Tugas Latihan

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.

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9.34

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