第4講統計的検定

尤度比検定による $F$ 検定の導出

X_{i j} \overset{i . i . d .}{\sim} N (μ_{i}, σ^{2}) (i = 1, \dots, a, j = 1, \dots, n)

$\begin{equation} X_{ij} \stackrel{i.i.d.}{\sim} N(\mu_i, \sigma^2)~~~~(i=1,\ldots,a,~j=1,\ldots,n) \end{equation}$

H_{0} : μ_{1} = μ_{2} = \dots = μ_{a} = μ

$\begin{equation} H_{0}~:~\mu_1 = \mu_2 = \cdots = \mu_a = \mu \end{equation}$

H_{1} : H_{0} の否定

$\begin{equation} H_{1}~:~H_0 \text{の否定} \end{equation}$

尤度比検定の棄却域は，

\frac{max_{μ_{1}, \dots, μ_{a}, σ^{2}} f (x; μ_{1}, \dots, μ_{a}, σ^{2})}{max_{μ, σ^{2}} f (x; μ, σ^{2})} \geq c

$\begin{equation} \dfrac{\ds\max_{\mu_1, \ldots, \mu_a, \sigma^2}f(\bm{x}; \mu_1, \ldots, \mu_a, \sigma^2)}{\ds\max_{\mu, \sigma^2}f(\bm{x}; \mu, \sigma^2)} \ge c \end{equation}$

同時密度関数は，

\begin{aligned} f (x; μ_{1}, \dots, μ_{a},, σ^{2}) & = \prod_{i = 1}^{a} \prod_{j = 1}^{n} f (x_{i j}; μ_{i}, σ^{2}) \\ = {(\frac{1}{\sqrt{2 π σ^{2}}})}^{a n} \exp [- \sum_{i = 1}^{a} \sum_{j = 1}^{n} \frac{(x_{i j} - μ_{i})^{2}}{2 σ^{2}}] \\ = {(\frac{1}{\sqrt{2 π σ^{2}}})}^{a n} \exp [- \frac{1}{2 σ^{2}} (n \sum_{i = 1}^{a} ({\bar{x}}_{i} - μ_{i})^{2} + \sum_{i = 1}^{a} \sum_{j = 1}^{n} (x_{i j} - {\bar{x}}_{i})^{2})] \end{aligned}

$\begin{align} f(\bm{x}; \mu_1, \ldots, \mu_a,, \sigma^2) &= \prod_{i=1}^{a}\prod_{j=1}^{n}f(x_{ij}; \mu_{i}, \sigma^2) \\ &= \left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^{an}\exp\left[-\sum_{i=1}^{a}\sum_{j=1}^{n}\frac{(x_{ij} - \mu_{i})^2}{2\sigma^2}\right] \\ &= \left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^{an}\exp\left[-\frac{1}{2\sigma^2}\left(n\sum_{i=1}^{a}(\bar{x}_{i} - \mu_{i})^2 + \sum_{i=1}^{a}\sum_{j=1}^{n}(x_{ij} - \bar{x}_{i})^2\right)\right] \end{align}$

H_{1}

$H_1$ の下で，

max_{μ_{1}, \dots, μ_{a}, σ^{2}} f (x; μ_{1}, \dots, μ_{a}, σ^{2})

$\max_{\mu_1, \ldots, \mu_a, \sigma^2}f(\bm{x}; \mu_1, \ldots, \mu_a, \sigma^2)$ を最大にする

μ_{i}

$\mu_i$ と

σ^{2}

$\sigma^2$ は，

{\hat{μ}}_{i} = {\bar{x}}_{i} = \frac{1}{n} \sum_{j = 1}^{n} x_{i j}

$\begin{equation} \hat{\mu}_i = \bar{x}_i = \frac{1}{n}\sum_{j=1}^{n}x_{ij} \end{equation}$

{\hat{σ}}^{2} = \frac{1}{a n} \sum_{i = 1}^{a} \sum_{j = 1}^{n} (x_{i j} - {\bar{x}}_{i})^{2}

$\begin{equation} \hat{\sigma}^2 = \frac{1}{an}\sum_{i=1}^{a}\sum_{j=1}^{n}(x_{ij} - \bar{x}_i)^2 \end{equation}$

よって，

\begin{aligned} max_{μ_{1}, \dots, μ_{a}, σ^{2}} f (x; μ_{1}, \dots, μ_{a}, σ^{2}) & = f (x; {\hat{μ}}_{1}, \dots, \hat{μ_{a}}, {\hat{σ}}^{2}) \\ = {(\frac{1}{\sqrt{2 π {\hat{σ}}^{2}}})}^{a n} \exp [- \frac{a n}{2}] \end{aligned}

$\begin{align} \max_{\mu_1, \ldots, \mu_a, \sigma^2}f(\bm{x}; \mu_1, \ldots, \mu_a, \sigma^2) &= f(\bm{x}; \hat{\mu}_1, \ldots, \hat{\mu_a}, \hat{\sigma}^2) \\ &= \left(\frac{1}{\sqrt{2\pi\hat{\sigma}^2}}\right)^{an}\exp\left[-\frac{an}{2}\right] \end{align}$

一方，

H_{0}

$H_0$ の下で，

max_{μ, σ^{2}} f (x; μ, σ^{2})

$\max_{\mu, \sigma^2}f(\bm{x}; \mu, \sigma^2)$ を最大にする

μ

$\mu$ と

σ^{2}

$\sigma^2$ は，

\tilde{μ} = \bar{x} = \frac{1}{a n} \sum_{j = 1}^{n} \sum_{j = 1}^{n} x_{i j}

$\begin{equation} \tilde{\mu} = \bar{x} = \frac{1}{an}\sum_{j=1}^{n}\sum_{j=1}^{n}x_{ij} \end{equation}$

{\tilde{σ}}^{2} = \frac{1}{a n} \sum_{i = 1}^{a} \sum_{j = 1}^{n} (x_{i j} - \bar{x})^{2}

$\begin{equation} \tilde{\sigma}^2 = \frac{1}{an}\sum_{i=1}^{a}\sum_{j=1}^{n}(x_{ij} - \bar{x})^2 \end{equation}$

よって，

\begin{aligned} max_{σ^{2}} f (x; μ, σ^{2}) & = f (x; \tilde{μ}, {\tilde{σ}}^{2}) \\ = {(\frac{1}{\sqrt{2 π {\tilde{σ}}^{2}}})}^{a n} \exp [- \frac{a n}{2}] \end{aligned}

$\begin{align} \max_{\sigma^2}f(\bm{x}; \mu, \sigma^2) &= f(\bm{x}; \tilde{\mu}, \tilde{\sigma}^2) \\ &= \left(\frac{1}{\sqrt{2\pi\tilde{\sigma}^2}}\right)^{an}\exp\left[-\frac{an}{2}\right] \end{align}$

よって，棄却域は，

{(\frac{\sum_{i = 1}^{a} \sum_{j = 1}^{n} (x_{i j} - \bar{x})^{2}}{\sum_{i = 1}^{a} \sum_{j = 1}^{n} (x_{i j} - {\bar{x}}_{i})^{2}})}^{\frac{a n}{2}} \geq c

$\begin{equation} \left(\frac{\sum_{i=1}^{a}\sum_{j=1}^{n}(x_{ij} - \bar{x})^2}{\sum_{i=1}^{a}\sum_{j=1}^{n}(x_{ij} - \bar{x}_i)^2}\right)^{\frac{an}{2}} \ge c \end{equation}$

ここで，

\sum_{i = 1}^{a} \sum_{j = 1}^{n} (x_{i j} - \bar{x})^{2} = \sum_{i = 1}^{a} \sum_{j = 1}^{n} (x_{i j} - {\bar{x}}_{i})^{2} + n \sum_{i = 1}^{a} ({\bar{x}}_{i} - \bar{x})^{2}

$\begin{equation} \sum_{i=1}^{a}\sum_{j=1}^{n}(x_{ij} - \bar{x})^2 = \sum_{i=1}^{a}\sum_{j=1}^{n}(x_{ij} - \bar{x}_i)^2 + n\sum_{i=1}^{a}(\bar{x}_i - \bar{x})^2 \end{equation}$

を用いれば，

{(1 + \frac{n \sum_{i = 1}^{a} ({\bar{x}}_{i} - \bar{x})^{2}}{\sum_{i = 1}^{a} \sum_{j = 1}^{n} (x_{i j} - {\bar{x}}_{i})^{2}})}^{\frac{a n}{2}} \geq c

$\begin{equation} \left(1 + \frac{n\sum_{i=1}^{a}(\bar{x}_i - \bar{x})^2}{\sum_{i=1}^{a}\sum_{j=1}^{n}(x_{ij} - \bar{x}_i)^2}\right)^{\frac{an}{2}} \ge c \end{equation}$

ここで，

F = \frac{\frac{n \sum_{i = 1}^{a} ({\bar{x}}_{i} - \bar{x})^{2}}{a - 1}}{\frac{\sum_{i = 1}^{a} \sum_{j = 1}^{n} (x_{i j} - {\bar{x}}_{i})^{2}}{a (n - 1)}} \sim F (a - 1, a (n - 1))

$\begin{equation} F = \dfrac{\dfrac{n\sum_{i=1}^{a}(\bar{x}_i - \bar{x})^2}{a-1}}{\dfrac{\sum_{i=1}^{a}\sum_{j=1}^{n}(x_{ij} - \bar{x}_i)^2}{a(n-1)}} \sim F(a-1,~a(n-1)) \end{equation}$

とおくと，

{(1 + \frac{a (n - 1)}{a - 1} F)}^{\frac{a n}{2}} \geq c

$\begin{equation} \left(1 + \frac{a(n-1)}{a-1}F\right)^{\frac{an}{2}} \ge c \end{equation}$

左辺は

F

$F$ に関して単調増加であることから，棄却域は次式で与えられることになる．

F \geq c^{'}

$\begin{equation} F \ge c' \end{equation}$

第4講 統計的検定

尤度比検定による FF 検定の導出

第4講統計的検定

尤度比検定による $F$ 検定の導出