統計検定 1級過去問解答/解答例と解説
2017年11月26日 (日) 試験

統計応用問1 [3]

設問の要約

尤度方程式から次が導かれることを示せ．
$ψ (α) - \log α = \log \frac{{\tilde{x}}_{n}}{{\bar{x}}_{n}}$
$\begin{equation} \psi(\alpha) - \log \alpha = \log\frac{\tilde{x}_n}{\bar{x}_n} \end{equation}$

$β = \frac{{\bar{x}}_{n}}{α}$
$\begin{equation} \beta = \frac{\bar{x}_n}{\alpha} \end{equation}$
ただし，
${\bar{x}}_{n} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$
$\begin{equation} \bar{x}_n = \frac{1}{n}\sum_{i=1}^n x_i \end{equation}$

${\tilde{x}}_{n} = {(\prod_{i = 1}^{n} x_{i})}^{\frac{1}{n}}$
$\begin{equation} \tilde{x}_n = \left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}} \end{equation}$
$\psi(\alpha)$ : ディガンマ関数 (対数ガンマ関数の1次導関数)

解答例

l (α, β; x) = - n α \log β - n \log Γ (α) + (α - 1) \sum_{i = 1}^{n} \log x_{i} - \frac{1}{β} \sum_{i = 1}^{n} x_{i}

$\begin{equation} l(\alpha, \beta; \bm{x}) = -n\alpha\log\beta -n\log \Gamma(\alpha) + (\alpha - 1)\sum_{i=1}^{n}\log{x_i} - \frac{1}{\beta}\sum_{i=1}^{n}x_i \end{equation}$

l (α, β; x)

$l(\alpha, \beta; \bm{x})$ を

β

$\beta$ で偏微分して 0 とおくと，

\begin{aligned} \frac{\partial}{\partial α} l (α, β; x) & = - n α \frac{1}{β} + \frac{1}{β^{2}} \sum_{i = 1}^{n} x_{i} \\ = - \frac{n}{β} (\frac{1}{β} \frac{1}{n} \sum_{i = 1}^{n} x_{i} - α) \\ = - \frac{n}{β} (\frac{1}{β} {\bar{x}}_{n} - α) \\ = 0 \end{aligned}

$\begin{align} \frac{\partial}{\partial \alpha}l(\alpha, \beta; \bm{x}) &= -n\alpha\frac{1}{\beta} + \frac{1}{\beta^2}\sum_{i=1}^{n}x_i \\ &= -\frac{n}{\beta}\left(\frac{1}{\beta}\frac{1}{n}\sum_{i=1}^{n}x_i - \alpha\right) \\ &= -\frac{n}{\beta}\left(\frac{1}{\beta}\bar{x}_n - \alpha\right) \\ &= 0 \end{align}$

ガンマ関数の定義より，

α > 0

$\alpha > 0$ なので，

\frac{1}{β} {\bar{x}}_{n} = α

$\begin{equation} \frac{1}{\beta}\bar{x}_n = \alpha \end{equation}$

β = \frac{{\bar{x}}_{n}}{α}

$\begin{equation} \beta = \frac{\bar{x}_n}{\alpha} \end{equation}$

l (α, β; x)

$l(\alpha, \beta; \bm{x})$ を

α

$\alpha$ で偏微分して 0 とおくと，

\begin{aligned} \frac{\partial}{\partial α} l (α, β; x) & = - n \log β - n ψ (α) + \sum_{i = 1}^{n} \log x_{i} \\ = - n \log β - n ψ (α) + \log \prod_{i = 1}^{n} x_{i} \\ = - n \log β - n ψ (α) + n \log {(\prod_{i = 1}^{n} x_{i})}^{\frac{1}{n}} \\ = - n \log β - n ψ (α) + n \log {\tilde{x}}_{n} \\ = - n (\log β + ψ (α) - \log {\tilde{x}}_{n}) \\ = 0 \end{aligned}

$\begin{align} \frac{\partial}{\partial \alpha}l(\alpha, \beta; \bm{x}) &= -n\log\beta - n\psi(\alpha) + \sum_{i=1}^{n}\log{x_i} \\ &= -n\log\beta - n\psi(\alpha) + \log\prod_{i=1}^{n}{x_i} \\ &= -n\log\beta - n\psi(\alpha) + n \log\left(\prod_{i=1}^{n}{x_i}\right)^{\frac{1}{n}} \\ &= -n\log\beta - n\psi(\alpha) + n \log\tilde{x}_n \\ &= -n(\log\beta + \psi(\alpha) - \log\tilde{x}_n) \\ &= 0 \end{align}$

\log β + ψ (α) - \log {\tilde{x}}_{n} = 0

$\begin{equation} \log\beta + \psi(\alpha) - \log\tilde{x}_n = 0 \end{equation}$

\begin{aligned} ψ (α) & = \log {\tilde{x}}_{n} - \log β \\ = \log {\tilde{x}}_{n} - \log \frac{{\bar{x}}_{n}}{α} \\ = \log {\tilde{x}}_{n} - (\log {\bar{x}}_{n} - \log α) \end{aligned}

$\begin{align} \psi(\alpha) &= \log\tilde{x}_n - \log\beta \\ &= \log\tilde{x}_n - \log\frac{\bar{x}_n}{\alpha} \\ &= \log\tilde{x}_n - (\log\bar{x}_n - \log{\alpha}) \end{align}$

\begin{aligned} ψ (α) - \log α & = \log {\tilde{x}}_{n} - \log {\bar{x}}_{n} \\ = \log \frac{{\tilde{x}}_{n}}{{\bar{x}}_{n}} \end{aligned}

$\begin{align} \psi(\alpha) - \log{\alpha} &= \log\tilde{x}_n - \log\bar{x}_n \\ &= \log\frac{\tilde{x}_n}{\bar{x}_n} \end{align}$

統計検定 1級 過去問 解答/解答例と解説 2017年11月26日 (日) 試験

統計応用 問1 [3]

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
2017年11月26日 (日) 試験

統計応用問1 [3]