(1) 共分散の定義より s x y = 1 n ∑ k = 1 n ( x k − x ˉ ) ( y k − y ˉ ) . \displaystyle
s_{xy}=\frac1n\sum\limits _{k=1}^n(x_k-\bar{x})(y_k-\bar{y}). s x y = n 1 k = 1 ∑ n ( x k − x ˉ ) ( y k − y ˉ ) . 1 n ∑ k = 1 n x k = x ˉ \displaystyle \frac1n\sum\limits _{k=1}^n x_k=\bar{x} n 1 k = 1 ∑ n x k = x ˉ 1 n ∑ k = 1 n y k = y ˉ \displaystyle \frac1n\sum\limits _{k=1}^n y_k=\bar{y} n 1 k = 1 ∑ n y k = y ˉ 1 n ∑ k = 1 n x k y k = z ˉ \displaystyle \frac1n\sum\limits _{k=1}^n x_ky_k=\bar{z} n 1 k = 1 ∑ n x k y k = z ˉ s x y = 1 n ∑ k = 1 n x k y k − y ˉ 1 n ∑ k = 1 n x k − x ˉ 1 n ∑ k = 1 n y k + x ˉ y ˉ = z ˉ − x ˉ y ˉ − x ˉ y ˉ + x ˉ y ˉ = z ˉ − x ˉ y ˉ . \displaystyle
\begin{aligned}
s_{xy}
&=\frac1n\sum\limits _{k=1}^n x_ky_k-\bar{y}\frac1n\sum\limits _{k=1}^n x_k-\bar{x}\frac1n\sum\limits _{k=1}^n y_k+\bar{x}\bar{y}\\
&=\bar{z}-\bar{x}\bar{y}-\bar{x}\bar{y}+\bar{x}\bar{y}
=\bar{z}-\bar{x}\bar{y}.
\end{aligned} s x y = n 1 k = 1 ∑ n x k y k − y ˉ n 1 k = 1 ∑ n x k − x ˉ n 1 k = 1 ∑ n y k + x ˉ y ˉ = z ˉ − x ˉ y ˉ − x ˉ y ˉ + x ˉ y ˉ = z ˉ − x ˉ y ˉ .
(2) q = n / l q=n/l q = n / l m = n m=n m = n k = 1 , … , n k=1,\ldots,n k = 1 , … , n y k = k y_k=k y k = k k = j l + t k=jl+t k = j l + t j = 0 , … , q − 1 j=0,\ldots,q-1 j = 0 , … , q − 1 t = 1 , … , l t=1,\ldots,l t = 1 , … , l x k = t x_k=t x k = t y k = j l + t y_k=jl+t y k = j l + t x x x 1 , 2 , … , l 1,2,\ldots,l 1 , 2 , … , l q q q y y y 1 , 2 , … , n 1,2,\ldots,n 1 , 2 , … , n
1 , 2 , … , N 1,2,\ldots,N 1 , 2 , … , N ∑ a = 1 N a = N ( N + 1 ) / 2 \displaystyle \sum\limits _{a=1}^N a=N(N+1)/2 a = 1 ∑ N a = N ( N + 1 ) /2 ∑ a = 1 N a 2 = N ( N + 1 ) ( 2 N + 1 ) / 6 \displaystyle \sum\limits _{a=1}^N a^2=N(N+1)(2N+1)/6 a = 1 ∑ N a 2 = N ( N + 1 ) ( 2 N + 1 ) /6 N + 1 2 , 1 N ∑ a = 1 N ( a − N + 1 2 ) 2 = N 2 − 1 12 \displaystyle
\frac{N+1}{2},\qquad \frac1N\sum\limits _{a=1}^N\left(a-\frac{N+1}{2}\right)^2=\frac{N^2-1}{12} 2 N + 1 , N 1 a = 1 ∑ N ( a − 2 N + 1 ) 2 = 12 N 2 − 1 x ˉ = l + 1 2 , y ˉ = n + 1 2 , s x 2 = l 2 − 1 12 , s y 2 = n 2 − 1 12 . \displaystyle
\bar{x}=\frac{l+1}{2},\quad \bar{y}=\frac{n+1}{2},\quad s_x^2=\frac{l^2-1}{12},\quad s_y^2=\frac{n^2-1}{12}. x ˉ = 2 l + 1 , y ˉ = 2 n + 1 , s x 2 = 12 l 2 − 1 , s y 2 = 12 n 2 − 1 . s x , s y s_x,s_y s x , s y x , y x,y x , y
また y k − y ˉ = ( t − l + 1 2 ) + l ( j − q − 1 2 ) , x k − x ˉ = t − l + 1 2 . \displaystyle
y_k-\bar{y}=\left(t-\frac{l+1}{2}\right)+l\left(j-\frac{q-1}{2}\right),\qquad x_k-\bar{x}=t-\frac{l+1}{2}. y k − y ˉ = ( t − 2 l + 1 ) + l ( j − 2 q − 1 ) , x k − x ˉ = t − 2 l + 1 . j j j ∑ t = 1 l ( t − ( l + 1 ) / 2 ) = 0 \displaystyle \sum\limits _{t=1}^l(t-(l+1)/2)=0 t = 1 ∑ l ( t − ( l + 1 ) /2 ) = 0 s x y = 1 q l ∑ j = 0 q − 1 ∑ t = 1 l ( t − l + 1 2 ) [ ( t − l + 1 2 ) + l ( j − q − 1 2 ) ] = 1 l ∑ t = 1 l ( t − l + 1 2 ) 2 = l 2 − 1 12 . \displaystyle
\begin{aligned}
s_{xy}
&=\frac1{ql}\sum\limits _{j=0}^{q-1}\sum\limits _{t=1}^l\left(t-\frac{l+1}{2}\right)\left[\left(t-\frac{l+1}{2}\right)+l\left(j-\frac{q-1}{2}\right)\right]\\
&=\frac1l\sum\limits _{t=1}^l\left(t-\frac{l+1}{2}\right)^2
=\frac{l^2-1}{12}.
\end{aligned} s x y = q l 1 j = 0 ∑ q − 1 t = 1 ∑ l ( t − 2 l + 1 ) [ ( t − 2 l + 1 ) + l ( j − 2 q − 1 ) ] = l 1 t = 1 ∑ l ( t − 2 l + 1 ) 2 = 12 l 2 − 1 . r = s x y s x s y = ( l 2 − 1 ) / 12 ( ( l 2 − 1 ) / 12 ) ( ( n 2 − 1 ) / 12 ) = l 2 − 1 n 2 − 1 . \displaystyle
r=\frac{s_{xy}}{s_xs_y}
=\frac{(l^2-1)/12}{\sqrt{((l^2-1)/12)((n^2-1)/12)}}
=\sqrt{\frac{l^2-1}{n^2-1}}. r = s x s y s x y = (( l 2 − 1 ) /12 ) (( n 2 − 1 ) /12 ) ( l 2 − 1 ) /12 = n 2 − 1 l 2 − 1 . l ≧ 2 l\ge2 l ≧ 2 n ≧ 2 n\ge2 n ≧ 2 n = l n=l n = l r = 1 r=1 r = 1 x k = y k = k x_k=y_k=k x k = y k = k
(3) k − 1 = q l + i k-1=ql+i k − 1 = q l + i q = 0 , 1 , … , l q=0,1,\ldots,l q = 0 , 1 , … , l i = 0 , 1 , … , l − 1 i=0,1,\ldots,l-1 i = 0 , 1 , … , l − 1 x k = i + 1 , y k = R l + 1 ( q l + i ) + 1 x_k=i+1,\qquad y_k=R_{l+1}(ql+i)+1 x k = i + 1 , y k = R l + 1 ( q l + i ) + 1 i i i q , q ′ q,q' q , q ′ l + 1 l+1 l + 1 l ( q − q ′ ) l(q-q') l ( q − q ′ ) l , l + 1 l,l+1 l , l + 1 l + 1 l+1 l + 1 q − q ′ q-q' q − q ′ 0 < ∣ q − q ′ ∣ ≦ l 0<|q-q'|\le l 0 < ∣ q − q ′ ∣ ≦ l q = 0 , 1 , … , l q=0,1,\ldots,l q = 0 , 1 , … , l y k y_k y k 1 , 2 , … , l + 1 1,2,\ldots,l+1 1 , 2 , … , l + 1 ( x k , y k ) (x_k,y_k) ( x k , y k ) 1 ≦ x ≦ l 1\le x\le l 1 ≦ x ≦ l 1 ≦ y ≦ l + 1 1\le y\le l+1 1 ≦ y ≦ l + 1
したがって x ˉ = l + 1 2 , y ˉ = l + 2 2 , z ˉ = 1 l ( l + 1 ) ( ∑ a = 1 l a ) ( ∑ b = 1 l + 1 b ) = ( l + 1 ) ( l + 2 ) 4 = x ˉ y ˉ . \displaystyle
\bar{x}=\frac{l+1}{2},\quad \bar{y}=\frac{l+2}{2},\quad
\bar{z}=\frac{1}{l(l+1)}\left(\sum\limits _{a=1}^l a\right)\left(\sum\limits _{b=1}^{l+1}b\right)
=\frac{(l+1)(l+2)}4
=\bar{x}\bar{y}. x ˉ = 2 l + 1 , y ˉ = 2 l + 2 , z ˉ = l ( l + 1 ) 1 ( a = 1 ∑ l a ) ( b = 1 ∑ l + 1 b ) = 4 ( l + 1 ) ( l + 2 ) = x ˉ y ˉ . s x y = 0 s_{xy}=0 s x y = 0 x x x 1 , … , l 1,\ldots,l 1 , … , l y y y 1 , … , l + 1 1,\ldots,l+1 1 , … , l + 1 l ≧ 2 l\ge2 l ≧ 2 r = s x y s x s y = 0. \displaystyle
r=\frac{s_{xy}}{s_xs_y}=0. r = s x s y s x y = 0.
