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%
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\title[Stat 609 Lecture 15]{Stat 609: Mathematical Statistics \\ Lecture 15}
\author[UW-Madison]{Jun Shao}
\institute[Statistics]{%
%  Department of Statistics\\
%  University of Wisconsin\\
%  Madison, WI 53706, USA\\
%~~~\\
  }
\date[2014]{}

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\AtBeginSection[]{\frame{\frametitle{Outline}\tableofcontents[current]}}
\def\back{\!\!\!\!\!}

\begin{document}


%\frame{\titlepage}

	

\begin{block}{Example}
	Let $F(x)$ and $G(y)$ be univariate cdf's with pdf's $f(x)$ and $g(y)$, respectively, and 
	$\alpha$ be a constant with $|\alpha | \leq 1$. 
	
	We want to show that
	\vspace{-2mm} 
	\[ H(x,y) =  F(x)G(y) \{ 1+\alpha [1-F(x)][1-G(y)] \} \quad\quad (x,y) \in {\cal R}^2 \vspace{-2mm} \]
	is a 2-dimensional cdf with pdf \vspace{-2mm} 
	\[  h(x,y) = f(x)g(y) \{ 1+\alpha [2F(x)-1][2G(y)-1] \} \quad\quad (x,y) \in {\cal R}^2  \]
\end{block}}
\frame{\vspace{-2mm} 
	\begin{block}{}
		Since \vspace{-3mm}
		\[ | \alpha [2F(x)-1][2G(y)-1] |  \leq |  [2F(x)-1][2G(y)-1]| \leq 1 \vspace{-2mm} \]
		we know that $h(x,y) \geq 0$. 
		
		\def\back{\!\!\!\!\!} 
		
		For any $(x,y) \in {\cal R}^2$, \vspace{-2mm} 
		\beqrs  
		\back && \back  \int_{-\infty}^x\int_{-\infty}^y h(u,v)dvdu \\
		\back  & =\back  & \int_{-\infty}^x  \int_{-\infty}^y f(u)g(v) \{ 1+\alpha [2F(u)-1][2G(v)-1] \}
		dvdu  \\
		\back & = & \back
		F(x) G(y)  + \alpha  \int_{-\infty}^x f(u)[2F(u)-1] du  \int_{-\infty}^y g(v)[2G(v)-1]dv \\
		\back  & = & \back 
		F(x) G(y) + \alpha [2F^2(x)-F(x)]  [G^2(y)-G(y)] \\
		\back & = & \back F(x)G(y) \{ 1 + \alpha [1-F(x)][1-G(y)] \} \\
		\back & = & \back H(x,y) 
		\eeqrs  
		$~$ \vspace{-7mm}
		
		This shows that $h(x,y)$ is a pdf of $H(x,y)$, if $H(x,y)$ is a cdf. 
		
		$H(x,y)$ is indeed a cdf, because \vspace{-2mm} 
		\[ \lim_{x\rightarrow \infty, y \rightarrow \infty} H(x,y) = 
		\lim_{x\rightarrow \infty, y \rightarrow \infty} F(x)G(y) = 1 \]
	\end{block}
	
	
					\frame{
						\begin{block}{}
							As an exercise, we now use convolution to derive the same result.
							\vspace{1mm}
							
							By the convolution formula, the pdf of $U+V$ is \vspace{-2mm}
							\[ f_{U+V}(z) = \int_{-\infty}^\infty \frac{1}{\pi (1+w^2)} \ \frac{1}{\pi [1+(z-w)^2]} dw \vspace{-2mm} \]
							After some calculation, we obtain that \vspace{-2mm}
							\beqrs
							\frac{1}{\pi^2 (1+w^2) [1+(z-w)^2]}
							&=& \frac{2}{\pi^2 (z^2+4)z} \left[ \frac{w}{1+w^2} - \frac{w-z}{1+(z-w)^2} \right]  \\
							&&  + \frac{1}{\pi^2(z^2+4)} \left[ \frac{1}{1+w^2} + \frac{1}{1+(z-w)^2} \right]
							\eeqrs
							For every fixed $z$, when $|w|$ is large enough,  \vspace{-2mm}
							\[ \left| \frac{w}{1+w^2} - \frac{w-z}{1+(z-w)^2} \right| = \left| \frac{2(1-w^2+zw) }{(1+w^2)[1+(z-w)^2]} \right|
							\leq \frac{4}{1+(z-w)^2}\vspace{-2mm} \]
							Hence the first term on the left hand side of the above expression is integrable and, by symmetry,  
						\end{block}}
						\frame{
							\begin{block}{}
								\[ \int_{-\infty}^\infty \left[ \frac{w}{1+w^2} - \frac{w-z}{1+(z-w)^2} \right] dw =0 \vspace{-2mm}
								\]
								Then, \vspace{-2mm}
								\beqrs
								f_{U+V}(z) &=&  \frac{1}{\pi^2(z^2+4)} \int_{-\infty}^\infty \left[ \frac{1}{1+w^2} + \frac{1}{1+(z-w)^2} \right] dw \\
								& = &  \frac{1}{\pi^2(z^2+4)} (\pi + \pi ) = \frac{1}{2\pi [1+(z/2)^2]}
								\eeqrs
								This is the pdf of $Cauchy(0,2)$ distribution.
								\vspace{1mm}
								
								
							\end{block}
							
						}
						\frame{
							\begin{block}{}
								\[ U_i = \sum_{j=1}^n a_{ij} X_j , \quad i=1,...,k, \quad V_r = \sum_{j=1}^n b_{rj} X_j , \quad r=1,...,m \vspace{-4mm}\]
								\begin{itemize}
									\item[a.]
									The random variables $U_i$ and $V_r$ are independent iff Cov$(U_i,V_r) =0$. Furthermore,
									Cov$(U_i,V_r) =\sum_{j=1}^n a_{ij}b_{rj} \sigma_j^2$.\vspace{1mm}
									\item[b.]
									The random vectors $(U_1,...,U_k)$ and $(V_1,...,V_m)$ are independent iff $U_i$ and $V_r$ are independent for all pairs $(i,r)$.
								\end{itemize}
							\end{block}\vspace{-2mm}
							\begin{block}{Proof.}
								The claims about independence can be shown by using transformations, or directly applying the results for multivariate normal distributions in Chapter 4.
								\vspace{1mm}
								
								The result for the covariance can be shown by direct calculation
								\vspace{-2mm}
								\beqrs
								{\rm Cov}(U_i,V_r) &=& {\rm Cov} \left(
								\sum_{j=1}^n a_{ij}X_j , \sum_{l=1}^n b_{rl} X_l \right) =
								\sum_{j=1}^n  \sum_{l=1}^n {\rm Cov} (a_{ij}X_j , b_{rl} X_l) \\
								&=&\sum_{j=1}^n  a_{ij} b_{rj} {\rm Var} (X_j) +
								\sum_{j \neq l}  a_{ij} b_{rj} {\rm Cov} (X_j, X_l)
								=  \sum_{j=1}^n a_{ij}b_{rl} \sigma_j^2
								\eeqrs
							\end{block}
						}
						\frame{\vspace{-2mm}
							\begin{alertblock}{Other proofs of the independence of $\bar{X}$ and $S^2$}
								Previously we have shown that $S^2$ is a function of $(X_2-\bar{X},...,X_n-\bar{X})$.
								
								Hence, by Lemma 5.3.3, we only need to show that $\bar{X}$ and $X_j-\bar{X}$ are uncorrelated for $j=2,...,n$.
								
								Note that \vspace{-2mm}
								\[ \bar{X} = \sum_{i=1}^n \left(\frac{1}{n} \right)  X_j, \quad\quad X_j - \bar{X} = \sum_{i=1}^n \left(\delta_{ij} - \frac{1}{n} \right)X_i ,
								\quad \delta_{ij} = \left\{ \br{ll}
								1 & \quad i=j \\
								0 & \quad i \neq j \er \right. \vspace{-2mm}\]
								By Lemma 5.3.3, for $j=2,...,n$, \vspace{-2mm}
								\[ {\rm Cov} (X_1 , X_j - \bar{X} )
								= \sum_{i=1}^n \frac{1}{n} \left(\delta_{ij} - \frac{1}{n} \right)\sigma^2 = \sigma^2 \left[ \frac{1}{n} \left(1-\frac{1}{n}\right)
								- \frac{n-1}{n^2}\right]  = 0 \vspace{-1mm} \]
								We can also prove the independence of $\bar{X}^2$ and $S^2$ using Cochran's theorem as follows.
								\vspace{1mm}
								
								Let $X$ be the $n$-dimensional vector whose $i$th component is $X_i$ and $J$ be the $n$-dimensional vector whose $i$th component is $n^{-1}$,
								$i=1,...,n$.
								\vspace{-3mm}
								\[ X'X = \sum_{i=1}^n X_i^2 , \quad\quad J' X = \sum_{i=1}^n n^{-1} X_i = \bar{X}, \quad \quad J'J = \sum_{i=1}^n n^{-2} = n^{-1}  \]
							\end{alertblock}
						}
						\frame{\vspace{-2mm}
							\begin{alertblock}{}Let $A_1 = nJJ'$, which is an $n\times n$ symmetric matrix of rank 1. \vspace{-2mm}
								\[ X'A_1X = X' nJJ' X = n (J'X)' (J'X) = n \bar{X}^2 \vspace{-2mm} \]
								Let $A_2 = I_n - nJJ'$, where $I_n$ is the identity matrix of order $n$. \vspace{-2.5mm}
								\[ X' A_2 X = X'(I_n-nJJ'X) = X'X - n \bar{X}^2 = \sum_{i=1}^n X_i^2 - n \bar{X}^2  = (n-1)S^2 \vspace{-2.5mm} \]
								Hence \vspace{-2mm}
								\[ X'X = n\bar{X} + X'X - n \bar{X}^2 = X'A_1X + X'A_2X \vspace{-2mm} \]
								What is the rank of $A_2$?
								\vspace{-2mm}
								\[ A_2^2 = (I_n-nJJ')(I_n-nJJ') = I_n -2nJJ' +n^2 JJ'JJ' = I_n - nJJ' = A_2 \vspace{-2mm} \]
								$A_2$ is a projection matrix with rank
								\vspace{-2mm}
								\[ {\rm trace}(A_2) = {\rm trace}(I_n) - {\rm trace}(nJJ') = n - nJ'J = n-1 \vspace{-2mm} \]
								Since rank of $A_1 + $ rank of $A_2 = 1 +n-1 = n$, by  Cochran's theorem,
								$X'A_1X = n \bar{X}^2$ and $X'A_2X = (n-1)S^2$ are independent.
								\vspace{1mm}
								
								Hence $\bar{X}^2$ and $S^2$ are independent.
								\vspace{1mm}
								
								By Cochran's Theorem, we can also show that the distribution of $(n-1)S^2 /\sigma^2$ is the central chi-square with degrees of freedom $n-1$.
							\end{alertblock}}
							\frame{
								\begin{block}{Sample moments from a normal population}
									If $X_1,...,X_n$ are iid $N(0,1)$, we want to show that \vspace{-2mm} 
									\[ E \left( \frac{M_3^j}{M_2^{3j/2}}\right) = \frac{E(M_3^j)}{E(M_2^{3j/2})} , \quad\quad 
									E \left( \frac{M_4^j}{M_2^{2j}}\right) = \frac{E(M_4^j)}{E(M_2^{2j})} , \quad\quad \quad j=1,2,... \vspace{-2mm} \]
									\[ E \left( \frac{\tilde M_3^j}{\tilde M_2^{3j/2}}\right) = \frac{E(\tilde M_3^j)}{E(\tilde M_2^{3j/2})} , \quad\quad
									E \left( \frac{\tilde M_4^j}{\tilde M_2^{2j}}\right) = \frac{E(\tilde M_4^j)}{E(\tilde M_2^{2j})} , \quad\quad \quad j=1,2,... \vspace{-1mm} \]
									From the result in lecture 19, \vspace{-2mm} 
									\[ M_2 \ \ \mbox{and} \ \ X_i^2/M_2 \ \ \mbox{are independent}, \ i=1,...,n  \vspace{-2mm} \]
									Repeatedly using the fact that functions of independent random variables are independent, we obtain\vspace{-2mm} 
									\[  \begin{aligned}
									M_2^{3/2} & \ \ \mbox{and} \ \ X_i^3/M_2^{3/2} \ \ \mbox{are independent}, \ i=1,...,n \\
									M_2^{3/2} & \ \ \mbox{and} \ \ \sum_{i=1}^n X_i^3/M_2^{3/2} = M_3/M_2^{3/2} \ \ \mbox{are independent}  \\
									M_2^{3j/2} & \ \ \mbox{and} \ \ M_3^j/M_2^{3j/2} \ \ \mbox{are independent} 
									\end{aligned} \]
								\end{block}}
								\frame{
									\begin{block}{}
										Then  
										\[ E(M_3^j ) = E \left( M_2^{3j/2} \frac{M_3^j}{M_2^{3j/2}}\right) = 
										E \left( M_2^{3j/2} \right) E\left( \frac{M_3^j}{M_2^{3j/2}}\right) \vspace{-2mm} \]
										and the first result follows. 
										\vspace{1.5mm}
										
										The proof of the second result is similar. 
										\vspace{1.5mm}
										
										The third and fourth results can be similarly derived if we can show that 
										\[ \tilde M_2 \ \ \mbox{and} \ \ (X_i-\bar{X})^2/ \tilde M_2 \ \ \mbox{are independent}, \ i=1,...,n  \vspace{-2mm} \]
										This result can be established using the transformation $Y_i = X_i - \bar{X}$ and the same argument in the proof of the independence of $M_2$ and 
										$X_i^2/M_2$, $i=1,...,n$. 
										\vspace{1.5mm}
										
										It can also be established using  Basu's theorem introduced in Chapter 6.  
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