UA MATH 571B 回歸 QE練習(xí)題 一元線性回歸理論

• 2015/1/5
• 2015/5/5
• 2016/5/6
• 2017/1/5
• 2017/5/6

這是2015年1月第五題，2015年5月第五題，2016年5月第六題，2017年1月第五題，2017年5月第六題。QE methodology的理論題目范圍就是一元線性回歸的相關(guān)理論，具體可以參考UA MATH571A 一元線性回歸I 模型設(shè)定與估計(jì)、UA MATH571A 一元線性回歸II 統(tǒng)計(jì)推斷1、UA MATH571A 一元線性回歸II 統(tǒng)計(jì)推斷2、UA MATH571A 一元線性回歸III 方差分析與相關(guān)性分析

2015/1/5

注意這道題題干有錯(cuò)，$\bar{Y}$后面應(yīng)該是減號(hào)。

Want to show $E[b_0]={\beta }_0$. Let’s first analyze the origin of randomness in $b_0$. Define
$$k_i=\frac{X_i?\bar{X}}{{\sum }_{i=1}^n(X_i?\bar{X}{)}^2}$$

so we can rewrite $b_0$ as
$$\begin{gathered} b_0=\bar{Y}?\bar{X}\sum _{i=1}^n{k}_i{Y}_i=\sum _{i=1}^n\left(\frac{1}{n}?k_i\bar{X}\right)Y_i=\sum _{i=1}^n\left(\frac{1}{n}?k_i\bar{X}\right)({\beta }_0+{\beta }_1{X}_i+{?}_i) \\ ={\beta }_0\sum _{i=1}^n\left(\frac{1}{n}?k_i\bar{X}\right)+{\beta }_1\sum _{i=1}^n{X}_i\left(\frac{1}{n}?k_i\bar{X}\right)+\sum _{i=1}^n\left(\frac{1}{n}?k_i\bar{X}\right){?}_i \end{gathered}$$

The first two terms are constant, and for the third term
$$E\sum _{i=1}^n\left(\frac{1}{n}?k_i\bar{X}\right){?}_i=\sum _{i=1}^n\left(\frac{1}{n}?k_i\bar{X}\right)E{?}_i=\sum _{i=1}^n\left(\frac{1}{n}?k_i\bar{X}\right)\times 0=0$$

Now simplify the first two terms.
$$\begin{gathered} \sum _{i=1}^n\left(\frac{1}{n}?k_i\bar{X}\right)=\sum _{i=1}^n\frac{1}{n}?\bar{X}\sum _{i=1}^n{k}_i=1 \\ \sum _{i=1}^n{X}_i\left(\frac{1}{n}?k_i\bar{X}\right)=\bar{X}?\bar{X}\sum _{i=1}^n{k}_i{X}_i=0 \end{gathered}$$

So $E[b_0]={\beta }_0$.

注意$k_i$是一個(gè)非常常用的構(gòu)造，它的求和有幾個(gè)很重要的性質(zhì)，可以參考上面的博文。

2015/5/5

暫時(shí)放棄這個(gè)題，兩周后再做

2016/5/6

Part a
Define
$$k_i=\frac{X_i}{{\sum }_{i=1}^n{X}_i^2}$$

and we can rewrite $b_1$ as
$$b_1=\sum _{i=1}^n{k}_i{Y}_i$$

Let’s analyze the randomness of $b_1$, note that $Y_i={\beta }_1{X}_i+{?}_i$,
$$b_1=\sum _{i=1}^n{k}_i({\beta }_1{X}_i+{?}_i)={\beta }_1\sum _{i=1}^n{k}_i{X}_i+\sum _{i=1}^n{k}_i{?}_i$$

Compute
$$\begin{gathered} \sum _{i=1}^n{k}_i{X}_i=\sum _{i=1}^n\frac{X_i^2}{{\sum }_{j=1}^n{X}_j^2}=1 \\ ?b_1={\beta }_1+\sum _{i=1}^n{k}_i{?}_i～N({\beta }_1,{\sigma }^2\sum _{i=1}^n{k}_i^2) \end{gathered}$$

Compute
$$\sum _{i=1}^n{k}_i^2=\sum _{i=1}^n\frac{X_i^2}{[{\sum }_{j=1}^n{X}_j^2{]}^2}=\frac{1}{{\sum }_{i=1}^n{X}_i^2}?b_1～N({\beta }_1,\frac{{\sigma }^2}{{\sum }_{i=1}^n{X}_i^2})$$

Part b
$${\hat{Y}}_h=\sum _{i=1}^n{k}_i{Y}_i{X}_h$$

Part c
$$Var({\hat{Y}}_h)=\frac{{\sigma }^2{X}_h^2}{{\sum }_{i=1}^n{X}_i^2}$$

2017/1/5

Part a
$$E[Y]={\beta }_0+{\beta }_1\xi =0?\xi =?\frac{{\beta }_0}{{\beta }_1}$$

Part b
$$\hat{\xi }=?\frac{{\hat{\beta }}_0}{{\hat{\beta }}_1}=?\frac{{\sum }_{i=1}^n(\frac{1}{n}?k_i\bar{X})Y_i}{{\sum }_{i=1}^n{k}_i{Y}_i}$$

where ${\hat{\beta }}_0$ and ${\hat{\beta }}_1$ are MLE of ${\beta }_0$ and ${\beta }_1$, and
$$k_i=\frac{X_i?\bar{X}}{{\sum }_{i=1}^n(X_i?\bar{X}{)}^2}$$

Part c
Let $\hat{\xi }=g({\hat{\beta }}_0,{\hat{\beta }}_1)=?\frac{{\hat{\beta }}_0}{{\hat{\beta }}_1}$. Notice $E[{\hat{\beta }}_0]={\beta }_0$ and $E[{\hat{\beta }}_1]={\beta }_1$ and compute
$$g_{{\hat{\beta }}_0}^{\prime }=?\frac{1}{{\beta }_1},g_{{\hat{\beta }}_1}^{\prime }=\frac{{\beta }_0}{{\beta }_1^2}$$

(i)
$$E[\hat{\xi }]\approx ?\frac{{\beta }_0}{{\beta }_1}+0+0=\xi$$

(ii)
$$\begin{gathered} Var(\hat{\xi })\approx \frac{{\sigma }^2}{{\beta }_1^2{\sum }_{i=1}^n(X_i?\bar{X}{)}^2}+{\sigma }^2\frac{{\beta }_0^2}{{\beta }_1^4}(\frac{1}{n}+\frac{{\bar{X}}^2}{{\sum }_{i=1}^n(X_i?\bar{X}{)}^2})? \\ \frac{2{\beta }_0}{{\beta }_1^3}Cov({\hat{\beta }}_0,{\hat{\beta }}_1) \end{gathered}$$

Compute
$$\begin{gathered} Cov({\hat{\beta }}_0,{\hat{\beta }}_1)=Cov({\beta }_0+\sum _{i=1}^n(1/n?k_i\bar{X})Y_i,{\beta }_1+\sum _{i=1}^n{k}_i{Y}_i) \\ =(\frac{1}{n}\sum _{i=1}^n{k}_i?\bar{X}\sum _{i=1}^n{k}_i^2){\sigma }^2=?\frac{\bar{X}{\sigma }^2}{{\sum }_{i=1}^n(X_i?\bar{X}{)}^2} \end{gathered}$$

2017/5/6

注意！這道題用相關(guān)結(jié)論就非常簡(jiǎn)單，可以口算寫(xiě)出答案，但標(biāo)準(zhǔn)答案是從頭開(kāi)始推的，LS估計(jì)的分布要自己推！
Part a
$$\begin{gathered} E[Y]=1+{\beta }_1\xi =0?\xi =?\frac{1}{{\beta }_1} \\ \hat{\xi }=g(\widehat{{\beta }_1})??\frac{1}{{\hat{\beta }}_1}=?\frac{1}{{\sum }_{i=1}^n{k}_i{Y}_i} \end{gathered}$$

Part b
$$E[\hat{\xi }]\approx ?\frac{1}{{\beta }_1}=\xi$$

Part c
$$E[\hat{\xi }]=\xi +\frac{{\sigma }^2{\xi }^3}{{\sum }_{i=1}^n(X_i?\bar{X})}$$

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