第3講統計的推定

順序統計量

一様分布に従う確率変数の順序統計量の分布

$X_1, X_2, \ldots, X_n$ が独立に一様分布 $U(0, 1)$ に従うとき，これらを大きい順に並び替えたとき， $k$ 番目の確率変数を $X_{(k)}$ と書くと，

X_{(1)} < X_{(2)} < \dots < X_{(k)} < \dots < X_{(n)}

$\begin{equation} X_{(1)} < X_{(2)} < \cdots < X_{(k)} < \cdots < X_{(n)} \end{equation}$

である．このとき，

X_{(k)}

$X_{(k)}$ はベータ分布に従う．

X_{(k)} \sim B e (k, n - k + 1)

$\begin{equation} X_{(k)} \sim Be(k, n-k+1) \end{equation}$

※証明は後述

順序統計量の分布

$X_1,\ldots,X_n$ を分布 $F_X(x)$ と確率密度関数 $f_X(x)$ を持つ連続型分布からの標本とする． $X_{(1)},\ldots,X_{(n)}$ を順序統計量としたとき， $X_{(k)}$ の分布関数と確率密度関数は次のようになる．

F_{X_{(k)}} (x) = \sum_{i = k}^{n}_{n} C_{i} F_{X} (x)^{i} (1 - F_{X} (x))^{n - i}

$\begin{equation} F_{X_{(k)}}(x) = \sum_{i=k}^{n}{}_{n}C_{i}F_X(x)^{i}(1 - F_X(x))^{n-i} \end{equation}$

f_{X_{(k)}} (x) = \frac{n!}{(k - 1)! (n - k)!} f_{X} (x) F_{X} (x)^{k - 1} (1 - F_{X} (x))^{n - k}

$\begin{equation} f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!}f_X(x)\,F_X(x)^{k-1}(1 - F_X(x))^{n-k} \end{equation}$

[証明]

\begin{aligned} F_{X_{(k)}} (x) & = P (X_{(k)} \leq x) \\ = \sum_{i = k}^{n} P (X_{(1)} \leq x, \dots, X_{(i)} \leq x, X_{(i + 1)} > x, \dots, X_{(n)} > x) \\ = \sum_{i = k}^{n} P (X_{(1)} \leq x) \dots P (X_{(i)} \leq x) P (X_{(i + 1)} > x) \dots P (X_{(n)} > x) \\ = \sum_{i = k}^{n}_{n} C_{i} {F_{X} (x)}^{i} (1 - F_{X} (x))^{n - i} \end{aligned}

$\begin{align} F_{X_{(k)}}(x) &= P(X_{(k)} \le x) \\ &= \sum_{i = k}^{n}P(X_{(1)} \le x, \ldots, X_{(i)} \le x, X_{(i+1)} > x, \ldots, X_{(n)} > x) \\ &= \sum_{i = k}^{n}P(X_{(1)} \le x) \cdots P(X_{(i)} \le x) \, P(X_{(i+1)} > x) \cdots P(X_{(n)} > x) \\ &= \sum_{i = k}^{n}{}_nC_{i} {F_X(x)}^{i}(1-F_X(x))^{n-i} \\ \end{align}$

ここで，

p = F_{X} (x)

$p = F_X(x)$ とし，

p (k, n)

$p(k,n)$ を

B i n (n, p)

$Bin(n,p)$ の確率関数とする．

p (k, n) = \frac{n!}{k! (n - k)!} p^{k} (1 - p)^{n - k}

$\begin{equation} p(k,n) = \frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} \end{equation}$

p (k - 1, n - 1) = \frac{(n - 1)!}{(k - 1)! (n - k)!} p^{k - 1} (1 - p)^{n - k}

$\begin{equation} p(k-1,n-1) = \frac{(n-1)!}{(k-1)!\,(n-k)!}p^{k-1}(1-p)^{n-k} \end{equation}$

p (k, n - 1) = \frac{(n - 1)!}{k! (n - k - 1)!} p^{k} (1 - p)^{n - k - 1}

$\begin{equation} p(k,n-1) = \frac{(n-1)!}{k!\,(n-k-1)!}p^k(1-p)^{n-k-1} \end{equation}$

p (n, n - 1) = 0

$\begin{equation} p(n,n-1) = 0 \end{equation}$

F (x)

$F(x)$ を

x

$x$ で微分して密度関数

f (x)

$f(x)$ を求めてみる．

\begin{aligned} f_{X_{(k)}} (x) & = \frac{d}{d p} F_{X_{(k)}} (x) \\ = \frac{d}{d p} \sum_{i = k}^{n}_{n} C_{i} p^{i} (1 - p)^{n - i} \\ = \sum_{i = k}^{n} {\frac{n!}{(i - 1)! (n - i)!} p^{i - 1} (1 - p)^{n - i} f_{X} (x) - \frac{n!}{i! (n - i - 1)!} p^{i} (1 - p)^{n - i - 1} f_{X} (x)} \\ = n f_{X} (x) \sum_{i = k}^{n} {p (i - 1, n - 1) - p (i, n - 1)} \\ = n f_{X} (x) {p (k - 1, n - 1) - p (k, n - 1) + p (k, n - 1) - p (k + 1, n - 1) + \dots + p (n - 1, n - 1) - p (n, n - 1)} \\ = n f_{X} (x) {p (k - 1, n - 1) - p (n, n - 1)} \\ = n f_{X} (x) p (k - 1, n - 1) \\ = n f_{X} (x) \frac{(n - 1)!}{(k - 1)! (n - k)!} F_{X} (x)^{k - 1} (1 - F_{X} (x))^{n - k} \\ = \frac{n!}{(k - 1)! (n - k)!} f_{X} (x) F_{X} (x)^{k - 1} (1 - F_{X} (x))^{n - k} \end{aligned}

$\begin{align} f_{X_{(k)}}(x) &= \frac{d}{dp}F_{X_{(k)}}(x) \\ &= \frac{d}{dp}\sum_{i=k}^n{}_nC_{i}p^i(1-p)^{n-i} \\ &= \sum_{i=k}^n\left\{\frac{n!}{(i-1)!\,(n-i)!}p^{i-1}(1-p)^{n-i}f_X(x) - \frac{n!}{i!\,(n-i-1)!}p^{i}(1-p)^{n-i-1}f_X(x)\right\} \\ &= n f_X(x)\sum_{i=k}^n\{p(i-1,n-1) - p(i,n-1)\} \\ &= n f_X(x)\left\{p(k-1,n-1) - p(k,n-1) + p(k,n-1) - p(k+1,n-1) + \cdots + p(n-1,n-1) - p(n,n-1)\right\} \\ &= n f_X(x)\left\{p(k-1,n-1) - p(n,n-1)\right\} \\ &= n f_X(x)\,p(k-1,n-1) \\ &= n f_X(x)\frac{(n-1)!}{(k-1)!\,(n-k)!}F_X(x)^{k-1}(1-F_X(x))^{n-k} \\ &= \frac{n!}{(k-1)!\,(n-k)!}f_X(x)F_X(x)^{k-1}(1-F_X(x))^{n-k} \end{align}$

ここで，母集団分布

F_{X} (x)

$F_X(x)$ が一様分布

U [0, 1]

$U[0,1]$ ならば，

F_{X} (x) = x, f_{X} (x) = 1

$F_X(x) = x,~~f_X(x) = 1$ であるので，

\begin{aligned} f_{X_{(k)}} (x) & = \frac{n!}{(k - 1)! (n - k)!} x^{k - 1} (1 - x)^{n - k} \\ = \frac{Γ (n + 1)}{Γ (k) Γ (n - k + 1)} x^{k - 1} (1 - x)^{n - k} \\ = \frac{1}{B (k, n - k + 1)} x^{k - 1} (1 - x)^{n - k} \end{aligned}

$\begin{align} f_{X_{(k)}}(x) &= \frac{n!}{(k-1)!\,(n-k)!}x^{k-1}(1-x)^{n-k} \\ &= \frac{\varGamma(n+1)}{\varGamma(k)\,\varGamma(n-k+1)}x^{k-1}(1-x)^{n-k} \\ &= \frac{1}{B(k,~n-k+1)}x^{k-1}(1-x)^{n-k} \end{align}$

となり，

X_{(k)} \sim B e (k, n - k + 1)

$\begin{equation} X_{(k)} \sim Be(k, n-k+1) \end{equation}$

となることがわかる．

順序統計量の性質

確率変数 $X_{(n)} = \max X_i$ に関しては，

F_{max X_{(i)}} (x) = P (max X_{i} \leq x) = F (x)^{n}

$\begin{equation} F_{\max X_{(i)}}(x) = P(\max X_i \le x) = F(x)^n \end{equation}$

f_{max X_{(i)}} (x) = n f (x) F (x)^{n - 1}

$\begin{equation} f_{\max X_{(i)}}(x) = nf(x)\,F(x)^{n-1} \end{equation}$

確率変数

X_{(1)} = min X_{i}

$X_{(1)} = \min X_i$ に関しては，

F_{min X_{(i)}} (x) = P (min X_{i} \leq x) = 1 - (1 - F (x))^{n}

$\begin{equation} F_{\min X_{(i)}}(x) = P(\min X_i \le x) = 1 - (1-F(x))^n \end{equation}$

f_{min X_{(i)}} (x) = n f (x) (1 - F (x))^{n - 1}

$\begin{equation} f_{\min X_{(i)}}(x) = nf(x)\,(1-F(x))^{n-1} \end{equation}$

順序統計量の同時確率密度関数 (1)

$i < j$ として， $X_{(i)}$ と $X_{(j)}$ の同時密度関数は次のようになる．

f_{X_{(i)}, X_{(j)}} (x, y) = \frac{n!}{(i - 1)! (j - i - 1)! (n - j)!} f (x) f (y) F (x)^{i - 1} (F (y) - F (x))^{j - i - 1} (1 - F (y))^{n - j}

$\begin{equation} f_{X_{(i)},X_{(j)}}(x,y) = \frac{n!}{(i-1)!\,(j-i-1)!\,(n-j)!}f(x)f(y)F(x)^{i-1}(F(y) - F(x))^{j-i-1}(1-F(y))^{n-j} \end{equation}$

順序統計量の同時確率密度関数 (2)

$X_1,\ldots,X_n$ を分布 $F_X(x)$ と確率密度関数 $f_X(x)$ を持つ連続型分布からの標本とする． $X_{(1)},\ldots,X_{(n)}$ を順序統計量としたとき， $(X_{(1)}, \ldots, X_{(n)})$ の同時確率密度関数は次のようになる．

f_{X_{(1)}, \dots, X_{(n)}} (x_{1}, \dots, x_{n}) = n! f_{X} (x_{1}) \dots f_{X} (x_{n})

$\begin{equation} f_{X_{(1)}, \ldots, X_{(n)}}(x_1, \ldots, x_n) = n!\,f_X(x_1)\cdots\,f_X(x_n) \end{equation}$