独立なガウス分布\(N(\mu_i,\sigma_i^2)\)に従う誤差\(\epsilon_i\)を含む回帰モデル\(y_i=\beta x_i+\alpha+\epsilon_i\)が\(n\)個の同時確率を考える
準備
\begin{eqnarray}
\bar{x}&=&\frac{1}{n}\sum_{i=1}^n x_i
\\\bar{y}&=&\frac{1}{n}\sum_{i=1}^n y_i
\\\sum_{i=1}^n\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)
&=&\sum_{i=1}^n\left(x_iy_i-x_i\bar{y}-\bar{x}y_i+\bar{x}\bar{y}\right)
\\&=&\sum_{i=1}^nx_iy_i-\bar{y}\sum_{i=1}^nx_i-\bar{x}\sum_{i=1}^ny_i+\bar{x}\bar{y}\sum_{i=1}^n1
\\&=&\sum_{i=1}^nx_iy_i-\bar{y}n\bar{x}-\bar{x}n\bar{y}+\bar{x}\bar{y}n
\\&=&\sum_{i=1}^nx_iy_i-2n\bar{x}\bar{y}+n\bar{x}\bar{y}
\\&=&\sum_{i=1}^nx_iy_i-n\bar{x}\bar{y}\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/10/sxy.html}{S_{xy}の式展開}
\\\sum_{i=1}^n(x_i-\bar{x})^2
&=&\sum_{i=1}^n(x_i^2-2x_i\bar{x}+\bar{x}^2)
\\&=&\sum_{i=1}^nx_i^2-2\bar{x}\sum_{i=1}^nx_i+\bar{x}^2\sum_{i=1}^n1
\\&=&\sum_{i=1}^nx_i^2-2\bar{x}n\bar{x}+\bar{x}^2n
\\&=&\sum_{i=1}^nx_i^2-2n\bar{x}^2+n\bar{x}^2
\\&=&\sum_{i=1}^nx_i^2-n\bar{x}^2\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/10/sxx.html}{S_{xx}の式展開}
\end{eqnarray}
同時確率を考える
\begin{eqnarray}
p(\epsilon_1,\epsilon_2,\cdots,\epsilon_n)
&=&\frac{1}{\sqrt{2\pi{\sigma_1}^2}}e^{-\frac{\left(\epsilon_1-\mu_1\right)^2}{2{\sigma_1}^2}}
\times\frac{1}{\sqrt{2\pi{\sigma_2}^2}}e^{-\frac{\left(\epsilon_2-\mu_2\right)^2}{2{\sigma_2}^2}}
\times\cdots\times\frac{1}{\sqrt{2\pi{\sigma_n}^2}}e^{-\frac{\left(\epsilon_n-\mu_n\right)^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{\left(\epsilon_i-\mu_i\right)^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{\left(y_i-\beta x_i-\alpha -\mu_i\right)^2}{2{\sigma_i}^2}}}\right)
\;\cdots\;y_i=\beta x_i+\alpha+\epsilon_i,\;\epsilon_i=y_i-\beta x_i-\alpha
\end{eqnarray}
負の対数によって情報量を求める
\begin{eqnarray}
\\-\ln(p)&=&-\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}\left(y_i-\beta x_i-\alpha -\mu_i\right)^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}\left(y_i-\beta x_i-\alpha -\mu_i\right)^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}\left(y_i-\beta x_i-\alpha -\mu_i\right)^2}}
\right)
\\&=&-\sum_{i=1}^n
\left\{-\frac{1}{2}\ln\left(
2\pi{\sigma_i}^2
\right)
\right\}
-\sum_{i=1}^n\left(-\frac{1}{2{\sigma_i}^2}\left(y_i-\beta x_i-\alpha -\mu_i\right)^2
\right)
\\&=&\sum_{i=1}^n
\left\{\frac{1}{2}\ln\left(
2\pi{\sigma_i}^2
\right)
\right\}
+\sum_{i=1}^n\frac{1}{2{\sigma_i}^2}\left(y_i-\beta x_i-\alpha -\mu_i\right)^2
\\&=&\sum_{i=1}^n
\left\{\frac{1}{2}\ln\left(
2\pi{\sigma_i}^2
\right)
\right\}
+\sum_{i=1}^n\frac{1}{2{\sigma_i}^2}\left(
y_i^2
+\beta^2x_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
\right)
\end{eqnarray}
回帰モデルの係数で微分する
\(\beta\)について．
\begin{eqnarray}
\frac{\partial}{\partial \beta}\left(-\ln(p)\right)&=&0
+\sum_{i=1}^n\frac{1}{\cancel{2}{\sigma_i}^2}\left(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\right)
\\&=&\sum_{i=1}^n\frac{1}{{\sigma_i}^2}\left({\beta}x_i^2-x_iy_i+\alpha x_i+x_i\mu_i\right)
\\&=&\beta\sum_{i=1}^n\frac{x_i^2}{{\sigma_i}^2}
-\sum_{i=1}^n\frac{x_i\left(y_i-\alpha-\mu_i\right)}{{\sigma_i}^2}
\end{eqnarray}
\(\alpha\)について．
\begin{eqnarray}
\frac{\partial}{\partial \alpha}\left(-\ln(p)\right)&=&0
+\sum_{i=1}^n\frac{1}{\cancel{2}{\sigma_i}^2}\left(0+0+\cancel{2}\alpha+0-0-\cancel{2}y_i
-0 +\cancel{2}\beta x_i+0+\cancel{2}\mu_i\right)
\\&=&\sum_{i=1}^n\frac{1}{{\sigma_i}^2}\left(\alpha -y_i+\beta x_i+\mu_i\right)
\\&=&\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{eqnarray}
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{eqnarray}
\\0&=&\frac{\partial}{\partial \beta}\left(-\ln(p)\right)
\\&=&\hat{\beta}\sum_{i=1}^n\frac{x_i^2}{{\sigma_i}^2}
-\sum_{i=1}^n\frac{x_i\left(y_i-\hat{\alpha}-\mu_i\right)}{{\sigma_i}^2}
\\\hat{\beta}\sum_{i=1}^n\frac{x_i^2}{{\sigma_i}^2}&=&
\sum_{i=1}^n\frac{x_i\left(y_i-\hat{\alpha}-\mu_i\right)}{{\sigma_i}^2}
\\\hat{\beta}&=&\frac{
\sum_{i=1}^n\frac{x_i\left(y_i-\hat{\alpha}-\mu_i\right)}{{\sigma_i}^2}
}{\sum_{i=1}^n\frac{x_i^2}{{\sigma_i}^2}}
\end{eqnarray}
\(\hat{\alpha}\)について．
\begin{eqnarray}
\\0&=&\frac{\partial}{\partial \alpha}\left(-\ln(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{eqnarray}
誤差の発生が全て同一モデル\(\epsilon_i \sim N(0,\sigma^2)\)から発生しているとする
\(\hat{\alpha}\)について．
\begin{eqnarray}
\hat{\alpha}&=&\frac{\frac{1}{{\sigma}^2} \sum_{i=1}^n \left(y_i-\hat{\beta}x_i-0\right)
}{\frac{n}{{\sigma}^2}}
\;\cdots\;\epsilon_i \sim N(0,\sigma^2)
\\&=&\frac{1}{n}\sum_{i=1}^n \left(y_i-\hat{\beta}x_i\right)
\\&=&\frac{\sum_{i=1}^n y_i}{n}-\hat{\beta}\frac{\sum_{i=1}^n x_i}{n}
\\&=&\bar{y}-\hat{\beta}\bar{x}
\end{eqnarray}
\(\hat{\beta}\)について．
\begin{eqnarray}
\\\hat{\beta}&=&\frac{
\frac{1}{{\sigma}^2} \sum_{i=1}^nx_i\left(y_i-\hat{\alpha}-0\right)
}{
\frac{1}{{\sigma}^2}\sum_{i=1}^nx_i^2
}
\;\cdots\;\epsilon_i \sim N(0,\sigma^2)
\\&=&\frac{
\sum_{i=1}^nx_i y_i-\hat{\alpha}\sum_{i=1}^nx_i
}{
\sum_{i=1}^nx_i^2
}
\\&=&\frac{\sum_{i=1}^nx_i y_i-\hat{\alpha}n\bar{x}}{\sum_{i=1}^nx_i^2}
\\&=&\frac{\sum_{i=1}^nx_i y_i-\left(\bar{y}-\hat{\beta}\bar{x}\right)n\bar{x}}{\sum_{i=1}^nx_i^2}
\;\cdots\;\hat{\alpha}=\bar{y}-\hat{\beta}\bar{x},\;\hat{\alpha}での結果を代入する
\\&=&\frac{\sum_{i=1}^nx_i y_i-n\bar{x}\bar{y}+\hat{\beta}n\bar{x}^2}{\sum_{i=1}^nx_i^2}
\\\hat{\beta}-\frac{\hat{\beta}n\bar{x}^2}{\sum_{i=1}^nx_i^2}
&=&\frac{\sum_{i=1}^nx_i y_i-n\bar{x}\bar{y}}{\sum_{i=1}^nx_i^2}
\\\hat{\beta}\left(\frac{\sum_{i=1}^nx_i^2-n\bar{x}^2}{\cancel{\sum_{i=1}^nx_i^2}}\right)
&=&\frac{\sum_{i=1}^nx_i y_i-n\bar{x}\bar{y}}{\cancel{\sum_{i=1}^nx_i^2}}
\\\hat{\beta}&=&\frac{\sum_{i=1}^nx_i y_i-n\bar{x}\bar{y}}{\sum_{i=1}^nx_i^2-n\bar{x}^2}
\\&=&\frac{\sum_{i=1}^n\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)}{\sum_{i=1}^n(x_i-\bar{x})^2}\;\cdots\;準備
より
\end{eqnarray}
まとめ
\begin{cases}\begin{eqnarray}
\hat{\beta}&=&\frac{\sum_{i=1}^nx_i
y_i-n\bar{x}\bar{y}}{\sum_{i=1}^nx_i^2-n\bar{x}^2}=\frac{\sum_{i=1}^n\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)}{\sum_{i=1}^n(x_i-\bar{x})^2}
\\\hat{\alpha}&=&\bar{y}-\hat{\beta}\bar{x}
\end{eqnarray}\end{cases}
0 件のコメント:
コメントを投稿