統計検定 1級過去問解答/解答例と解説
2016年11月27日 (日) 試験

統計応用問3 [5]

設問の要約

$E[\hat{C}_P] = a C_P$ と表される． $a$ の値を求めよ．
$Y \sim \chi^2(\nu)$ の確率密度関数
$f_{Y} (y) = \frac{1}{2 Γ (\frac{ν}{2})} {(\frac{y}{2})}^{\frac{ν}{2} - 1} e^{- \frac{y}{2}}$
$\begin{equation} f_Y(y) = \dfrac{1}{2\varGamma\left(\dfrac{\nu}{2}\right)}\left(\frac{y}{2}\right)^{\frac{\nu}{2} - 1}e^{-\frac{y}{2}} \end{equation}$

解答例

全確率 $\ds\int_{0}^{\infty}f_Y(y)dy = 1$ より、

\begin{aligned} \int_{0}^{\infty} \frac{1}{2 Γ (\frac{ν}{2})} {(\frac{y}{2})}^{\frac{ν}{2} - 1} e^{- \frac{y}{2}} d y = 1 \end{aligned}

$\begin{align} \int_{0}^{\infty}\dfrac{1}{2\varGamma\left(\dfrac{\nu}{2}\right)}\left(\frac{y}{2}\right)^{\frac{\nu}{2} - 1}e^{-\frac{y}{2}}dy = 1 \end{align}$

\begin{aligned} \int_{0}^{\infty} {(\frac{y}{2})}^{\frac{ν}{2} - 1} e^{- \frac{y}{2}} d y = 2 Γ (\frac{ν}{2}) \end{aligned}

$\begin{align} \int_{0}^{\infty}\left(\frac{y}{2}\right)^{\frac{\nu}{2} - 1}e^{-\frac{y}{2}}dy = 2\varGamma\left(\dfrac{\nu}{2}\right) \end{align}$

Y = \frac{(n - 1) S^{2}}{σ^{2}} \sim χ^{2} (n - 1)

$Y = \dfrac{(n-1)S^2}{\sigma^2} \sim \chi^2(n-1)$ とおくと、

\begin{aligned} E [{\hat{C}}_{P}] & = E [\frac{S_{U} - S_{L}}{6 S}] \\ = \frac{S_{U} - S_{L}}{6 σ} \cdot \sqrt{n - 1} \cdot E [\frac{σ}{S \sqrt{n - 1}}] \\ = C_{P} \cdot \sqrt{n - 1} \cdot E [{(\frac{σ^{2}}{S^{2} (n - 1)})}^{\frac{1}{2}}] \\ = C_{P} \cdot \sqrt{n - 1} \cdot E [Y^{- \frac{1}{2}}] \\ = C_{P} \cdot \sqrt{n - 1} \cdot \int_{0}^{\infty} y^{- \frac{1}{2}} f_{Y} (y) d y \\ = C_{P} \cdot \sqrt{n - 1} \cdot \int_{0}^{\infty} y^{- \frac{1}{2}} \frac{1}{2 Γ (\frac{n - 1}{2})} {(\frac{y}{2})}^{\frac{n - 1}{2} - 1} e^{- \frac{y}{2}} d y \\ = C_{P} \cdot \sqrt{n - 1} \cdot \frac{1}{2 \sqrt{2} Γ (\frac{n - 1}{2})} \cdot \int_{0}^{\infty} {(\frac{y}{2})}^{- \frac{1}{2}} {(\frac{y}{2})}^{\frac{n - 1}{2} - 1} e^{- \frac{y}{2}} d y \\ = C_{P} \cdot \sqrt{n - 1} \cdot \frac{1}{2 \sqrt{2} Γ (\frac{n - 1}{2})} \cdot \int_{0}^{\infty} {(\frac{y}{2})}^{\frac{n - 2}{2} - 1} e^{- \frac{y}{2}} d y \\ = C_{P} \cdot \sqrt{n - 1} \cdot \frac{1}{2 \sqrt{2} Γ (\frac{n - 1}{2})} \cdot 2 Γ (\frac{n - 2}{2}) \\ = C_{P} \cdot \frac{\sqrt{n - 1} Γ (\frac{n - 2}{2})}{\sqrt{2} Γ (\frac{n - 1}{2})} \end{aligned}

$\begin{align} E[\hat{C}_P] &= E\left[\frac{S_U - S_L}{6S}\right] \\ &= \frac{S_U - S_L}{6\sigma} \cdot \sqrt{n-1} \cdot E\left[\frac{\sigma}{S\sqrt{n-1}}\right] \\ &= C_P \cdot \sqrt{n-1} \cdot E\left[\left(\frac{\sigma^2}{S^2(n-1)}\right)^{\frac{1}{2}}\right] \\ &= C_P \cdot \sqrt{n-1} \cdot E\left[Y^{-\frac{1}{2}}\right] \\ &= C_P \cdot \sqrt{n-1} \cdot \int_{0}^{\infty}y^{-\frac{1}{2}}f_Y(y)dy \\ &= C_P \cdot \sqrt{n-1} \cdot \int_{0}^{\infty}y^{-\frac{1}{2}}\dfrac{1}{2\varGamma\left(\dfrac{n-1}{2}\right)}\left(\frac{y}{2}\right)^{\frac{n-1}{2} - 1}e^{-\frac{y}{2}}dy \\ &= C_P \cdot \sqrt{n-1} \cdot \dfrac{1}{2\sqrt{2}\varGamma\left(\dfrac{n-1}{2}\right)} \cdot \int_{0}^{\infty}\left(\frac{y}{2}\right)^{-\frac{1}{2}}\left(\frac{y}{2}\right)^{\frac{n-1}{2} - 1}e^{-\frac{y}{2}}dy \\ &= C_P \cdot \sqrt{n-1} \cdot \dfrac{1}{2\sqrt{2}\varGamma\left(\dfrac{n-1}{2}\right)} \cdot \int_{0}^{\infty}\left(\frac{y}{2}\right)^{\frac{n-2}{2} - 1}e^{-\frac{y}{2}}dy \\ &= C_P \cdot \sqrt{n-1} \cdot \dfrac{1}{2\sqrt{2}\varGamma\left(\dfrac{n-1}{2}\right)} \cdot 2\varGamma\left(\dfrac{n-2}{2}\right) \\ &= C_P \cdot \dfrac{\sqrt{n-1}\varGamma\left(\dfrac{n-2}{2}\right)}{\sqrt{2}\varGamma\left(\dfrac{n-1}{2}\right)} \\ \end{align}$

よって、

\begin{aligned} a = \frac{\sqrt{n - 1} Γ (\frac{n - 2}{2})}{\sqrt{2} Γ (\frac{n - 1}{2})} \end{aligned}

$\begin{align} a = \dfrac{\sqrt{n-1}\varGamma\left(\dfrac{n-2}{2}\right)}{\sqrt{2}\varGamma\left(\dfrac{n-1}{2}\right)} \end{align}$

統計検定 1級 過去問 解答/解答例と解説 2016年11月27日 (日) 試験

統計応用 問3 [5]

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
2016年11月27日 (日) 試験

統計応用問3 [5]