UA MATH566 统计理论7 另一个例子：二项检验
UA MATH566 統計理論7 另一個例子：二項檢驗
假設X1,X2,?,Xn～Ber(p)X_1,X_2,\cdots,X_n \sim Ber(p)X1?,X2?,?,Xn?～Ber(p)，想根據這組樣本做如下檢驗：
H0:p=p0Ha:p≠p0H_0:p=p_0 \\ H_a:p \ne p_0H0?:p=p0?Ha?:p?=p0?
參數空間為
Θ0={p=p0}Θa={p≠p0}\Theta_0 = \{p=p_0\} \\ \Theta_a = \{p \ne p_0\}Θ0?={p=p0?}Θa?={p?=p0?}
根據Karlin-Rubin定理，這個檢驗的UMP拒絕域為
C={X:λ(X)≤kα}C = \{X:\lambda(X) \le k_{\alpha}\}C={X:λ(X)≤kα?}
先計算樣本的似然函數：
L(p∣X)=∏i=1npXi(1?p)1?Xi=p∑i=1nXi(1?p)n?∑i=1nXiL(p|X) = \prod_{i=1}^n p^{X_i} (1-p)^{1-X_i} = p^{\sum_{i=1}^n X_i} (1-p)^{n-\sum_{i=1}^n X_i}L(p∣X)=i=1∏n?pXi?(1?p)1?Xi?=p∑i=1n?Xi?(1?p)n?∑i=1n?Xi?
根據Neyman-Fisher因子定理，定義T(X)=∑i=1nXi=nXˉT(X)=\sum_{i=1}^n X_i=n\bar{X}T(X)=∑i=1n?Xi?=nXˉ，則T(X)T(X)T(X)是充分統計量。似然函數可以寫成
L(p∣X)=∏i=1npXi(1?p)1?Xi=pT(X)(1?p)n?T(X)?ln?L(p∣X)?p=??p[T(X)ln?p+(n?T(X))ln?(1?p)]=T(X)p?n?T(X)1?p=0L(p|X) = \prod_{i=1}^n p^{X_i} (1-p)^{1-X_i} = p^{T(X)} (1-p)^{n-T(X)} \\ \frac{\partial \ln L(p|X)}{\partial p} = \frac{\partial }{\partial p} [T(X)\ln p + (n-T(X))\ln(1-p)] \\ = \frac{T(X)}{p} - \frac{n-T(X)}{1-p} = 0L(p∣X)=i=1∏n?pXi?(1?p)1?Xi?=pT(X)(1?p)n?T(X)?p?lnL(p∣X)?=?p??[T(X)lnp+(n?T(X))ln(1?p)]=pT(X)??1?pn?T(X)?=0
因此p^=Xˉ\hat{p}=\bar{X}p^?=Xˉ是ppp的最大似然估計。計算似然比，
L(p0∣X)L(p^∣X)=(np0T(X))T(X)(n?np0n?T(X))n?T(X)\frac{L(p_0|X)}{L(\hat{p}|X)} = \left( \frac{np_0}{T(X)} \right)^{T(X)} \left( \frac{n-np_0}{n-T(X)} \right)^{n-T(X)}L(p^?∣X)L(p0?∣X)?=(T(X)np0??)T(X)(n?T(X)n?np0??)n?T(X)
記這個似然比的對數為g(T(X))g(T(X))g(T(X))，則
g(T(X))=T(X)(ln?np0?ln?T(X))+(n?T(X))(ln?(n?np0)?ln?(n?T(X)))dg(T(X))dT(X)=(ln?np0?ln?T(X))?1?(ln?(n?np0)?ln?(n?T(X)))+1=ln?np0(n?T(X))(n?np0)T(X)g(T(X)) = T(X)(\ln np_0 - \ln T(X)) + (n-T(X)) (\ln (n-np_0) - \ln (n-T(X))) \\ \frac{dg(T(X))}{dT(X)} = (\ln np_0 - \ln T(X)) - 1 - (\ln (n-np_0) - \ln (n-T(X))) + 1 \\ = \ln \frac{np_0 (n-T(X))}{(n-np_0)T(X)} g(T(X))=T(X)(lnnp0??lnT(X))+(n?T(X))(ln(n?np0?)?ln(n?T(X)))dT(X)dg(T(X))?=(lnnp0??lnT(X))?1?(ln(n?np0?)?ln(n?T(X)))+1=ln(n?np0?)T(X)np0?(n?T(X))?
考慮一個特殊情況，如果這個導數為正，則λ(X)<kα\lambda(X)<k_{\alpha}λ(X)<kα?等價于T(X)<cα,?cαT(X)<c_{\alpha},\exists c_{\alpha}T(X)<cα?,?cα?。拒絕域為
C={X:T(X)≤cα}C=\{X:T(X) \le c_{\alpha}\}C={X:T(X)≤cα?}
原假設下T(X)～Binom(n,p0)T(X) \sim Binom(n,p_0)T(X)～Binom(n,p0?)，因此可以取cαc_{\alpha}cα?為Binom(n,p0)Binom(n,p_0)Binom(n,p0?)的左側α\alphaα分位點，這個檢驗也由此得名二項檢驗（binomial test）。（R語言中可以用binom.test）
在這個特殊情況下，如果樣本數量nnn足夠大，根據中心極限定理
Z=Xˉ?pp(1?p)/n→dN(0,1)Z = \frac{\bar{X}-p}{\sqrt{p(1-p)/n}} \to_d N(0,1)Z=p(1?p)/n?Xˉ?p?→d?N(0,1)
Xˉ\bar{X}Xˉ是ppp的最大似然估計，也是充分統計量。可以根據ZZZ構造拒絕域：
C={X:∣Xˉ?p0p0(1?p0)/n∣≥zα/2}C = \{X:|\frac{\bar{X}-p_0}{\sqrt{p_0(1-p_0)/n}}| \ge z_{\alpha/2} \}C={X:∣p0?(1?p0?)/n?Xˉ?p0??∣≥zα/2?}
其中zα/2z_{\alpha/2}zα/2?是標準正態分布的α/2\alpha/2α/2上分位點，這個檢驗叫做比例檢驗（proportion test），（R語言中可以用prop.test）。它的勢函數為
Power=P(∣Z∣≤zα/2)=P(∣Xˉ?pp(1?p)/n∣≤zα/2)Power = P(|Z| \le z_{\alpha/2}) = P(|\frac{\bar{X}-p}{\sqrt{p(1-p)/n}}| \le z_{\alpha/2}) Power=P(∣Z∣≤zα/2?)=P(∣p(1?p)/n?Xˉ?p?∣≤zα/2?)
考慮
?zα/2≤Xˉ?p0p0(1?p0)/n≤zα/2p0?zα/2p0(1?p0)/n≤Xˉ≤p0+p0(1?p0)/np0?pp0(1?p0)/n?zα/2p0(1?p0)p(1?p)≤Z≤p0?pp0(1?p0)/n+zα/2p0(1?p0)p(1?p)-z_{\alpha/2} \le \frac{\bar{X}-p_0}{\sqrt{p_0(1-p_0)/n}} \le z_{\alpha/2} \\ p_0-z_{\alpha/2}\sqrt{p_0(1-p_0)/n} \le \bar{X} \le p_0 + \sqrt{p_0(1-p_0)/n} \\ \frac{p_0-p}{\sqrt{p_0(1-p_0)/n}} - z_{\alpha/2} \sqrt{\frac{p_0(1-p_0)}{p(1-p)}} \le Z \le \frac{p_0-p}{\sqrt{p_0(1-p_0)/n}} +z_{\alpha/2} \sqrt{\frac{p_0(1-p_0)}{p(1-p)}}?zα/2?≤p0?(1?p0?)/n?Xˉ?p0??≤zα/2?p0??zα/2?p0?(1?p0?)/n?≤Xˉ≤p0?+p0?(1?p0?)/n?p0?(1?p0?)/n?p0??p??zα/2?p(1?p)p0?(1?p0?)??≤Z≤p0?(1?p0?)/n?p0??p?+zα/2?p(1?p)p0?(1?p0?)??
所以
Power=Φ(p0?pp0(1?p0)/n+zα/2p0(1?p0)p(1?p))?Φ(p0?pp0(1?p0)/n?zα/2p0(1?p0)p(1?p))Power = \Phi(\frac{p_0-p}{\sqrt{p_0(1-p_0)/n}} +z_{\alpha/2} \sqrt{\frac{p_0(1-p_0)}{p(1-p)}}) - \Phi(\frac{p_0-p}{\sqrt{p_0(1-p_0)/n}} - z_{\alpha/2} \sqrt{\frac{p_0(1-p_0)}{p(1-p)}})Power=Φ(p0?(1?p0?)/n?p0??p?+zα/2?p(1?p)p0?(1?p0?)??)?Φ(p0?(1?p0?)/n?p0??p??zα/2?p(1?p)p0?(1?p0?)??)
總結
以上是生活随笔為你收集整理的UA MATH566 统计理论7 另一个例子：二项检验的全部內容，希望文章能夠幫你解決所遇到的問題。
- 上一篇： UA MATH564 概率论II 连续型
- 下一篇： UA MATH566 统计理论5 假设检