Tugas Latihan

9.28

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’)^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

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’) = (2\pi\theta_2’)^{-n/2} \exp\left(-\frac{1}{2\theta_2’}\sum{i=1}^n (x_i-\bar{x})^2\right)$$

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

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

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

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’}\right)^{-n/2} \exp\left(\frac{n}{2}\left(1 - \frac{S^2}{n\theta_2’}\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

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

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}, \quad \theta \geq \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 $X_{(n)}=\max (X_1,\dots ,X_n)$.