[1]
\(\bar{y}\)
\(\displaystyle \bar{y}=\frac{n_1\bar{y_1}+n_2\bar{y_2}}{n}\)
\(s^2\)
\(\displaystyle s^2= \frac{1}{n-1}\sum_{i=1}^n(y_i-\bar{y})^2\)
\(\displaystyle (n-1)s^2= \sum_{i=1}^{n_1} (y_i-\bar{y})^2+\sum_{i=n_1+1}^{n} (y_i-\bar{y})^2\)
右辺の第1項について、
\(\displaystyle \sum_{i=1}^{n_1} (y_i-\bar{y})^2\)\(\displaystyle = \sum_{i=1}^{n_1} ((y_i-\bar{y_1})-(\bar{y}-\bar{y_1}))^2\)
\(\displaystyle = \sum_{i=1}^{n_1} (y_i-\bar{y_1})^2 \)\(\displaystyle-2\sum_{i=1}^{n_1}(y_i-\bar{y_1})(\bar{y}-\bar{y_1})\)\(\displaystyle+\sum_{i=1}^{n_1}(\bar{y}-\bar{y_1})^2\)
\(\displaystyle = \sum_{i=1}^{n_1} (y_i-\bar{y_1})^2 \)\(\displaystyle-2\sum_{i=1}^{n_1}(y_i-\bar{y_1})(\bar{y}-\bar{y_1})\)\(\displaystyle+\sum_{i=1}^{n_1}(\bar{y}-\bar{y_1})^2\)
\(\displaystyle = (n_1-1)s_1^2 \)\(\displaystyle+n_1(\bar{y}-\bar{y_1})^2\)
同様に、
\(\displaystyle \sum_{i=n_1+1}^{n} (y_i-\bar{y})^2\)\(\displaystyle = (n_2-1)s_2^2 \)\(\displaystyle+n_2(\bar{y}-\bar{y_2})^2\)
以上より、
\( (n-1)s^2\)\(\displaystyle = (n_1-1)s_1^2 \)\(\displaystyle+n_1(\bar{y}-\bar{y_1})^2\)\(\displaystyle + (n_2-1)s_2^2 \)\(\displaystyle+n_2(\bar{y}-\bar{y_2})^2\)
\( = (n_1-1)s_1^2\)\(\displaystyle + (n_2-1)s_2^2 \)\(\displaystyle +\frac{n_1n_2}{n}(\bar{y_1}-\bar{y_2})^2\)
\(s^2 =\displaystyle\frac{1}{n-1}( (n_1-1)s_1^2\)\(\displaystyle + (n_2-1)s_2^2 \)\(\displaystyle +\frac{n_1n_2}{n}(\bar{y_1}-\bar{y_2})^2)\)
[2]
組標本を\((x,y)\)とすると、\(x\)の不偏分散\(s_x^2\)は、[1]の\(s^2\)で\(s_1^2=0\)\(,s_2^2=0\)\(,\bar{y_1}=a\)\(,\bar{y_2}=-a\)で置き換えて、
\(s_x^2=\displaystyle \frac{4n_1n_2a^2}{n(n-1)}\)
不偏共分散\(s_{xy}\)は、
\(\displaystyle s_{xy} = \frac{1}{n-1}\sum_{i=1}^n (x_i -\bar{x})(y_i -\bar{y})\)
\(\displaystyle (n-1)s_{xy} = \sum_{i=1}^n x_iy_i -n\bar{x }\bar{y }\)
\(\displaystyle (n-1)s_{xy} = n_1a\bar{y_1}-n_2a\bar{y_2}\)\( \displaystyle -n\frac{n_1a+n_2(-a)}{n}\cdot\frac{n_1\bar{y_1}+n_2\bar{y_2}}{n}\)
\(\displaystyle =\frac{2n_1n_2a(\bar{y_1}-\bar{y_2})}{n}\)
以上より、
\(\displaystyle s_{xy}=\frac{2n_1n_2a(\bar{y_1}-\bar{y_2})}{n(n-1)}\)
相関係数\(\rho\)は、
\(\rho = \displaystyle \frac{s_{xy}}{\sqrt{s_x^2}\sqrt{s^2}}\)
\(= \displaystyle \frac{\frac{2n_1n_2a(\bar{y_1}-\bar{y_2})}{n(n-1)}}{\sqrt{\frac{4n_1n_2a^2}{n(n-1)}}\sqrt{s^2}}\)\(\displaystyle =\frac{\sqrt{n_1n_2}(\bar{y_1}-\bar{y_2})}{\sqrt{n(n-1)s^2}}\)
[3]
[1],[2]で求めた式に代入して値を得る。
[4]
\(f(z|z \geq 0) =\displaystyle \frac{f(z)}{\int^\infty_0 f(z)dz}\)\(=\displaystyle 2 f(z)\)
\(E[Z|Z \geq 0] = \displaystyle \int^\infty_0 zf(z|z \geq 0) dz\)
\( = \displaystyle 2\int^\infty_0 z\frac{1}{\sqrt{2\pi}}e^{-z^2/2} dz\)\(\displaystyle =\sqrt{\frac{2}{\pi}}\)
[5]
\(P(X \geq 0) = P(X<0) =0.5\)より、
\(P(T=t)=0.5 \quad (t= a,-a)\)
なので、
\(E[T]=0.5a +0.5(-a)=0\)
\(V[T]=0.5a^2+0.5(-a)^2=a^2\)
[6]
\(E[Y|X \geq 0]\)
\(X\)を定数で固定した場合の条件付確率変数\(Y|X\)は条件より、
\(Y|X=x \sim N(\rho x, 1-\rho^2)\)
なので、
\(E[Y|X \geq 0] =E[\rho X|X \geq 0] \)\(=\rho E[X|X \geq 0] \)\(\displaystyle =\sqrt{\frac{2}{\pi}}\rho\) ([3]を利用)
\(\xi\)
\(\displaystyle \xi = \frac{Cov[T,Y]}{\sqrt{V[T]}\sqrt{V[Y]}}\)
\(\displaystyle = \frac{E[TY]-E[T]E[Y]}{\sqrt{a^2}\sqrt{1}}\)\(\displaystyle = \frac{E[TY]}{a}\)
\(E[TY] \)\(\displaystyle = \int^{\infty}_{-\infty} \sum_{i=1}^2 t_iy f(t_i,y)dy\) (ただし、\(t_1 = a,t_2 = -a\))
\(T\)が決まると\(X\)の正負が決まり、\(X\)の正負が決まると\(Y\)の条件付き分布が決まるので\(Y|T\)を考えていくと、
\(\displaystyle = \int^{\infty}_{-\infty} \sum_{i=1}^2 t_iy f(y|t_i) f(t_i)dy\)
\(\displaystyle = \frac{1}{2}\int^{\infty}_{-\infty} \{ay f(y|t=a)-ayf(y|t=-a)\} dy\)
\(\displaystyle = \frac{1}{2}\int^{\infty}_{-\infty} \{ay f(y|x \geq 0)-ayf(y| x<0 )\} dy\)
\(\displaystyle = \frac{1}{2}aE[Y|X \geq0]- \frac{1}{2}aE[Y|X <0]\)
\(\displaystyle = \sqrt{\frac{2}{\pi}}\rho a\)
以上より、
\(\displaystyle \xi = \sqrt{\frac{2}{\pi}}\rho \)
[3]との関連
\(\xi\)は組標本\((x_1,y_1),…,(x_n,y_n)\)に対して、\(x_i\)が\(0\)以上の時\(a\)、\(x_i\)が\(0\)以下の時\(-a\)で置き換えて標本の相関係数を計算したものである。
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