ex7.B.2 フィッシャーの直接法

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入門・演習数理統計 [ 野田一雄 ]
価格:3780円(税込、送料無料) (2018/4/3時点)



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ex7.B.2

分割表の$(i,j)$セルの値をそれぞれ$N_{ij}$とし, $\bm{p} = (p_{11},p_{12},p_{21},p_{22})$と表すことにする.また, $n$を合計とする.

また, $p_{+1}$のように添字に$+$が付くものは見やすさのために$p_{*1}$のように$*$に置き換えて記述する.

(i)

$\bm{N} = (N_{11},N_{12},N_{21},N_{22})$は多項分布に従うから, $H_0$の下での確率密度関数$f$は,

\begin{align}
f(\bm{N};\bm{p}) &= \frac{n!}{n_{11}!n_{12}!n_{21}!n_{22}!}\cdot \underline{{p_{11}}^{n_{11}}{p_{12}}^{n_{12}}{p_{21}}^{n_{21}}{p_{22}}^{n_{22}}} \label{eq-7b2-1}
\end{align}

となる.ここで帰無仮説より $p_{ij} = p_{i*}p_{*j}$だから, 下線部は
\begin{align}
{p_{11}}^{n_{11}}&{p_{12}}^{n_{12}}{p_{21}}^{n_{21}}{p_{22}}^{n_{22}} \\
&= {\left(p_{1*}p_{*1}\right)}^{n_{11}}{\left(p_{1*}p_{*2}\right)}^{n_{12}}{\left(p_{2*}p_{*1}\right)}^{n_{21}}{\left(p_{2*}p_{*2}\right)}^{n_{22}}\\
&= p_{1*}^{(n_{11}+n_{12})}p_{2*}^{(n_{21}+n_{22})}p_{*1}^{(n_{11}+n_{21})}p_{*2}^{(n_{12}+n_{22})}\\
&=p_{1*}^{\ n_{1*}}p_{2*}^{\ n_{2*}}p_{*1}^{\ n_{*1}}p_{*2}^{\ n_{*2}}
\end{align}

となる. 従って$\eqref{eq-7b2-1}$は,
\begin{align}
&h(\bm{N}) = \frac{n!}{n_{11}!n_{12}!n_{21}!n_{22}!}\lnl
&T(\bm{N}) = (N_{11}+N_{12},N_{21}+N_{22},N_{11}+N_{21},N_{12}+N_{22}) = (N_{1*},N_{2*},N_{*1},N_{*2})\lnl
&g(T(\bm{N}),\bm{p}) = p_{1*}^{\ n_{1*}}p_{2*}^{\ n_{2*}}p_{*1}^{\ n_{*1}}p_{*2}^{\ n_{*2}}
\end{align}

とおけば因子分解定理により, $T(\bm{N})=(N_{1*},N_{2*},N_{*1},N_{*2})$が$\bm{p}=(p_{1*}, p_{2*}, p_{*1}, p_{*2})$の十分統計量であることがわかる.

(ii)

条件付き確率の定義より,

\begin{align}
P(N_{11}=&n_{11},\cdots,N_{22}=n_{22}|N_{1*}=n_{1*},\cdots,N_{*2}=n_{*2})\lnl
&= \frac{P(N_{11}=n_{11},\cdots,N_{22}=n_{22} , N_{1*}=n_{1*},\cdots,N_{*2}=n_{*2})}{P(N_{1*}=n_{1*},\cdots,N_{*2}=n_{*2})}\label{eq-7b2-7}
\end{align}

である.分子は$\eqref{eq-7b2-1}$の$f(\bm{N},\bm{p})$に他ならない.分母を$f_2$とおくことにする.

$N_{1*}=n_{1*},\cdots,N_{*2}=n_{*2}$を満たす全ての非負整数の組$(n_{11},\cdots,n_{22})$の集合を$A$とする.つまり,

\begin{align}
A = \Big\{(n_{11},n_{12},n_{21},n_{22}) \in \mathbb{Z_{+}}^4; n_{11}+n_{12}=n_{1*},\cdots , n_{12}+n_{22} = n_{*2} \Big \}
\end{align}

$f_2(n_{1*},\cdots,n_{*2})$は$A$の全ての元に対する確率の和である. 帰無仮説$H_0$の下,
\begin{align}
f_2(n_{1*},\cdots,n_{*2}) &= \sum_{(n_{11},\cdots,n_{22})\in A}f\big(n_{11},\cdots,n_{22};\bm{p}\big) \lnl
&=\sum_{(n_{11},\cdots,n_{22})\in A} \frac{n!}{n_{11}!n_{12}!n_{21}!n_{22}!} p_{1*}^{\ n_{1*}}p_{2*}^{\ n_{2*}}p_{*1}^{\ n_{*1}}p_{*2}^{\ n_{*2}} \label{eq-7b2-5}
\end{align}

となる. ここで,
\begin{align}
\sum_{(n_{11},\cdots,n_{22})\in A} \frac{1}{n_{11}!n_{12}!n_{21}!n_{22}!}
\end{align}

を$n_{1*},\cdots,n_{*2}$を用いて表すことを考える. そのために, 変数$x_1,x_2$に対して次の恒等式を考える.
\begin{align}
(x_1+x_2)^n = (x_1+x_2)^{n_{*1}}(x_1+x_2)^{n_{*2}} \label{eq-7b2-8}
\end{align}

両辺の${x_1}^{n_{1*}}{x_2}^{n_{2*}}$の係数を比べる. $\eqref{eq-7b2-8}$の左辺は二項定理より,
\begin{align}
\binom{n}{n_{1*}} = \frac{n!}{n_{1*}!(n-n_{1*})!} = \frac{n!}{n_{1*}!n_{2*}!} \label{eq-7b2-2}
\end{align}

となる.$\eqref{eq-7b2-8}$の右辺の2つの式の乗数は各列の列和となっている. ${x_1}^{n_{1*}}{x_2}^{n_{2*}}$の係数は, その条件の下で$x_1,x_2$の乗数が行和になるようにしたときの総数である. つまり,${x_1}^{n_{1*}}{x_2}^{n_{2*}}$の係数は全ての$(n_{11},\cdots,n_{22})\in A$を動くことになるから,
\begin{align}
\sum_{(n_{11},\cdots,n_{22})\in A}\binom{n_{*1}}{n_{11}}\cdot \binom{n_{*2}}{n_{12}} = \sum_{(n_{11},\cdots,n_{22})\in A} \frac{n_{*1}!}{n_{11}!n_{21}!}\frac{n_{*2}!}{n_{12}!n_{22}!}\label{eq-7b2-3}
\end{align}

となる. $\eqref{eq-7b2-2}, \eqref{eq-7b2-3}$が等しいので,
\begin{align}
&\sum_{(n_{11},\cdots,n_{22})\in A} \frac{n_{*1}!}{n_{11}!n_{21}!}\frac{n_{*2}!}{n_{12}!n_{22}!} = \frac{n!}{n_{1*}!n_{2*}!}\lnl
\Longleftrightarrow &\sum_{(n_{11},\cdots,n_{22})\in A} \frac{1}{n_{11}!n_{12}!n_{21}!n_{22}!} = \frac{n!}{n_{1*}!n_{2*}!n_{*1}!n_{*2}!}\label{eq-7b2-4}
\end{align}

$\eqref{eq-7b2-5}$に$\eqref{eq-7b2-4}$を代入して,
\begin{align}
f_2(n_{1*},\cdots,n_{*2}) &= \sum_{(n_{11},\cdots,n_{22})\in A} \frac{n!}{n_{11}!n_{12}!n_{21}!n_{22}!} p_{1*}^{\ n_{1*}}p_{2*}^{\ n_{2*}}p_{*1}^{\ n_{*1}}p_{*2}^{\ n_{*2}} \lnl
&= \frac{(n!)^2}{n_{1*}!n_{2*}!n_{*1}!n_{*2}!} p_{1*}^{\ n_{1*}}p_{2*}^{\ n_{2*}}p_{*1}^{\ n_{*1}}p_{*2}^{\ n_{*2}}\label{eq-7b2-6}
\end{align}

となる.$\eqref{eq-7b2-7}$に$\eqref{eq-7b2-1}$と$\eqref{eq-7b2-6}$を代入して,
\begin{align}
P(N_{11}=&n_{11},\cdots,N_{22}=n_{22}|N_{1*}=n_{1*},\cdots,N_{*2}=n_{*2})\lnl
&= \frac{f(\bm{N},\bm{p})}{f_2(n_{1*},\cdots,n_{*2})}\lnl
&= \frac{\cfrac{n!}{n_{11}!n_{12}!n_{21}!n_{22}!}p_{1*}^{\ n_{1*}}p_{2*}^{\ n_{2*}}p_{*1}^{\ n_{*1}}p_{*2}^{\ n_{*2}}} {\cfrac{(n!)^2}{n_{1*}!n_{2*}!n_{*1}!n_{*2}!} p_{1*}^{\ n_{1*}}p_{2*}^{\ n_{2*}}p_{*1}^{\ n_{*1}}p_{*2}^{\ n_{*2}}}\lnl
&= \frac{\left(\cfrac{n_{1*}!}{n_{11}!n_{12}!}\right) \left(\cfrac{n_{2*}!}{n_{21}!n_{22}!}\right)}{ \left(\cfrac{n!}{n_{*1}!n_{*2}!}\right)}\lnl
&= \frac{\left(\cfrac{n_{1*}!}{n_{11}!(n_{1*}-n_{11})!}\right) \left(\cfrac{n_{2*}!}{n_{21}!(n_{2*}-n_{21})!}\right)}{ \left(\cfrac{n!}{n_{*1}!(n-n_{*1})!}\right)}\lnl
&= \frac{\displaystyle \binom{n_{1*}}{n_{11}}\binom{n_{2*}}{n_{21}}}{\displaystyle \binom{n}{n_{*1}}}
\end{align}

となり示された.

(iii)

欠番?

(iv)

帰無仮説$H_0$の下での$\bm{p}=(p_{11},\cdots,p_{22})$を$\bm{p}_0$, 対立仮説${H_1}^*$の下での$\bm{p}$を$\bm{p}_1$とする.
ネイマン・ピアソン補助定理より, 次のような検定関数$\varphi(\bm{N})$を持つ検定は有意水準$\alpha$の最強力検定である.

\begin{align}
&E_{\bm{p}_0}\Big[\varphi(\bm{N})\Big] = \alpha \label{eq-7b2-9}\lnl
&\varphi(\bm{n}) = \begin{cases} 1 &\qquad f(\bm{n};\bm{p}_1) > k f(\bm{n};\bm{p}_0)\\
0&\qquad f(\bm{n};\bm{p}_1) < k f(\bm{n};\bm{p}_0)\end{cases} \end{align}
ただし$k$は$\eqref{eq-7b2-9}$を満たすように選んだ正の定数. 尤度比は,
\begin{align} \frac{ f(\bm{n};\bm{p}_1)}{ f(\bm{n};\bm{p}_0)} &= \frac{\cfrac{n!}{n_{11}!\cdots{n_{22}!}} {p_{11}^*}^{n_{11}} {p_{12}^*}^{n_{12}} {p_{21}^*}^{n_{21}} {p_{22}^*}^{n_{22}} }{\cfrac{n!}{n_{11}!\cdots{n_{22}!}} {p_{1*}}^{n_{1*}} {p_{2*}}^{n_{2*}} {p_{*1}}^{n_{*1}} {p_{*2}}^{n_{*2}} }\lnl &=\frac{{p_{11}^*}^{n_{11}} {p_{12}^*}^{n_{12}} {p_{21}^*}^{n_{21}} {p_{22}^*}^{n_{22}} }{{p_{1*}}^{n_{1*}} {p_{2*}}^{n_{2*}} {p_{*1}}^{n_{*1}} {p_{*2}}^{n_{*2}} } \end{align}
ここで分子は定数だから, $K$を定数として
\begin{align} \frac{ f(\bm{n};\bm{p}_1)}{ f(\bm{n};\bm{p}_0)} &= K{p_{11}^*}^{n_{11}} {p_{12}^*}^{n_{12}} {p_{21}^*}^{n_{21}} {p_{22}^*}^{n_{22}} \lnl &=K\left(\frac{ p_{11}^*p_{22}^*}{p_{12}^*p_{21}^*}\right)^{n_{11}} {p_{12}^*}^{n_{1*}}{p_{21}^*}^{n_{*1}}{p_{22}^*}^{n-n_{1*}-n_{*1}} \end{align}
と表せる.結局棄却域は, $\displaystyle \left(\frac{ p_{11}^*p_{22}^*}{p_{12}^*p_{21}^*}\right)^{n_{11}} > K’$と表せ, これは$n_{11}$だけに依存する.

$p_{11}^* > p_{1*}^*p_{*1}^*$のとき,

\begin{align}
\frac{ p_{11}^*p_{22}^*}{p_{12}^*p_{21}^*} &= \frac{ p_{11}^*(1-p_{11}^*-p_{12}^*-p_{21}^*)}{(p_{1*}^*-p_{11}^*)(p_{*1}^*-p_{11}^*)}\lnl
&=\frac{ p_{11}^*(1-p_{11}^*-(p_{1*}^*-p_{11}^*)-(p_{*1}^*-p_{11}^*))}{(p_{1*}^*-p_{11}^*)(p_{*1}^*-p_{11}^*)}\lnl
&=\frac{ p_{11}^*(1+p_{11}^*-p_{1*}^*-p_{*1}^*)}{(p_{1*}^*-p_{11}^*)(p_{*1}^*-p_{11}^*)}\lnl
&>\frac{ p_{1*}^*p_{*1}^*(1+p_{1*}^*p_{*1}^*-p_{1*}^*-p_{*1}^*)}{(p_{1*}^*-p_{11}^*)(p_{*1}^*-p_{11}^*)}\lnl
&=\frac{ p_{1*}^*p_{*1}^*(1-p_{1*}^*)(1-p_{*1}^*)}{(p_{1*}^*-p_{11}^*)(p_{*1}^*-p_{11}^*)}\lnl
&=\frac{ (p_{*1}^*-p_{*1}^*p_{1*}^*)(p_{1*}^*-p_{1*}^*p_{*1}^*)}{(p_{1*}^*-p_{11}^*)(p_{*1}^*-p_{11}^*)}\lnl
&>\frac{ (p_{*1}^*-p_{11}^*)(p_{1*}^*-p_{11}^*)}{(p_{1*}^*-p_{11}^*)(p_{*1}^*-p_{11}^*)}\lnl
&=1
\end{align}

となる. 同様にして$p_{11}^* < p_{1*}^*p_{*1}^*$のとき, $\cfrac{ p_{11}^*p_{22}^*}{p_{12}^*p_{21}^*} < 1$となる. 従って$p_{11}^* > p_{1*}^*p_{*1}^*$のとき$n_{11} \ge c_0$なら$H_0$を棄却. $p_{11}^* < p_{1*}^*p_{*1}^*$のとき$n_{11} \le c_0'$なら$H_0$を棄却する検定を考えればよい.

(v)

問題ではないので省略.