Chap5.tex

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\begin{document}
\author{Zhizhong Pu}
\title{Solutions - Chapter 5}
\date{May 2023}
\maketitle

\thispagestyle{empty}

%% 5.1
\begin{Exercise} Show that if $Q\sim \chi^2_r$, then $\E[Q]=r$ and $Var[Q]=2r$ 

    First write $Q=\sum_{i=1}^rZ_i$ with $Z_i\sim\mathcal{N}(0,1)$ i.i.d.. Then $\E[Z_i^2]=1$ and $\E[Z_i^4]=3$ are easy to calculate.
    
    Then $\E[Q]=r$ and
    \[
    Var(Q) = r Var(Z_i^2) = r (\E[Z_i^4] - \E[Z_i^2]^2) = 2r
    \]
\end{Exercise}

%% 5.2
\begin{Exercise} Show that if $e \sim \mathcal{N}(0,\I_n\sigma^2)$ and $H'H=\I_n$ then $u=H'e \sim \mathcal{N}(0,\I_n\sigma^2)$

By Theorem 5.3.3, we have 
\[
u = H'e \sim \mathcal{N}(H'0,H'\I_n\sigma^2H) = \mathcal{N}(0,H'H\sigma^2) = \mathcal{N}(0,\I_n\sigma^2)
\]
\end{Exercise}

%% 5.3
\begin{Exercise} Show that if $e \sim \mathcal{N}(0,AA')$ then $u=A^{-1}e \sim \mathcal{N}(0,\I_n)$

By Theorem 5.3.3, we have 
\[
u = A^{-1} e \sim \mathcal{N}(A^{-1} 0,A^{-1}AA' (A^{-1})' = \mathcal{N}(0,\I_n)
\]
\end{Exercise}

%% 5.4
\begin{Exercise}\end{Exercise}

%% 5.5
\begin{Exercise} Show that $\hat{Y}_i|X \sim \mathcal{N}(X'_i\beta,\sigma^2h_i)$ where are the leverage values (3.40).
    \[
    \hat{Y} = PY = X(X'X)^{-1}X'(X\beta+e) = X\beta + X(X'X)^{-1}X'e =  X\beta + Pe
    \]
    Then
    \[
    \hat{Y} \sim \mathcal{N}(X\beta, PPVar(e)) \equiv \mathcal{N}(X\beta, \sigma^2P)
    \]
    Since $h_i$ specifies the $i$th diagonal element of $P$, then this shows that $\hat{Y}_i|X \sim \mathcal{N}(X'_i\beta,\sigma^2h_i)$

\end{Exercise}

%% 5.6
\begin{Exercise} \end{Exercise}

%% 5.7
\begin{Exercise} In the normal regression model show that the robust covariance matrices $V_{\hat{\beta}}^{HCi}$ are independent of the OLS estimator $\hat{\beta}$, conditional on $X$. 

Recall
    \[
    V_{\hat{\beta}}^{HC0} = (X'X)^{-1} (\sum_{i=1}^n X_iX_i' \widehat{e_i^2} ) (X'X^{-1}) 
    \]
    \[
    V_{\hat{\beta}}^{HC1} = \frac{n}{n-k} (X'X)^{-1} (\sum_{i=1}^n X_iX_i' \widehat{e_i^2} ) (X'X^{-1}) 
    \]
    \[
    V_{\hat{\beta}}^{HC2} = (X'X)^{-1} (\sum_{i=1}^n (1-h_i)^{-1} X_iX_i' \widehat{e_i^2} ) (X'X)^{-1}
    \]
    \[
    V_{\hat{\beta}}^{HC3} = (X'X)^{-1} (\sum_{i=1}^n (1-h_i)^{-2} X_iX_i' \widehat{e_i^2} ) (X'X)^{-1}
    \]
Then it can be seen that the only stochastic term in each of the variance estimators is $\widehat{e_i^2}$ after conditioning on $X$. By Theorem 5.6, we have $\widehat{e_i^2}$ being independent of $\hat{\beta}$ conditional on $X$. 
\end{Exercise}

%% 5.8
\begin{Exercise} Let $F(u)$ be the distribution function of a random variable $X$ whose density is symmetric about $0$. Show that $F(-u) = 1-F(u)$.

    Let $f(u)$ be the p.d.f. of the distribution specified by $F(u)$. We know that $f(u)$ is symmetric about $0$, then $F(0)=0.5$ and for any $x$ on the support of $f$, we have:
    \[
    \int_{-u}^0f(x) = \int_0^uf(x) \Rightarrow F(0) - F(-u) = F(u) - F(0) \Rightarrow F(-u) = 1-F(u)
    \]
\end{Exercise}

%% 5.9
\begin{Exercise} Let $\hat{C}_\beta=[L,U]$ be a $1-\alpha$ confidence interval for $\beta$,and consider the transformation where $g()$ is monotonically increasing. Consider the confidence interval $\hat{C}_\theta = [g(L),g(U)]$ for $\theta$. Show that $Pr(\theta \in \hat{C}_\theta) = Pr(\beta \in \hat{C}_\beta)$.

    Note that since $g()$ is monotonically increasing, then $g(x) \le g(L) \Leftrightarrow x \le L$. Then
    \[
    Pr(\beta \in \hat{C}_\beta) = Pr(\beta \le U) - Pr(\beta \le L) = Pr(g(\beta) \le g(U)) - Pr(g(\beta) \le g(L)) = Pr(\theta \in \hat{C}_\theta) 
    \]

\end{Exercise}

%% 5.10
\begin{Exercise} \end{Exercise}

%% 5.11
\begin{Exercise} \end{Exercise}

%% 5.12
\begin{Exercise} In the normal regression model let $s^2$ be the unbiased estimator of the error variance $\sigma^2$ from (4.31). Show that $Var(s^2) = 2\sigma^4 / (n-k)$ and that it is strictly larger than the Cramér-Rao Lower Bound for $\sigma^2$

    By Theorem 5.7 we have
    \[
    \frac{(n-k)s^2}{\sigma^2} \sim \chi^2_{n-k}
    \]
    Note that the variance of LHS is $2(n-k)$, and $(n-k)$ and $\sigma$ are non-stochastic, so 
    \[
    \frac{(n-k)^2}{\sigma^4} Var(s^2) = 2(n-k)
    \]
    This shows that $Var(s^2) = 2\sigma^4 / (n-k)$. Note that $k>0$ and the Cramér-Rao Lower Bound for $\sigma^2$ is $2\sigma^4/n$ from (5.20), so $Var(s^2) = 2\sigma^4 / (n-k) > 2\sigma^4/n $

\end{Exercise}


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