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Drink H · 2020年02月15日

问一道题：NO.PZ202001080100002601

* 问题详情，请查看题干

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.

这题计算量那么大，考试应该不会考吧

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orange品职答疑助手 · 2020年02月15日

同学你好，是的，我也觉得应该不会考，因为咱们计算器只能算一元线性回归的系数。这是原版书的题啦，原版书有些题是有点偏

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2024-05-17 12:58 1 · 回答

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2024-05-17 12:16 1 · 回答

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