目次
- はじめに
- ex5.2.1
- (i)二項分布の十分統計量
- (ii)幾何分布の十分統計量
- (iii)負の二項分布の十分統計量
- (iv)ガンマ分布の十分統計量(パラメータ$\alpha$:既知 , $\beta$:未知)
- (v)ガンマ分布の十分統計量(パラメータ$\alpha$:未知 , $\beta$:既知)
- (vi)ガンマ分布の十分統計量(パラメータ$\alpha$:未知 , $\beta$:未知)
- (vii)ベータ分布の十分統計量(パラメータ$\alpha$:未知 , $\beta$:既知)
- (viii)ベータ分布の十分統計量(パラメータ$\alpha$:既知 , $\beta$:未知)
- (ix)ベータ分布の十分統計量(パラメータ$\alpha$:未知 , $\beta$:未知)
- (x)一様分布の十分統計量(下限:既知, 上限:未知)
- (xi)一様分布の十分統計量(下限:未知, 上限:未知)
- (xii)一様分布の十分統計量(平均:未知 , 区間の幅:既知)
はじめに
「入門・演習 数理統計」の演習問題の自作解答を紹介します。
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スポンサーリンク
$\newcommand{\lnl}{\\[8pt]}$ $\newcommand{\Lnl}{\\[18pt]}$ $\newcommand{\delt}{\mathrm{d}}$ $\newcommand{\comb}{\mathrm{C}}$ $\DeclareMathOperator*{\ssum}{\Sigma}$ $\DeclareMathOperator*{\sprod}{\Pi}$
ex5.2.1
(i)二項分布の十分統計量
各$X_i$の確率関数は,
\begin{align}
P(X_i = x) = \begin{cases}\displaystyle \binom{m}{x}p^x(1-p)^{m-x}&x=0,1,\cdots\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
P(X_i = x) = \begin{cases}\displaystyle \binom{m}{x}p^x(1-p)^{m-x}&x=0,1,\cdots\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
である.
$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $, $T=t=x_1+x_2+\cdots+x_n$となる確率は
\begin{align}
P(\bm{X}=\bm{x}) &= P(X_1=x_1)\cdot P(X_2=x_2) \cdots P(X_n=x_n)\lnl
&=\prod_{i=1}^n \binom{m}{x_i}p^{x_i}(1-p)^{m-x_i}\lnl
&=p^{x_1+x_2+\cdots+x_n}(1-p)^{(m-x_1)+(m-x_2)+\cdots+(m-x_n)} \prod_{i=1}^n\binom{m}{x_i}\lnl
&=p^t(1-p)^{nm-t} \prod_{i=1}^n\binom{m}{x_i}
\end{align}
P(\bm{X}=\bm{x}) &= P(X_1=x_1)\cdot P(X_2=x_2) \cdots P(X_n=x_n)\lnl
&=\prod_{i=1}^n \binom{m}{x_i}p^{x_i}(1-p)^{m-x_i}\lnl
&=p^{x_1+x_2+\cdots+x_n}(1-p)^{(m-x_1)+(m-x_2)+\cdots+(m-x_n)} \prod_{i=1}^n\binom{m}{x_i}\lnl
&=p^t(1-p)^{nm-t} \prod_{i=1}^n\binom{m}{x_i}
\end{align}
また, $T$は二項分布の再生性より$\mathrm{B}(nm,p)$に従うから,
\begin{align}
P(T=t) = \binom{nm}{t}p^t(1-p)^{nm-t}
\end{align}
P(T=t) = \binom{nm}{t}p^t(1-p)^{nm-t}
\end{align}
となる.従って$T=t$が与えられたときの$\bm{X}=\bm{x}$の条件付き確率は
\begin{align}
P(\bm{X}=\bm{x}|T=t) &= \frac{P(\bm{X} = \bm{x})}{P(T=t)}\lnl
&=\frac{\displaystyle p^t(1-p)^{nm-t} \prod_{i=1}^n\binom{m}{x_i}}{\displaystyle \binom{nm}{t}p^t(1-p)^{nm-t}} \lnl
&=\left\{\prod_{i=1}^n\binom{m}{x_i} \right\}\bigg/ \binom{nm}{t}
\end{align}
P(\bm{X}=\bm{x}|T=t) &= \frac{P(\bm{X} = \bm{x})}{P(T=t)}\lnl
&=\frac{\displaystyle p^t(1-p)^{nm-t} \prod_{i=1}^n\binom{m}{x_i}}{\displaystyle \binom{nm}{t}p^t(1-p)^{nm-t}} \lnl
&=\left\{\prod_{i=1}^n\binom{m}{x_i} \right\}\bigg/ \binom{nm}{t}
\end{align}
これはパラメータ$p$に依存しないので$T=\displaystyle \sum_{i=1}^n X_i$は$p$の十分統計量である.
(ii)幾何分布の十分統計量
各$X_i$の確率関数は,
\begin{align}
P(X_i = x) = \begin{cases}p(1-p)^x &x=0,1,\cdots\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
P(X_i = x) = \begin{cases}p(1-p)^x &x=0,1,\cdots\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
である.
$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $, $T=t=x_1+x_2+\cdots+x_n$となる確率は
\begin{align}
P(\bm{X}=\bm{x}) &= P(X_1=x_1)\cdot P(X_2=x_2) \cdots P(X_n=x_n)\lnl
&=\prod_{i=1}^n p(1-p)^{x_i}\lnl
&=p^n(1-p)^{x_1+x_2+\cdots+x_n}\lnl
&=p^n(1-p)^{t}
\end{align}
P(\bm{X}=\bm{x}) &= P(X_1=x_1)\cdot P(X_2=x_2) \cdots P(X_n=x_n)\lnl
&=\prod_{i=1}^n p(1-p)^{x_i}\lnl
&=p^n(1-p)^{x_1+x_2+\cdots+x_n}\lnl
&=p^n(1-p)^{t}
\end{align}
また, $T$はテキスト(3.4.6)よりパラメータ$n,p$の負の二項分布に従うから,
\begin{align}
P(T=t) = \binom{n+t-1}{t}p^n(1-p)^{t}
\end{align}
P(T=t) = \binom{n+t-1}{t}p^n(1-p)^{t}
\end{align}
となる.従って$T=t$が与えられたときの$\bm{X}=\bm{x}$の条件付き確率は
\begin{align}
P(\bm{X}=\bm{x}|T=t) &= \frac{P(\bm{X} = \bm{x})}{P(T=t)}\lnl
&=\frac{\displaystyle p^n(1-p)^t}{\displaystyle \binom{n+t-1}{t}p^n(1-p)^{t}} \lnl
&=\binom{n+t-1}{t}^{-1}
\end{align}
P(\bm{X}=\bm{x}|T=t) &= \frac{P(\bm{X} = \bm{x})}{P(T=t)}\lnl
&=\frac{\displaystyle p^n(1-p)^t}{\displaystyle \binom{n+t-1}{t}p^n(1-p)^{t}} \lnl
&=\binom{n+t-1}{t}^{-1}
\end{align}
これはパラメータ$p$に依存しないので$T=\displaystyle \sum_{i=1}^n X_i$は$p$の十分統計量である.
(iii)負の二項分布の十分統計量
各$X_i$の確率関数は,
\begin{align}
P(X_i = x) = \begin{cases}\displaystyle \binom{r+x-1}{x}p^r(1-p)^x &x=0,1,\cdots\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
P(X_i = x) = \begin{cases}\displaystyle \binom{r+x-1}{x}p^r(1-p)^x &x=0,1,\cdots\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
である.
$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $, $T=t=x_1+x_2+\cdots+x_n$となる確率は
\begin{align}
P(\bm{X}=\bm{x}) &= P(X_1=x_1)\cdot P(X_2=x_2) \cdots P(X_n=x_n)\lnl
&=\prod_{i=1}^n \binom{r+x-1}{x}p^r(1-p)^x\lnl
&=p^{nr}(1-p)^{x_1+x_2+\cdots+x_n} \prod_{i=1}^n \binom{r+x-1}{x}\lnl
&=p^{nr}(1-p)^{t}\prod_{i=1}^n \binom{r+x-1}{x}
\end{align}
P(\bm{X}=\bm{x}) &= P(X_1=x_1)\cdot P(X_2=x_2) \cdots P(X_n=x_n)\lnl
&=\prod_{i=1}^n \binom{r+x-1}{x}p^r(1-p)^x\lnl
&=p^{nr}(1-p)^{x_1+x_2+\cdots+x_n} \prod_{i=1}^n \binom{r+x-1}{x}\lnl
&=p^{nr}(1-p)^{t}\prod_{i=1}^n \binom{r+x-1}{x}
\end{align}
また, $T$は負の二項分布の再生性より$\mathrm{NB}(nr,p)$に従うから,
\begin{align}
P(T=t) = \binom{nr+t-1}{t}p^{nr}(1-p)^{t}
\end{align}
P(T=t) = \binom{nr+t-1}{t}p^{nr}(1-p)^{t}
\end{align}
となる.従って$T=t$が与えられたときの$\bm{X}=\bm{x}$の条件付き確率は
\begin{align}
P(\bm{X}=\bm{x}|T=t) &= \frac{P(\bm{X} = \bm{x})}{P(T=t)}\lnl
&=\frac{\displaystyle p^{nr}(1-p)^{t}\prod_{i=1}^n \binom{r+x-1}{x}}{\displaystyle \binom{nr+t-1}{t}p^{nr}(1-p)^{t}} \lnl
&=\left(\prod_{i=1}^n \binom{r+x-1}{x}\right) \bigg/ \binom{nr+t-1}{t}
\end{align}
P(\bm{X}=\bm{x}|T=t) &= \frac{P(\bm{X} = \bm{x})}{P(T=t)}\lnl
&=\frac{\displaystyle p^{nr}(1-p)^{t}\prod_{i=1}^n \binom{r+x-1}{x}}{\displaystyle \binom{nr+t-1}{t}p^{nr}(1-p)^{t}} \lnl
&=\left(\prod_{i=1}^n \binom{r+x-1}{x}\right) \bigg/ \binom{nr+t-1}{t}
\end{align}
これはパラメータ$p$に依存しないので$T=\displaystyle \sum_{i=1}^n X_i$は$p$の十分統計量である.
(iv)ガンマ分布の十分統計量(パラメータ$\alpha$:既知 , $\beta$:未知)
各$X_i$の確率密度関数は,
\begin{align}
f(x) = \begin{cases}\displaystyle \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} &x>0\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
f(x) = \begin{cases}\displaystyle \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} &x>0\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
である.
$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $の結合確率密度関数$f_X(\bm{x})$は, $T=t=x_1+x_2+\cdots+x_n$としたとき,
\begin{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{\beta^\alpha}{\Gamma(\alpha)}{x_i}^{\alpha-1}e^{-\beta x_i} \lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} e^{-\beta(x_1+x_2+\cdots+x_n)}\lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} e^{-\beta t}\lnl
\end{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{\beta^\alpha}{\Gamma(\alpha)}{x_i}^{\alpha-1}e^{-\beta x_i} \lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} e^{-\beta(x_1+x_2+\cdots+x_n)}\lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} e^{-\beta t}\lnl
\end{align}
ここで,
\begin{align}
h(\bm{x}) = \frac{1}{\Gamma(\alpha)^n}\left(\prod_{i=1}^n x_i\right)^{\alpha-1} ,\quad g(t,\beta) = \beta^{n\alpha} e^{-\beta t}
\end{align}
h(\bm{x}) = \frac{1}{\Gamma(\alpha)^n}\left(\prod_{i=1}^n x_i\right)^{\alpha-1} ,\quad g(t,\beta) = \beta^{n\alpha} e^{-\beta t}
\end{align}
とおけば$f(\bm{x};\beta) = h(\bm{x})g(t,\beta)$と表せるので因子分解定理より$T=\displaystyle \sum_{i=1}^n X_i$は$\beta$の十分統計量となる.
(v)ガンマ分布の十分統計量(パラメータ$\alpha$:未知 , $\beta$:既知)
各$X_i$の確率密度関数は,
\begin{align}
f(x) = \begin{cases}\displaystyle \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} &x>0\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
f(x) = \begin{cases}\displaystyle \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} &x>0\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
である.
$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $の結合確率密度関数$f_X(\bm{x})$は, $T=t=x_1\cdot x_2 \cdots x_n$としたとき,
\begin{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{\beta^\alpha}{\Gamma(\alpha)}{x_i}^{\alpha-1}e^{-\beta x_i} \lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} e^{-\beta(x_1+x_2+\cdots+x_n)}\lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} t^{\alpha-1} \exp\left(-\beta \sum_{i=1}^n x_i \right)\lnl
\end{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{\beta^\alpha}{\Gamma(\alpha)}{x_i}^{\alpha-1}e^{-\beta x_i} \lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} e^{-\beta(x_1+x_2+\cdots+x_n)}\lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} t^{\alpha-1} \exp\left(-\beta \sum_{i=1}^n x_i \right)\lnl
\end{align}
ここで,
\begin{align}
h(\bm{x}) = \exp\left(-\beta \sum_{i=1}^n x_i \right) ,\quad g(t,\alpha) = \frac{\beta^{n\alpha}}{\Gamma(\alpha)^n}t^{\alpha-1}
\end{align}
h(\bm{x}) = \exp\left(-\beta \sum_{i=1}^n x_i \right) ,\quad g(t,\alpha) = \frac{\beta^{n\alpha}}{\Gamma(\alpha)^n}t^{\alpha-1}
\end{align}
とおけば$f(\bm{x};\alpha) = h(\bm{x})g(t,\alpha)$と表せるので因子分解定理より$T=\displaystyle \prod_{i=1}^n X_i$は$\alpha$の十分統計量となる.
(vi)ガンマ分布の十分統計量(パラメータ$\alpha$:未知 , $\beta$:未知)
各$X_i$の確率密度関数は,
\begin{align}
f(x) = \begin{cases}\displaystyle \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} &x>0\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
f(x) = \begin{cases}\displaystyle \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} &x>0\lnl
\quad 0&\text{その他}
\end{cases}
\end{align}
である.
\begin{align}
T_1 = t_1 = \prod_{i=1}^n x_i ,\quad T_2 = t_2 = \sum_{i=1}^n x_i
\end{align}
T_1 = t_1 = \prod_{i=1}^n x_i ,\quad T_2 = t_2 = \sum_{i=1}^n x_i
\end{align}
としたとき,$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $の結合確率密度関数$f_X(\bm{x})$は,
\begin{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{\beta^\alpha}{\Gamma(\alpha)}{x_i}^{\alpha-1}e^{-\beta x_i} \lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} e^{-\beta(x_1+x_2+\cdots+x_n)}\lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} t_1^{\alpha-1} \exp\left(-\beta t_2 \right)\lnl
\end{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{\beta^\alpha}{\Gamma(\alpha)}{x_i}^{\alpha-1}e^{-\beta x_i} \lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} e^{-\beta(x_1+x_2+\cdots+x_n)}\lnl
&=\frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} t_1^{\alpha-1} \exp\left(-\beta t_2 \right)\lnl
\end{align}
ここで,
\begin{align}
h(\bm{x}) = 1 ,\quad g((t_1,t_2),\alpha,\beta) = \frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} t_1^{\alpha-1} \exp\left(-\beta t_2 \right)
\end{align}
h(\bm{x}) = 1 ,\quad g((t_1,t_2),\alpha,\beta) = \frac{\beta^{n\alpha}}{\Gamma(\alpha)^n} t_1^{\alpha-1} \exp\left(-\beta t_2 \right)
\end{align}
とおけば$f(\bm{x};\alpha,\beta) = h(\bm{x})g((t_1,t_2),\alpha,\beta)$と表せるので因子分解定理より$(T_1,T_2)=\displaystyle\left(\prod_{i=1}^n X_i,\sum_{i=1}^n X_i\right)$は$\alpha,\beta$の結合十分統計量となる.
(vii)ベータ分布の十分統計量(パラメータ$\alpha$:未知 , $\beta$:既知)
各$X_i$の確率密度関数は,
\begin{align}
f(x) = \begin{cases}\displaystyle \frac{1}{B(\alpha,\beta)} x^{\alpha-1}(1-x)^{\beta-1} &0 < x < 1\lnl \quad 0&\text{その他} \end{cases}\end{align}
である.
$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $の結合確率密度関数$f_X(\bm{x})$は, $T=t=x_1\cdot x_2\cdots x_n$としたとき,
f(x) = \begin{cases}\displaystyle \frac{1}{B(\alpha,\beta)} x^{\alpha-1}(1-x)^{\beta-1} &0 < x < 1\lnl \quad 0&\text{その他} \end{cases}\end{align}
\begin{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{1}{B(\alpha,\beta)} {x_i}^{\alpha-1}(1-x_i)^{\beta-1} \lnl
&=\frac{1}{B(\alpha,\beta)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} \left(\prod_{i=1}^n (1-x_i)\right)^{\beta-1}\lnl
&=\frac{1}{B(\alpha,\beta)^n} t^{\alpha-1} \left(\prod_{i=1}^n (1-x_i)\right)^{\beta-1}
\end{align}
ここで,
\begin{align}
h(\bm{x}) = \left(\prod_{i=1}^n (1-x_i)\right)^{\beta-1} ,\quad g(t,\alpha) = \frac{1}{B(\alpha,\beta)^n} t^{\alpha-1}
\end{align}
とおけば$f(\bm{x};\alpha) = h(\bm{x})g(t,\alpha)$と表せるので因子分解定理より$T=\displaystyle \prod_{i=1}^n X_i$は$\alpha$の十分統計量となる.
(viii)ベータ分布の十分統計量(パラメータ$\alpha$:既知 , $\beta$:未知)
各$X_i$の確率密度関数は,
\begin{align}
f(x) = \begin{cases}\displaystyle \frac{1}{B(\alpha,\beta)} x^{\alpha-1}(1-x)^{\beta-1} &0 < x < 1\lnl \quad 0&\text{その他} \end{cases} \end{align}
である.
$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $の結合確率密度関数$f_X(\bm{x})$は, $T=t=(1-x_1)(1-x_2)\cdots(1-x_n)$としたとき,
f(x) = \begin{cases}\displaystyle \frac{1}{B(\alpha,\beta)} x^{\alpha-1}(1-x)^{\beta-1} &0 < x < 1\lnl \quad 0&\text{その他} \end{cases} \end{align}
\begin{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{1}{B(\alpha,\beta)} {x_i}^{\alpha-1}(1-x_i)^{\beta-1} \lnl
&=\frac{1}{B(\alpha,\beta)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} \left(\prod_{i=1}^n (1-x_i)\right)^{\beta-1}\lnl
&=\frac{1}{B(\alpha,\beta)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} t^{\beta-1}
\end{align}
ここで,
\begin{align}
h(\bm{x}) = \left(\prod_{i=1}^n x_i\right)^{\alpha-1} ,\quad g(t,\beta) = \frac{1}{B(\alpha,\beta)^n} t^{\beta-1}
\end{align}
とおけば$f(\bm{x};\beta) = h(\bm{x})g(t,\beta)$と表せるので因子分解定理より$T=\displaystyle \prod_{i=1}^n (1-X_i)$は$\beta$の十分統計量となる.
(ix)ベータ分布の十分統計量(パラメータ$\alpha$:未知 , $\beta$:未知)
各$X_i$の確率密度関数は,
\begin{align}
f(x) = \begin{cases}\displaystyle \frac{1}{B(\alpha,\beta)} x^{\alpha-1}(1-x)^{\beta-1} &0 < x < 1\lnl \quad 0&\text{その他} \end{cases} \end{align}
である.
f(x) = \begin{cases}\displaystyle \frac{1}{B(\alpha,\beta)} x^{\alpha-1}(1-x)^{\beta-1} &0 < x < 1\lnl \quad 0&\text{その他} \end{cases} \end{align}
\begin{align}
T_1 = t_1= \prod_{i=1}^n x_i ,\quad T_2 = t_2= \prod_{i=1}^n (1-x_i)
\end{align}
$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $の結合確率密度関数$f_X(\bm{x})$は,
\begin{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{1}{B(\alpha,\beta)} {x_i}^{\alpha-1}(1-x_i)^{\beta-1} \lnl
&=\frac{1}{B(\alpha,\beta)^n} \left(\prod_{i=1}^n x_i\right)^{\alpha-1} \left(\prod_{i=1}^n (1-x_i)\right)^{\beta-1}\lnl
&=\frac{1}{B(\alpha,\beta)^n} {t_1}^{\alpha-1} {t_2}^{\beta-1}
\end{align}
ここで,
\begin{align}
h(\bm{x}) = 1 ,\quad g((t_1,t_2),\alpha,\beta) = \frac{1}{B(\alpha,\beta)^n} {t_1}^{\alpha-1} {t_2}^{\beta-1}
\end{align}
とおけば$f(\bm{x};\alpha,\beta) = h(\bm{x})g((t_1,t_2),\alpha,\beta)$と表せるので因子分解定理より$(T_1,T_2)=\displaystyle\left(\prod_{i=1}^n X_i,\prod_{i=1}^n (1-X_i)\right)$は$\alpha,\beta$の結合十分統計量となる.
(x)一様分布の十分統計量(下限:既知, 上限:未知)
$\varphi(x)$を,
\begin{align}
\varphi(x) = \begin{cases}0 & x < 0\\ 1& x\ge 1\end{cases} \end{align}
と定義する.下限が既知($0$)なので,各$X_i$の確率密度関数は,
\varphi(x) = \begin{cases}0 & x < 0\\ 1& x\ge 1\end{cases} \end{align}
\begin{align}
f(x) = \begin{cases}\displaystyle \frac{1}{\theta} &0\le x \le \theta\lnl
0&\text{その他}
\end{cases} = \frac{\varphi(\theta - x)}{\theta}
\end{align}
である.
$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $の結合確率密度関数$f_X(\bm{x})$は, $\displaystyle T=t=\max_{i} x_i$とすると,
\begin{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{\varphi(\theta - x_i)}{\theta} \lnl
&=\frac{1}{\theta^n} \varphi(\theta - \max_{i}x_i) \lnl
&=\frac{1}{\theta^n} \varphi(\theta - t)
\end{align}
ここで,
\begin{align}
h(\bm{x}) = 1 ,\quad g(t,\theta) = \frac{1}{\theta^n} \varphi(\theta - t)
\end{align}
とおけば$f(\bm{x};\theta) = h(\bm{x})g(t,\theta)$と表せるので因子分解定理より$T=\displaystyle \max_{i} X_i$は$\theta$の十分統計量となる.
(xi)一様分布の十分統計量(下限:未知, 上限:未知)
$\varphi(x)$は(x)と同様に定義する.
各$X_i$の確率密度関数は,
\begin{align}
f(x) &= \begin{cases}\displaystyle \frac{1}{\theta_2 – \theta_1} &\theta_1 \le x \le \theta_2\lnl
0&\text{その他}
\end{cases}\lnl
&= \frac{1}{\theta_2-\theta_1}\varphi(\theta_2 – x)\varphi(x-\theta_1)
\end{align}
f(x) &= \begin{cases}\displaystyle \frac{1}{\theta_2 – \theta_1} &\theta_1 \le x \le \theta_2\lnl
0&\text{その他}
\end{cases}\lnl
&= \frac{1}{\theta_2-\theta_1}\varphi(\theta_2 – x)\varphi(x-\theta_1)
\end{align}
である.
\begin{align}
T_1 = t_1 = \min_{i} x_i , &\quad T_2 = t_2 = \max_{i} x_i
\end{align}
T_1 = t_1 = \min_{i} x_i , &\quad T_2 = t_2 = \max_{i} x_i
\end{align}
とすると,$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $の結合確率密度関数$f_X(\bm{x})$は,
\begin{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{1}{\theta_2-\theta_1}\varphi(\theta_2 – x_i)\varphi(x_i-\theta_1) \lnl
&=\frac{1}{(\theta_2-\theta_1)^n} \varphi(\min_{i}x_i-\theta_1)\varphi(\theta_2 – \max_{i}x_i) \lnl
&=\frac{1}{(\theta_2-\theta_1)^n} \varphi(t_1 – \theta_1) \varphi(\theta_2 – t_2)
\end{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \frac{1}{\theta_2-\theta_1}\varphi(\theta_2 – x_i)\varphi(x_i-\theta_1) \lnl
&=\frac{1}{(\theta_2-\theta_1)^n} \varphi(\min_{i}x_i-\theta_1)\varphi(\theta_2 – \max_{i}x_i) \lnl
&=\frac{1}{(\theta_2-\theta_1)^n} \varphi(t_1 – \theta_1) \varphi(\theta_2 – t_2)
\end{align}
ここで,
\begin{align}
h(\bm{x}) = 1 ,\quad g((t_1,t_2),\theta_1,\theta_2) = \frac{1}{(\theta_2-\theta_1)^n} \varphi(t_1 – \theta_1) \varphi(\theta_2 – t_2)
\end{align}
h(\bm{x}) = 1 ,\quad g((t_1,t_2),\theta_1,\theta_2) = \frac{1}{(\theta_2-\theta_1)^n} \varphi(t_1 – \theta_1) \varphi(\theta_2 – t_2)
\end{align}
とおけば$f(\bm{x};\theta_1,\theta_2) = h(\bm{x})g((t_1,t_2),\theta_1,\theta_2)$と表せるので因子分解定理より$(T_1,T_2)$は$\theta_1,\theta_2$の結合十分統計量となる.
(xii)一様分布の十分統計量(平均:未知 , 区間の幅:既知)
$\varphi(x)$は(x)と同様に定義する.
各$X_i$の確率密度関数は,
\begin{align}
f(x) &= \begin{cases}\displaystyle 1 &\theta-0.5 \le x \le \theta+0.5\lnl
0&\text{その他}
\end{cases}\lnl
&= \varphi(x-\theta-0.5)\varphi(\theta-0.5-x)
\end{align}
f(x) &= \begin{cases}\displaystyle 1 &\theta-0.5 \le x \le \theta+0.5\lnl
0&\text{その他}
\end{cases}\lnl
&= \varphi(x-\theta-0.5)\varphi(\theta-0.5-x)
\end{align}
である.
\begin{align}
T_1 = t_1 = \min_{i} x_i , &\quad T_2 = t_2 = \max_{i} x_i
\end{align}
T_1 = t_1 = \min_{i} x_i , &\quad T_2 = t_2 = \max_{i} x_i
\end{align}
とすると,$\bm{X} = \bm{x} = (x_1,x_2,\cdots,x_n) $の結合確率密度関数$f_X(\bm{x})$は,
\begin{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \varphi(x_i-\theta-0.5)\varphi(\theta-0.5-x_i)\lnl
&=\varphi(\min_{i}x_i-\theta-0.5)\varphi(\theta-0.5-\max_{i}x_i) \lnl
&=\varphi(t_1-\theta-0.5)\varphi(\theta-0.5-t_2)
\end{align}
f_{X}(\bm{x}) &= f(x_1)\cdot f(x_2) \cdots f(x_n)\lnl
&=\prod_{i=1}^n \varphi(x_i-\theta-0.5)\varphi(\theta-0.5-x_i)\lnl
&=\varphi(\min_{i}x_i-\theta-0.5)\varphi(\theta-0.5-\max_{i}x_i) \lnl
&=\varphi(t_1-\theta-0.5)\varphi(\theta-0.5-t_2)
\end{align}
ここで,
\begin{align}
h(\bm{x}) = 1 ,\quad g((t_1,t_2),\theta) = \varphi(t_1-\theta-0.5)\varphi(\theta-0.5-t_2)
\end{align}
h(\bm{x}) = 1 ,\quad g((t_1,t_2),\theta) = \varphi(t_1-\theta-0.5)\varphi(\theta-0.5-t_2)
\end{align}
とおけば$f(\bm{x};\theta) = h(\bm{x})g(t,\theta)$と表せるので因子分解定理より$(T_1,T_2)$は$\theta$の結合十分統計量となる.