\documentclass[12pt]{article}

\newcommand{\CN}{Mathematical Statistics}
\newcommand{\Ti}{Homework 1}
\newcommand{\Pf}{Dr. Didier}
\newcommand{\FN}{Zehao}
\newcommand{\LN}{Wang}



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    \textsc{\Huge \CN}\\[0.5cm]
    \textsc{\large \Pf}\\[1.0cm]
    
    \textsc{\LARGE \Ti}\\[0.5cm]
    \textsc{\large \LN, \FN}\\
    {Master student in Statistics of Math Dept.}
    
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    {\Large \emph{\today}}
    
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    \section*{\Ti}

    \begin{solution}
        Let $f(t)$ be the distribution of $T$ and $T=m(U)$. Then $f^*(u)=f(m(U))m'(U)$. For $T$, if it is complete, then for any function $g(T)$, 
        \[
            \int g(t)f(t)\der t=0 \Rightarrow \Pr(g(t)=0)=1. 
        \]
        Substitute $t$ with $m(u)$. Then, 
        \[
            \begin{aligned}
                &\quad\,\int g(m(u))f(m(u))\der (m(u))\\
                &=\int g(m(u))f(m(u))m'(u)\der u\\
                &=\int g(m(u))f^*(u)\der u=0\Rightarrow \Pr(g(m(u))=0)=1. 
            \end{aligned}
        \]
        And because the arbitrary of $g$, $g(m(u))$ can be taken as any function of $u$. So, $U$ is also complete. 
    \end{solution}
    
    \begin{exercise}
        Let \(X_{1}, \ldots, X_{n} \stackrel{i.i.d. }{\sim} \mathcal{N}\left(\mu, \sigma^{2}\right)\), \(\mu \in \mathbb{R}, \sigma^{2}>0 .\) Show that \(\bar{X}\) and \(S^{2}\) are independent. 
        
        (\rmk{2}: this classical result can be shown by means of characteristic functions. However, in the context of our course, a much shorter proof can be provided). 
    \end{exercise}
    
    \begin{solution}
        We know that $(\bar{X}, S^2)$ are sufficient. And $\bar{X}\sim N(\mu,\sigma^2/n)$. So, $\bar{X}$ is complete for $\mu$ (Because $N(\mu,\sigma^2/n)$ is full rank exponential family). 
            
        And $S^2$ only depends on $\sigma^2$ (Because $\frac{(n-1)S^2}{\sigma^2}=\sum_{i=1}^n(\frac{X-\bar{X}}{\sigma})^2\sim\chi_n^2$), it is an ancillary statistic. So, from Basu's theorem, they are independent. 
    \end{solution}
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