ポアンカレ計量による距離ではX軸に沿った移動では線分が最短にならない

計量による距離

$$\begin{array}{rcl}{g}_{(x,y)}(\mathit{v},\mathit{w})& =& \frac{⟨\mathit{v},\mathit{w}⟩}{y^2}\cdots \mathrm{ポ}\mathrm{ア}\mathrm{ン}\mathrm{カ}\mathrm{レ}\mathrm{計}\mathrm{量},⟨\mathit{x},\mathit{y}⟩=\sum _{i=1}^n{x}_i{y}_i\mathrm{は}\mathrm{標}\mathrm{準}\mathrm{内}\mathrm{積}\\ \gamma \left(t\right)& =& (x(t),y(t))\cdots \mathrm{曲}\mathrm{線}\gamma ,t\mathrm{に}\mathrm{よ}\mathrm{る}\mathrm{媒}\mathrm{介}\mathrm{表}\mathrm{示},\mathrm{始}\mathrm{点}\gamma (a),\mathrm{終}\mathrm{点}\gamma (b)\\ {\gamma }^{\prime }\left(t\right)& =& (x^{\prime }(t),y^{\prime }(t))\cdots (x(t),y(t))\mathrm{で}\mathrm{の}\mathrm{接}\mathrm{ベ}\mathrm{ク}\mathrm{ト}\mathrm{ル}\\ L\left(\gamma \right)& =& {\int }_a^b\sqrt{g({\gamma }^{\prime }\left(t\right),{\gamma }^{\prime }\left(t\right))}\cdots \gamma \mathrm{の}\mathrm{計}\mathrm{量}\mathrm{に}\mathrm{よ}\mathrm{る}\mathrm{距}\mathrm{離}\\ & =& {\int }_a^b\sqrt{\frac{⟨{\gamma }^{\prime }\left(t\right),{\gamma }^{\prime }\left(t\right)⟩}{y^2}}\mathrm{d}t\\ & =& {\int }_a^b\frac{\sqrt{⟨{\gamma }^{\prime }\left(t\right),{\gamma }^{\prime }\left(t\right)⟩}}{\sqrt{y^2}}\mathrm{d}t\\ & =& {\int }_a^b\frac{||{\gamma }^{\prime }\left(t\right)||}{y}\mathrm{d}t\cdots ||x||=\sqrt{⟨\mathit{x},\mathit{x}⟩}\end{array}$$

$(\cos \left(\theta \right),\sin \left(\theta \right))$から $(-\cos \left(\theta \right),\sin \left(\theta \right))$までの線分で移動(ただし$\theta \in (0,\frac{\pi }{2})$)

$$\begin{array}{rcl}{\gamma }_3(t)& =& ((x_1-x_0)t+x_0,(y_1-y_0)t+y_0)\\ & =& ((-\cos \left(\theta \right)-\cos \left(\theta \right))t+\cos \left(\theta \right),(\sin \left(\theta \right)-\sin \left(\theta \right))t+\sin \left(\theta \right))\\ & =& ((1-2t)\cos \left(\theta \right),\sin \left(\theta \right))\\ {\gamma }_3^{\prime }(t)& =& (-2\cos \left(\theta \right),0)\\ L_3=L\left({\gamma }_3\right)& =& {\int }_0^1\frac{||{\gamma }_3^{\prime }\left(t\right)||}{y(t)}\mathrm{d}t\\ & =& {\int }_0^1\frac{\sqrt{{(-2\cos \left(\theta \right))}^2+0^2}}{\sin \left(\theta \right)}\mathrm{d}t\\ & =& {\int }_0^1\frac{2\cos \left(\theta \right)}{\sin \left(\theta \right)}\mathrm{d}t\\ & =& {\int }_0^1\frac{2}{\tan \left(\theta \right)}\mathrm{d}t\\ & =& \frac{2}{\tan \left(\theta \right)}{\int }_0^1\mathrm{d}t\\ & =& \frac{2}{\tan \left(\theta \right)}[t{]}_0^1\\ & =& \frac{2}{\tan \left(\theta \right)}[1-0]\\ & =& \frac{2}{\tan \left(\theta \right)}\cdot 1\\ & =& \frac{2}{\tan \left(\theta \right)}\end{array}$$

$(\cos \left(\theta \right),\sin \left(\theta \right))$から $(-\cos \left(\theta \right),\sin \left(\theta \right))$まで，原点を中心とした半径1の円弧で移動(ただし $\theta \in (0,\frac{\pi }{2})$)

$$\begin{array}{rcl}{\gamma }_4(t)& =& (\cos (\theta +(\pi -2\theta )t),\sin (\theta +(\pi -2\theta )t))\\ {\gamma }_4^{\prime }(t)& =& (-(\pi -2\theta )\sin (\theta +(\pi -2\theta )t),(\pi -2\theta )\cos (\theta +(\pi -2\theta )t))\\ L_4=L\left({\gamma }_4\right)& =& {\int }_0^1\frac{||{\gamma }_4^{\prime }\left(t\right)||}{y(t)}\mathrm{d}t\\ & =& {\int }_0^1\frac{\sqrt{{(-(\pi -2\theta )\sin (\theta +(\pi -2\theta )t))}^2+{((\pi -2\theta )\cos (\theta +(\pi -2\theta )t))}^2}}{\sin (\theta +(\pi -2\theta )t)}\mathrm{d}t\\ & =& {\int }_0^1\frac{\sqrt{(\pi -2\theta {)}^2({\sin }^2(\theta +(\pi -2\theta )t)+{\cos }^2(\theta +(\pi -2\theta )t))}}{\sin (\theta +(\pi -2\theta )t)}\mathrm{d}t\\ & =& {\int }_0^1\frac{\sqrt{(\pi -2\theta {)}^2\cdot 1}}{\sin (\theta +(\pi -2\theta )t)}\mathrm{d}t\\ & =& {\int }_0^1\frac{\pi -2\theta }{\sin (\theta +(\pi -2\theta )t)}\mathrm{d}t\\ & =& {\int }_0^1\frac{\alpha }{\sin (\beta +\alpha t)}\mathrm{d}t\\ & =& {\int }_{\beta }^{\alpha +\beta }\frac{\cancel{\alpha }}{\sin \left(u\right)}\frac{1}{\cancel{\alpha }}\mathrm{d}u\\ & =& {[\ln |\tan \left(\frac{u}{2}\right)|]}_{\beta }^{\alpha +\beta }\cdots \int \frac{1}{\sin \left(x\right)}\mathrm{d}x=\ln |\tan \left(\frac{x}{2}\right)|+C\\ & =& {[\ln |\tan \left(\frac{u}{2}\right)|]}_{\theta }^{\pi -2\theta +\theta }\\ & =& {[\ln |\tan \left(\frac{u}{2}\right)|]}_{\theta }^{\pi -\theta }\\ & =& \ln |\tan \left(\frac{\pi -\theta }{2}\right)|-\ln |\tan \left(\frac{\theta }{2}\right)|\\ & =& \ln \left|\frac{\tan \left(\frac{\pi -\theta }{2}\right)}{\tan \left(\frac{\theta }{2}\right)}\right|\\ & =& \ln \left|\frac{\cot \left(\frac{\theta }{2}\right)}{\tan \left(\frac{\theta }{2}\right)}\right|\\ & =& \ln \left|\frac{1}{{\tan }^2\left(\frac{\theta }{2}\right)}\right|\\ & =& \ln |{\tan }^{-2}\left(\frac{\theta }{2}\right)|\\ & =& -2\ln |\tan \left(\frac{\theta }{2}\right)|\end{array}$$

$L_3-L_4$

$$\begin{array}{rcl}{L}_3-L_4& =& \left\{\frac{2}{\tan \left(\theta \right)}\right\}-\{-2\ln (\tan \left(\frac{\theta }{2}\right))\}\\ & =& 2\{\frac{1}{\tan \left(\theta \right)}+\ln (\tan \left(\frac{\theta }{2}\right))\}\\ \frac{\mathrm{d}}{\mathrm{d}\theta }(L_3-L_4)& =& 2\{\frac{-1}{{\sin }^2\left(\theta \right)}+\frac{1}{2\cos \left(\frac{\theta }{2}\right)\sin \left(\frac{\theta }{2}\right)}\}\\ & =& 2\{\frac{-1}{{\sin }^2\left(\theta \right)}+\frac{1}{\sin \left(\theta \right)}\}\\ & =& 2\{\frac{-1}{{\sin }^2\left(\theta \right)}+\frac{1}{\sin \left(\theta \right)}\frac{\sin \left(\theta \right)}{\sin \left(\theta \right)}\}\\ & =& 2\{\frac{-1}{{\sin }^2\left(\theta \right)}+\frac{\sin \left(\theta \right)}{{\sin }^2\left(\theta \right)}\}\\ & =& 2\left\{\frac{\sin \left(\theta \right)-1}{{\sin }^2\left(\theta \right)}\right\}\le 0\cdots \mathrm{分}\mathrm{母}\mathrm{は}\mathrm{常}\mathrm{に}\mathrm{正}，\mathrm{分}\mathrm{子}\mathrm{は}\theta \in (0,\frac{\pi }{2})\mathrm{な}\mathrm{の}\mathrm{で}(-1,0)\mathrm{で}\mathrm{あ}\mathrm{り}，\mathrm{全}\mathrm{体}\mathrm{で}\mathrm{は}\mathrm{常}\mathrm{に}\mathrm{負}\mathrm{と}\mathrm{な}\mathrm{る}．\\ \underset{\theta \to \frac{\pi }{2}-0}{lim}{L}_3-L_4& =& 0\\ & & \cdots \underset{\theta \to \frac{\pi }{2}-0}{lim}\frac{1}{\tan \left(\theta \right)}=0\\ & & \cdots \underset{\theta \to \frac{\pi }{2}-0}{lim}\ln (\tan \left(\frac{\theta }{2}\right))=0\\ \mathrm{よ}\mathrm{っ}\mathrm{て}\\ L_3-L_4& \ge & 0\cdots ，\theta \in (0,\frac{\pi }{2})\mathrm{で}\mathrm{は}\mathrm{微}\mathrm{分}\mathrm{が}\mathrm{常}\mathrm{に}\mathrm{負}(\mathrm{減}\mathrm{少}\mathrm{関}\mathrm{数})\mathrm{で}\mathrm{あ}\mathrm{り}，\theta =\frac{\pi }{2}\mathrm{で}0\mathrm{な}\mathrm{の}\mathrm{で}，\mathrm{範}\mathrm{囲}\mathrm{内}\mathrm{で}\mathrm{は}\mathrm{常}\mathrm{に}\mathrm{正}．\end{array}$$

まとめ

ポアンカレ計量では水平の線分より，$x$軸から離れる側となる円弧に沿った経路の方が短いことになる．
これは境界$x$軸$y=0$に近いほど計量中にある$\frac{1}{y^2}$の効果で値が大きくなり，距離としては長くなることが影響した結果．

ポアンカレ計量による距離でもY軸に沿った移動なら線分が最短になる

計量による距離

$$\begin{array}{rcl}{g}_{(x,y)}(\mathit{v},\mathit{w})& =& \frac{⟨\mathit{v},\mathit{w}⟩}{y^2}\cdots \mathrm{ポ}\mathrm{ア}\mathrm{ン}\mathrm{カ}\mathrm{レ}\mathrm{計}\mathrm{量},⟨\mathit{x},\mathit{y}⟩=\sum _{i=1}^n{x}_i{y}_i\mathrm{は}\mathrm{標}\mathrm{準}\mathrm{内}\mathrm{積}\\ \gamma \left(t\right)& =& (x(t),y(t))\cdots \mathrm{曲}\mathrm{線}\gamma ,t\mathrm{に}\mathrm{よ}\mathrm{る}\mathrm{媒}\mathrm{介}\mathrm{表}\mathrm{示},\mathrm{始}\mathrm{点}\gamma (a),\mathrm{終}\mathrm{点}\gamma (b)\\ {\gamma }^{\prime }\left(t\right)& =& (x^{\prime }(t),y^{\prime }(t))\cdots (x(t),y(t))\mathrm{で}\mathrm{の}\mathrm{接}\mathrm{ベ}\mathrm{ク}\mathrm{ト}\mathrm{ル}\\ L\left(\gamma \right)& =& {\int }_a^b\sqrt{g({\gamma }^{\prime }\left(t\right),{\gamma }^{\prime }\left(t\right))}\cdots \gamma \mathrm{の}\mathrm{計}\mathrm{量}\mathrm{に}\mathrm{よ}\mathrm{る}\mathrm{距}\mathrm{離}\\ & =& {\int }_a^b\sqrt{\frac{⟨{\gamma }^{\prime }\left(t\right),{\gamma }^{\prime }\left(t\right)⟩}{y^2}}\mathrm{d}t\\ & =& {\int }_a^b\frac{\sqrt{⟨{\gamma }^{\prime }\left(t\right),{\gamma }^{\prime }\left(t\right)⟩}}{\sqrt{y^2}}\mathrm{d}t\\ & =& {\int }_a^b\frac{||{\gamma }^{\prime }\left(t\right)||}{y}\mathrm{d}t\cdots ||x||=\sqrt{⟨\mathit{x},\mathit{x}⟩}\end{array}$$

$(0,y_0)$から$(0,y_1)$までの線分${\gamma }_1$

$$\begin{array}{rcl}{\gamma }_1\left(t\right)& =& (0,(y_1-y_0)t+y_0)\cdots t\in [0,1]\\ {\gamma }_1^{\prime }\left(t\right)& =& (0,y_1-y_0)\\ L_1=L\left({\gamma }_1\right)& =& {\int }_0^1\frac{||{\gamma }_1^{\prime }\left(t\right)||}{y(t)}\mathrm{d}t\\ & =& {\int }_0^1\frac{\sqrt{0^2+(y_1-y_0{)}^2}}{(y_1-y_0)t+y_0}\mathrm{d}t\\ & =& {\int }_0^1\frac{y_1-y_0}{(y_1-y_0)t+y_0}\mathrm{d}t\\ & =& {\int }_{y_0}^{y_1}\frac{\cancel{y_1-y_0}}{u}\frac{1}{\cancel{y_1-y_0}}\mathrm{d}u\\ & =& {\int }_{y_0}^{y_1}\frac{1}{u}\mathrm{d}u\\ & =& {[\log \left(u\right)]}_{y_0}^{y_1}\\ & =& \log \left(y_1\right)-\log \left(y_0\right)\\ & =& \log \left(\frac{y_1}{y_0}\right)\end{array}$$

$(0,y_0)$から$(0,y_1)$までの任意の曲線${\gamma }_2$

$$\begin{array}{rcl}{\gamma }_2\left(t\right)& =& (x(t),y(t))\cdots t\in [0,1]\\ {\gamma }_2^{\prime }\left(t\right)& =& (x^{\prime }(t),y^{\prime }(t))\\ L_2=L\left({\gamma }_2\right)& =& {\int }_0^1\frac{||{\gamma }_2^{\prime }\left(t\right)||}{y(t)}\mathrm{d}t\\ & =& {\int }_0^1\frac{\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}}{y(t)}\mathrm{d}t\\ & \ge & {\int }_0^1\frac{\sqrt{y^{\prime }(t{)}^2}}{y(t)}\mathrm{d}t\cdots x^{\prime }(t{)}^2(\ge 0)\mathrm{を}\mathrm{取}\mathrm{り}\mathrm{除}\mathrm{く}\mathrm{の}\mathrm{で}\mathrm{等}\mathrm{し}\mathrm{い}\mathrm{か}\mathrm{小}\mathrm{さ}\mathrm{く}\mathrm{な}\mathrm{る}\\ & =& {\int }_0^1\frac{|y^{\prime }(t)|}{y(t)}\mathrm{d}t\\ & =& {[\log (y(t))]}_0^1\\ & =& \log (y(1))-\log (y(0))\\ & =& \log \left(y_1\right)-\log \left(y_0\right)\\ & =& \log \left(\frac{y_1}{y_0}\right)=L_1\\ L_2& \ge & L_1\cdots y\mathrm{軸}\mathrm{方}\mathrm{向}\mathrm{で}\mathrm{は}\mathrm{線}\mathrm{分}\mathrm{が}\mathrm{一}\mathrm{番}\mathrm{短}\mathrm{い}\end{array}$$

独立なガウス分布に従う誤差を含む回帰モデルがn個の同時確率を考える

独立なガウス分布$N({\mu }_i,{\sigma }_i^2)$に従う誤差${ϵ}_i$を含む回帰モデル$y_i=\beta x_i+\alpha +{ϵ}_i$が$n$個の同時確率を考える

準備

$$\begin{array}{rcl}\overline{x}& =& \frac{1}{n}\sum _{i=1}^n{x}_i\\ \overline{y}& =& \frac{1}{n}\sum _{i=1}^n{y}_i\\ \sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})& =& \sum _{i=1}^n(x_i{y}_i-x_i\overline{y}-\overline{x}{y}_i+\overline{x}\overline{y})\\ & =& \sum _{i=1}^n{x}_i{y}_i-\overline{y}\sum _{i=1}^n{x}_i-\overline{x}\sum _{i=1}^n{y}_i+\overline{x}\overline{y}\sum _{i=1}^{n}1\\ & =& \sum _{i=1}^n{x}_i{y}_i-\overline{y}n\overline{x}-\overline{x}n\overline{y}+\overline{x}\overline{y}n\\ & =& \sum _{i=1}^n{x}_i{y}_i-2n\overline{x}\overline{y}+n\overline{x}\overline{y}\\ & =& \sum _{i=1}^n{x}_i{y}_i-n\overline{x}\overline{y}\cdots S_{xy}\mathrm{の}\mathrm{式}\mathrm{展}\mathrm{開}\\ \sum _{i=1}^n(x_i-\overline{x}{)}^2& =& \sum _{i=1}^n(x_i^2-2x_i\overline{x}+{\overline{x}}^2)\\ & =& \sum _{i=1}^n{x}_i^2-2\overline{x}\sum _{i=1}^n{x}_i+{\overline{x}}^2\sum _{i=1}^{n}1\\ & =& \sum _{i=1}^n{x}_i^2-2\overline{x}n\overline{x}+{\overline{x}}^{2}n\\ & =& \sum _{i=1}^n{x}_i^2-2n{\overline{x}}^2+n{\overline{x}}^2\\ & =& \sum _{i=1}^n{x}_i^2-n{\overline{x}}^2\cdots S_{xx}\mathrm{の}\mathrm{式}\mathrm{展}\mathrm{開}\end{array}$$

同時確率を考える

$$\begin{array}{rcl}p({ϵ}_1,{ϵ}_2,\cdots ,{ϵ}_n)& =& \frac{1}{\sqrt{2\pi {{\sigma }_1}^2}}{e}^{-\frac{{({ϵ}_1-{\mu }_1)}^2}{2{{\sigma }_1}^2}}\times \frac{1}{\sqrt{2\pi {{\sigma }_2}^2}}{e}^{-\frac{{({ϵ}_2-{\mu }_2)}^2}{2{{\sigma }_2}^2}}\times \cdots \times \frac{1}{\sqrt{2\pi {{\sigma }_n}^2}}{e}^{-\frac{{({ϵ}_n-{\mu }_n)}^2}{2{{\sigma }_n}^2}}\\ & =& \left(\prod _{i=1}^n\frac{1}{\sqrt{2\pi {{\sigma }_i}^2}}\right)\left(\prod _{i=1}^n{e}^{-\frac{{({ϵ}_i-{\mu }_i)}^2}{2{{\sigma }_i}^2}}\right)\\ & =& \left(\prod _{i=1}^n{\left(2\pi {{\sigma }_i}^2\right)}^{-\frac{1}{2}}\right)\left(\prod _{i=1}^n{e}^{-\frac{{(y_i-\beta x_i-\alpha -{\mu }_i)}^2}{2{{\sigma }_i}^2}}\right)\cdots y_i=\beta x_i+\alpha +{ϵ}_i,{ϵ}_i=y_i-\beta x_i-\alpha \end{array}$$

負の対数によって情報量を求める

$$\begin{array}{r}\\ -\ln \left(p\right)& =& -\ln \left(\left(\prod _{i=1}^n{\left(2\pi {{\sigma }_i}^2\right)}^{-\frac{1}{2}}\right)\left(\prod _{i=1}^n{e}^{-\frac{1}{2{{\sigma }_i}^2}{(y_i-\beta x_i-\alpha -{\mu }_i)}^2}\right)\right)\\ & =& -\ln \left(\prod _{i=1}^n{\left(2\pi {{\sigma }_i}^2\right)}^{-\frac{1}{2}}\right)-\ln \left(\prod _{i=1}^n{e}^{-\frac{1}{2{{\sigma }_i}^2}{(y_i-\beta x_i-\alpha -{\mu }_i)}^2}\right)\\ & =& -\sum _{i=1}^n\ln \left({\left(2\pi {{\sigma }_i}^2\right)}^{-\frac{1}{2}}\right)-\sum _{i=1}^n\ln \left(e^{-\frac{1}{2{{\sigma }_i}^2}{(y_i-\beta x_i-\alpha -{\mu }_i)}^2}\right)\\ & =& -\sum _{i=1}^n\{-\frac{1}{2}\ln \left(2\pi {{\sigma }_i}^2\right)\}-\sum _{i=1}^n(-\frac{1}{2{{\sigma }_i}^2}{(y_i-\beta x_i-\alpha -{\mu }_i)}^2)\\ & =& \sum _{i=1}^n\{\frac{1}{2}\ln \left(2\pi {{\sigma }_i}^2\right)\}+\sum _{i=1}^n\frac{1}{2{{\sigma }_i}^2}{(y_i-\beta x_i-\alpha -{\mu }_i)}^2\\ & =& \sum _{i=1}^n\{\frac{1}{2}\ln \left(2\pi {{\sigma }_i}^2\right)\}+\sum _{i=1}^n\frac{1}{2{{\sigma }_i}^2}(y_i^2+{\beta }^2{x}_i^2+{\alpha }^2+{\mu }_i^2-2\beta x_i{y}_i-2\alpha y_i-2y_i{\mu }_i+2\alpha \beta x_i+2\beta x_i{\mu }_i+2\alpha {\mu }_i)\end{array}$$

回帰モデルの係数で微分する

$\beta$について． $$\begin{array}{rcl}\frac{\partial }{\partial \beta }(-\ln \left(p\right))& =& 0+\sum _{i=1}^n\frac{1}{\cancel{2}{{\sigma }_i}^2}(0+\cancel{2}\beta x_i^2+0+0-\cancel{2}{x}_i{y}_i-0-0+\cancel{2}\alpha x_i+\cancel{2}{x}_i{\mu }_i+0)\\ & =& \sum _{i=1}^n\frac{1}{{{\sigma }_i}^2}(\beta x_i^2-x_i{y}_i+\alpha x_i+x_i{\mu }_i)\\ & =& \beta \sum _{i=1}^n\frac{x_i^2}{{{\sigma }_i}^2}-\sum _{i=1}^n\frac{x_i(y_i-\alpha -{\mu }_i)}{{{\sigma }_i}^2}\end{array}$$ $\alpha$について． $$\begin{array}{rcl}\frac{\partial }{\partial \alpha }(-\ln \left(p\right))& =& 0+\sum _{i=1}^n\frac{1}{\cancel{2}{{\sigma }_i}^2}(0+0+\cancel{2}\alpha +0-0-\cancel{2}{y}_i-0+\cancel{2}\beta x_i+0+\cancel{2}{\mu }_i)\\ & =& \sum _{i=1}^n\frac{1}{{{\sigma }_i}^2}(\alpha -y_i+\beta x_i+{\mu }_i)\\ & =& \alpha \sum _{i=1}^n\frac{1}{{{\sigma }_i}^2}-\sum _{i=1}^n\frac{y_i-\beta x_i-{\mu }_i}{{{\sigma }_i}^2}\end{array}$$

微分が0となる$\hat{\alpha },\hat{\beta }$,つまり極値となるパラメータを求める

$\hat{\beta }$について． $$\begin{array}{r}\\ 0& =& \frac{\partial }{\partial \beta }(-\ln \left(p\right))\\ & =& \hat{\beta }\sum _{i=1}^n\frac{x_i^2}{{{\sigma }_i}^2}-\sum _{i=1}^n\frac{x_i(y_i-\hat{\alpha }-{\mu }_i)}{{{\sigma }_i}^2}\\ \hat{\beta }\sum _{i=1}^n\frac{x_i^2}{{{\sigma }_i}^2}& =& \sum _{i=1}^n\frac{x_i(y_i-\hat{\alpha }-{\mu }_i)}{{{\sigma }_i}^2}\\ \hat{\beta }& =& \frac{\sum _{i=1}^n\frac{x_i(y_i-\hat{\alpha }-{\mu }_i)}{{{\sigma }_i}^2}}{\sum _{i=1}^n\frac{x_i^2}{{{\sigma }_i}^2}}\end{array}$$ $\hat{\alpha }$について． $$\begin{array}{r}\\ 0& =& \frac{\partial }{\partial \alpha }(-\ln \left(p\right))\\ & =& \hat{\alpha }\sum _{i=1}^n\frac{1}{{{\sigma }_i}^2}-\sum _{i=1}^n\frac{y_i-\hat{\beta }{x}_i-{\mu }_i}{{{\sigma }_i}^2}\\ \hat{\alpha }\sum _{i=1}^n\frac{1}{{{\sigma }_i}^2}& =& \sum _{i=1}^n\frac{y_i-\hat{\beta }{x}_i-{\mu }_i}{{{\sigma }_i}^2}\\ \hat{\alpha }& =& \frac{\sum _{i=1}^n\frac{y_i-\hat{\beta }{x}_i-{\mu }_i}{{{\sigma }_i}^2}}{\sum _{i=1}^n\frac{1}{{{\sigma }_i}^2}}\end{array}$$

誤差の発生が全て同一モデル${ϵ}_i\sim N(0,{\sigma }^2)$から発生しているとする

$\hat{\alpha }$について． $$\begin{array}{rcl}\hat{\alpha }& =& \frac{\frac{1}{{\sigma }^2}\sum _{i=1}^n(y_i-\hat{\beta }{x}_i-0)}{\frac{n}{{\sigma }^2}}\cdots {ϵ}_i\sim N(0,{\sigma }^2)\\ & =& \frac{1}{n}\sum _{i=1}^n(y_i-\hat{\beta }{x}_i)\\ & =& \frac{\sum _{i=1}^n{y}_i}{n}-\hat{\beta }\frac{\sum _{i=1}^n{x}_i}{n}\\ & =& \overline{y}-\hat{\beta }\overline{x}\end{array}$$ $\hat{\beta }$について． $$\begin{array}{r}\\ \hat{\beta }& =& \frac{\frac{1}{{\sigma }^2}\sum _{i=1}^n{x}_i(y_i-\hat{\alpha }-0)}{\frac{1}{{\sigma }^2}\sum _{i=1}^n{x}_i^2}\cdots {ϵ}_i\sim N(0,{\sigma }^2)\\ & =& \frac{\sum _{i=1}^n{x}_i{y}_i-\hat{\alpha }\sum _{i=1}^n{x}_i}{\sum _{i=1}^n{x}_i^2}\\ & =& \frac{\sum _{i=1}^n{x}_i{y}_i-\hat{\alpha }n\overline{x}}{\sum _{i=1}^n{x}_i^2}\\ & =& \frac{\sum _{i=1}^n{x}_i{y}_i-(\overline{y}-\hat{\beta }\overline{x})n\overline{x}}{\sum _{i=1}^n{x}_i^2}\cdots \hat{\alpha }=\overline{y}-\hat{\beta }\overline{x},\hat{\alpha }\mathrm{で}\mathrm{の}\mathrm{結}\mathrm{果}\mathrm{を}\mathrm{代}\mathrm{入}\mathrm{す}\mathrm{る}\\ & =& \frac{\sum _{i=1}^n{x}_i{y}_i-n\overline{x}\overline{y}+\hat{\beta }n{\overline{x}}^2}{\sum _{i=1}^n{x}_i^2}\\ \hat{\beta }-\frac{\hat{\beta }n{\overline{x}}^2}{\sum _{i=1}^n{x}_i^2}& =& \frac{\sum _{i=1}^n{x}_i{y}_i-n\overline{x}\overline{y}}{\sum _{i=1}^n{x}_i^2}\\ \hat{\beta }\left(\frac{\sum _{i=1}^n{x}_i^2-n{\overline{x}}^2}{\cancel{\sum _{i=1}^n{x}_i^2}}\right)& =& \frac{\sum _{i=1}^n{x}_i{y}_i-n\overline{x}\overline{y}}{\cancel{\sum _{i=1}^n{x}_i^2}}\\ \hat{\beta }& =& \frac{\sum _{i=1}^n{x}_i{y}_i-n\overline{x}\overline{y}}{\sum _{i=1}^n{x}_i^2-n{\overline{x}}^2}\\ & =& \frac{\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})}{\sum _{i=1}^n(x_i-\overline{x}{)}^2}\cdots \mathrm{準}\mathrm{備}\mathrm{よ}\mathrm{り}\end{array}$$

まとめ

$$\{\begin{array}{l}\begin{array}{rcl}\hat{\beta }& =& \frac{\sum _{i=1}^n{x}_i{y}_i-n\overline{x}\overline{y}}{\sum _{i=1}^n{x}_i^2-n{\overline{x}}^2}=\frac{\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})}{\sum _{i=1}^n(x_i-\overline{x}{)}^2}\\ \hat{\alpha }& =& \overline{y}-\hat{\beta }\overline{x}\end{array}\end{array}$$