1번
\(X_1,X_2,\dots, X_n\)이 \(N(\mu, \sigma^2)\) (단 \(\mu\):unknown)로부터의 랜덤샘플일 때 다음 가설에 대한 가능도비 검정의 유의수준 \(\alpha\)기각역을 구하시오.
\[H_0: \sigma^2 = \sigma_0^2 \ vs \ H_1: \sigma^2 \neq \sigma_0^2\]
\((\hat \mu, \hat \sigma^2) = (\bar X, \frac{\sum_{i=1}^n (X_i-\bar X)^2}{n})\)
\((\hat \mu_0, \hat \sigma^2_0) = (\bar X, \sigma_0^2)\)
\(L(\mu, \sigma^2 |x)=\Pi_{i=1}^n \frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{(X_i-\mu)^2}{2\sigma^2}}\)
\(\gamma(X)=\dfrac{(\frac{1}{2\pi \hat \sigma_0^2})^{2/n} exp(-\frac{\sum_{i=1}^n(X_i-\bar X)^2}{2\hat \sigma_0^2})}{(\frac{1}{2\pi\hat \sigma^2})^{2/n}exp(-\frac{\sum_{i=1}^n(X_i-\bar X)^2}{2\hat \sigma^2})}\)
\(=(\frac{\hat \sigma^2}{\sigma_0^2})^{2/n}exp(-\frac{\sum_{i=1}^n (X_i-\bar X)^2}{2\sigma_0^2}+\frac{n}{2})\)
\(=(\frac{\hat \sigma^2}{ \sigma_0^2})^{2/n}exp((-\frac{n}{2}) (\frac{\hat \sigma^2}{\sigma_0^2})+\frac{n}{2})\)
\(P(\gamma(X) \leq C |H_0) = \alpha\) 를 찾자.
\(e^{n/2}\)식은 상수이므로 곱해서 없애주면
\((\frac{\hat \sigma^2}{\sigma_0^2})^{2/n} exp(-\frac{\hat \sigma^2}{\sigma_0^2}) \leq C\)
\(\because \frac{n\hat \sigma^2}{\sigma_0^2} = \frac{\sum_{i=1}^n (X_i-\bar X)^2}{\sigma_0^2}\) ~ \(\chi^2(n-1))\)
\(\rightarrow P((\frac{n\hat \sigma^2}{\sigma_0^2})^{2/n} exp(-\frac{n\hat \sigma^2}{\sigma_0^2}) \leq C |H_0) = \alpha\)
\(\therefore \dfrac{\sum_{i=1}^n (X_i -\bar X)^2}{\sigma_0^2} \geq \chi^2_{\alpha/2}(n-1)\) or \(\dfrac{\sum_{i=1}^n (X_i -\bar X)^2}{\sigma_0^2} \leq \chi^2_{1-\alpha/2}(n-1)\)
2번
\(X_k\) ~ \(N(k\theta. 1), k=1,2,\dots,n\)이고 서로 독립인 확률변수이다. 다음 가설에 대한 가능도비 검정의 유의수준 \(\alpha\) 기각역을 구하시오.
\[H_0:\theta=0 \ vs \ H_1: \theta \neq 0\]
위 확률변수는 평균이 다 다르므로 랜덤표본이 아니다.
\(L(\theta: x_1, \dots, x_n)= f(x_1,\dots,x_n;\theta)=\Pi_{k=1}^n f_k (x_k ; \theta) = \Pi_{k=1}^n \frac{1}{\sqrt{2\pi}} e^{-\frac{(X_k-k\theta)^2}{2}}=(\frac{1}{2\pi})^{n/2}e^{-\frac{\sum_{i=1}^n(X_k-k\theta)^2}{2}}\)
\(h: \{ \theta : -\infty < \theta < \infty \}\)
\(h_0: \{ \theta : \theta=0 \}\)
\(l(\theta)=\frac{n}{2}log 2\pi - \frac{\sum_{k=1}^n (X_k - k\theta)^2}{2}\)
\(l'(\theta) = \sum_{k=1}^n k(X_k - k\theta) = 0\)
\(\hat \theta^{MLE} = \dfrac{\sum_{k=1}^n kX_k}{\sum_{k=1}^n k^2}\)
\(\gamma(X) = \dfrac{(\frac{1}{2\pi})^{2/n} e^{-\frac{\sum_{k=1}^n (X_k)^2}{2}}}{(\frac{1}{2\pi})^{2/n} e^{-\frac{\sum_{k=1}^n (X_k-k\hat \theta^{MLE})^2}{2}}}\)
\(=e^{-\frac{\sum X_k^2}{2}+\frac{\sum(X_k-k\hat \theta^{MLE})^2}{2}}\)
\(=exp(-(\sum kX_k) \frac{\sum kX_k}{\sum k^2} + \frac{1}{2}\sum k^2 \frac{(\sum kX_k)^2}{(\sum k^2)^2})\)
\(=exp(-\dfrac{(\sum kX_k)^2}{2\sum k^2}) \leq C\)
위 식에서 log를 취하면
\(\dfrac{(\sum kX_k)^2}{\sum k^2} \geq C\)
\(kX_k\) ~ \(N(k^2 \theta, k^2)\)
\(\sum_{k=1}^n k X_k\) ~ \(N(\sum_{k=1}^n k^2 \theta, \sum_{k=1}^n k^2)\)
under \(H_0\),
\(\sum_{k=1}^n k X_k\) ~ \(N(0,\sum_{k=1}^n k^2)\)
표준화하면,
\(\dfrac{\sum_{k=1}^n k X_k - 0}{\sqrt{\sum_{k=1}^n k^2}}\) ~ \(N(0,1)\)
제곱하면,
\(\dfrac{(\sum_{k=1}^n k X_k)^2}{\sum_{k=1}^n k^2}\) ~ \(\chi^2(1)\)
즉,
\(P \{ (X_1, \dots, X_n): \dfrac{(\sum_{k=1}^n kX_k)^2}{\sum_{k=1}^n k^2} \geq C \} (\theta_0)=\alpha\)
\(\dfrac{(\sum_{k=1}^n kX_k)^2}{\sum_{k=1}^n k^2} \geq \chi^2_\alpha(1)\)