統計検定 1級過去問解答/解答例と解説
2014年11月30日 (日) 試験

統計数理問4 [4]

設問の要約

$H_1$ の下での $F$ 統計量の非心度 $\lambda$ を求めよ．
$Z_i,~i=1,\ldots,f$ : 互いに独立に $N(\mu_i,~1)$ に従う
$Y = {Z_1}^2 + \cdots + {Z_f}^2$ の分布 : 非心度 $\lambda = {\mu_1}^2 + \cdots + {\mu_f}^2$ ，自由度が $f$ の非心カイ二乗分布
$Y_1$ と $Y_2$ は互いに独立
$Y_1$ : 非心度 $\lambda$ で自由度 $f_1$ の非心カイ二乗分布
$Y_2$ : 自由度 $f_2$ のカイ二乗分布
$F$ 統計量 $F= (Y_1/f_1)/(Y_2/f_2)$ の非心度は $\lambda$

解答例

$\bm{\theta}_0 = (\alpha, \alpha, \alpha, \alpha, \alpha)'$ とする．問題 [2] で求めた (1) と (2) の場合の分散をそれぞれ $s^2_{(1)}, s^2_{(2)}$ とする．

$F$ 統計量は次のように書ける．

F = \frac{\frac{1}{s_{(i)}^{2}} ‖ X \hat{θ} - X θ_{0} ‖^{2} / (5 - 1)}{\frac{1}{s_{(i)}^{2}} ‖ x - X \hat{θ} ‖^{2} / 5} \sim F (4, 5) (k = 1, 2)

$\begin{equation} F = \frac{\frac{1}{s^2_{(i)}}\|X \hat{\bm{\theta}} - X\bm{\theta}_0\|^2 / (5 - 1)}{\frac{1}{s^2_{(i)}}\|\bm{x} - X \hat{\bm{\theta}}\|^2 / 5} \sim F(4, 5)~~~~(k = 1,2) \end{equation}$

(1)のとき

まず，分母が従う確率分布が自由度 5 の $\chi^2$ 分布に従うことを証明する．

x - X \hat{θ} = (\begin{matrix} x_{1} - {\hat{θ}}_{1} \\ x_{2} - {\hat{θ}}_{1} \\ x_{3} - {\hat{θ}}_{2} \\ x_{4} - {\hat{θ}}_{2} \\ x_{5} - {\hat{θ}}_{3} \\ x_{6} - {\hat{θ}}_{3} \\ x_{7} - {\hat{θ}}_{4} \\ x_{8} - {\hat{θ}}_{4} \\ x_{9} - {\hat{θ}}_{5} \\ x_{10} - {\hat{θ}}_{5} \end{matrix})

$\begin{equation} \bm{x} - X\hat{\bm{\theta}} = \begin{pmatrix} x_1 - \hat{\theta}_1 \\ x_2 - \hat{\theta}_1 \\ x_3 - \hat{\theta}_2 \\ x_4 - \hat{\theta}_2 \\ x_5 - \hat{\theta}_3 \\ x_6 - \hat{\theta}_3 \\ x_7 - \hat{\theta}_4 \\ x_8 - \hat{\theta}_4 \\ x_9 - \hat{\theta}_5 \\ x_{10} - \hat{\theta}_5 \end{pmatrix} \end{equation}$

\begin{aligned} \frac{1}{s_{(1)}^{2}} {‖ x - X \hat{θ} ‖}^{2} & = \frac{(x_{1} - {\hat{θ}}_{1})^{2}}{s_{(1)}^{2}} + \frac{(x_{2} - {\hat{θ}}_{1})^{2}}{s_{(1)}^{2}} + \dots + \frac{(x_{9} - {\hat{θ}}_{5})^{2}}{s_{(1)}^{2}} + \frac{(x_{10} - {\hat{θ}}_{5})^{2}}{s_{(1)}^{2}} \\ = \frac{({\hat{θ}}_{1} - x_{1})^{2}}{s_{(1)}^{2}} + \frac{({\hat{θ}}_{1} - x_{2})^{2}}{s_{(1)}^{2}} + \dots + \frac{({\hat{θ}}_{5} - x_{9})^{2}}{s_{(1)}^{2}} + \frac{({\hat{θ}}_{5} - x_{10})^{2}}{s_{(1)}^{2}} \end{aligned}

$\begin{align} \frac{1}{s^2_{(1)}}\left\| \bm{x} - X\hat{\bm{\theta}} \right\|^2 &= \frac{(x_1 - \hat{\theta}_1)^2}{s^2_{(1)}} + \frac{(x_2 - \hat{\theta}_1)^2}{s^2_{(1)}} + \cdots + \frac{(x_9 - \hat{\theta}_5)^2}{s^2_{(1)}} + \frac{(x_{10} - \hat{\theta}_5)^2}{s^2_{(1)}} \\ &= \frac{(\hat{\theta}_1 - x_1)^2}{s^2_{(1)}} + \frac{(\hat{\theta}_1 - x_2)^2}{s^2_{(1)}} + \cdots + \frac{(\hat{\theta}_5 - x_9)^2}{s^2_{(1)}} + \frac{(\hat{\theta}_5 - x_{10})^2}{s^2_{(1)}} \\ \end{align}$

ここで，

\begin{aligned} E [x - X \hat{θ}] & = E [x] - E [X \hat{θ}] \\ = E [X θ + ϵ] - E [X \hat{θ}] \\ = E [X θ] + E [ϵ] - E [X \hat{θ}] \\ = X θ + 0 - X θ \\ = 0 \end{aligned}

$\begin{align} E[\bm{x} - X\hat{\bm{\theta}}] &= E[\bm{x}] - E[X\hat{\bm{\theta}}] \\ &= E[X\bm{\theta} + \bm{\epsilon}] - E[X\hat{\bm{\theta}}] \\ &= E[X\bm{\theta}] + E[\bm{\epsilon}] - E[X\hat{\bm{\theta}}] \\ &= X\bm{\theta} + \bm{0} - X\bm{\theta} \\ &= \bm{0} \end{align}$

より，

E [{\hat{θ}}_{i} - x_{j}] = 0

$\begin{equation} E[\hat{\theta}_i - x_j] = 0 \end{equation}$

であるので，

\frac{{\hat{θ}}_{i} - α}{\sqrt{s_{(1)}^{2}}} \sim N (0, 1)

$\begin{equation} \frac{\hat{\theta}_i - \alpha}{\sqrt{s^2_{(1)}}} \sim N(0, 1) \end{equation}$

よって，

\frac{({\hat{θ}}_{1} - α)^{2}}{s_{(1)}^{2}} + \frac{({\hat{θ}}_{1} - α)^{2}}{s_{(1)}^{2}} + \dots + \frac{({\hat{θ}}_{5} - α)^{2}}{s_{(1)}^{2}} + \frac{({\hat{θ}}_{5} - α)^{2}}{s_{(1)}^{2}} \sim χ^{2} (5)

$\begin{equation} \frac{(\hat{\theta}_1 - \alpha)^2}{s^2_{(1)}} + \frac{(\hat{\theta}_1 - \alpha)^2}{{s}^2_{(1)}} + \cdots + \frac{(\hat{\theta}_5 - \alpha)^2}{s^2_{(1)}} + \frac{(\hat{\theta}_5 - \alpha)^2}{s^2_{(1)}} \sim \chi^2(5) \end{equation}$

次に，分子の従う確率分布が自由度 4 の非心

χ^{2}

$\chi^2$ 分布に従うことを証明し，非心度

λ

$\lambda$ を求める．

X \hat{θ} - X θ_{0} = (\begin{matrix} {\hat{θ}}_{1} - α \\ {\hat{θ}}_{1} - α \\ {\hat{θ}}_{2} - α \\ {\hat{θ}}_{2} - α \\ {\hat{θ}}_{3} - α \\ {\hat{θ}}_{3} - α \\ {\hat{θ}}_{4} - α \\ {\hat{θ}}_{4} - α \\ {\hat{θ}}_{5} - α \\ {\hat{θ}}_{5} - α \end{matrix})

$\begin{equation} X\hat{\bm{\theta}} - X\bm{\theta}_0 = \begin{pmatrix} \hat{\theta}_1 - \alpha \\ \hat{\theta}_1 - \alpha \\ \hat{\theta}_2 - \alpha \\ \hat{\theta}_2 - \alpha \\ \hat{\theta}_3 - \alpha \\ \hat{\theta}_3 - \alpha \\ \hat{\theta}_4 - \alpha \\ \hat{\theta}_4 - \alpha \\ \hat{\theta}_5 - \alpha \\ \hat{\theta}_5 - \alpha \end{pmatrix} \end{equation}$

\frac{1}{s_{(1)}^{2}} {‖ X \hat{θ} - X θ_{0} ‖}^{2} = \frac{({\hat{θ}}_{1} - α)^{2}}{s_{(1)}^{2}} + \frac{({\hat{θ}}_{1} - α)^{2}}{s_{(1)}^{2}} + \dots + \frac{({\hat{θ}}_{5} - α)^{2}}{s_{(1)}^{2}} + \frac{({\hat{θ}}_{5} - α)^{2}}{s_{(1)}^{2}}

$\begin{equation} \frac{1}{s^2_{(1)}}\left\| X\hat{\bm{\theta}} - X\bm{\theta}_0 \right\|^2 = \frac{(\hat{\theta}_1 - \alpha)^2}{s^2_{(1)}} + \frac{(\hat{\theta}_1 - \alpha)^2}{{s}^2_{(1)}} + \cdots + \frac{(\hat{\theta}_5 - \alpha)^2}{s^2_{(1)}} + \frac{(\hat{\theta}_5 - \alpha)^2}{s^2_{(1)}} \end{equation}$

ここで，

E [{\hat{θ}}_{i} - α] = E [{\hat{θ}}_{i}] - α = θ_{i} - α

$\begin{equation} E[\hat{\theta}_i - \alpha] = E[\hat{\theta}_i] - \alpha = \theta_i - \alpha \end{equation}$

であるので，

\frac{{\hat{θ}}_{i} - α}{\sqrt{s_{(1)}^{2}}} \sim N (θ_{i} - α, 1)

$\begin{equation} \frac{\hat{\theta}_i - \alpha}{\sqrt{s^2_{(1)}}} \sim N(\theta_i - \alpha, 1) \end{equation}$

よって，非心度を

λ

$\lambda$ とすれば，

\frac{({\hat{θ}}_{1} - α)^{2}}{s_{(1)}^{2}} + \frac{({\hat{θ}}_{1} - α)^{2}}{s_{(1)}^{2}} + \dots + \frac{({\hat{θ}}_{5} - α)^{2}}{s_{(1)}^{2}} + \frac{({\hat{θ}}_{5} - α)^{2}}{s_{(1)}^{2}} \sim χ^{2} (4, λ)

定義より，非心

χ^{2}

$\chi^2$ 分布の非心度

λ

$\lambda$ は，

\begin{aligned} λ & = (θ_{1} - α)^{2} + (θ_{i} - α)^{2} + \dots + (θ_{5} - α)^{2} + (θ_{5} - α)^{2} \\ = 2 \sum_{i = 1}^{5} (θ_{i} - α)^{2} \end{aligned}

$\begin{align} \lambda &= (\theta_1 - \alpha)^2 + (\theta_i - \alpha)^2 + \cdots + (\theta_5 - \alpha)^2 + (\theta_5 - \alpha)^2 \\ &= 2\sum_{i=1}^5(\theta_i - \alpha)^2 \end{align}$

定義より，この非心度

λ

$\lambda$ は，

F

$F$ 統計量の非心度となる．

(2)のとき

分母が $\chi^2$ 分布，分子が非心 $\chi^2$ 分布に従うことの証明は省略し，非心度のみを求める．

E [X \hat{θ} - X θ_{0}] = E (\begin{matrix} {\hat{θ}}_{1} + {\hat{θ}}_{2} - (α + α) \\ {\hat{θ}}_{1} + {\hat{θ}}_{3} - (α + α) \\ {\hat{θ}}_{1} + {\hat{θ}}_{4} - (α + α) \\ {\hat{θ}}_{1} + {\hat{θ}}_{5} - (α + α) \\ {\hat{θ}}_{2} + {\hat{θ}}_{3} - (α + α) \\ {\hat{θ}}_{2} + {\hat{θ}}_{4} - (α + α) \\ {\hat{θ}}_{2} + {\hat{θ}}_{5} - (α + α) \\ {\hat{θ}}_{3} + {\hat{θ}}_{4} - (α + α) \\ {\hat{θ}}_{3} + {\hat{θ}}_{5} - (α + α) \\ {\hat{θ}}_{4} + {\hat{θ}}_{5} - (α + α) \end{matrix}) = (\begin{matrix} (θ_{1} - α) + (θ_{2} - α) \\ (θ_{1} - α) + (θ_{3} - α) \\ (θ_{1} - α) + (θ_{4} - α) \\ (θ_{1} - α) + (θ_{5} - α) \\ (θ_{2} - α) + (θ_{3} - α) \\ (θ_{2} - α) + (θ_{4} - α) \\ (θ_{2} - α) + (θ_{5} - α) \\ (θ_{3} - α) + (θ_{4} - α) \\ (θ_{3} - α) + (θ_{5} - α) \\ (θ_{4} - α) + (θ_{5} - α) \end{matrix})

$\begin{equation} E[X\hat{\bm{\theta}} - X\bm{\theta}_0] = E\begin{pmatrix} \hat{\theta}_1 + \hat{\theta}_2 - (\alpha + \alpha) \\ \hat{\theta}_1 + \hat{\theta}_3 - (\alpha + \alpha) \\ \hat{\theta}_1 + \hat{\theta}_4 - (\alpha + \alpha) \\ \hat{\theta}_1 + \hat{\theta}_5 - (\alpha + \alpha) \\ \hat{\theta}_2 + \hat{\theta}_3 - (\alpha + \alpha) \\ \hat{\theta}_2 + \hat{\theta}_4 - (\alpha + \alpha) \\ \hat{\theta}_2 + \hat{\theta}_5 - (\alpha + \alpha) \\ \hat{\theta}_3 + \hat{\theta}_4 - (\alpha + \alpha) \\ \hat{\theta}_3 + \hat{\theta}_5 - (\alpha + \alpha) \\ \hat{\theta}_4 + \hat{\theta}_5 - (\alpha + \alpha) \\ \end{pmatrix} = \begin{pmatrix} (\theta_1 - \alpha) + (\theta_2 - \alpha) \\ (\theta_1 - \alpha) + (\theta_3 - \alpha) \\ (\theta_1 - \alpha) + (\theta_4 - \alpha) \\ (\theta_1 - \alpha) + (\theta_5 - \alpha) \\ (\theta_2 - \alpha) + (\theta_3 - \alpha) \\ (\theta_2 - \alpha) + (\theta_4 - \alpha) \\ (\theta_2 - \alpha) + (\theta_5 - \alpha) \\ (\theta_3 - \alpha) + (\theta_4 - \alpha) \\ (\theta_3 - \alpha) + (\theta_5 - \alpha) \\ (\theta_4 - \alpha) + (\theta_5 - \alpha) \\ \end{pmatrix} \end{equation}$

よって，非心度

λ

$\lambda$ は，

\begin{aligned} λ & = {(θ_{1} - α) + (θ_{2} - α)}^{2} + \dots + {(θ_{4} - α) + (θ_{5} - α)}^{2} \\ = 4 \sum_{i = 1}^{5} (θ_{i} - α)^{2} + 2 \sum_{i = 1}^{4} \sum_{j = i + 1}^{5} (θ_{i} - α) (θ_{j} - α) \end{aligned}

$\begin{align} \lambda &= \{(\theta_1 - \alpha) + (\theta_2 - \alpha)\}^2 + \cdots + \{(\theta_4 - \alpha) + (\theta_5 - \alpha)\}^2 \\ &= 4\sum_{i=1}^{5}(\theta_i - \alpha)^2 + 2\sum_{i=1}^{4}\sum_{j=i+1}^{5}(\theta_i - \alpha)(\theta_j - \alpha) \end{align}$

ここで，

θ_{1} + θ_{2} + θ_{3} + θ_{4} + θ_{5} = 5 α

$\theta_1 + \theta_2 + \theta_3 + \theta_4 + \theta_5 = 5\alpha$ であることに注意すると，

\begin{aligned} 2 \sum_{i = 1}^{4} \sum_{j = i + 1}^{5} (θ_{i} - α) (θ_{j} - α) \\ = \sum_{i = 1}^{5} \sum_{j = 1}^{5} (θ_{i} - α) (θ_{j} - α) - \sum_{i = 1}^{5} (θ_{i} - α)^{2} \\ = \sum_{i = 1}^{5} (θ_{i} - α) \sum_{j = 1}^{5} (θ_{j} - α) - \sum_{i = 1}^{5} (θ_{i} - α)^{2} \\ = \sum_{i = 1}^{5} (θ_{i} - α) {(θ_{1} - α) + (θ_{2} - α) + (θ_{3} - α) + (θ_{4} - α) + (θ_{5} - α)} - \sum_{i = 1}^{5} (θ_{i} - α)^{2} \\ = \sum_{i = 1}^{5} (θ_{i} - α) (θ_{1} + θ_{2} + θ_{3} + θ_{4} + θ_{5} - 5 α) - \sum_{i = 1}^{5} (θ_{i} - α)^{2} \\ = 0 - \sum_{i = 1}^{5} (θ_{i} - α)^{2} \\ = - \sum_{i = 1}^{5} (θ_{i} - α)^{2} \end{aligned}

$\begin{align} &2 \sum_{i=1}^{4}\sum_{j=i+1}^{5}(\theta_i - \alpha)(\theta_j - \alpha) \\ &= \sum_{i=1}^{5}\sum_{j=1}^{5}(\theta_i - \alpha)(\theta_j - \alpha) - \sum_{i=1}^{5}(\theta_i - \alpha)^2 \\ &= \sum_{i=1}^{5}(\theta_i - \alpha)\sum_{j=1}^{5}(\theta_j - \alpha) - \sum_{i=1}^{5}(\theta_i - \alpha)^2 \\ &= \sum_{i=1}^{5}(\theta_i - \alpha)\{(\theta_1 - \alpha) + (\theta_2 - \alpha) + (\theta_3 - \alpha) + (\theta_4 - \alpha) + (\theta_5 - \alpha)\} - \sum_{i=1}^{5}(\theta_i - \alpha)^2 \\ &= \sum_{i=1}^{5}(\theta_i - \alpha)(\theta_1 + \theta_2 + \theta_3 + \theta_4 + \theta_5 - 5\alpha) - \sum_{i=1}^{5}(\theta_i - \alpha)^2 \\ &= 0 - \sum_{i=1}^{5}(\theta_i - \alpha)^2 \\ &= - \sum_{i=1}^{5}(\theta_i - \alpha)^2 \end{align}$

となるなので，非心度

λ

$\lambda$ は，

λ = 3 \sum_{i = 1}^{5} (θ_{i} - α)^{2}

$\begin{equation} \lambda = 3\sum_{i=1}^{5}(\theta_i - \alpha)^2 \end{equation}$

統計検定 1級 過去問 解答/解答例と解説 2014年11月30日 (日) 試験

統計数理 問4 [4]

設問の要約

解答例

統計検定 1級過去問解答/解答例と解説
2014年11月30日 (日) 試験

統計数理問4 [4]