Chap7.tex

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\begin{document}
\author{Zhizhong Pu}
\title{Solutions - Chapter 7}
\date{April 2023}
\maketitle

\thispagestyle{empty}

%% 4.1
\begin{Exercise} Question see book. 

   Given the DGS: $Y=X_1'\beta_1+X_2'\beta_2+e$ with $\E[Xe]=0$, regressing $Y$ on $X_1$ only yields 
   \[\begin{split}
     \hat{\beta_1} & = (X_1X_1')^{-1}X_1Y \\ 
    & = (X_1X_1')^{-1}X_1(X_1'\beta_1+X_2'\beta_2+e) \\ 
    & = \beta_1 + (X_1X_1')^{-1}(X_1X_2'\beta_2) + (X_1X_1')^{-1}(X_1e) \\
    & = \beta_1 + (\frac{1}{n} \sum_{i=1}^n X_{1i} X_{1i}')^{-1} (\frac{1}{n} \sum_{i=1}^n X_{1i} X_{2i}' )\beta_2 + (\frac{1}{n} \sum_{i=1}^n X_{1i} X_{1i}')^{-1} (\frac{1}{n} \sum_{i=1}^n X_{1i} e_i)
    \end{split}\]
   Note that by WLLN, we have 

   \begin{enumerate}
        \item $(\frac{1}{n} \sum_{i=1}^n X_{1i} X_{1i}')^{-1} \rightarrow \E[X_1X_1']^{-1}$
        
        \item $\frac{1}{n} \sum_{i=1}^n X_{1i} e_i \rightarrow \E[X_1e] = 0$
        
        \item $\frac{1}{n} \sum_{i=1}^n X_{1i} X_{2i}' \rightarrow \E[X_1X_2']$ if $\E[X_1X_2']<\infty$
        
   \end{enumerate}
   
   Then $\hat{\beta_1}$ is generally not consistent, and it is consistent iff $\E[X_1X_2'] = 0$ or $\beta_2 = 0$
    
\end{Exercise}

%% 7.2
\begin{Exercise}\end{Exercise}

%% 7.3
\begin{Exercise}\end{Exercise}

%% 7.4
\begin{Exercise}\end{Exercise}

%% 7.5
\begin{Exercise}\end{Exercise}

%% 7.6
\begin{Exercise} Full problem see book. Find the method of moments estimators $(\hat{\beta},\hat{\Omega})$ for $(\beta,\Omega)$ where $\Omega = \E[XX'e^2]$.

    Since $\hat{\beta} = \E[XX']\E[XY]$, then we could substitute in the sample moment for population moment: $\hat{\beta}_{MoM} = (\frac{1}{n} \sum_{i=1}^n X_i X_i')^{-1} (\frac{1}{n} \sum_{i=1}^n X_i Y_i') $. And we can do the same for $\Omega$:
    \[
        \hat{\Omega}_{MoM} = (\frac{1}{n} \sum_{i=1}^n X_i X_i' e_i^2)
    \]
    

\end{Exercise}

%% 7.7
\begin{Exercise}\end{Exercise}

%% 7.8
\begin{Exercise} Find the asymptotic distribution of $\sqrt{n}(\widehat{\sigma^2} - \sigma^2)$ as $n \rightarrow \infty$

    First, we can re-write:
   \[\begin{split}
     \sqrt{n}(\widehat{\sigma^2} - \sigma^2) & = \sqrt{n}(\widehat{\sigma^2} - \frac{n-k}{n} \sigma^2 - \frac{k}{n} \sigma^2) \\ 
    & = \sqrt{n}(\widehat{\sigma^2} - \frac{n-k}{n} \sigma^2) - \frac{k}{\sqrt{n}} \sigma^2
    \end{split}\]

    Note that $\widehat{\sigma^2} \equiv n^{-1} \sum_{i=1}^n \hat{e}'\hat{e} $ by definition and that in Chapter 4.11 we have established that 
    \[
    \E[\widehat{\sigma^2}] = \frac{n-k}{n} \sigma^2
    \]
    Then we can view $\widehat{\sigma^2}$ and $\frac{n-k}{n}$ as a sample mean and its population mean, respectively. This warrants applying the CLT, which states that 
    \[
    \sqrt{n}(\widehat{\sigma^2} - \frac{n-k}{n} \sigma^2) \xrightarrow[]{d} \mathcal{N}(0,V)
    \]
    where $V=\E[(\widehat{\sigma^2} - \frac{n-k}{n} \sigma^2)^2]$. Lastly, as $- \frac{k}{\sqrt{n}} \sigma^2$ is a constant that converges to $0$, adding it is equivalent to a mean-shift of size $0$. Then
    \[
    \sqrt{n}(\widehat{\sigma^2} - \sigma^2) \xrightarrow[]{d} \mathcal{N}(0,V)
    \]

\end{Exercise}

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