线性回归分析-HW2

Prob-1

β^1OLS=∑i=1N(xi−xˉ)(yi−yˉ)∑i=1N(xi−xˉ)2\hat{\beta}_1^{OLS}=\frac{\sum_{i=1}^{N}(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^{N} (x_i-\bar{x})^{2}}

β^0OLS=yˉ−β^1xˉ\hat{\beta}_0^{OLS}=\bar{y}-\hat{\beta}_1\bar{x}

Prob-2

β^0ML,β^1ML=β^0LS,β^1LSσ^2,ML=∑ei2n\begin{align} \hat{\beta}_{0}^{ML},\hat{\beta}_{1}^{ML} &= \hat{\beta}_{0}^{LS},\hat{\beta}_{1}^{LS} \\ \hat{\sigma}^{2,ML}&=\frac{\sum e_{i}^{2}}{n} \end{align}

Prob-3

1.29

It means that the expected value of YY is 0 given X=0X=0. The curve of the function passes (0,0)(0,0) on a graph.

1.41

a.

Solve

argmin⁡β^1∑i=1n(yi−β^1xi)2\operatorname*{argmin}_{\hat{\beta}_1}\sum_{i=1}^{n} (y_i-\hat{\beta}_1x_i)^{2}

we get

β^1OLS=∑i=1nxiyi∑i=1nxi2\hat{\beta}_1^{OLS}=\frac{\sum_{i=1}^{n}x_iy_i}{\sum_{i=1}^{n} x_i^{2}}

b.

L(β1∣X,Y)=∏i=1nfYi(yi∣β1,σ2,X)=⋯exp⁡(−12σ2∑i=1n(yi−β1xi)2)\begin{align} L(\beta_1\mid \boldsymbol{X},\boldsymbol{Y}) & =\prod_{i=1}^{n} f_{Y_i}(y_i\mid \beta_1,\sigma^{2},\boldsymbol{X}) \\ & = \cdots \exp \left( -\frac{1}{2\sigma^{2}}\sum_{i=1}^{n} (y_i-\beta_1x_i)^{2} \right) \end{align}

β^1ML=argmax⁡β1L(β1∣X,Y)=argmin⁡β1∑i=1n(yi−β1xi)2=β^1OLS\hat{\beta}^{ML}_1=\operatorname{argmax}_{\beta_1}L(\beta_1\mid \boldsymbol{X},\boldsymbol{Y})=\operatorname{argmin}_{\beta_1}\sum_{i=1}^{n} (y_i-\beta_1x_i)^{2}=\hat{\beta}_1^{OLS}

It's the same as the least square estimator.

c.

Eβ^1ML=E[∑i=1nxiyi∑i=1nxi2]=1∑i=1nxi2E[∑i=1nxi(β1x1+εi)]=1∑i=1nxi2∑i=1n(β1xi2+xiEεi)=β1\begin{align} \mathbb{E}\hat{\beta}_1^{ML} &= \mathbb{E}\left[ \frac{\sum_{i=1}^{n}x_iy_i}{\sum_{i=1}^{n} x_i^{2}} \right] \\ & = \frac{1}{\sum_{i=1}^{n} x_i^{2}}\mathbb{E}\left[ \sum_{i=1}^{n} x_i(\beta_1x_1+\varepsilon_i) \right] \\ & = \frac{1}{\sum_{i=1}^{n} x_i^{2}}\sum_{i=1}^{n} \left( \beta_1x_i^{2}+x_i\mathbb{E}\varepsilon_i \right) \\ & = \beta_1 \end{align}

Prob-4

a.

No. I don't know the exact distribution of ε\varepsilon.

b.

Y=100+20⋅5+εY=100+20\cdot 5+\varepsilon where ε∼N(0,25)\varepsilon\sim N(0,25), so

P(195<Y<205)=P(−5<ε<5)=Φ(1)−Φ(−1)P(195<Y<205)=P(-5<\varepsilon<5)=\Phi(1)-\Phi(-1)

where Φ(⋅)\Phi(\cdot) is the CDF of standard normal distribution.

Prob-5

a.

The data obtained in the study were observational data because no experiment was made.

b.

It is not very valid. Relationship does not mean causality.

c.

Age, occupation

d.

Add more variables into the regression so that the error term has zero expectation. Or use instrumental variables to get the real effect.

Prob-6

a.

β^0=2.11, β^1=0.039\hat{\beta}_0=2.11,\, \hat{\beta}_1=0.039

The estimated regression function is

y^=2.11+0.039x\hat{y}=2.11+0.039x

b.


It seems good.

c.

3.278863

d.

0.038.


线性回归分析-HW2
https://wtx17.github.io/2026/09/26/线性回归分析-HW2/
作者
wtx
发布于
2026年9月26日
更新于
2026年11月6日
许可协议