統計検定 1級過去問解答/解答例と解説
2015年11月29日 (日) 試験

統計数理問3 [4]

設問の要約

$x_{1}, x_{2}$ を用いる重回帰モデルは正しいとする．
説明変数を $x_{1}$ だけとした単回帰分析の $x_1$ の回帰係数の最小二乗推定量 $\tilde{\beta}_1$ は，
${\tilde{β}}_{1} = \frac{S_{1 y}}{S_{11}}$
$\begin{equation} \tilde{\beta}_1 = \frac{S_{1y}}{S_{11}} \end{equation}$
$V[\hat{\beta}_1]$ と $V[\tilde{\beta}_1]$ の差を $r_{12}$ を用いて表わせ．
$E[\tilde{\beta}_1]$ を求めよ．
$\tilde{\beta}_1$ の平均二乗誤差 (MSE) を求めよ．
$\tilde{\beta}_1$ と $\hat{\beta}_1$ の MSE を比較せよ．

解答例

単回帰モデルの $x_1$ の回帰係数の分散は，次で求めることができる．

V [{\tilde{β}}_{1}] = \frac{σ^{2}}{S_{11}}

$\begin{equation} V[\tilde{\beta}_1] = \frac{\sigma^2}{S_{11}} \end{equation}$

これを用いれば，

\begin{aligned} V [{\hat{β}}_{1}] - V [{\tilde{β}}_{1}] & = \frac{σ^{2}}{S_{11}} \frac{1}{1 - {r_{12}}^{2}} - \frac{σ^{2}}{S_{11}} \\ = \frac{σ^{2}}{S_{11}} (\frac{1}{1 - {r_{12}}^{2}} - 1) \\ = \frac{σ^{2}}{S_{11}} \frac{{r_{12}}^{2}}{1 - {r_{12}}^{2}} \end{aligned}

$\begin{align} V[\hat{\beta}_1] - V[\tilde{\beta}_1] &= \frac{\sigma^2}{S_{11}}\frac{1}{1 - {r_{12}}^2} - \frac{\sigma^2}{S_{11}} \\ &= \frac{\sigma^2}{S_{11}}\left(\frac{1}{1 - {r_{12}}^2} - 1\right) \\ &= \frac{\sigma^2}{S_{11}}\frac{{r_{12}}^2}{1 - {r_{12}}^2} \end{align}$

\begin{aligned} E [{\tilde{β}}_{1}] & = E [\frac{S_{1 y}}{S_{11}}] \\ = \frac{1}{S_{11}} E [S_{1 y}] \\ = \frac{1}{S_{11}} E [\sum_{i = 1}^{n} x_{1 i} y_{i}] \\ = \frac{1}{S_{11}} \sum_{i = 1}^{n} E [x_{1 i} y_{i}] \\ = \frac{1}{S_{11}} \sum_{i = 1}^{n} x_{1 i} E [y_{i}] \\ = \frac{1}{S_{11}} \sum_{i = 1}^{n} x_{1 i} E [β_{0} + β_{1} x_{1 i} + β_{2} x_{2 i} + ϵ_{i}] \\ = \frac{1}{S_{11}} \sum_{i = 1}^{n} x_{1 i} (E [β_{0}] + E [β_{1} x_{1 i}] + E [β_{2} x_{2 i}] + E [ϵ_{i}]) \\ = \frac{1}{S_{11}} \sum_{i = 1}^{n} x_{1 i} (β_{0} + β_{1} x_{1 i} + β_{2} x_{2 i}) \\ = \frac{1}{S_{11}} \sum_{i = 1}^{n} (β_{0} x_{1 i} + β_{1} {x_{1 i}}^{2} + β_{2} x_{1 i} x_{2 i}) \\ = \frac{1}{S_{11}} (β_{0} \sum_{i = 1}^{n} x_{1 i} + β_{1} \sum_{i = 1}^{n} {x_{1 i}}^{2} + β_{2} \sum_{i = 1}^{n} x_{1 i} x_{2 i}) \\ = \frac{1}{S_{11}} (0 + β_{1} S_{11} + β_{2} S_{12}) \\ = β_{1} + \frac{S_{12}}{S_{11}} β_{2} \end{aligned}

$\begin{align} E[\tilde{\beta}_1] &= E\left[\frac{S_{1y}}{S_{11}}\right] \\ &= \frac{1}{S_{11}}E\left[S_{1y}\right] \\ &= \frac{1}{S_{11}}E\left[\sum_{i=1}^{n}x_{1i}y_{i}\right] \\ &= \frac{1}{S_{11}}\sum_{i=1}^{n}E\left[x_{1i}y_{i}\right] \\ &= \frac{1}{S_{11}}\sum_{i=1}^{n}x_{1i} E\left[y_{i}\right] \\ &= \frac{1}{S_{11}}\sum_{i=1}^{n}x_{1i} E\left[\beta_0 + \beta_1 x_{1i} + \beta_2 x_{2i} + \epsilon_i\right] \\ &= \frac{1}{S_{11}}\sum_{i=1}^{n}x_{1i} \left(E[\beta_0] + E[\beta_1 x_{1i}] + E[\beta_2 x_{2i}] + E[\epsilon_i]\right) \\ &= \frac{1}{S_{11}}\sum_{i=1}^{n}x_{1i} \left(\beta_0 + \beta_1 x_{1i} + \beta_2 x_{2i}\right) \\ &= \frac{1}{S_{11}}\sum_{i=1}^{n}\left(\beta_0 x_{1i} + \beta_1 {x_{1i}}^2 + \beta_2 x_{1i} x_{2i}\right) \\ &= \frac{1}{S_{11}}\left(\beta_0\sum_{i=1}^{n}x_{1i} + \beta_1\sum_{i=1}^{n}{x_{1i}}^2 + \beta_2\sum_{i=1}^{n}x_{1i} x_{2i}\right) \\ &= \frac{1}{S_{11}}\left(0 + \beta_1S_{11} + \beta_2 S_{12}\right) \\ &= \beta_1 + \frac{S_{12}}{S_{11}}\beta_2 \end{align}$

\begin{aligned} M S E [{\tilde{β}}_{1}] & = E [({\tilde{β}}_{1} - β_{1})^{2}] \\ = E [{\tilde{β}}_{1}^{2} - 2 β_{1} {\tilde{β}}_{1} + {β_{1}}^{2}] \\ = E [{\tilde{β}}_{1}^{2}] - 2 β_{1} E [{\tilde{β}}_{1}] + {β_{1}}^{2} \\ = V [{\tilde{β}}_{1}] + {(E [{\tilde{β}}_{1}])}^{2} - 2 β_{1} E [{\tilde{β}}_{1}] + {β_{1}}^{2} \\ = V [{\tilde{β}}_{1}] + {(β_{1} + \frac{S_{12}}{S_{11}} β_{2})}^{2} - 2 β_{1} (β_{1} + \frac{S_{12}}{S_{11}} β_{2}) + {β_{1}}^{2} \\ = V [{\tilde{β}}_{1}] + {β_{1}}^{2} + 2 \frac{S_{12}}{S_{11}} β_{1} β_{2} + {(\frac{S_{12}}{S_{11}} β_{2})}^{2} - 2 {β_{1}}^{2} - 2 \frac{S_{12}}{S_{11}} β_{1} β_{2} + {β_{1}}^{2} \\ = V [{\tilde{β}}_{1}] + {(\frac{S_{12}}{S_{11}})}^{2} {β_{2}}^{2} \end{aligned}

$\begin{align} MSE[\tilde{\beta}_1] &= E[(\tilde{\beta}_1 - \beta_1)^2] \\ &= E[{\tilde{\beta}_1}^2 - 2\beta_1\tilde{\beta}_1 + {\beta_1}^2] \\ &= E[{\tilde{\beta}_1}^2] - 2\beta_1E[\tilde{\beta}_1] + {\beta_1}^2 \\ &= V[{\tilde{\beta}_1}] + \left(E[{\tilde{\beta}_1}]\right)^2 - 2\beta_1E[\tilde{\beta}_1] + {\beta_1}^2 \\ &= V[{\tilde{\beta}_1}] + \left(\beta_1 + \frac{S_{12}}{S_{11}}\beta_2\right)^2 - 2\beta_1\left(\beta_1 + \frac{S_{12}}{S_{11}}\beta_2\right) + {\beta_1}^2 \\ &= V[{\tilde{\beta}_1}] + {\beta_1}^2 + 2\frac{S_{12}}{S_{11}}\beta_1\beta_2 + \left(\frac{S_{12}}{S_{11}}\beta_2\right)^2 - 2{\beta_1}^2 - 2\frac{S_{12}}{S_{11}}\beta_1\beta_2 + {\beta_1}^2 \\ &= V[{\tilde{\beta}_1}] + \left(\frac{S_{12}}{S_{11}}\right)^2{\beta_2}^2 \end{align}$

\begin{aligned} M S E [{\hat{β}}_{1}] & = E [({\hat{β}}_{1} - β_{1})^{2}] \\ = E [{\hat{β}}_{1}^{2} - 2 β_{1} {\hat{β}}_{1} + {β_{1}}^{2}] \\ = E [{\hat{β}}_{1}^{2}] - 2 β_{1} E [{\hat{β}}_{1}] + {β_{1}}^{2} \\ = V [{\hat{β}}_{1}] + {(E [{\hat{β}}_{1}])}^{2} - 2 β_{1} E [{\hat{β}}_{1}] + {β_{1}}^{2} \\ = V [{\hat{β}}_{1}] + {β_{1}}^{2} - 2 {β_{1}}^{2} + {β_{1}}^{2} \\ = V [{\hat{β}}_{1}] \end{aligned}

$\begin{align} MSE[\hat{\beta}_1] &= E[(\hat{\beta}_1 - \beta_1)^2] \\ &= E[{\hat{\beta}_1}^2 - 2\beta_1\hat{\beta}_1 + {\beta_1}^2] \\ &= E[{\hat{\beta}_1}^2] - 2\beta_1E[\hat{\beta}_1] + {\beta_1}^2 \\ &= V[{\hat{\beta}_1}] + \left(E[{\hat{\beta}_1}]\right)^2 - 2\beta_1E[\hat{\beta}_1] + {\beta_1}^2 \\ &= V[{\hat{\beta}_1}] + {\beta_1}^2 - 2{\beta_1}^2 + {\beta_1}^2 \\ &= V[{\hat{\beta}_1}] \\ \end{align}$

\begin{aligned} M S E [{\hat{β}}_{1}] - M S E [{\tilde{β}}_{1}] & = V [{\hat{β}}_{1}] - {V [{\tilde{β}}_{1}] + {(\frac{S_{12}}{S_{11}})}^{2} {β_{2}}^{2}} \\ = V [{\hat{β}}_{1}] - V [{\tilde{β}}_{1}] - {(\frac{S_{12}}{S_{11}})}^{2} {β_{2}}^{2} \end{aligned}

$\begin{align} MSE[\hat{\beta}_1] - MSE[\tilde{\beta}_1] &= V[\hat{\beta}_1] - \left\{V[\tilde{\beta}_1] + \left(\frac{S_{12}}{S_{11}}\right)^2{\beta_2}^2\right\} \\ &= V[\hat{\beta}_1] - V[\tilde{\beta}_1] - \left(\frac{S_{12}}{S_{11}}\right)^2{\beta_2}^2 \\ \end{align}$

ところで，

\begin{aligned} V [{\hat{β}}_{1}] - V [{\tilde{β}}_{1}] & = \frac{σ^{2}}{S_{11}} \frac{{r_{12}}^{2}}{1 - {r_{12}}^{2}} \\ = {r_{12}}^{2} \frac{S_{22}}{S_{11}} \frac{σ^{2}}{S_{22}} \frac{1}{1 - {r_{12}}^{2}} \\ = \frac{{S_{12}}^{2}}{S_{11} S_{22}} \frac{S_{22}}{S_{11}} \frac{σ^{2}}{S_{22}} \frac{1}{1 - {r_{12}}^{2}} \\ = {(\frac{S_{12}}{S_{11}})}^{2} \frac{σ^{2}}{S_{22}} \frac{1}{1 - {r_{12}}^{2}} \\ = {(\frac{S_{12}}{S_{11}})}^{2} V [{\hat{β}}_{2}] \end{aligned}

$\begin{align} V[\hat{\beta}_1] - V[\tilde{\beta}_1] &= \frac{\sigma^2}{S_{11}}\frac{{r_{12}}^2}{\ds 1 - {r_{12}}^2} \\ &= {r_{12}}^2\frac{S_{22}}{S_{11}}\frac{\sigma^2}{S_{22}}\frac{1}{\ds 1 - {r_{12}}^2} \\ &= \frac{{S_{12}}^2}{S_{11}S_{22}}\frac{S_{22}}{S_{11}}\frac{\sigma^2}{S_{22}}\frac{1}{\ds 1 - {r_{12}}^2} \\ &= \left(\frac{S_{12}}{S_{11}}\right)^2\frac{\sigma^2}{S_{22}}\frac{1}{\ds 1 - {r_{12}}^2} \\ &= \left(\frac{S_{12}}{S_{11}}\right)^2 V[\hat{\beta}_2] \\ \end{align}$

これを用いて，

\begin{aligned} M S E [{\hat{β}}_{1}] - M S E [{\tilde{β}}_{1}] & = {(\frac{S_{12}}{S_{11}})}^{2} V [{\hat{β}}_{2}] - {(\frac{S_{12}}{S_{11}})}^{2} {β_{2}}^{2} \\ = {(\frac{S_{12}}{S_{11}})}^{2} (V [{\hat{β}}_{2}] - {β_{2}}^{2}) \end{aligned}

$\begin{align} MSE[\hat{\beta}_1] - MSE[\tilde{\beta}_1] &= \left(\frac{S_{12}}{S_{11}}\right)^2 V[\hat{\beta}_2] - \left(\frac{S_{12}}{S_{11}}\right)^2{\beta_2}^2 \\ &= \left(\frac{S_{12}}{S_{11}}\right)^2 \left(V[\hat{\beta}_2] - {\beta_2}^2\right) \end{align}$

よって，

V [{\hat{β}}_{2}] - {β_{2}}^{2} \geq 0

$V[\hat{\beta}_2] - {\beta_2}^2 \ge 0$ のとき，

M S E [{\hat{β}}_{1}] \geq M S E [{\tilde{β}}_{1}]

$\begin{equation} MSE[\hat{\beta}_1] \ge MSE[\tilde{\beta}_1] \end{equation}$

V [{\hat{β}}_{2}] - {β_{2}}^{2} < 0

$V[\hat{\beta}_2] - {\beta_2}^2 < 0$ のとき，

M S E [{\hat{β}}_{1}] < M S E [{\tilde{β}}_{1}]

$\begin{equation} MSE[\hat{\beta}_1] < MSE[\tilde{\beta}_1] \end{equation}$

統計検定 1級 過去問 解答/解答例と解説 2015年11月29日 (日) 試験

統計数理 問3 [4]

設問の要約

解答例

統計検定 1級過去問解答/解答例と解説
2015年11月29日 (日) 試験

統計数理問3 [4]