第4講統計的検定

尤度比検定による $t$ 検定の導出

X_{i} \sim N (μ, σ^{2}), i . i . d . (i = 1, \dots, n)

$\begin{equation} X_i \sim N(\mu, \sigma^2),~i.i.d.~~~~(i=1,\ldots,n) \end{equation}$

H_{0} : μ = μ_{0}

$\begin{equation} H_{0}~:~\mu = \mu_{0} \end{equation}$

H_{1} : μ \neq μ_{0}

$\begin{equation} H_{1}~:~\mu \ne \mu_{0} \end{equation}$

とする．同時密度関数は，

\begin{aligned} f (x; μ, σ^{2}) & = \prod_{i = 1}^{n} f (x_{i}; μ, σ^{2}) \\ = {(\frac{1}{\sqrt{2 π σ^{2}}})}^{n} \exp [- \sum_{i = 1}^{n} \frac{(x_{i} - μ)^{2}}{2 σ^{2}}] \end{aligned}

$\begin{align} f(\bm{x}; \mu, \sigma^2) &= \prod_{i=1}^{n}f(x_i; \mu, \sigma^2) \\ &= \left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^{n}\exp\left[-\sum_{i=1}^{n}\frac{(x_i - \mu)^2}{2\sigma^2}\right] \end{align}$

尤度比検定の棄却域は，

\frac{max_{μ, σ^{2}} f (x; μ, σ^{2})}{max_{σ^{2}} f (x; μ_{0}, σ^{2})} \geq c

$\begin{equation} \dfrac{\ds\max_{\mu, \sigma^2}f(\bm{x}; \mu, \sigma^2)}{\ds\max_{\sigma^2}f(\bm{x}; \mu_0, \sigma^2)} \ge c \end{equation}$

H_{1}

$H_1$ の下で，分子

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，

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

$\begin{equation} \hat{\mu} = \bar{x} = \sum_{i=1}^{n}x_i \end{equation}$

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

$\begin{equation} \hat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 \end{equation}$

よって，

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

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

一方，

H_{0}

$H_0$ の下で，分母

max_{σ^{2}} f (x; μ_{0}, σ^{2})

$\max_{\sigma^2}f(\bm{x}; \mu_0, \sigma^2)$ を最大にする

σ^{2}

$\sigma^2$ は，

{\tilde{σ}}^{2} = \frac{1}{n} \sum_{i = 1}^{n} (x_{i} - μ_{0})^{2}

$\begin{equation} \tilde{\sigma}^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \mu_0)^2 \end{equation}$

よって，

\begin{aligned} max_{σ^{2}} f (x; μ_{0}, σ^{2}) & = f (x; μ_{0}, {\tilde{σ}}^{2}) \\ = {(\frac{1}{\sqrt{2 π {\tilde{σ}}^{2}}})}^{n} \exp [- \frac{n}{2}] \end{aligned}

$\begin{align} \max_{\sigma^2}f(\bm{x}; \mu_0, \sigma^2) &= f(\bm{x}; \mu_0, \tilde{\sigma}^2) \\ &= \left(\frac{1}{\sqrt{2\pi\tilde{\sigma}^2}}\right)^{n}\exp\left[-\frac{n}{2}\right] \end{align}$

よって，棄却域は，

\frac{{(\frac{1}{\sqrt{2 π {\hat{σ}}^{2}}})}^{n} \exp [- \frac{n}{2}]}{{(\frac{1}{\sqrt{2 π {\tilde{σ}}^{2}}})}^{n} \exp [- \frac{n}{2}]} \geq c

$\begin{equation} \dfrac{\left(\dfrac{1}{\sqrt{2\pi\hat{\sigma}^2}}\right)^{n}\exp\left[-\dfrac{n}{2}\right]}{\left(\dfrac{1}{\sqrt{2\pi\tilde{\sigma}^2}}\right)^{n}\exp\left[-\dfrac{n}{2}\right]} \ge c \end{equation}$

⟺ \frac{{(\frac{1}{\sqrt{{\hat{σ}}^{2}}})}^{n}}{{(\frac{1}{\sqrt{{\tilde{σ}}^{2}}})}^{n}} \geq c

$\begin{equation} \Longleftrightarrow~\dfrac{\left(\dfrac{1}{\sqrt{\hat{\sigma}^2}}\right)^n}{\left(\dfrac{1}{\sqrt{\tilde{\sigma}^2}}\right)^n} \ge c \end{equation}$

⟺ {(\frac{{\tilde{σ}}^{2}}{{\hat{σ}}^{2}})}^{\frac{n}{2}} \geq c

$\begin{equation} \Longleftrightarrow~\left(\frac{\tilde{\sigma}^2}{\hat{\sigma}^2}\right)^{\frac{n}{2}} \ge c \end{equation}$

⟺ {(\frac{\sum_{i = 1}^{n} (x_{i} - μ_{0})^{2}}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}})}^{\frac{n}{2}} \geq c

$\begin{equation} \Longleftrightarrow~\left(\frac{\sum_{i=1}^{n}(x_i - \mu_0)^2}{\sum_{i=1}^{n}(x_i - \bar{x})^2}\right)^{\frac{n}{2}} \ge c \end{equation}$

これを

\sum (x_{i} - μ_{0})^{2} = \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2} + n (\bar{x} - μ_{0})^{2}

$\begin{equation} \sum(x_i - \mu_0)^2 = \sum_{i=1}^{n}(x_i - \bar{x})^2 + n(\bar{x} - \mu_0)^2 \end{equation}$

を用いて変形すると，

{(\frac{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2} + n (\bar{x} - μ_{0})^{2}}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}})}^{\frac{n}{2}} \geq c

$\begin{equation} \left(\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2 + n(\bar{x} - \mu_0)^2}{\sum_{i=1}^{n}(x_i - \bar{x})^2}\right)^{\frac{n}{2}} \ge c \end{equation}$

⟺ {(1 + \frac{n (\bar{x} - μ_{0})^{2}}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}})}^{\frac{n}{2}} \geq c

$\begin{equation} \Longleftrightarrow~\left(1 + \frac{n(\bar{x} - \mu_0)^2}{\sum_{i=1}^{n}(x_i - \bar{x})^2}\right)^{\frac{n}{2}} \ge c \end{equation}$

ここで，

t = \frac{\bar{x} - μ_{0}}{\sqrt{\frac{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}}{n}}} \sim t (n - 1)

$\begin{equation} t = \dfrac{\bar{x} - \mu_0}{\sqrt{\dfrac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n}}} \sim t(n-1) \end{equation}$

とおくと，

{(1 + t^{2})}^{\frac{n}{2}} \geq c

$\begin{equation} \left(1 + t^2\right)^{\frac{n}{2}} \ge c \end{equation}$

左辺は

t

$t$ 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．

| t | \geq c^{'}

$\begin{equation} |t| \ge c' \end{equation}$

第4講 統計的検定

尤度比検定による tt 検定の導出

第4講統計的検定

尤度比検定による $t$ 検定の導出