統計検定 1級 2016年統計数理問3 解答解説

[1]
1. \(b_0\)
2. \(b_1\)
[2]
[3]
参考サイト

[1]

\(b_0\)

\(\displaystyle E[b_0]=E\left[\frac{1}{n}\sum_{i=1}^n\frac{Y_i}{x_i}\right]\)

\(\displaystyle =E\left[\frac{1}{n}\sum_{i=1}^n\left(\beta+\frac{\varepsilon_i}{x_i}\right)\right]\)

\(\displaystyle = \beta+\frac{1}{n}\sum_{i=1}^n E\left[\frac{\varepsilon_i}{x_i}\right]\)

\(=\beta\)

\(b_1\)

\(\displaystyle E[b_1]=E\left[\frac{\sum_{i=1}^n Y_i}{\sum_{i=1}^n x_i}\right]\)

\(\displaystyle = \frac{1}{\sum_{i=1}^n x_i}E \left[\sum_{i=1}^n (\beta x_i+\varepsilon_i) \right]\)

\(\displaystyle = \frac{1}{\sum_{i=1}^n x_i}\left(\beta \sum_{i=1}^n x_i +\sum_{i=1}^n E[\varepsilon_i]\right)\)

\(=\beta\)

[2]

\(\displaystyle S=\sum_{i=1}^n (Y_i-\beta x_i^2)\)

とすると、\(Y=(Y_1, \cdots , Y_n)^T\)\(,x=(x_1, \cdots , x_n)^T\)として、

\(S=(Y-\beta x)^T(Y-\beta x)\)

\(= (Y^T -\beta x^T)(Y-\beta x)\)

\(=Y^TY -\beta x^T Y -\beta Y^T x + \beta^2 x^Tx\)

\(=Y^TY -2\beta x^T Y + \beta^2 x^Tx\)(\(x^TY=Y^T x\)のため)

\(\displaystyle \frac{\partial S}{\partial \beta}=-2 x^T Y+ 2\beta x^Tx\)\(=0\)

として、

\(\displaystyle \hat{\beta}=\frac{x^T Y}{x^Tx}\)

\(\displaystyle =\frac{\sum_{i=1}^n x_iY_i}{\sum_{i=1}^n x_i^2}\)\(=b_2\)

また、

\(\displaystyle E[b_2]=E\left[\frac{\sum_{i=1}^n x_iY_i}{\sum_{i=1}^n x_i^2}\right]\)

\(\displaystyle =\frac{1}{\sum_{i=1}^n x_i^2}E\left[\sum_{i=1}^n x_i(\beta x_i + \varepsilon_i)\right]\)

\(\displaystyle =\frac{1}{\sum_{i=1}^n x_i^2}\left( \beta \sum_{i=1}^n x_i^2 +\sum_{i=1}^n x_iE[\varepsilon_i] \right)\)

\(=\beta\)

[3]

\(V[b_0]\)

\(\displaystyle V[b_0]=V\left[ \frac{1}{n}\sum_{i=1}^n\frac{Y_i}{x_i}\right]\)

\(\displaystyle =V\left[ \frac{1}{n}\sum_{i=1}^n\left(\beta+\frac{\varepsilon_i}{x_i}\right)\right]\)

\(\displaystyle =\frac{1}{n^2}V\left[\sum_{i=1}^n \frac{\varepsilon_i}{x_i}\right]\)

\(\displaystyle =\frac{1}{n^2}\sum_{i=1}^n V\left[ \frac{\varepsilon_i}{x_i}\right]\)

\(\displaystyle =\frac{1}{n^2}\sum_{i=1}^n \frac{1}{x_i^2}V[ \varepsilon_i]\)

\(\displaystyle =\frac{1}{n^2}\sum_{i=1}^n \frac{1}{x_i^2}\sigma^2\)

\(\displaystyle =\frac{\sigma^2}{n^2}\sum_{i=1}^n \frac{1}{x_i^2}\)

\(V[b_1]\)

\(\displaystyle V[b_1]=V\left[\frac{\sum_{i=1}^n Y_i}{\sum_{i=1}^n x_i}\right]\)

\(\displaystyle = \frac{1}{\left(\sum_{i=1}^n x_i\right)^2}V\left[\sum_{i=1}^n (\beta x_i + \varepsilon_i)\right]\)

\(\displaystyle = \frac{1}{\left(\sum_{i=1}^n x_i\right)^2}\sum_{i=1}^nV\left[ \beta x_i + \varepsilon_i \right]\)

\(\displaystyle = \frac{1}{\left(\sum_{i=1}^n x_i\right)^2}\sum_{i=1}^nV\left[ \varepsilon_i \right]\)

\(\displaystyle = \frac{1}{\left(\sum_{i=1}^n x_i\right)^2}\sum_{i=1}^n\sigma^2\)

\(\displaystyle = \frac{n\sigma^2}{\left(\sum_{i=1}^n x_i\right)^2}\)

\(V[b_2]\)

\(\displaystyle V[b_1]=V\left[\frac{\sum_{i=1}^n x_iY_i}{\sum_{i=1}^n x_i^2}\right]\)

\(\displaystyle =\frac{1}{\left(\sum_{i=1}^n x_i^2\right)^2}V\left[\sum_{i=1}^n x_iY_i\right]\)

\(\displaystyle =\frac{1}{\left(\sum_{i=1}^n x_i^2\right)^2}V\left[\sum_{i=1}^n x_i(\beta x_i +\varepsilon_i)\right]\)

\(\displaystyle =\frac{1}{\left(\sum_{i=1}^n x_i^2\right)^2}V\left[\sum_{i=1}^n x_i\varepsilon_i\right]\)

\(\displaystyle =\frac{1}{\left(\sum_{i=1}^n x_i^2\right)^2}\sum_{i=1}^n x_i^2V\left[ \varepsilon_i\right]\)

\(\displaystyle =\frac{1}{\left(\sum_{i=1}^n x_i^2\right)^2}\sum_{i=1}^n x_i^2\sigma^2\)

\(\displaystyle =\frac{\sigma^2}{\sum_{i=1}^n x_i^2}\)

大小比較

\(V[b_0]\)\(,V[b_1]\)\(,V[b_2]\)に\(x_1=1\)\(,x_2=2\)\(,n=2\)を代入すると、

\(V[b_0]=\displaystyle \frac{5}{16}\sigma^2\)\(,V[b_1]=\displaystyle \frac{2}{9}\sigma^2\)\(,V[b_2]=\displaystyle \frac{1}{5}\sigma^2\)なので、\(V[b_2]\)\(\leq V[b_1]\)\(\leq V[b_0]\)だと予想出来る。

これで答えとして良いが以下証明する。

\(V[b_2]\leq V[b_1]\)

\(\Leftrightarrow \displaystyle \frac{\sigma^2}{\sum_{i=1}^n x_i^2}\)\(\leq \displaystyle \frac{n\sigma^2}{\left(\sum_{i=1}^n x_i\right)^2}\)

\(\Leftrightarrow \displaystyle \left(\sum_{i=1}^n x_i\right)^2\)\(\leq \displaystyle n\sum_{i=1}^n x_i^2\)

\(\Leftrightarrow (1\cdot x_1+\cdots +1\cdot x_n)^2\)\(\leq (1^2 +\cdots+ 1^2)(x_1^2 +\cdots+ x_n^2)\)

これはシュワルツの不等式により成り立つことが分かる。

\(V[b_1]\leq V[b_0]\)

\(\Leftrightarrow \displaystyle \frac{n\sigma^2}{\left(\sum_{i=1}^n x_i\right)^2}\)\(\leq \displaystyle \frac{\sigma^2}{n^2}\sum_{i=1}^n \frac{1}{x_i^2}\)

\(\Leftrightarrow \displaystyle n^3 \)\(\leq \displaystyle \left(\sum_{i=1}^n x_i\right)^2\sum_{i=1}^n \frac{1}{x_i^2}\)

これもシュワルツの不等式を使って証明しそうな感じがするので変形していく。

\((右辺)\)\(=\displaystyle \left(\sum_{i=1}^n x_i\right)^2\sum_{i=1}^n \frac{1}{x_i^2}\)

\(= \displaystyle \left(\sum_{i=1}^n x_i\right)^2\)\(\displaystyle \left(\frac{1}{x_1^2}+\cdots +\frac{1}{x_n^2}\right)\)\((1^2 +\cdots+ 1^2)\)\(\displaystyle \times \frac{1}{n}\)

\(\geq \displaystyle \left(\sum_{i=1}^n x_i\right)^2\)\(\displaystyle \left(\frac{1}{x_1}+\cdots +\frac{1}{x_n}\right)^2\)\(\displaystyle \times \frac{1}{n}\)(シュワルツの不等式を使った)

\(= \displaystyle (x_1 + \cdots + x_n)^2\)\(\displaystyle \left(\frac{1}{x_1}+\cdots +\frac{1}{x_n}\right)^2\)\(\displaystyle \times \frac{1}{n}\)

\(= \displaystyle ((\sqrt{x_1})^2 + \cdots + (\sqrt{x_n})^2)^2\)\(\displaystyle \left(\left(\sqrt{\frac{1}{x_1}}\right)^2+\cdots +\left(\sqrt{\frac{1}{x_n}}\right)^2\right)^2\)\(\displaystyle \times \frac{1}{n}\)

\(\geq \displaystyle \left(\sqrt{x_1} \cdot \sqrt{\frac{1}{x_1}} + \cdots \sqrt{x_n} \cdot \sqrt{\frac{1}{x_n}}\right)^4 \)\(\displaystyle \times \frac{1}{n}\) (シュワルツの不等式を使った)

\(= \displaystyle n^4 \)\(\displaystyle \times \frac{1}{n}\) \(=n^3\)