【推导结果】如何得到 回归均方误差 & 估计系数的标准误 |
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【推导结果】如何得到 回归均方误差 & 估计系数的标准误
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对线性回归模型系数标准差标准误的理解
1.生成数据
yxe3.610.633.42-1.387.631.017.44-1.0111.651.3811.46-0.63
y = β 0 + β 1 x + ϵ y = \beta_{0}+\beta_{1}x+\epsilon y=β0+β1x+ϵ y i = β 0 + β 1 x i + e i y_{i}=\beta_{0}+\beta_{1} x_{i}+e_{i} yi=β0+β1xi+ei reg y x Source | SS df MS Number of obs = 6 -------------+---------------------------------- F(1, 4) = 34.60 Model | 57.422285 1 57.422285 Prob > F = 0.0042 Residual | 6.63771505 4 1.65942876 R-squared = 0.8964 -------------+---------------------------------- Adj R-squared = 0.8705 Total | 64.0600001 5 12.812 Root MSE = 1.2882 ------------------------------------------------------------------------------ y | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- x | 1.811429 .3079359 5.88 0.004 .9564615 2.666396 _cons | 1.16 1.199238 0.97 0.388 -2.169618 4.489618 ------------------------------------------------------------------------------ 3.计算回归的标准误差 （1）SSE\SSR\SSTS S E SSE SSE: Sum of Squares Error, S S E = ∑ i = 1 n ( y i ^ − y i ) 2 = ∑ i = 1 n ( e i − e ˉ ) 2 SSE= \sum_{i=1}^{n}(\hat{y_{i}}-y_{i})^2 = \sum_{i=1}^{n}(e_{i}-\bar{e})^2 SSE=i=1∑n(yi^−yi)2=i=1∑n(ei−eˉ)2 在本示例中， S S E = ( 3.6 − 2.97 ) 2 + ( 3.4 − 4.78 ) 2 + ( 7.6 − 6.95 ) 2 + ( 7.4 − 8.41 ) 2 + ( 11.6 − 10.22 ) 2 + ( 11.4 − 12.03 ) 2 = 6.637713 SSE=(3.6-2.97)^2+(3.4-4.78)^2+(7.6-6.95)^2+(7.4-8.41)^2+(11.6-10.22)^2+(11.4-12.03)^2 = 6.637713 SSE=(3.6−2.97)2+(3.4−4.78)2+(7.6−6.95)2+(7.4−8.41)2+(11.6−10.22)2+(11.4−12.03)2=6.637713 S S R SSR SSR: Sum of Squares of the Regression S S R = ∑ i = 1 n ( y i ^ − y ˉ ) 2 SSR= \sum_{i=1}^{n}(\hat{y_{i}}-\bar{y})^2 SSR=i=1∑n(yi^−yˉ)2 S S T SST SST: Total Sum of Squares S S T = ∑ i = 1 n ( y i − y ˉ ) 2 SST= \sum_{i=1}^{n}(y_{i}-\bar{y})^2 SST=i=1∑n(yi−yˉ)2 （2）MSE回归的标准误差为： s 2 = M S E = S S E n − K = ∑ i = 1 n ( e i − e ˉ ) 2 n − K s^{2}=MSE=\frac{SSE}{n-K}=\frac{\sum_{i=1}^{n}(e_{i}-\bar{e})^2}{n-K} s2=MSE=n−KSSE=n−K∑i=1n(ei−eˉ)2 s = M S E s=\sqrt{MSE} s=MSE s 2 = 6.637713 6 − 2 = 1.6594282 ; s = 1.288188 s^2 = \frac{6.637713}{6 - 2}=1.6594282; \ \ \ \ \ \ \ s=1.288188 s2=6−26.637713=1.6594282; s=1.288188 （3）SES β ^ = s 2 ∑ i = 1 n ( x i − x ˉ ) S_{\hat{\beta}} = \sqrt{\frac{s^2}{{\sum_{i=1}^{n}(x_{i}-\bar{x})}}} Sβ^=∑i=1n(xi−xˉ)s2 S β ^ = 1 n − 2 ∑ i = 1 n e 2 ^ ∑ i = 1 n ( x i − x ˉ ) S_{\hat{\beta}} = \sqrt{\frac{\frac{1}{n-2}\sum_{i=1}^{n} \hat{e^{2}}}{{\sum_{i=1}^{n}(x_{i}-\bar{x})}}} Sβ^=∑i=1n(xi−xˉ)n−21∑i=1ne2^ S β ^ = 1 4 × 6.637713 ( 1 − 3.5 ) 2 + ( 2 − 3.5 ) 2 + ( 3 − 3.5 ) 2 + ( 4 − 3.5 ) 2 + ( 5 − 3.5 ) 2 + ( 6 − 3.5 ) 2 S_{\hat{\beta}} = \sqrt{\frac{\frac{1}{4} \times 6.637713}{(1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2}} Sβ^=(1−3.5)2+(2−3.5)2+(3−3.5)2+(4−3.5)2+(5−3.5)2+(6−3.5)241×6.637713 SE为何会很大？ 样本少，分母可能大极端值多X没有足够的variation Appendix 1. simulation code clear set obs 6 gen y = 3.6 in 1 replace y = 3.4 in 2 replace y = 7.6 in 3 replace y = 7.4 in 4 replace y = 11.6 in 5 replace y = 11.4 in 6 gen x = _n reg y x predict xb gen e = y - xb format %9.2f xb format %9.2f e egen addtext_mean = rowmean(y xb) forv i = 1/6{ su add in `i',d global y`i' = r(mean) su e in `i',d global e`i' = r(mean) } tw (scatter y x, mlab(y) mlabp(1)) /// (lfit y x) /// (scatter xb x, mlab(xb) mlabp(1)) /// (rspike y xb x) ,legend(off) /// text($y1 0.9 "0.63",size(vsmall) color(red)) /// text($y2 1.9 "-1.38",size(vsmall) color(red)) /// text($y3 2.9 "1.01",size(vsmall) color(red)) /// text($y4 3.9 "-1.01",size(vsmall) color(red)) /// text($y5 4.9 "1.38",size(vsmall) color(red)) /// text($y6 5.9 "-0.63",size(vsmall) color(red)) 2.序列相关 同方差 or 异方差对于①参数线性②不存在“严格多重共线性”③随机抽样④严格外生性⑤“球形扰动项”(条件同方差+不存在自相关)五个假定均能够满足时 OLS估计量为BLUE，最优无偏线性估计量 此时，x的协方差矩阵为： V a r ( β 1 ^ ∣ x ) = V a r ( β 1 + ∑ ( x i − x ˉ ) e i ∑ ( x i − x ˉ ) ∣ x ) Var(\hat{\beta_{1}}|x)=Var({\beta_{1}+\frac{\sum(x_{i}-\bar{x})e_{i}}{\sum(x_{i}-\bar{x})}}|x) Var(β1^∣x)=Var(β1+∑(xi−xˉ)∑(xi−xˉ)ei∣x) V a r ( β 1 ^ ∣ x ) = V a r ( ∑ ( x i − x ˉ ) e i ∣ x ) [ ∑ ( x i − x ˉ ) 2 ] 2 Var(\hat{\beta_{1}}|x)=\frac{Var(\sum(x_{i}-\bar{x})e_{i}|x)}{[\sum(x_{i}-\bar{x})^2]^2} Var(β1^∣x)=[∑(xi−xˉ)2]2Var(∑(xi−xˉ)ei∣x) 倘若序列无关，那么和的方差即等价于方差的和，假设 V a r ( e i ∣ x ) = σ 2 Var(e_i|x)=\sigma^2 Var(ei∣x)=σ2KaTeX parse error: Unknown column alignment: * at position 71: … \begin{array}{*̲*lr**} … 序列相关：σ 2 ^ = ∑ e i 2 n − k − 1 \hat{\sigma^2}=\frac{\sum e_{i}^2}{n-k-1} σ2^=n−k−1∑ei2 3.calculate SE in matlab sqrt(inv(X'*X)*1.6594282) |
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