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\begin{document}

\title{�₳����
�m���p�����g���b�N���v}

\author{���c����(��t��w�j}

\date{today}

\maketitle 

%================ section 1  ============================

���v�̗��_�ł͒��S�Ɍ��藝�����Ƃɐ��K���z�������Ƃ�
�d�v�Ȉʒu���߂Ă���B���������̂悤�ȕ��z���ʂ�����
�W�{�f�[�^�Ƃ��ē���ꂽ��W�c�̕��z�Ƃ��Đ^�̂��̂��ǂ����A
���邢�͐^�łȂ��Ƃ��\���ɂ悢�ߎ����Ă����
���Ȃ��邾�낤���H�@
���܂܂ŏq�ׂĂ��������M����Ԃɂ����ẮA���̏�����
���肳��āA���̌��ʂ�����ꂫ���B�������K���z�Ƃ����O�������
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�ΏۂƂ������z�̏�񂪂킩���Ă���Ȃ�΁A���̏���
�p�����ق����悢�B�@�����҂��邢�͊ϑ��҂����Ƃɂ��Ă��镪�z�̌`��
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�˂�Ȃ����@���]�܂����B
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���̕��ς╪�U�̃p�����[�^�ɂ��Ď�舵���Ă����B�����ł�
�u�m���p�����g���b�N(nonparametric)�v���邢��
�u���z�`�ɂ��Ȃ�(distribution-free)�v�Ƃ�΂����@��
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\section{��{�̊T�O}
���ꂩ��q�ׂ鐄��⌟��ɂ́A���̂悤�ȕ��z�����肵�Ȃ����߁A
�悭�p������T�O�́A���o�����W�{�f�[�^(�傫���̏��ɂȂ��Ă��Ȃ�)��
�傫���̏��ɕ��ёւ��邱�ƁA�������v�ʂ��d�v�ȓ���������B
�����������𒲂ׂĂ݂�B

$\bullet$
���o�����f�[�^�F
$X_1, X_2, X_3, \cdots, X_n$ (�傫���̏��ɂȂ��Ă��Ȃ�)

$\bullet$
�傫���̏��ɕ��ׂ��f�[�^�F
       $Z_1 \leq Z_2 \leq Z_3 \leq \cdots \leq Z_n$


{\gt (I) �������v�ʂ̕��z�G}
�^����ꂽ�m���ϐ�$X_1,X_2,\cdots,X_n$�͘A���^���z�ł��̖��x$f(x)$������
����B�傫���̏��ɕ��ёւ����m���ϐ�$Z_1 \leq Z_2 \leq \cdots <Z_n$��
���z�����߂悤�B�����ł̉���Ƃ��āA�^����ꂽ���z�͘A���^�ŁA���x�֐���
���Ƃ��Ă��邩��A�����̋N����ꍇ�͊m���[���ł��邩��A�����Ȃ���
�s�����̏ꍇ�݂̂��l����B

�Ⴆ��
3�̊m���ϐ�������ꍇ�ɂ́A������
�Ɨ��Ɖ��肵�A�������z�ɏ]���Ƃ���ƁA�����i�����j���z�͐ςɓ������B����������
$g(x_1,x_2,x_3)=f(x_1)f(x_2)f(x_3),$ �������傫���̑傫���̏��ł͂Ȃ��A
$-\infty <x_1 < \infty,-\infty <x_2 < \infty,-\infty <x_3 < \infty$�B
����$x_1< x_2< x_3$�Ȃ�΁A
���ʓ��v�ʂ̒�`���� $z_1=x_1, z_2=x_2, z_3=x_3$�ƂȂ�B
���l��,
�����f�[�^�̑傫���̏����A$x_2< x_1< x_3$�Ȃ�΁A
$z_1=x_2, z_2=x_1, z_3=x_3$�ȂǂƂƂȂ�B
���̂Ƃ��̓����i�����j���z�́A����̕��z�Ƃ��邩��, 
$x_2 <x_1 <x_3$�Ȃ��
$h(z_1,z_2,z_3)=f(x_2)f(x_1)f(x_3)=f(x_1)f(x_2)f(x_3)=f(z_1)f(z_2)f(z_3)$
�ŗ^������B���̂悤�ɏꍇ���������Ȃ�΁A
$\{x_2 <x_1 <x_3\} =\{z_1 <z_2 <z_3\} $
�ɓ���������A3�̃f�[�^�����ׂčl����� 
$\{1,2,3\}$����̕��ёւ�$(i,j,k) , i<j<k$�A
$\{x_i <x_j <x_k\} =\{z_1 <z_2 <z_3\}$�A $3!=6$�ʂ�ɂ��ē����ł��邩��A
$h(z_1,z_2,z_3)=3! f(z_1)f(z_2)f(z_3), \quad z_1 <z_2 <z_3$�ŗ^������B

�藝�F
��ʂ� $n$�̏ꍇ�ɂ�
\begin{equation}                               \label{orderdist}
  h(z_1,z_2,\cdots,z_n)=n! f(z_1)f(z_2)\cdots f(z_n), \quad 
z_1 <z_2 <\cdots <z_n
\end{equation}
�ŗ^������B

���̒藝�����Ƃɂ��Ă������̐��������߂�B

{\gt (II) �ő�ŏ��̌������x}
$n$�̂����̍ő�$z_n$�ƍŏ�$z_1$�̓���(����)���z�����߂�B
�ϐ���$n$�����āA�ő�ƍŏ��̕��z�����߂�ɂ́A�c���$n-2$��
$z_2,\cdots,z_{n-1}$��
�ϕ����āA���ӕ��z�����߂�΂悢�B����������
\begin{displaymath}
 \begin{array}{ll}
   g(z_1,z_n) 
& = \underbrace{\int_{-\infty}^{\infty}\cdots
    \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}}_{\mbox{$n-2$�̐ϕ�}}
   h(z_1,z_2,\cdots,z_n)dz_2 dz_3\cdots dz_{n-1}\\[3.5ex]
& = n(n-1)[F(z_n)-F(z_1)]^{n-2}f(z_1)f(z_n), \quad -\infty < z_1 < z_n < \infty
 \end{array}
\end{displaymath}


%$z_1 < z_n$�ɑ΂���
%\[
%\begin{array}{ll}
% g(z_1,z_n) &
%  = \displaystyle
%  \int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
%  h(z_1,z_2,\cdots,z_n)dz_2 dz_3\cdots dz_{n-1} \\
%&  = \displaystyle
%  \int_{z_1}^{z_n}\cdots\int_{z_1}^{z_4} \int_{z_1}^{z_3}
%  h(z_1,z_2,\cdots,z_n)dz_2 dz_3\cdots dz_{n-1} \\
%  & 
%  = \displaystyle
%  n! f(z_1) f(z_n) \int_{z_1}^{z_n}\cdots\int_{z_1}^{z_4} \int_{z_1}^{z_3}
%   f(z_2)\cdots f(z_{n-1})dz_2 dz_3\cdots dz_{n-1}  
% \end{array}
%\]
\noindent
�Ȃ��Ȃ�΁A�ݎ�(����)�ϕ��́A���z�֐�
 $\displaystyle\int_{a}^{b} f(t) dt = F(b)-F(a)$,
 $\displaystyle\int_{a}^{b} [F(t) - F(a)]^k f(t) dt 
  = \dfrac{[F(b)-F(a)]^{k+1}}{k+1}$�Ȃ�
���������Čv�Z�ł���B
$ z_1 < z_2 < z_3 < \cdots < z_n$ ������ $z_2$ �Őϕ�����΁A
$ \displaystyle\int_{z_1}^{z_3} f(z_2) dz_2 = F(z_3) - F(z_1),$ 
������$ z_1 <  z_3  <z_4 < \cdots < z_n$�ɒ��ӂ���B ����$z_3$ �Őϕ�����΁A
$ \displaystyle\int_{z_1}^{z_4} [F(z_3) - F(z_1)]f(z_3) dz_3 
= \dfrac{1}{2}[F(z_4) - F(z_1)]^2,$
$\, z_1 <  z_4 < \cdots < z_n$�ƂȂ�B
������J��Ԃ���
$ \displaystyle\int_{z_1}^{z_5} \dfrac{1}{2}[F(z_4) - F(z_1)]^2f(z_4) dz_4 
= \dfrac{1}{3!}[F(z_5) - F(z_1)]^2 $�𓾂�B
����ɂ���𑱂��āA����
\begin{equation} \label{maxandmin}
 g(z_1,z_n) 
  = n(n-1)[F(z_n)-F(z_1)]^{n-2}f(z_1)f(z_n), \quad -\infty < z_1 < z_n < \infty
\end{equation}
�𓾂邱�Ƃ��ł���B

{\gt (III) �ő�̖��x�֐��ƍŏ��̖��x�֐�}�G

��ŋ��߂�(\ref{maxandmin})�ɂ�������ӕ��z�����߂邱�ƂŁA�ő�
���邢�͍ŏ��̕��z���v�Z�ł���B
���ꂼ��̖��x�֐���$h_{(n)}$, $h_{(1)}$�ƕ\���΁A���̎��ŗ^������B

\begin{equation} \label{maxdist}
  h_{(n)} (z_n) = n [F(z_n)]^{n-1} f(z_n), \, -\infty < z_n < \infty
\end{equation}
\begin{equation} \label{mindist}
  h_{(1)} (z_1) = n [1-F(z_1)]^{n-1} f(z_1), \, -\infty < z_1 < \infty
\end{equation}

{\gt (IV) ��ʂ�$i$�Ԗڂ̖��x�֐�$(i=1,2,\cdots n)$}�G
\begin{equation} \label{ith-density}
 \begin{array}{ll}
  h_{(i,i)}(z_{i})
   = i \displaystyle\binom{n}{i}[F(z_i)]^{i-1}[1-F(z_i)]^{n-i} f(z_i)
 \end{array}
\end{equation}

�藝�̎�(\ref{orderdist})�����Ƃɂ��āA�ォ������Ǝ���
���z�����߁A����ɉ���������ӕ��z�����߂Ă����B
���̖��x�֐���$h_{(i,j)}(z_i,z_{i+1}, \cdots, z_{j-1},z_{j}) $
�Ƃ����B
�܂� $z_1 < z_2 < \cdots < z_{n-1}$ �ɂ���
\[
 \begin{array}{ll}
  h_{(1,n-1)}(z_1,z_2, \cdots, z_{n-2},z_{n-1})
  & = \displaystyle\int_{z_{n-1}}^{\infty} 
    h(z_1,z_2, \cdots, z_{n-1},z_{n}) dz_n \\
  & = n! f(z_1) f(z_2) \cdots f(z_{n-1}) [1- F(z_{n-1})]
 \end{array}
\]
�����
$
   \displaystyle\int_{z_{n-2}}^{\infty} 
     [1 - F(z_{n-1})] f(z_{n-1}) dz_{n-1}
   = \dfrac{[1 - F(z_{n-2})]^2}{2} 
$
�ł��邩��A
$z_1 < z_2 < \cdots < z_{n-2}$ �ɂ���
\[
 \begin{array}{ll}
  h_{(1,n-2)}(z_1,z_2, \cdots, z_{n-2})
  & = \displaystyle\int_{z_{n-1}}^{\infty} 
    h_{(1,n-1)}(z_1,z_2, \cdots, z_{n-1}) dz_{n-1} \\
  & = n! f(z_1) f(z_2) \cdots f(z_{n-2})\dfrac{[1 - F(z_{n-2})]^2}{2}
 \end{array}
\]
�����J��Ԃ���
$z_1 < z_2 < \cdots < z_{i}$ �ɂ���
\[
 \begin{array}{ll}
  h_{(1,i)}(z_1,z_2, \cdots, z_{i})
  & = \displaystyle\int_{z_{i}}^{\infty} 
    h_{(1,i+1)}(z_1,z_2, \cdots, z_{i+1}) dz_{i+1} \\
  & = n! f(z_1) f(z_2) \cdots f(z_{i})
    \dfrac{[1 - F(z_{i})]^{n-i}}{(n-i)!}
 \end{array}
\]
����ǂ�$z_1,z_2,\cdots, z_{i-1}$�̕ϐ��ɂ��Đϕ����Ă����B
$ z_2 < \cdots < z_{i}$ �ɂ���
\[
 \begin{array}{ll}
  h_{(2,i)}(z_2, \cdots, z_{i})
  & = \displaystyle\int_{-\infty}^{z_2} 
    h_{(1,i)}(z_1,z_2, \cdots, z_{i}) dz_1 \\
  & = n! \int_{-\infty}^{z_2} f(z_1) dz_1 f(z_2) \cdots f(z_{i})
    \dfrac{[1 - F(z_{i})]^{n-i}}{(n-i)!}\\
  & = n! F(z_2) f(z_2) \cdots f(z_{i})
    \dfrac{[1 - F(z_{i})]^{n-i}}{(n-i)!}\\
 \end{array}
\]
���l��
$ z_k < \cdots < z_{i}$ �ɂ���
\[
 \begin{array}{ll}
  h_{(k,i)}(z_{k}, \cdots, z_{i})
  & = \displaystyle\int_{-\infty}^{z_k} 
    h_{(k-1,i)}(z_{k-1}, \cdots, z_{i}) dz_{k-1} \\
  & = n! \dfrac{[F(z_k)]^{k-1}}{(k-1)!} f(z_k) \cdots f(z_{i})
    \dfrac{[1 - F(z_{i})]^{n-i}}{(n-i)!}\\
 \end{array}
\]
�Ƃ���$k=i$�܂ŌJ��Ԃ��΁A$-\infty < z_i < \infty$�ɂ���
\[
 \begin{array}{ll}
  h_{(i,i)}(z_{i})
  & = \displaystyle\int_{-\infty}^{z_i} 
    h_{(i-1,i)}(z_{i-1}, z_{i}) dz_{i-1} 
   = n! \dfrac{[F(z_i)]^{i-1}}{(i-1)!} f(z_i)
    \dfrac{[1 - F(z_{i})]^{n-i}}{(n-i)!}\\
  & = \dfrac{n!}{(i-1)!(n-i)!} [F(z_i)]^{i-1}[1-F(z_i)]^{n-i} f(z_i) \\
  & = i \displaystyle\binom{n}{i}[F(z_i)]^{i-1}[1-F(z_i)]^{n-i} f(z_i)
 \end{array}
\]

���ܓƗ��œ��ꕪ�z�ɏ]���Ƃ��āA���̕��z����l���z
$F(x) = P(X_i \leq x) = x , i=1,2,\cdots$�Ƃ����
\begin{equation} \label{ith-expectation}
  E(Z_i) = \dfrac{i}{n+1}, i=1,2,\cdots ,n
\end{equation}
�����藧�B���ɂ��Ȃ킿�ŏ��l$z_1$�̊��Ғl�� $\dfrac{1}{n+1}$,
�ő�l$z_n$�̊��Ғl�� $\dfrac{n}{n+1}$�ł���B
��ʐ����������ƂȂ��A��l���z���l���闝�R�͂��̕ϊ����\�ł��邩��B
�^����ꂽ�m���ϐ�$X$ �Ƃ��̕��z�֐� $F(x)= P(X \leq x)$ ����
$Z= F(X)$ �Ƃ����ƁA$Z$�̕��z$F_Z$�͈�l���z�ɂ��������B�܂�
$F_Z(z) = P(Z \leq z) = P(F(X) \leq z ) 
= P( X \leq F^{-1}(z) ) = F(F^{-1}(z)) = z$
�ƂȂ邩��B
�㎮(\ref{ith-expectation})�͎��̂悤�ɂ��ċ��߂���B
��(\ref{ith-density})�ɂ�����,��l���z�̏ꍇ$F(x) = x, f(x) = 1, 0<x<1$ ��
�l����΁A
\[
  \begin{array}{ll}
   E[u_i] & = \int_{0}^{1} u \times i \binom{n}{i} u^{i-1} (1-u)^{n-i} du \\
          & = i \binom{n}{i} \int_{0}^{1} u^i (1-u)^{n-i} du \\
          & = i \binom{n}{i} 
                 \dfrac{\Gamma(i+1)\Gamma(n-i+1)}{\Gamma(i+1 + n-i+1)}\\
          & = i \binom{n}{i} \dfrac{i! (n-i)!}{ (n+1)!}\\
          & = \dfrac{i}{n+1}
  \end{array}
\]
�����Őϕ�$\int_{0}^{1} u^{p-1} (1-u)^{q-1} du, \, p>0,q>0$�̓x�[�^�ϕ�
$B(p,q) = \dfrac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$�Ƃ�΂�A
�K���}�֐�$\Gamma(p) = \int_0^{\infty} x^{p-1} e^{-x} dx, \, p>0$
���g���ĕ\�����Ƃ��ł��A����ɐ��̐����ɑ΂���J���}�֐���$\Gamma(k) = (k-1)!$
�ƂȂ邱�Ƃ𗘗p�����B



{\gt (V) �͈�(�ő傩��ŏ��܂�)�̕��z}

�ő�ƍŏ��̓���(����)���z��
\[
   h_{(1,n)}(z_1,z_n) = n(n-1)(z_n - z_1)^{n-2}, 
   \quad 0 < z_1 < z_n < 1 
\]
�ŗ^����ꂽ����A���ꂩ��͈� $R = Z_n - Z_1$ �̕��z�����߂邱�Ƃ�
�ł���B
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% STR 2 0 3 0
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% $z_n$
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\end{picture}%
\caption{�ϕ��̗̈�}
\label{fig:range}
\end{wrapfigure}

���z�֐� 
$
F_R (r) = P(R \leq r) = P(Z_n - Z_1 \leq r )
= \int\!\!\int_{\{z_n - z_1 \leq r\}} h_{(1,n)}(z_1,z_n) dz_1 dz_n
$
���v�Z����΂悢�B�}�ɂ���悤�ɂ���2�d�ϕ��ɂ�����
�ϕ��͈͂�2�ɕ�����΁A
\[
 \begin{array}{ll}
  F_R (r) 
   & = \biggl(\displaystyle\int_0^{1-r} dz_1 \int_{z_1}^{r+z_1} dz_n
       + \int_{1-r}^1 dz_1 \int_{z_1}^1 dz_n
       \biggr) h_{(1,n)}(z_1,z_n) \\
   & = n r^{n-1} \displaystyle\int_0^{1-r}dz_1 + n \int_{1-r}^1 (1-z_1)^{n-1}dz_1 \\
    & = n r^{n-1}(1-r) + r^n = n r^{n-1} -(n-1)r^n, \quad 0 < r < 1 
 \end{array}
\]
���������Ĕ���������΁A���x�֐�
\[
  f_R(r) = n(n-1) r^{n-2}(1-r) , \quad 0 < r < 1
\]
��������B


\fbox{���}
�J�[�h��$n$�������āA�ԍ�$\{1,2,\cdots,n\}$���L����Ă���B���ꂩ��
$m$��̌��ɖ߂�����1���Â��o����(�񕜌����o�j�B
���̎��o���ꂽ�ԍ���
\[X_{(1)} < X_{(2)} < \cdots < X_{(m)}\]
�Ƃ���B���̂Ƃ�, �e$i=1,2,\cdots,m$�ɂ���
\[
  P(X_{(i)} = k) = \dfrac{\binom{k-1}{i-1} \binom{n-k}{m-i}}{\binom{n}{m}},
  \quad, k=1,2,\cdots, n
\]
�����藧�B

�Ȃ��Ȃ�A����$\{X_{(i)} = k\}$�Ƃ�

\[
 \{X_{(i)} = k\}
   \Leftrightarrow
 \left\{
   \begin{array}{l}
     X_{(1)} < X_{(2)} < \cdots < X_{(i-1)}
      : \{1,2,\cdots,k-1\} ���� i-1 �� \\
     X_{(i+1)} < X_{(i+2)} < \cdots < X_{(n)}
      : \{k+1,k+2,\cdots,n\} ���� m-i ��
   \end{array}
 \right\}
\]
�ƂȂ�΂悢�B���ꂼ���$\binom{k-1}{i-1}$, 
$\binom{n-k}{m-i}$ �̏ꍇ�̒ʂ肪����B
�S�̂�$\binom{n}{m}$�ł��邩��A��L�̊m���ƂȂ�B

���Ƃ��΁A$n$���̒�����$m=3$�������o���ƁA
\begin{enumerate}
\item �ő�l $\max_i X_i = X_{(3)}$ �́A$m=3,i=3$ ����
\[
   P(X_{(3)} = k) = \dfrac{\binom{k-1}{3-1}\binom{n-k}{3-3}}{\binom{n}{3}}
                  = \binom{k-1}{2}{\binom{n}{3}}^{-1}
\]
����Ɋ��Ғl�����߂�ƁA
\[
   E(X_{(3)})  = \sum_{k=3}^n k \dfrac{\binom{k-1}{2}}{\binom{n}{3}} 
               = {\binom{n}{3}}^{-1}  \sum_{k=3}^n 3 \binom{k}{3} 
               = {\binom{n}{3}}^{-1} 3 \binom{n+1}{4} 
               = \dfrac{3}{4} (n+1)
\]
\item �����l$\rm{med} \{X_i\} = X_{(2)}$�̊��Ғl��$m=3,i=2$ ����
\[
\begin{array}{ll}
   E(X_{(2)}) 
 & = \sum_{k=2}^n k \dfrac{\binom{k-1}{2-1}\binom{n-k}{3-2}}{\binom{n}{3}}
   = {\binom{n}{3}}^{-1} 
        \sum_{k=2}^{n-1} 2 \binom{k}{2} (n-k) \\
 & = {\binom{n}{3}}^{-1} \left\{
        2n \sum_{k=2}^{n-1} \binom{k}{2} 
         - 2 \sum_{k=2}^{n-1} k \binom{k}{2} \right\} \\
 &  = {\binom{n}{3}}^{-1} \left\{
         2n \binom{n}{3} 
         - 2 \left[\sum_{k=2}^{n-1} (k+1) \binom{k}{2} 
              - \sum_{k=2}^{n-1}  \binom{k}{2} \right]
         \right\} \\
 &  = {\binom{n}{3}}^{-1} \left\{
         2n \binom{n}{3} 
         - 2 [ 3 \binom{n+1}{4} - \binom{n}{3}] 
         \right\} 
    = {\binom{n}{3}}^{-1} \binom{n}{3} \left\{
         2n 
         - 2 \cdot 3 \dfrac{n+1}{4} - 2 
         \right\}\\
 &  = \dfrac{n+1}{2}
\end{array}
\]
\item �ŏ��l$\min_i X_i = X_{(1)}$�̊��Ғl��$m=3,i=3$ ����A���l�Ȍv�Z�ɂ����
\[
   E(X_{(1)}) = \dfrac{n+1}{4}
\]
�𓾂�B���邢�� $E\left[\dfrac{X_{(1)} + X_{(2)} + X_{(3)}}{3}\right] = 
 \dfrac{1}{n}(1+2+\cdots+n) = \dfrac{n+1}{2}$������v�Z�����B

\end{enumerate}




\section{U���v�ʂƐ���̕s�ΐ�}

��W�c���z�Ɋ܂܂��ꐔ (�Ⴆ�Ε���,���U���̃��[�����g)�̐���ʂƂ���, 
U-���v�ʂ��L���m���Ă���. U-���v�ʂƂ͐�ΘA���ȕ��z�̃N���X�̒���, 
�Ή�����ꐔ�̈�l�ŏ����U�s�ΐ���ʂł���. Halmos(1947)�ɂ�蓱�����ꂽ��, 
�����̑傫�Ȕ��W��Hoeffding(1948�Ƃ���ȍ~)�ɂ��. 
�Q�l���Ƃ��ẮA

\begin{itemize}
  \item W.Hoeffding, 1948 Ann. Math. Stat. {\bf 19} 293-325.
  \item M.Denker, 1985 Fr Viewieg \& Sohn, Wiesbaden.
  \item A.J.Lee, 1990 Marcel Dekker, New York.
  \item R.J.Serfing, 1980 John Wiley \& Sons, New York.
  \item T.S.Ferguson, Notes for Statistics, UCLA 200B. 
\end{itemize}

���̑Q�ߓI�����Ƃ���, ���S�Ɍ��藝, �s�ό�����, �Q�ߓI�������������. 



�����ł͕��U�̐���̂��߂̕s�ΐ����l����B

�܂�{\cal P}�𕪕z���Ƃ��āA$\theta$ �� $P \in {\cal P}$ �̃p�����[�^�Ƃ���B
��`�Ƃ���
$\theta$ ������\�ł���Ƃ́A���鐔 $m$ ��
�Ɨ����ꕪ�z $X_1,X_2,\cdots,X_m$���������֐�$h(X_1,X_2,\cdots,X_m)$��
\[
  E[h(X_1,X_2,\cdots,X_m)] = \theta
\] 
�𖞂����Ƃ��A����$m$�Ő���\�Ƃ����B
�܂��֐�$h$���j(�J�[�l��)�Ƃ�΂��B
���� $\mu$ �ɑ΂��Ă� $E[X_1] = \mu$ �ŁA$h(x) = x, m=1$ �Ő���\�ł���B

$U$���v�ʂƂ͊j�֐�$h$����A
\[
   U_n = \dfrac{1}{n!/(n-m)!} \sum_{{\cal P}_{m,n}} 
           h(X_{i_1}, X_{i_2}, \cdots, X_{i_m})
       = \dfrac{1}{\binom{n}{m}} \sum_{{\cal C}_{m,n}} 
           h(X_{i_1}, X_{i_2}, \cdots, X_{i_m})
\]
������${\cal P}_{m,n}$(permutation)��$n$��$\{1,2,\cdots,n\}$����I��
����$(i_1, i_2,\cdots,i_m)$�ɂ킽��a���Ӗ�����B
�܂�${\cal C}_{m,n}$(combination)��$n$��$\{1,2,\cdots,n\}$����I��
�g�ݍ��킹$(i_1, i_2,\cdots,i_m)$�ɂ킽��a���Ӗ�����B
�L��
$\dfrac{n!}{(n-m)!} = (n)_m = n(n-1)\cdots(n-m+1), 
  \displaystyle\binom{n}{m} = \dfrac{n!}{m! (n-m)!} = \dfrac{n (n-1) \cdots (n-m+1)}{m!}$
���p����B








\section{
    Wilcoxon�̕������ʌ���
}

A.J.Lee; U-Statistics -- Theory and Proctice
Statistics:Textbook and Monograph vol.110 
Marcel Dekker, Inc. ISBN:0-8247-8253-4

Wilcoxson one-sample statistics 

$X_1, X_2, \cdots, X_n$ �͓Ɨ�����ŁA���A���^��
���̕��z�֐���$F$,���x�֐���$f$�Ƃ���B
$R_i^+$ the rank of $|X_i|$, $(i=1,2,\cdots,n)$.
$R_i^+$ the position of $|X_i|$ when r.v.  $|X_1|, |X_2|, \cdots, |X_n|$.

Wilcoxson
  $ W_n^+$ = sum of the qualities $R_i$ corresponding to the position $X_i$.

Walsh average $\dfrac{1}{2}(X_i + X_j)$, $1 \ge i \ge j \ge n$(��������j

\( i < j ,
   I\{X_i + X_j > 0\} = I\{X_i > 0\} I\{ |X_j| < X_i \}  
                       + I\{X_j > 0\} I\{ |X_i| < X_j \} 
\)


  $ \begin{array}{ll}
    W_n^+ & = \sum_{1 \le i \le j \le n} I\{\dfrac{1}{2}(X_i + X_j) > 0\} \\
          &  = \sum_{1 \le i < j \le n} I\{X_i > 0 \} I\{ |X_j| < X_i \}
           + \sum_{1 \le i < j \le n} \{X_j > 0 \} I\{ |X_i| < X_j \}
           + \sum_{i=1}^{n} I\{ X_i > 0 \}  \\
          & = \sum_{i < j} I\{ X_i + X_j > 0 \} + 
               \sum_{i=1}^{n} I\{ X_i > 0 \} \\
          & = \binom{n}{2}  U_n^{(2)} + n U_n^{(2)} \\
          & = \sum_{i=1}^n R_i^+ I\{X_i > 0\} 
    \end{array}
  $
������
\(
  U_n^(1) = \dfrac{1}{n}\sum_{i=1}^{n} I(X_i > 0),
  U_n^(2) = \dfrac{1}{\binom{n}{2}}\sum_{i < j} I(X_i + X_j > 0).
\)



�������z���[���̎���őΏ̂Ȃ�΁A���̐��������藧�F
\[\forall x; f(x) = f(-x),  \, F(x) = 1 - F(-x), \,  F(0)=1/2\]
���ꂩ��m���̌v�Z������B
\begin{enumerate}
  \item[(1)] $p_1 = P(X_1 > 0) = 1/2$
  \item[(2)] $p_2 = P(X_1 + X_2 > 0) = 1/2$
  \item[(3)] $p_3 = P(X_1 + X_2 > 0, X_1 >0) = 3/8$
  \item[(4)] $p_4 = P(X_1 + X_2 > 0, X_2 + X_3 > 0) = 1/3$
\end{enumerate}
(�ؖ�)
(1)  $P(X_1 > 0) = \displaystyle\int_0^{\infty} f(x) dx 
           = \dfrac{1}{2} \Big\{\int_0^{\infty} f(x) dx 
               + \int_0^{\infty} f(x) dx \Big\}
           = \dfrac{1}{2} \Big\{\int_0^{\infty} f(x) dx
               + \int_{-\infty}^0 f(x) dx \Big\}
           = 1/2 $

\noindent
(2) $P(X_1 + X_2 > 0 ) 
          = \displaystyle\int\!\!\int_{\{(x,y): x+y>0\}} \hspace{-8ex} f(x) f(y) dx dy
          = \int_{-\infty}^{\infty} f(y) dy \int_{-y}^{\infty} f(x) dx
          = \int_{-\infty}^{\infty} f(y) \{ 1- F(-y)  \}  dy
          = \int_{-\infty}^{\infty} f(y)  F(y)  dy
          = \int_{-\infty}^{\infty} F(y)  dF(y)
          = \int_{0}^{1} z  dz 
          = 1/2 $

\noindent
(3) $P(X_1 + X_2 > 0, X_1 >0 ) 
         = \displaystyle\int\!\!\int_{\{(x,y): x+y>0, x>0\}} \hspace{-8ex} f(x) f(y) dx dy
         = \int_{0}^{\infty} f(x) dx \int_{-x}^{\infty} f(y) dy
         = \int_{0}^{\infty} f(x) \{ 1- F(-x)  \}  dx
         = \int_{0}^{\infty} f(x)  F(x)  dx
         = \int_{0}^{\infty} F(x)  dF(x)
         = \int_{1/2}^{1} z  dz 
         = \dfrac{1}{2} - \dfrac{1}{2}\Big(\dfrac{1}{2}\Big)^2 
         = 3/8 $

\noindent
(4) $P(X_1 + X_2 > 0, X_1 + X_2 >0 ) 
        = \displaystyle
          \int\!\!\int\!\!\int_{\{(x,y): x+y>0, y+z>0\}} \hspace{-8ex} f(x) f(y) dx dy
        = \int_{\infty}^{\infty} f(x)dx 
            \int_{-x}^{\infty} f(y)dy \int_{-y}^{\infty} f(z) dz
        = \int_{-\infty}^{\infty} f(x)dx 
            \int_{-x}^{\infty} f(y) \{ 1- F(-y)  \} dy  
        = \int_{-\infty}^{\infty} f(x)dx 
            \int_{-x}^{\infty} f(y)  F(y) dy 
        = \int_{-\infty}^{\infty} f(x)dx \int_{-x}^{\infty} F(y) dF(y)
        = \int_{-\infty}^{\infty} f(x) 
            \Big[ \dfrac{w^2}{2} \Big]_{w=F(-x)}^{1} dx 
        = \int_{-\infty}^{\infty} f(x) \dfrac{1-F(-x)^2}{2} dx 
        = \int_{-\infty}^{\infty} f(-x) \dfrac{1-F(x)^2}{2} dx 
        = \int_{-\infty}^{\infty} f(x) \dfrac{1-F(x)^2}{2} dx 
        = \dfrac{1}{2} (1-E[F(X_1)^2]) $
������
    $P(F(X_1) \le w ) = P(X_1 \le F^{-1}(w)) 
         = \displaystyle\int_{-\infty}^{F^{-1}(w)} f(x)dx
         = F(t)\Big|_{t=-\infty}^{t=F^{-1}(w)}
         = F(F^{-1}(w)) - F(-\infty) = w $
���A$F(X_1) $ �͒P�ʋ��$[0,1]$��̈�l���z�ɂ��������B
$E[F(X_1)^2] = \displaystyle\int_{0}^{1} z^2 dz = 1/3 $
����āA
$
   P(X_1 + X_2 > 0, X_1 + X_2 >0 ) = \dfrac{1}{2} (1-E[F(X_1)^2]) 
        = \dfrac{1}{2} (1- \dfrac{1}{3}) = 1/3
$





�m���ϐ� $Y_1, Y_2, \ldots, Y_n$ �͓Ɨ��œ���Ȋm�����z�ɂ��������A
�A���^�őΏ̂ł���Ƃ���B���̖��x�֐��� $f_Y(y)$�Ƃ���B
�A�������ƑΗ��������\������B

Define the sign indecator $Z_i, i = 1, 2, \cdots , n$, as

$Z_i = Y_i -X_i$


The Wilcoxon test statistic is calculated as:



If the null hypothesis is true, the mean and variance of the Wilcoxon test statistic 
are given by :

For $n > 12$, the $z$ score of $W$ has an approximately 
standard normal distribution, i.e.,
Therefore, if $n$ is large, $H_o$ will be rejected in favor 
of $H_1$ if
\[
         Z_W < z_{a/2} \quad \mbox{or} \quad  Z_W > z_{1-a/2}
\]

\begin{verbatim}
http://davidmlane.com/hyperstat/dist_free.html
http://faculty.vassar.edu/lowry/webtext.html
http://lbm.ab.a.u-tokyo.ac.jp/~omori/kensyu/nonpara.htm
\end{verbatim}
%================ section   ============================


���

\begin{tabular}{cccccccc}
   ��1 & ��2 & ��3 & ��4 & ��5 & ��6 & ��7 \\
�ԍ�& $X_A$ & $X_B$ & $X_A-X_B$ & ��Βl $|X_A-X_B|$ &
   ��Βl�̏���  & �����t������ \\
1 &  78& 78& 0 & 0& --& -- \\
2 & 24 & 24& 0& 0& --& --\\
3 & 64& 62& $+2$& 2& 1& $+1$\\
4 & 45 & 48& $-3$& 3& 2& $-2$\\
5 & 64& 68& $-4$& 4& 3.5& $-3.5$\\
6 & 52& 56& $-4$& 4& 3.5& $-3.5$\\
7 & 30& 25& $+5$& 5& 5& $+5$\\
8 & 50& 44& $+6$& 6& 6& $+6$\\
9 & 64& 56& $+8$& 8& 7& $+7$\\
10& 50& 40& $+10$& 10& 8.5&  $+8.5$\\
11& 78& 68& $+10$& 10& 8.5& $+8.5$\\
12& 22& 36& $-14$& 14& 10& $-10$\\
13& 84& 68& $+16$& 16& 11& $+11$\\
14& 40& 20& $+20$& 20& 12& $+12$\\
15& 90& 58& $+32$& 32& 13& $+13$\\
16& 72& 32& $+40$& 40& 14& $+14$\\
  &   &   &    &   &   &  W = 67.0 
TN = 14  \\

\end{tabular}

 To illustrate, suppose that 16 students in an introductory statistics course are presented with a number of questions (of the sort you encountered in Chapters 5 and 6) concerning basic probabilities. In each instance, the question takes the form "What is the probability of such-and-such?" However, the students are not allowed to perform calculations. Their answers must be immediate, based only on their raw intuitions. They are instructed to frame each answer in terms of a zero to 100 percent rating scale, with 0\% corresponding to P=0.0, 27\% corresponding to P=.27, and so forth. They are also told that they can give non-integer answers if they wish to make really fine-grained distinctions; for example, 49.0635...\%. (As it turns out, none do.) 

   The instructor of the course is particularly interested in student's responses to two of the questions, which we will designate as question A and question B. He reasons that if students have developed a good, solid understanding of the basic concepts, they will tend to give higher probability ratings for question A than for question B; whereas, if they were sleeping through that portion of the course, their answers will be mere shots in the dark and there will be no overall tendency one way or the other. The instructor's hypothesis is of course directional: he expects his students have mastered the concepts well enough to sense, if only intuitively, that the event described in question A has the higher probability. The following table shows the probability ratings of the 16 subjects for each of the two questions. 

Voila! The observed results are consistent with the hypothesis. The probability ratings do on average end up higher for question A than for question B. Now to determine whether the degree of the observed difference reflects anything more than some lucky guessing.

The sum of the signed ranks in column 7 is a quantity symbolized as W, which for the present example is equal to 67. Two of the original 16 subjects were removed from consideration because of the zero difference they produced in columns 4 and 5, so our observed value of W is based on a sample of size N=14.


\end{document}
 

Overview of Distribution-Free Tests (1 of 2)

Most (*)inferential statistics assume (*)normal distributions. 
Although these statistical tests work well even if the assumption
of normality is violated, extreme deviations from normality can 
distort the results. Usually the effect of violating the 
assumption of normality is to decrease the (*)Type I error rate. 
Although this may sound like a good thing, it often is accompanied 
by a substantial decrease in (*)power. There is a collection of tests 
called distribution-free tests that do not make any assumptions 
about the distribution from which the numbers were sampled. 
Thus the name, "distribution-free." The main advantage of 
distribution-free tests is that they provide more power than traditional 
tests when the samples are from highly-(*)skewed distributions. 
Since alternative means of dealing with skewed distributions such 
as taking logarithms or square roots of the data are available, 
distribution-free tests have not achieved a high level of popularity. 
Distribution-free tests are nonetheless a worthwhile alternative.
Moreover, computer analysis has made possible new and promising 
variations of these tests.

	

Overview of Distribution-Free Tests (2 of 2)

Distribution-free tests are sometimes referred to as nonparametric 
tests because, strictly speaking, they do not test hypotheses about 
population (*)parameters. Nonethless, they do provide a basis for 
making inferences about populations and are therefore classified 
as inferential statistics. Some researchers take the position that 
measurement on at least an (*)interval scale is necessary for techniques 
such as the analysis of variance that result in inferences about means. 
These researchers use distribution- free tests as alternatives 
to the usual parametric tests. As discussed in another section, 
most researchers do not accept the argument that tests such as 
the analysis of variance can only be done with interval-level data.



%================ section   ============================

Randomization Tests (1 of 6)

Most distribution-free tests are based on the principle of randomization. 
The best way to understand the principle of randomization 
is in terms of a specific example of a randomization test. 
Assume that four numbers are sampled from each of two populations. 
The numbers are shown below.

\begin{verbatim}
         Group 1    Group 2
           11          2
           14          9
            7          0
            8          5
      Mean 10          4
\end{verbatim}

The first step is to compute the difference between means. 
For these data, the difference is six. The second step is to compute 
the number of ways these eight numbers could be divided into two groups 
of four. The general formula is:
\[ 
  W = N! / (n_1! n_2!)
\] 
where $W$ is the number of ways, $N$ is the total number of numbers 
(8 in this case), 
$n_1$ is the size of the first group (4 in this case) and $n_2$ is 
the size of the 
second group (4 in this case). Therefore, $W = 8!/(4! 4!) = 70$. 

Randomization Tests (2 of 6)

This means that there are 70 ways in which eight numbers can be divided 
into two groups of four.
\begin{verbatim}
                            Mean 
 Group 1  ->  11  14  9  8   10.5
 Group 2  ->   2   7  0  5    3.5
\end{verbatim}
\begin{verbatim}
                            Mean 
 Group 1  ->  11  14  9  7   10.25
 Group 2  ->   2   8  0  5    3.75
\end{verbatim}

The third step is to determine how many of these $W$ ways of dividing 
the data result in differences between the means as large or larger 
than the difference obtained in the actual data. An examination of 
the data shows that there are only two ways the eight numbers can be 
divided into two groups of four with a larger difference 
between means than the difference of six found in the actual arrangement. 
These two ways are shown on the right side of this page. Thus, including the 
original data, there are three ways in which the eight numbers can be arranged so 
that the difference between means is six or more. 



Randomization Tests (3 of 6)

To compute the (*)probability value for a (*)one-tailed test of 
the difference between groups, divide this value of three by 
the $W$ of ways of dividing the data into two groups of four. 
The probability value is therefore: $p = 3/70 = .0429$. 
For a two-tailed test, 
the three cases in which Group 2 had a mean that was greater than Group 1 
by six 
or more would be considered. This would make the two-tailed probability: 
$p = 6/70 = .0857$. 
In summary, a randomization test proceeds from the data actually collected. 
It compares a computed statistic (the difference between means in this example) 
with the value of that statistic for other arrangements of the data. 
The probability value is simply the proportion of arrangements leading to 
a value of 
the statistic as large or larger than the value obtained from the actual data. 

Randomization Tests (4 of 6)

Consider one more example of a randomization test. Suppose a researcher 
wished to know 
whether or not there was a relationship between two variables: $X$ and $Y$. 
The most common 
way to test the relationship would be using (*)Pearson's $r$.
\begin{verbatim}
     X  Y
     4  3
     8  5
     2  1
    10  7
     9  8
\end{verbatim}

The test based on the principle of randomization would proceed as follows. 
First, Pearson's 
correlation would be computed for the data as they stand. The value is: 
$r = .9556$. Next, 
the number of ways the $X$ and $Y$ numbers could be paired is calculated 
(Note that $X$'s do not 
become $Y$'s and $Y$'s do not become $X$'s. It is the pairings that change.) 
The formula for the number of ways that the numbers can be paired is simply: 
$W = N!$ where 
$N$ is the number of pairs of numbers. For this example, $N = 5$ and $W = 120$. 
This means there 
are 120 ways the numbers can be paired. Of these pairings, only one would 
produce a higher 
correlation than $.9556$. It is shown on the next page. Therefore, there 
are two ways of 
arranging the data that result in correlations of $.9556$ or higher.

Randomization Tests (5 of 6)
\begin{verbatim}
      X   Y
      4   3
      8   5
      2   1
     10   8
      9   7
    r = .981
\end{verbatim}

This makes the one-tailed probability equal to : $p = 2/120 = .017$. 
Naturally, the two-tailed 
probability is $.034$. 
Randomization tests have only one major drawback: they are impractical 
to compute with moderate to large sample sizes. For example, 
the number of ways 45 scores 
can be equally divided among three groups is
\begin{verbatim}
 W = 45!/ (15! 15! 15!) = 5.34 * 10^19
\end{verbatim}
which is an astronomical number. Even high-speed computers are incapable 
of the necessary calculations. 

Randomization Tests (6 of 6)

Fortunately, there is an alternative computational method called "resampling." 
Instead of 
calculating all possible ways the data could be arranged, one can take 
numerous random 
samples from the set of possible arrangements. For the problem of dividing 
45 numbers into 
three groups of 15 each, one could randomly select $10,000$ of the possible 
arrangements and 
determine the proportion of these for which the effect size in the sample 
is exceeded. 
This proportion is then an accurate estimate of the probability level. 
This procedure is 
currently not in frequent use due, in part, to the general unavailability 
of programs to 
do this type of computation.

%================ section   ============================

Rank Randomization Tests (1 of 4)

The major problem with randomization tests is that they are very difficult to compute. 
Rank randomization tests are performed by first converting the scores to ranks and then 
computing a randomization test. The primary advantage of rank randomization tests is that 
there are tables that can be used to determine significance. The disadvantage is that some 
information is lost when the numbers are converted to ranks. Therefore, rank randomization 
tests are generally less powerful than randomization tests based on the original numbers. 
Consider the following data:
\begin{verbatim}
    Group 1     Group 2
      11          2
      14          9
       7          0
       8          5
\end{verbatim}
that were used as an example in the section on randomization tests. A rank randomization test 
on these data begins by converting the numbers to ranks. 
The converted data are:
\begin{verbatim}
    Group 1     Group 2
       7          2
       8          6
       4          1
       5          3
\end{verbatim}
The sum of the ranks of the first group is 24 and the sum of the ranks of the second group 
is 12. The difference is therefore 12. The problem is to figure out the number of ways that 
data can be arranged so that the difference between summed ranks is 12 or more. The bottom 
of this page shows the four arrangements of the eight ranks that lead to a difference of 12 
or more. Since there are $W = 8!/(4! 4!) = 70$ ways of arranging the data, the one-tailed 
probability is: $4/70 = .057$.
\begin{verbatim}
         8 4
         7 3
         6 2
         5 1
         
         8 5
         7 3
         6 2
         4 1

         8 6
         7 3
         5 2
         4 1

         8 5
         7 4
         6 2
         3 1
\end{verbatim}


The two-tailed probability is therefore: $8/70 = .114$.
Notice that this probability value is slightly higher than the $.0857$ obtained 
with the 
randomization test for these same data. This rank randomization test for differences between 
two groups has several names. It is most often called the Mann-Whitney U test or the Wilcoxon 
Rank Sum test. In practice, tables are available to look up the probability value based on 
the sum of the ranks of the group with the lower mean rank. For these data, the sum of the ranks 
is: $2+6+1+3 =12$. A table would show that for a one-tailed test to be significant at the .05 
level, the sum of the ranks must be $\le 11$. Since $12>11$, the test is not significant at this 
level. This agrees with the calculated probability value of $.057$. The purpose of this section 
is to present the concepts rather than the details of the computations. Therefore, if you wish 
to actually perform this test, you should find a textbook with appropriate tables.

Rank randomization tests are also available for correlation and for testing differences between 
more than two groups (analogous to one-way  ANOVA). The rank randomization test for correlation 
is called  Spearman's rho. The rank randomization test for differences among groups is called 
the Kruskal and Wallis Test. In summary, rank randomization tests are based on the principle of 
randomization. They are much easier to compute than randomization tests. However, they are less 
powerful than randomization tests and the more typically performed parameteric tests such as 
analysis of variance. For this reason, they are not as widely used.

%================ section   ============================
\section{��������}

Sign Test (1 of 2)

Suppose a researcher were interested in whether a drug helped people fall asleep. Subjects took 
the drug one night and a placebo another night with the order counterbalanced. The number of 
minutes it took each of 8 subjects to fall asleep is shown below.
\begin{verbatim}
       Drug       Placebo
        12          21
         9          16
        11           8
        21          36
        17          28
        22          20
        18          29
        11          22
\end{verbatim}

If the drug has no effect, then the population proportion (��)$\pi$ of 
people falling asleep faster 
in the drug condition is $.50$. Therefore, the null hypothesis is that 
$\pi  = .50$. Of the 8 subjects, 
7 fell asleep faster with the drug.



	
Sign Test (2 of 2)
Next chapter: Measuring effect size

Using the formula for the binomial distribution, the probability of obtaining 7 or more 
"successes" with $N = 8$ and $\pi  = .50$ can be calculated to be .0352 as follows:
\begin{verbatim}
P(7) = 8!/7!(8-7)! * 5^7 (1-.5)^(8-7)
     = 8 * .00391 = .03125

P(8) = 8!/8!(8-8)! * 5^8 (1-.5)^(8-8)
     = .00391
\end{verbatim}
Therefore, $P(7) + P(8) = .03125 + .00391 = .0352$. 
The one-tailed probability value is 
therefore .03516; The two-tailed probability value is $(2)(.03516) = .0732$. 
Technically speaking, 
the sign test is a test of whether the median is a particular value. In this example, the test 
is whether the median difference between conditions is zero. The sign test is generally less 
powerful than a t test of the difference between dependent means. For the example data, 
the t test would be: $t(7) = -3.21$. The two-tailed probability value for this t is 0.015.


%================ section   ============================

Chapter 17

\begin{enumerate}
 \item%
[1.] Explain the principle of randomization.
 \item% 
[2.] What is the reason that rank randomization tests rather than pure randomization tests are 
conducted.
 \item% 
[3.] Which is more powerful: randomization or rank randomization tests? Why?
 \item% 
[4.] Conduct a rank randomization test, a randomization test, and a $t$- test for the data shown 
below. Make all tests two-tailed.
\begin{verbatim}
Group 1 Group 2
   9      5
   7      2
   12     8
   16     4
   14    10 
\end{verbatim}
\end{enumerate}

(answer)
4.
Rank randomization test: $p = 0.063$.

Randomization test: $p = 0.040$

$t$-test: $t(8) = -2.675$, $p =0.028$.
\begin{verbatim}
 cf. http://davidmlane.com/hyperstat/index.html
\end{verbatim}



\end{document}
