开发者:上海品职教育科技有限公司隐私政策详情

应用版本:4.2.11(IOS)｜3.2.5(安卓)APP下载

学习体验
App下载
手机上的品职教育

随时随地学习课程，支持音视频下载！
- 扫码下载品职教育APP
进入课程
登录 | 注册

Brian邵彬 · 2024年05月17日

这里计算是不是不对？计算出来是-0.8909？

* 问题详情，请查看题干

NO.PZ202001080100002601

问题如下：

a. Apply OLS linear regression to find the parameters for the following equation:

$Y_i = \alpha + \beta_1X_{1i} + \beta_2X_{2i} + \epsilon_i$

选项：

解释：

Start by setting up the basic calculations:

$\mu_Y=-2.823, \mu_{X_1}=-0.91, \mu_{X_2}=-0.101,$

$\sigma_Y^2=7.58, \sigma_{X_1}^2=1.345, \sigma_{X_2}^2=0.911$

$\sigma_{X_1X_2}=-0.172, \sigma_Y\sigma_{X_1}=2.729, \sigma_Y\sigma{X_2}=-1.434$

First regressing on the last independent variable:

$X_{1i}=\delta_0+\delta_1X_{2i}+\widetilde{X_{1i}}$

$\delta_1=\frac{\sigma_{X_1X_2}}{\sigma_{X_2}^2}=-0.172/0.911=0.189$

$\delta_0=\mu_{X_1}-\delta_1\mu_{X_2}=-0.91-0.189*(-0.101)=-0.929$

$Y_i=\gamma_0+\gamma_1X_{2i}+\widetilde{Y_i}$

$\gamma_1=\sigma_{YX_2}/\sigma_{X_2}^2=-1.434/0.911=-1.574$

$\gamma_0=\mu_Y-\gamma_1\mu_{X_2}=-2.823-(-1.574*-0.101)=-2.982$

so $\beta_1=\frac{Cov()\widetilde{Y}, \widetilde{X_1}}{\sigma_{\widetilde{X_1}}^2}=2.459/1.313=1.873$

Repeating the process for $X_2$

$\beta_2=-1.085/0.889=-1.221$

Tying things together:

$Y_i-\beta_1X_{1i}-\beta_2X_{2i}=\alpha+\epsilon_i=-1.873X_{1i}+1.221X_{2i}$

So $\alpha = -1.242$ , and the variance of the residuals is 0.718.

Finally, asserting the presumption of normality:

$Y_i=-1.242+1.873*X_{1i}-1.221*X_{2i}+\epsilon_i, \epsilon_i~N(0, 0.718)$

Note that the variance for $\epsilon_i$ is biased and the unbiased figure is this figure multiplied by 10/(10 - 3) or 1.206.

这里计算是不是不对？计算出来是-0.8909？

添加评论

0
0

1 个答案

pzqa39 · 2024年05月17日

嗨，爱思考的PZer你好：

是的同学上一步𝛿1 应该是=-0.189，这样𝛿0=−0.929，感谢指正~

𝜇𝑋1=−0.91, 𝜇𝑋2=−0.101

𝛿1=𝜎𝑋1𝑋2/𝜎𝑋2^2=−0.172/0.911=-0.189

𝛿0=𝜇𝑋1−𝛿1𝜇𝑋2 =-0.91−(-0.189)∗(-0.101)=−0.929

----------------------------------------------
加油吧，让我们一起遇见更好的自己！

添加评论

0
0

1
回答
0
关注
150
浏览

我要回答关注问题

相关问题

NO.PZ202001080100002601 问题如下 Apply OLS lineregression to finthe parameters for the following equation: Yi=α+β1X1i+β2X2i+ϵiY_i = \alpha + \beta_1X_{1i} + \beta_2X_{2i} + \epsilon_iYi=α+β1X1i+β2X2i+ϵi Start setting up the basic calculations: μY=−2.823,μX1=−0.91,μX2=−0.101,\mu_Y=-2.823, \mu_{X_1}=-0.91, \mu_{X_2}=-0.101,μY=−2.823,μX1=−0.91,μX2=−0.101,σY2=7.58,σX12=1.345,σX22=0.911\sigma_Y^2=7.58, \sigma_{X_1}^2=1.345, \sigma_{X_2}^2=0.911σY2=7.58,σX12=1.345,σX22=0.911σX1X2=−0.172,σYσX1=2.729,σYσX2=−1.434\sigma_{X_1X_2}=-0.172, \sigma_Y\sigma_{X_1}=2.729, \sigma_Y\sigma{X_2}=-1.434σX1X2=−0.172,σYσX1=2.729,σYσX2=−1.434First regressing on the last inpennt variable: X1i=δ0+δ1X2i+X1i~X_{1i}=\lta_0+\lta_1X_{2i}+\witil{X_{1i}}X1i=δ0+δ1X2i+X1iδ1=σX1X2σX22=−0.172/0.911=0.189\lta_1=\frac{\sigma_{X_1X_2}}{\sigma_{X_2}^2}=-0.172/0.911=0.189δ1=σX22σX1X2=−0.172/0.911=0.189δ0=μX1−δ1μX2=−0.91−0.189∗(−0.101)=−0.929\lta_0=\mu_{X_1}-\lta_1\mu_{X_2}=-0.91-0.189*(-0.101)=-0.929δ0=μX1−δ1μX2=−0.91−0.189∗(−0.101)=−0.929Yi=γ0+γ1X2i+Yi~Y_i=\gamma_0+\gamma_1X_{2i}+\witil{Y_i}Yi=γ0+γ1X2i+Yiγ1=σYX2/σX22=−1.434/0.911=−1.574\gamma_1=\sigma_{YX_2}/\sigma_{X_2}^2=-1.434/0.911=-1.574γ1=σYX2/σX22=−1.434/0.911=−1.574γ0=μY−γ1μX2=−2.823−(−1.574∗−0.101)=−2.982\gamma_0=\mu_Y-\gamma_1\mu_{X_2}=-2.823-(-1.574*-0.101)=-2.982γ0=μY−γ1μX2=−2.823−(−1.574∗−0.101)=−2.982so β1=Cov()Y~,X1~σX1~2=2.459/1.313=1.873\beta_1=\frac{Cov()\witil{Y}, \witil{X_1}}{\sigma_{\witil{X_1}}^2}=2.459/1.313=1.873β1=σX12Cov()Y,X1=2.459/1.313=1.873Repeating the process for X2X_2X2β2=−1.085/0.889=−1.221\beta_2=-1.085/0.889=-1.221β2=−1.085/0.889=−1.221Tying things together:Yi−β1X1i−β2X2i=α+ϵi=−1.873X1i+1.221X2iY_i-\beta_1X_{1i}-\beta_2X_{2i}=\alpha+\epsilon_i=-1.873X_{1i}+1.221X_{2i}Yi−β1X1i−β2X2i=α+ϵi=−1.873X1i+1.221X2iSo α=−1.242\alpha = -1.242α=−1.242, anthe varianof the resials is 0.718.Finally, asserting the presumption of normality:Yi=−1.242+1.873∗X1i−1.221∗X2i+ϵi,ϵi N(0,0.718)Y_i=-1.242+1.873*X_{1i}-1.221*X_{2i}+\epsilon_i, \epsilon_i~N(0, 0.718)Yi=−1.242+1.873∗X1i−1.221∗X2i+ϵi,ϵi N(0,0.718)Note ththe varianfor ϵi\epsilon_iϵi is biaseanthe unbiasefigure is this figure multiplie10/(10 - 3) or 1.206. 请问红框中的是怎么计算出来的？按照上面的公式x残差数算出来对不上，y残差那一列是可以对的上的。

2024-05-17 12:58 1 · 回答

NO.PZ202001080100002601 问题如下 Apply OLS lineregression to finthe parameters for the following equation: Yi=α+β1X1i+β2X2i+ϵiY_i = \alpha + \beta_1X_{1i} + \beta_2X_{2i} + \epsilon_iYi=α+β1X1i+β2X2i+ϵi Start setting up the basic calculations: μY=−2.823,μX1=−0.91,μX2=−0.101,\mu_Y=-2.823, \mu_{X_1}=-0.91, \mu_{X_2}=-0.101,μY=−2.823,μX1=−0.91,μX2=−0.101,σY2=7.58,σX12=1.345,σX22=0.911\sigma_Y^2=7.58, \sigma_{X_1}^2=1.345, \sigma_{X_2}^2=0.911σY2=7.58,σX12=1.345,σX22=0.911σX1X2=−0.172,σYσX1=2.729,σYσX2=−1.434\sigma_{X_1X_2}=-0.172, \sigma_Y\sigma_{X_1}=2.729, \sigma_Y\sigma{X_2}=-1.434σX1X2=−0.172,σYσX1=2.729,σYσX2=−1.434First regressing on the last inpennt variable: X1i=δ0+δ1X2i+X1i~X_{1i}=\lta_0+\lta_1X_{2i}+\witil{X_{1i}}X1i=δ0+δ1X2i+X1iδ1=σX1X2σX22=−0.172/0.911=0.189\lta_1=\frac{\sigma_{X_1X_2}}{\sigma_{X_2}^2}=-0.172/0.911=0.189δ1=σX22σX1X2=−0.172/0.911=0.189δ0=μX1−δ1μX2=−0.91−0.189∗(−0.101)=−0.929\lta_0=\mu_{X_1}-\lta_1\mu_{X_2}=-0.91-0.189*(-0.101)=-0.929δ0=μX1−δ1μX2=−0.91−0.189∗(−0.101)=−0.929Yi=γ0+γ1X2i+Yi~Y_i=\gamma_0+\gamma_1X_{2i}+\witil{Y_i}Yi=γ0+γ1X2i+Yiγ1=σYX2/σX22=−1.434/0.911=−1.574\gamma_1=\sigma_{YX_2}/\sigma_{X_2}^2=-1.434/0.911=-1.574γ1=σYX2/σX22=−1.434/0.911=−1.574γ0=μY−γ1μX2=−2.823−(−1.574∗−0.101)=−2.982\gamma_0=\mu_Y-\gamma_1\mu_{X_2}=-2.823-(-1.574*-0.101)=-2.982γ0=μY−γ1μX2=−2.823−(−1.574∗−0.101)=−2.982so β1=Cov()Y~,X1~σX1~2=2.459/1.313=1.873\beta_1=\frac{Cov()\witil{Y}, \witil{X_1}}{\sigma_{\witil{X_1}}^2}=2.459/1.313=1.873β1=σX12Cov()Y,X1=2.459/1.313=1.873Repeating the process for X2X_2X2β2=−1.085/0.889=−1.221\beta_2=-1.085/0.889=-1.221β2=−1.085/0.889=−1.221Tying things together:Yi−β1X1i−β2X2i=α+ϵi=−1.873X1i+1.221X2iY_i-\beta_1X_{1i}-\beta_2X_{2i}=\alpha+\epsilon_i=-1.873X_{1i}+1.221X_{2i}Yi−β1X1i−β2X2i=α+ϵi=−1.873X1i+1.221X2iSo α=−1.242\alpha = -1.242α=−1.242, anthe varianof the resials is 0.718.Finally, asserting the presumption of normality:Yi=−1.242+1.873∗X1i−1.221∗X2i+ϵi,ϵi N(0,0.718)Y_i=-1.242+1.873*X_{1i}-1.221*X_{2i}+\epsilon_i, \epsilon_i~N(0, 0.718)Yi=−1.242+1.873∗X1i−1.221∗X2i+ϵi,ϵi N(0,0.718)Note ththe varianfor ϵi\epsilon_iϵi is biaseanthe unbiasefigure is this figure multiplie10/(10 - 3) or 1.206. 这题答案有错误把。。。。

2024-05-17 11:09 1 · 回答

二元的系数公式是什么？

2020-03-07 15:55 3 · 回答

这题计算量那么大，考试应该不会考吧

2020-02-15 00:19 1 · 回答