_0}& \ge k\\ 5^{\sum x_i}{e}^{-4}& \ge k\\ 5^{\sum x_i}& \ge k{e}^4\\ \sum x_i& \ge {\log }_5(k{e}^4)\end{array}$$

Karena ${\log }_5(k{e}^4)$ adalah konstanta positif critical region berbentuk ${\sum }_{i=1}^{10}{x}_i\ge c$ untuk suatu konstanta $c$, artinya, $C:{\sum }_{i=1}^{10}{x}_i\ge 3$ adalah best critical region.

Di bawah $H_0$, ${\sum }_{i=1}^{10}{X}_i\sim \text{Poisson}(10\times 0.1)=\text{Poisson}(1)$ sehingga

$$\begin{array}{rl}\alpha & =P_{H_0}\left[\sum _{i=1}^{10}{X}_i\ge 3\right]\\ & =1-P[\sum _{i=1}^{10}{X}_i\le 2]\\ & =1-\left(P[Y=0]+P[Y=1]+P[Y=2]\right)\\ & =1-\left(\frac{1^0{e}^{-1}}{0!}+\frac{1^1{e}^{-1}}{1!}+\frac{1^2{e}^{-1}}{2!}\right)\\ & =1-e^{-1}\left(1+1+\frac{1}{2}\right)\\ & =1-\frac{5e^{-1}}{2}\\ & =\boxed{1-\frac{5}{2e}\approx 0.0803}\end{array}$$

Di bawah $H_1$, ${\sum }_{i=1}^{10}{X}_i\sim \text{Poisson}(10\times 0.5)=\text{Poisson}(5)$. Misal $Y\sim \text{Poisson(5)}$,sehingga

$$\begin{array}{rl}\alpha & =P_{H_1}\left[\sum _{i=1}^{10}{X}_i\ge 3\right]\\ & =1-P[Y\le 2]\\ & =1-\left(\frac{5^0{e}^{-5}}{0!}+\frac{5^1{e}^{-5}}{1!}+\frac{5^2{e}^{-5}}{2!}\right)\\ & =1-e^{-5}\left(1+5+\frac{25}{2}\right)\\ & =1-\frac{37e^{-5}}{2}\\ & =\boxed{1-\frac{37}{2e^5}\approx 0.875}\end{array}$$

$\therefore$ Critical region $C:{\sum }_{i=1}^{10}{x}_i\ge 3$ adalah best critical region dengan significane level $\alpha \approx 0.0803$ dan power $\approx 0.875$

9.12

Let $X$ have a p.d.f. of the form $f(x;\theta )=1/\theta$, $0<x<\theta$, zero elsewhere. Let $Y_1<Y_2<Y_3<Y_4$ denote the order statistics of a random sample of size $4$ from this distribution. Let the observed value of $Y_4$ be $y_4$. We reject $H_0:\theta =1$ and accept $H_1:\theta \ne 1$ if either $y_4\le \frac{1}{2}$ or $y_4\ge 1$. Find the power function $K(\theta )$, $0<\theta$, of the test.

Answer

Diketahui $f(x;\theta )$ berasal dari distribusi uniform $U(0,\theta )$. CDF-nya adalah $F(x)=x/\theta$ untuk $0<x<\theta$.

CDF dari statistik terurut terbesar, $Y_4$, dari sampel ukuran $n=4$ adalah: $G(y_4)=[F(y_4){]}^4=(y_4/\theta {)}^4,0<y_4<\theta$

Daerah penolakan adalah $C=\{y_4:y_4\le 1/2\text{ atau }{y}_4\ge 1\}$. $K(\theta )=P(Y_4\le 1/2|\theta )+P(Y_4\ge 1|\theta )$

Kasus 1: $0<\theta \le 1/2$

• $P(Y_4\le 1/2)=1$
• $P(Y_4\ge 1)=0$, karena $Y_4<\theta \le 1/2$. Jadi, $K(\theta )=1$.

Kasus 2: $1/2<\theta <1$

• $P(Y_4\le 1/2)=G(1/2)=((1/2)/\theta {)}^4=1/(16{\theta }^4)$.
• $P(Y_4\ge 1)=0$, karena $Y_4<\theta <1$. Jadi, $K(\theta )=1/(16{\theta }^4)$.

Kasus 3: $\theta \ge 1$

• $P(Y_4\le 1/2)=G(1/2)=((1/2)/\theta {)}^4=1/(16{\theta }^4)$.
• $P(Y_4\ge 1)=1-P(Y_4<1)=1-G(1)=1-(1/\theta {)}^4$. Jadi, $K(\theta )=\frac{1}{16{\theta }^4}+1-\frac{1}{{\theta }^4}=1-\frac{15}{16{\theta }^4}$.

Menggabungkan semua kasus, fungsi power adalah: $K(\theta )=\{\begin{array}{ll}1& 0<\theta \le 1/2\\ 1/(16{\theta }^4)& 1/2<\theta <1\\ 1-15/(16{\theta }^4)& \theta \ge 1\end{array}$

$\therefore$ Fungsi power dari tes tersebut adalah $\boxed{K(\theta )=\{\begin{array}{ll}1& 0<\theta \le 1/2\\ 1/(16{\theta }^4)& 1/2<\theta <1\\ 1-15/(16{\theta }^4)& \theta \ge 1\end{array}}$.

9.14

Consider the two normal distributions $N({\mu }_1,400)$ and $N({\mu }_2,225)$. Let $\theta ={\mu }_1-{\mu }_2$. Let $\bar{x}$ and $\bar{y}$ denote the observed means of two independent random samples, each of size $n$, from these two distributions. We reject $H_0:\theta =0$ and accept $H_1:\theta >0$ if and only if $\bar{x}-\bar{y}\ge c$. If $K(\theta )$ is the power function of this test, find $n$ and $c$ so that $K(0)=0.05$ and $K(10)=0.90$, approximately.

Answer

Diketahui $\bar{X}\sim N({\mu }_1,400/n)$ dan $\bar{Y}\sim N({\mu }_2,225/n)$.

Misalkan $W=\bar{X}-\bar{Y}$. Karena $\bar{X}$ dan $\bar{Y}$ independen, $W$ berdistribusi normal dengan:

• $E[W]=E[\bar{X}]-E[\bar{Y}]={\mu }_1-{\mu }_2=\theta$
• $\text{Var}(W)=\text{Var}(\bar{X})+\text{Var}(\bar{Y})=\frac{400}{n}+\frac{225}{n}=\frac{625}{n}$ Sehingga, $W\sim N(\theta ,625/n)$.

Fungsi power $K(\theta )$ dari tes yang menolak $H_0$ jika $W\ge c$ adalah: $K(\theta )=P(W\ge c)=P\left(\frac{W-\theta }{\sqrt{625/n}}\ge \frac{c-\theta }{\sqrt{625/n}}\right)=P\left(Z\ge \frac{c-\theta }{25/\sqrt{n}}\right)$ dimana $Z\sim N(0,1)$.

Diberikan dua kondisi:

1. $K(0)=0.05$

   $P\left(Z\ge \frac{c}{25/\sqrt{n}}\right)=0.05$.

   Dari tabel normal standar, $z_{0.05}\approx 1.645$.

   Maka, $\frac{c}{25/\sqrt{n}}=1.645⟹c=1.645\frac{25}{\sqrt{n}}$ (1)

2. $K(10)=0.90$.

   $P\left(Z\ge \frac{c-10}{25/\sqrt{n}}\right)=0.90$.

   Dari tabel, nilai z yang sesuai adalah $z_{0.90}=-z_{0.10}\approx -1.282$.

   Maka, $\frac{c-10}{25/\sqrt{n}}=-1.282⟹c-10=-1.282\frac{25}{\sqrt{n}}$ (2)

Substitusikan (1) ke dalam (2):

$$\begin{array}{rl}1.645\frac{25}{\sqrt{n}}-10& =-1.282\frac{25}{\sqrt{n}}\\ (1.645+1.282)\frac{25}{\sqrt{n}}& =10\\ 2.927\frac{25}{\sqrt{n}}& =10\\ \sqrt{n}& =\frac{2.927\times 25}{10}=7.3175\\ n& =(7.3175{)}^2\approx 53.54\end{array}$$

Karena $n$ harus bilangan bulat, $n$ dapat dibulatkan ke atas menjadi $n=54$.

Lalu, $c$ dapat dicari menggunakan $n=54$ dalam persamaan (1): $c=1.645\frac{25}{\sqrt{54}}\approx 1.645\times \frac{25}{7.348}\approx 5.596$

$\therefore$ Nilai $n$ dan $c$ yang memenuhi adalah $\boxed{n=54}$ dan $\boxed{c\approx 5.596}$.

9.18

Let $X_1,X_2,\dots ,X_n$ denote a random sample from a normal distribution $N(\theta ,16)$. Find the sample size $n$ and a uniformly most powerful test of $H_0:\theta =25$ against $H_1:\theta <25$ with power function $K(\theta )$ so that approximately $K(25)=0.10$ and $K(23)=0.90$.

Answer

Diketahui $X_i\sim N(\theta ,16)$, sehingga mean sampel $\bar{X}\sim N(\theta ,16/n)$.

Berdasarkan Teorema Karlin-Rubin (karena keluarga distribusi normal memiliki Monotone Likelihood Ration (mlr)), uniformly most powerful test adalah dengan menolak $H_0$ jika statistik cukup $\bar{X}$ kurang dari suatu nilai kritis $c$.

Bentuk tes UMP: Tolak $H_0$ jika $\bar{X}\le c$.

Fungsi power $K(\theta )$ adalah: $K(\theta )=P(\bar{X}\le c|\theta )=P\left(\frac{\bar{X}-\theta }{4/\sqrt{n}}\le \frac{c-\theta }{4/\sqrt{n}}\right)=\Phi \left(\frac{c-\theta }{4/\sqrt{n}}\right)$ dimana $\Phi$ adalah CDF dari distribusi normal standar.

Diberikan dua kondisi:

1. $K(25)=0.10$ (tingkat signifikansi $\alpha$).

   $\Phi \left(\frac{c-25}{4/\sqrt{n}}\right)=0.10$.

   Dari tabel normal, $z_{0.10}\approx -1.282$.

   Maka, $\frac{c-25}{4/\sqrt{n}}=-1.282⟹c=25-1.282\frac{4}{\sqrt{n}}$ (1)

2. $K(23)=0.90$.

   $\Phi \left(\frac{c-23}{4/\sqrt{n}}\right)=0.90$.

   Dari tabel normal, $z_{0.90}\approx 1.282$.

   Maka, $\frac{c-23}{4/\sqrt{n}}=1.282⟹c=23+1.282\frac{4}{\sqrt{n}}$ (2)

Sehingga dapat diperoleh

$$\begin{array}{rl}25-1.282\frac{4}{\sqrt{n}}& =23+1.282\frac{4}{\sqrt{n}}\\ 2& =2\left(1.282\frac{4}{\sqrt{n}}\right)\\ 1& =\frac{5.128}{\sqrt{n}}\\ \sqrt{n}& =5.128\\ n& =(5.128{)}^2\approx 26.29\end{array}$$

Karena $n$ harus bilangan bulat, kita bulatkan ke atas menjadi $n=27$.

Cari nilai $c$ menggunakan $n=27$ dalam persamaan (2): $c=23+1.282\frac{4}{\sqrt{27}}\approx 23+1.282\frac{4}{5.196}\approx 23+0.987\approx 23.987$

$\therefore$ Ukuran sampel yang dibutuhkan adalah $\boxed{n=27}$, dan tes UMP adalah $\boxed{\text{Tolak }{H}_0\text{ jika }\bar{X}\le 24}$.

9.21

Let $X_1,X_2,\dots ,X_n$ be a random sample from a distribution with pdf $f(x;\theta )=\theta x^{\theta -1},0<x<\infty$, zero elsewhere, where $\theta >0$. Find a sufficient statistic for $\theta$ and show that a uniformly most powerful test of $H_0:\theta =6$ against $H_1:\theta <6$ is based on this statistic.

Jawab

Fungsi likelihood untuk sampel adalah:

$$\begin{array}{rl}L(\theta )& =\prod _{i=1}^n\theta x_i^{\theta -1}\\ & ={\theta }^n\prod _{i=1}^n{x}_i^{\theta -1}\\ & ={\theta }^n{\left(\prod _{i=1}^n{x}_i\right)}^{\theta -1}\end{array}$$

Log-likelihood:

$$\begin{array}{rl}\ln L(\theta )& =n\ln \theta +(\theta -1)\sum _{i=1}^n\ln x_i\\ & =n\ln \theta +(\theta -1)\ln \left(\prod _{i=1}^n{x}_i\right)\end{array}$$

Note

Untuk mencari UMPT, perlu dicari UMP Critical Region, perlu dicari Best Critical Region menggunakan Neyman-Pearson Theorem. Selain itu, dari Hogg&Craig ed8,

Perhatikan bahwa berdasarkan Neyman Theorem, $Y_1={\prod }_{i=1}^n{X}_i=u_1(\mathbf{X})$ adalah statistik cukup untuk $\theta$ dengan:

$$\begin{array}{rl}Y& ={\theta }^n{\left(\prod x_i\right)}^{\theta -1}\cdot 1\\ & =k_1[u_1(\mathbf{x});\theta ]\cdot k_2(\mathbf{x})\end{array}$$

Misalkan ${\theta }_1<6$. Rasio likelihood adalah:

$$\begin{array}{rl}\frac{L({\theta }_1)}{L(6)}& =\frac{{\theta }_1^n{\left(\prod x_i\right)}^{{\theta }_1-1}}{6^n{\left(\prod x_i\right)}^{6-1}}\\ & ={\left(\frac{{\theta }_1}{6}\right)}^n{\left(\prod x_i\right)}^{{\theta }_1-6}\end{array}$$

Rasio log:

$$\begin{array}{rl}\ln \left(\frac{L({\theta }_1)}{L(6)}\right)& =n\ln \left(\frac{{\theta }_1}{6}\right)+({\theta }_1-6)\ln \left(\prod x_i\right)\\ & =n\ln \left(\frac{{\theta }_1}{6}\right)+({\theta }_1-6)\sum _{i=1}^n\ln x_i\end{array}$$

Lalu:

1. Karena ${\theta }_1<6$, maka ${\theta }_1-6<0$.
2. Karena $\ln x_i$ negatif untuk $0<x_i<1$, maka $({\theta }_1-6)\sum \ln x_i>0$
3. Perhatikan bahwa $({\theta }_1-6)\sum \ln x_i$ meningkat ketika $\sum \ln x_i$ semakin negatif
4. Karena $\sum \ln x_i=\ln \left(\prod x_i\right)$ maka $\ln \left(\frac{L({\theta }_1)}{L(6)}\right)$ meningkat ketika $\prod x_i$ mengecil ($y$ mengecil maka $\ln (y)$ membesar)
5. Sehingga, $\forall {\theta }_1<6$, best critical region adalah ${\prod }_{i=1}^n{X}_i\le k$ untuk suatu konstanta $k$

Karena bentuk critical region sama untuk semua ${\theta }_1<6$ (hanya konstanta $k$ yang berbeda, tergantung pada significance level), maka test yang menolak $H_0$ ketika ${\prod }_{i=1}^n{X}_i\le k$ adalah uniformly most powerful test.

$\therefore$ Statistik cukup untuk $\theta$ adalah $Y_1={\prod }_{i=1}^n{X}_i$. UMPT untuk $H_0:\theta =6$ melawan $H_1:\theta <6$ yang berkaitan dengan $Y_1$ adalah menolak $H_0$ ketika ${\prod }_{i=1}^n{X}_i\le k$ untuk suatu konstanta $k$.

9.28

Let $X_1,X_2,\dots ,X_n$ be a random sample from the normal distribution $N(\theta ,1)$. Show that the likelihood ratio principle for testing $H_0:\theta ={\theta }^{\prime }$, where ${\theta }^{\prime }$ is specified, against $H_1:\theta \ne {\theta }^{\prime }$ leads to the inequality $|\bar{x}-{\theta }^{\prime }|\ge c$. Is this a uniformly most powerful test of $H_0$ against $H_1$?

Diketahui $X_i\sim N(\theta ,1)$. Ambil sembarang $\theta \in \Omega$. Fungsi likelihood untuk sampel adalah:

$$\begin{array}{rl}L(\theta )& =\prod _{i=1}^n\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{(x_i-\theta {)}^2}{2}\right)\\ & =(2\pi {)}^{-n/2}\exp \left(-\frac{1}{2}\sum _{i=1}^n(x_i-\theta {)}^2\right)\\ & \\ \frac{\partial \ln L(\theta )}{\partial \theta }& =\frac{\partial }{\partial \theta }\left\{\ln \left[(2\pi {)}^{-n/2}\exp \left(-\frac{1}{2}\sum _{i=1}^n(x_i-\theta {)}^2\right)\right]\right\}\\ & =\frac{\partial }{\partial \theta }\left\{\ln \left[(2\pi {)}^{-n/2}\right]+\ln \left[\exp \left(-\frac{1}{2}\sum _{i=1}^n(x_i-\theta {)}^2\right)\right]\right\}\\ & =\frac{\partial }{\partial \theta }\left\{-\frac{1}{2}\sum _{i=1}^n(x_i-\theta {)}^2\right\}\\ 0& =\sum _{i=1}^n(x_i-\theta )\\ 0& =\sum _{i=1}^n{x}_i-n\theta \\ \theta & =\bar{x}\end{array}$$

Sehingga diperoleh MLE $\hat{\theta }=\bar{x}$ dan: $L(\hat{\theta })=(2\pi {)}^{-n/2}\exp \left(-\frac{1}{2}{\sum }_{i=1}^n(x_i-\bar{x}{)}^2\right)$

Likelihood ratio-nya adalah:

$$\begin{array}{rl}\lambda & =\frac{L({\theta }^{\prime })}{L(\hat{\theta })}\\ & =\frac{\exp \left(-\frac{1}{2}{\sum }_{i=1}^n(x_i-{\theta }^{\prime }{)}^2\right)}{\exp \left(-\frac{1}{2}{\sum }_{i=1}^n(x_i-\bar{x}{)}^2\right)}\\ & =\exp \left(-\frac{1}{2}\left[\sum _{i=1}^n(x_i-{\theta }^{\prime }{)}^2-\sum _{i=1}^n(x_i-\bar{x}{)}^2\right]\right)\end{array}$$

Perhatikan bahwa:

$$\begin{array}{rl}\sum _{i=1}^n(x_i-{\theta }^{\prime }{)}^2-\sum _{i=1}^n(x_i-\bar{x}{)}^2& =\sum _{i=1}^n[(x_i-{\theta }^{\prime }{)}^2-(x_i-\bar{x}{)}^2]\\ & =\sum _{i=1}^n[x_i^2-2x_i{\theta }^{\prime }+({\theta }^{\prime }{)}^2-x_i^2+2x_i\bar{x}-{\bar{x}}^2]\\ & =\sum _{i=1}^n[2x_i(\bar{x}-{\theta }^{\prime })+({\theta }^{\prime }{)}^2-{\bar{x}}^2]\\ & =2(\bar{x}-{\theta }^{\prime })\sum _{i=1}^n{x}_i+n[({\theta }^{\prime }{)}^2-{\bar{x}}^2]\\ & =2n(\bar{x}-{\theta }^{\prime }{)}^2+n[({\theta }^{\prime }{)}^2-{\bar{x}}^2]\\ & =n(\bar{x}-{\theta }^{\prime }{)}^2\end{array}$$

Sehingga: $\lambda =\exp \left(-\frac{n(\bar{x}-{\theta }^{\prime }{)}^2}{2}\right)$

Ambil sembarang $k$ konstanta positif. Dapat diperoleh

$$\begin{array}{rl}\exp \left(-\frac{n(\bar{x}-{\theta }^{\prime }{)}^2}{2}\right)& \le k\\ -\frac{n(\bar{x}-{\theta }^{\prime }{)}^2}{2}& \le \ln k\\ (\bar{x}-{\theta }^{\prime }{)}^2& \ge -\frac{2\ln k}{n}\\ |\bar{x}-{\theta }^{\prime }|& \ge c\end{array}$$

dimana $c=\sqrt{-\frac{2\ln k}{n}}$

Test ini bukan UMPT, karena UMPT didefiniskan untuk $H_1$ composite hypothesis, sementara $H_1$ yang diberikan adalah simple hypothesis.

$\therefore$ Terbukti $|\bar{x}-{\theta }^{\prime }|\ge c$. Tetapi, test ini bukan UMPT karena $H_1$ adalah composite hypothesis.

9.29

Let $X_1,X_2,\dots ,X_n$ be a random sample from the normal distribution $N({\theta }_1,{\theta }_2)$.

Show that the likelihood ratio principle for testing $H_0:{\theta }_2={\theta }_2^{\prime }$ specified, and ${\theta }_1$ specified, against $H_1:{\theta }_2\ne {\theta }_2$, ${\theta }_1$ unspecified, leads to a test that rejects when ${\sum }_i^n(x_i-\bar{x}{)}^2\le c$ or ${\sum }_i^n(x_i-\bar{x}{)}^2\ge c_2$, where $c_1<c_2$ are selected appropriately

Diketahui $X_i\sim N({\theta }_1,{\theta }_2)$. Fungsi likelihood untuk sampel adalah: $L({\theta }_1,{\theta }_2)={\prod }_{i=1}^n\frac{1}{\sqrt{2\pi {\theta }_2}}\exp \left(-\frac{(x_i-{\theta }_1{)}^2}{2{\theta }_2}\right)=(2\pi {\theta }_2{)}^{-n/2}\exp \left(-\frac{1}{2{\theta }_2}{\sum }_{i=1}^n(x_i-{\theta }_1{)}^2\right)$

Di bawah $H_0$: ${\theta }_2={\theta }_2^{\prime }$ (specified), ${\theta }_1$ unspecified.

Untuk memaksimumkan $L({\theta }_1,{\theta }_2^{\prime })$, cari ${\hat{\theta }}_1$ dengan:

$$\begin{array}{rl}\frac{\partial }{\partial {\theta }_1}\ln L({\theta }_1,{\theta }_2^{\prime })& =\frac{\partial }{\partial {\theta }_1}\left[-\frac{1}{2{\theta }_2^{\prime }}\sum _{i=1}^n(x_i-{\theta }_1{)}^2\right]=0\\ \frac{1}{{\theta }_2^{\prime }}\sum _{i=1}^n(x_i-{\theta }_1)& =0\\ {\hat{\theta }}_1& =\bar{x}\end{array}$$

Sehingga: $L({\hat{\theta }}_1,{\theta }_2^{\prime })=(2\pi {\theta }_2^{\prime }{)}^{-n/2}\exp \left(-\frac{1}{2{\theta }_2^{\prime }}{\sum }_{i=1}^n(x_i-\bar{x}{)}^2\right)$

Di bawah $H_1$: ${\theta }_1$ dan ${\theta }_2$ keduanya unspecified.

Untuk memaksimumkan $L({\theta }_1,{\theta }_2)$:

$$\begin{array}{rl}\frac{\partial \ln L}{\partial {\theta }_1}& =0⟹{\hat{\theta }}_1=\bar{x}\\ \frac{\partial \ln L}{\partial {\theta }_2}& =\frac{\partial }{\partial {\theta }_2}\left[-\frac{n}{2}\ln {\theta }_2-\frac{1}{2{\theta }_2}\sum _{i=1}^n(x_i-\bar{x}{)}^2\right]=0\\ -\frac{n}{2{\theta }_2}+\frac{1}{2{\theta }_2^2}\sum _{i=1}^n(x_i-\bar{x}{)}^2& =0\\ {\hat{\theta }}_2& =\frac{1}{n}\sum _{i=1}^n(x_i-\bar{x}{)}^2\end{array}$$

Sehingga: $L({\hat{\theta }}_1,{\hat{\theta }}_2)=(2\pi {\hat{\theta }}_2{)}^{-n/2}\exp \left(-\frac{n}{2}\right)$

Likelihood ratio adalah:

$$\begin{array}{rl}\lambda & =\frac{L({\hat{\theta }}_1,{\theta }_2^{\prime })}{L({\hat{\theta }}_1,{\hat{\theta }}_2)}\\ & =\frac{(2\pi {\theta }_2^{\prime }{)}^{-n/2}\exp \left(-\frac{1}{2{\theta }_2^{\prime }}\sum (x_i-\bar{x}{)}^2\right)}{(2\pi {\hat{\theta }}_2{)}^{-n/2}\exp (-n/2)}\\ & ={\left(\frac{{\hat{\theta }}_2}{{\theta }_2^{\prime }}\right)}^{-n/2}\exp \left(-\frac{1}{2{\theta }_2^{\prime }}\sum (x_i-\bar{x}{)}^2+\frac{n}{2}\right)\\ & ={\left(\frac{{\hat{\theta }}_2}{{\theta }_2^{\prime }}\right)}^{-n/2}\exp \left(\frac{n}{2}-\frac{n{\hat{\theta }}_2}{2{\theta }_2^{\prime }}\right)\\ & ={\left(\frac{{\hat{\theta }}_2}{{\theta }_2^{\prime }}\right)}^{-n/2}\exp \left(\frac{n}{2}\left(1-\frac{{\hat{\theta }}_2}{{\theta }_2^{\prime }}\right)\right)\end{array}$$

Misalkan $S^2={\sum }_{i=1}^n(x_i-\bar{x}{)}^2=n{\hat{\theta }}_2$. Maka: $\lambda ={\left(\frac{S^2}{n{\theta }_2^{\prime }}\right)}^{-n/2}\exp \left(\frac{n}{2}\left(1-\frac{S^2}{n{\theta }_2^{\prime }}\right)\right)$

Test menolak $H_0$ ketika $\lambda \le k$ untuk konstanta $k$.

Perhatikan bahwa $\lambda$ adalah fungsi dari $S^2/{\theta }_2^{\prime }$. Fungsi $\lambda (S^2)$ mencapai maksimum ketika $S^2=n{\theta }_2^{\prime }$ dan menurun ketika $S^2$ menjauh dari $n{\theta }_2^{\prime }$ (baik lebih kecil atau lebih besar).

Karena $\lambda \le k$ ekuivalen dengan $S^2\le c_1$ atau $S^2\ge c_2$ untuk konstanta $c_1<n{\theta }_2^{\prime }<c_2$ yang dipilih sesuai dengan tingkat signifikansi.

$\therefore$ Test menolak $H_0$ ketika $\boxed{\sum _{i=1}^n(x_i-\bar{x}{)}^2\le c_1\text{ atau }\sum _{i=1}^n(x_i-\bar{x}{)}^2\ge c_2}$ dimana $c_1<c_2$ dipilih sesuai tingkat signifikansi.

9.32

Let $Y_1<Y_2<\cdots <Y_5$ be the order statistics of a random sample of size $n=5$ from a distribution with p.d.f. $f(x;\theta )=\frac{1}{2}{e}^{-|x-\theta |}$, $-\infty <x<\infty$, for all real $\theta$.

Find the likelihood ratio test $\lambda$ for testing $H_0={\theta }_0$ against $H_1:\theta \ne {\theta }_0$.

Fungsi likelihood untuk sampel adalah:

$$\begin{array}{rl}L(\theta )& =\prod _{i=1}^5\frac{1}{2}{e}^{-|x_i-\theta |}\\ & =\frac{1}{32}\exp \left(-\sum _{i=1}^5|x_i-\theta |\right)\end{array}$$

Untuk memaksimumkan $L(\theta )$, perlu meminimumkan ${\sum }_{i=1}^5|x_i-\theta |$. Diketahui bahwa median sampel meminimumkan jumlah deviasi absolut. Untuk $n=5$, median adalah $Y_3$.

Sehingga $\hat{\theta }=Y_3$ (statistik terurut ketiga).

Di bawah $H_0$: $\theta ={\theta }_0$ $L({\theta }_0)=\frac{1}{32}\exp \left(-{\sum }_{i=1}^5|x_i-{\theta }_0|\right)$

Di bawah $H_1$: $\theta$ unspecified $L(\hat{\theta })=L(Y_3)=\frac{1}{32}\exp \left(-{\sum }_{i=1}^5|x_i-Y_3|\right)$

Likelihood ratio adalah:

$$\begin{array}{rl}\lambda & =\frac{L({\theta }_0)}{L(\hat{\theta })}\\ & =\frac{\exp \left(-{\sum }_{i=1}^5|x_i-{\theta }_0|\right)}{\exp \left(-{\sum }_{i=1}^5|x_i-Y_3|\right)}\\ & =\exp \left(-\sum _{i=1}^5|x_i-{\theta }_0|+\sum _{i=1}^5|x_i-Y_3|\right)\\ & =\exp \left(\sum _{i=1}^5|x_i-Y_3|-\sum _{i=1}^5|x_i-{\theta }_0|\right)\end{array}$$

Perhatikan bahwa:

$$\begin{array}{rl}\sum _{i=1}^5|x_i-Y_3|& =\sum _{i=1}^5|Y_i-Y_3|\\ & =(Y_3-Y_1)+(Y_3-Y_2)+0+(Y_4-Y_3)+(Y_5-Y_3)\\ & =2Y_3-Y_1-Y_2+Y_4+Y_5-2Y_3\\ & =Y_4+Y_5-Y_1-Y_2\end{array}$$

$\therefore$ Likelihood ratio test adalah $\boxed{\lambda =\exp \left(Y_4+Y_5-Y_1-Y_2-\sum _{i=1}^5|Y_i-{\theta }_0|\right)}$

9.34

A random sample $X_1,X_2,\dots ,X_n$ arises from a distribution given by

\begin{align} H_{0} & : f(x;\theta) = \frac{1}{\theta} & 0<x< \theta ,\quad 0 \text{ elsewhere} \\ H_{1} & : f(x;\theta) = \frac{1}{\theta} e^{x/\theta} & 0<x< \theta ,\quad 0 \text{ elsewhere} \end{align}

Determine the likelihood ratio test $(\lambda )$ associated with the test of $H_0$ against $H_1$.

Di bawah $H_0$: $f(x;\theta )=\frac{1}{\theta }$ untuk $0<x<\theta$ (distribusi uniform)

Fungsi likelihood: $L_0(\theta )={\prod }_{i=1}^n\frac{1}{\theta }=\frac{1}{{\theta }^n},\theta \ge \max (x_i)=x_{(n)}$

Untuk memaksimumkan $L_0(\theta )$, pilih $\theta$ sekecil mungkin, yaitu ${\hat{\theta }}_0=x_{(n)}$.

Sehingga: $L_0({\hat{\theta }}_0)=\frac{1}{x_{(n)}^n}$

Di bawah $H_1$: $f(x;\theta )=\frac{1}{\theta }{e}^{-x/\theta }$ untuk $x>0$ (distribusi eksponensial)

Fungsi likelihood:

$$\begin{array}{rl}{L}_1(\theta )& =\prod _{i=1}^n\frac{1}{\theta }{e}^{-x_i/\theta }\\ & =\frac{1}{{\theta }^n}\exp \left(-\frac{1}{\theta }\sum _{i=1}^n{x}_i\right)\end{array}$$

Untuk memaksimumkan:

$$\begin{array}{rl}\frac{\partial \ln L_1}{\partial \theta }& =\frac{\partial }{\partial \theta }\left[-n\ln \theta -\frac{1}{\theta }\sum _{i=1}^n{x}_i\right]=0\\ -\frac{n}{\theta }+\frac{1}{{\theta }^2}\sum _{i=1}^n{x}_i& =0\\ {\hat{\theta }}_1& =\frac{1}{n}\sum _{i=1}^n{x}_i=\bar{x}\end{array}$$

Sehingga: $L_1({\hat{\theta }}_1)=\frac{1}{{\bar{x}}^n}\exp \left(-\frac{n\bar{x}}{\bar{x}}\right)=\frac{1}{{\bar{x}}^n}{e}^{-n}$

Likelihood ratio adalah:

$$\begin{array}{rl}\lambda & =\frac{L_0({\hat{\theta }}_0)}{L_1({\hat{\theta }}_1)}\\ & =\frac{1/x_{(n)}^n}{e^{-n}/{\bar{x}}^n}\\ & =\frac{{\bar{x}}^n}{x_{(n)}^n}{e}^{-n}\\ & ={\left(\frac{\bar{x}}{x_{(n)}}\right)}^n{e}^{-n}\end{array}$$

$\therefore$ Likelihood ratio test adalah $\boxed{\lambda ={\left(\frac{\bar{X}}{X_{(n)}}\right)}^n{e}^{-n}}$ dimana $X_{(n)}=\max (X_1,\dots ,X_n)$.

9.47

Consider a random sample $X_1,X_2,\dots ,X_n$ from a distribution with pdf $f(x;\theta )=\theta (1-x{)}^{\theta -1},0<x<1$, zero elsewhere, where $\theta >0$.

1. Find the form of the uniformly most powerful test of $H_0:\theta =1$ against $H_1:\theta >1$.
2. What is the likelihood ratio for $\lambda$ for testing $H_0:\theta =1$ against $H_1:\theta \ne 1$?

9.47.1

Fungsi likelihood untuk sampel adalah:

$$\begin{array}{rl}L(\theta )& =\prod _{i=1}^n\theta (1-x_i{)}^{\theta -1}\\ & ={\theta }^n\prod _{i=1}^n(1-x_i{)}^{\theta -1}\\ & ={\theta }^n{\left[\prod _{i=1}^n(1-x_i)\right]}^{\theta -1}\end{array}$$

Misalkan ${\theta }_1>1$. Rasio likelihood adalah:

$$\begin{array}{rl}\frac{L({\theta }_1)}{L(1)}& =\frac{{\theta }_1^n{\left[{\prod }_{i=1}^n(1-x_i)\right]}^{{\theta }_1-1}}{1^n{\left[{\prod }_{i=1}^n(1-x_i)\right]}^{1-1}}\\ & ={\theta }_1^n{\left[\prod _{i=1}^n(1-x_i)\right]}^{{\theta }_1-1}\\ & \\ \ln \left(\frac{L({\theta }_1)}{L(1)}\right)& =n\ln ({\theta }_1)+({\theta }_1-1)\ln \left[\prod _{i=1}^n(1-x_i)\right]\\ & =n\ln ({\theta }_1)+({\theta }_1-1)\sum _{i=1}^n\ln (1-x_i)\end{array}$$

Karena ${\theta }_1>1$, maka ${\theta }_1-1>0$. Untuk $0<x_i<1$, kita memiliki $0<1-x_i<1$ sehingga $\ln (1-x_i)<0$.

Rasio likelihood meningkat ketika ${\sum }_{i=1}^n\ln (1-x_i)$ meningkat (menjadi kurang negatif), yang terjadi ketika ${\prod }_{i=1}^n(1-x_i)$ meningkat.

Oleh karena itu, untuk setiap ${\theta }_1>1$, best critical region adalah: ${\prod }_{i=1}^n(1-X_i)\ge k⟺{\sum }_{i=1}^n\ln (1-X_i)\ge k^{\prime },\forall {\theta }_1>1$ untuk suatu konstanta $k^{\prime }$.

$\therefore$ UMPT untuk $H_0:\theta =1$ versus $H_1:\theta >1$ adalah:

$\boxed{\text{Tolak }{H}_0\text{ jika }\sum _{i=1}^n\ln (1-X_i)\ge k^{\prime }}$

9.47.2

Untuk $H_0:\theta =1$:

$$\begin{array}{rl}L(1)& =\prod _{i=1}^n(1-x_i{)}^{1-1}=1\\ \ln L(\theta )& =n\ln (\theta )+(\theta -1)\sum _{i=1}^n\ln (1-x_i)\\ \frac{\partial \ln L}{\partial \theta }& =\frac{n}{\theta }+\sum _{i=1}^n\ln (1-x_i)=0\end{array}$$

Sehingga: $\hat{\theta }=-\frac{n}{{\sum }_{i=1}^n\ln (1-x_i)}$

Likelihood maksimum:

$$\begin{array}{rl}L(\hat{\theta })& ={\hat{\theta }}^n{\left[\prod _{i=1}^n(1-x_i)\right]}^{\hat{\theta }-1}\\ & ={\left(-\frac{n}{{\sum }_{i=1}^n\ln (1-x_i)}\right)}^n{\left[\prod _{i=1}^n(1-x_i)\right]}^{-\frac{n}{{\sum }_{i=1}^n\ln (1-x_i)}-1}\end{array}$$

Likelihood ratio adalah: $\boxed{\lambda =\frac{L(1)}{L(\hat{\theta })}=\frac{1}{L(\hat{\theta })}=\frac{{\left({\sum }_{i=1}^n\ln (1-x_i)\right)}^n}{(-n{)}^n}{\left[\prod _{i=1}^n(1-x_i)\right]}^{\frac{n}{{\sum }_{i=1}^n\ln (1-x_i)}+1}}$

Footnotes

1. Normal distribution ↩