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\title[Stat 609 Lecture 17]{Stat 609: Mathematical Statistics \\ Lecture 17}
\author[UW-Madison]{Jun Shao}
\institute[Statistics]{%
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%  University of Wisconsin\\
%  Madison, WI 53706, USA\\
%~~~\\
  }
\date[2015]{}

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\AtBeginSection[]{\frame{\frametitle{Outline}\tableofcontents[current]}}

\begin{document}

%\frame{\titlepage}

\frame {\vspace{-2mm}
\frametitle{Chapter 5: Properties of a Random Sample\\
Lecture 17: Population, random sample, and statistics}
\begin{alertblock}{Populations, samples, and models}
\begin{itemize}
  \item One or a series of random experiments is performed.
  \item Some data from the experiment(s) are collected.
  \item Planning experiments and collecting data are not discussed in the textbook.
  \item Data analysis and inference: extract information from the data, interpret the results, and draw some conclusions.
  \item The data set is a realization of a
random vector defined on a sample space.
  \item The distribution of the random vector is called the {\bf population}.

  In some cases, a population may be a set of elements from which we draw a sample.
  \item The random vector that produces the
data is called a {\bf sample} from the population.
\end{itemize}
\end{alertblock}}
\frame{\vspace{-3mm}
\begin{alertblock}{}
\begin{itemize}
  \item The size of the data set is called the {\bf sample size}.
  \item A population is {\bf known} iff
the distribution is completely  known.
\item In a statistical problem, the population is at least partially unknown.
  \item We would like to deduce some properties of the population based on the
available sample.
  \item A {\bf statistical model} is a set of assumptions
on the population and is often postulated
to make the analysis possible or easy.
  \item Postulated models are often based on knowledge of the
problem under consideration.
\item
A statistical model or population is parametric if it can be indexed by a vector of fixed dimension.
Otherwise it is nonparametric.
\end{itemize}
\end{alertblock}\vspace{-2mm}
\begin{block}{Statistics and their distributions}
\begin{itemize}
  \item Our data set is a realization of a sample
(random vector) $X$ from an unknown population
  \item Statistic $T(X)$: A function $T$ of $X$;
$T(X)$ is a known value whenever $X$ is known.
\end{itemize}
\end{block}}
\frame{\vspace{-2mm}
\begin{block}{}
\begin{itemize}
\item Statistical analysis and inference is based on various statistics,
for various purposes.
  \item $X$ itself is a statistic, but it is a trivial statistic.
  \item The range of a nontrivial statistic $T(X)$
is usually simpler than that of $X$, i.e., $T(X)$ provides a ``reduction".
  \item For example, $X$ may be a random $n$-vector and
$T(X)$ may be a random $m$-vector with an $m$ much smaller
than $n$.
      \item A statistic $T(X)$ is a random vector (element).
      \item If the distribution of $X$ is unknown,
then the distribution of $T$ may also be unknown, although $T$ is a
known function.
       \item Finding the form of the distribution of $T$ is one of
the major problems in statistical inference.
\item Since $T$ is a transformation of $X$, tools we learn in Chapters 1-4 for
transformations may be useful in finding the distribution or
an approximation to the distribution of $T(X)$.
\item
Approximations are often given in terms of limits, i.e., the sample size $n $ increases to $\infty$.
\end{itemize}
\end{block}}
\frame{
\begin{alertblock}{Definition 5.1.1 (random sample)}
We say that a set of random vectors $X_1,...,X_n$ is a random sample (of size $n$) from a population
(a cdf $F$) iff \\
(a) $X_1,...,X_n$ are independent and \\
(b) the cdf of $X_i$  is $F$ for all
$i$.

When (a) and (b) hold, we also say that $X_1,...,X_n$ are iid (independent and identically distributed)
or $X_1,...,X_n$ is an iid sample.
 \end{alertblock} \vspace{-2mm}
\begin{alertblock}{}
\begin{itemize}
\item
The joint cdf of a random sample $X_1,...,X_n$ with cdf $F$ is \vspace{-2mm}
\[ F(x_1) \cdots F(x_n) = \prod_{i=1}^n F(x_i) , \quad\quad x_i \in {\cal R}^k, \ i=1,...,n, \vspace{-2mm} \]
where $k$ is the dimension of $X_i$.
\item
If $F$ in the previous expression has a pdf or pmf $f$, then the same expression holds with $F$ replaced by $f$.
\item
A random sample is viewed as sampling from an infinite population or from a finite population with replacement so that $X_i$'s
are independently observed.
\end{itemize}
\end{alertblock}}
\frame{\vspace{-2mm}
\begin{block}{Sampling without replacement from a finite population}
Sometimes we  consider sampling without replacement from a finite population; e.g.,
a survey of $n$ persons from a population of size $N$.
\begin{itemize}
\item
If each person in the population has characteristic $x_j$ (a $k$-dimensional vector), then a sample $X_1,...,X_n$
is $n$ random vectors and the range of each $X_i$ is $\{ x_1,...,x_N\}$.
\item
If sampling is without replacement, then $X_1,...,X_n$ can not be a random sample because, if $X_1 =x_k$, then $X_2$ can not be $x_k$
so that $X_1$ and $X_2$ are not independent.
\item
Is there a similar concept to ``random sample"?
\item
$X_1,...,X_n$ is called a simple random sample of size $n$ without replacement
from population $\{ x_1,...,x_N\}$ iff \vspace{-2mm}
\[ P( X_1 = x_{i_1} , ..., X_n = x_{i_n} ) = {N \choose n}^{-1} , \quad \quad \mbox{for any $\{ i_1 ,..., i_n \} \subset \{ 1,...,N \}$} \vspace{-2mm} \]
\item
In a simple random sample, $X_i$'s have the same distribution; however, they are not independent.
\item
The dependence becomes weak when $N$ is much larger than $n$.
\end{itemize}
\end{block}
}
\frame{\vspace{-2mm}
\begin{alertblock}{Example.}
The simplest finite population is the population with $N$ characteristics $x_1,...,x_N$ whose values
are either 0 or 1 (binary).
\vspace{0.5mm}

In such a case the number of ones, $M$, or the proportion $M/N$ is the only thing unknown in the population.
\vspace{0.5mm}


If $X_1,...,X_n$ is a simple random sample without replacement from this population and $Y = X_1+ \cdots + X_n $, then 
 \vspace{-2mm}
\[ P(Y=y) = \left\{ \begin{array}{ll}
\frac{{M \choose y} {N-M \choose n-y} }{{N \choose n}}
 & \quad  y=0,1,...,n \\
0 & \quad \mbox{otherwise} \end{array} \right. \vspace{-2mm} \]
assuming that $n < M$ and $n < N-M$.
\vspace{0.5mm}

But $X_1$ and $X_2$ are not independent, since \vspace{-3mm}
\[ P(X_2=1|X_1 =1) = \frac{M-1}{N-1} \neq \frac{M}{N} =P(X_2=1) \vspace{-3mm} \]
Suppose now that sampling is with replacement so that after $X_1$ is sampled, it does not affect sampling $X_2,...,X_n$.
\vspace{0.5mm}

Then, $X_1,...,X_n$ are $n$ independent Bernoulli random variables

In this case, $Y$ follows the $binomial(n,M/N)$ distribution.
\end{alertblock}
}
\frame{\vspace{-2mm}
\begin{block}{Some important statistics}
As we have defined earlier, a statistic is a function (possibly vector-valued) of a sample $X_1,...,X_n$ (not necessary a random sample or a simple random sample).

The following are some important statistics used in applications.
\begin{itemize}
\item
The {\bf sample mean} is the (simple) average of $X_1,...,X_n$, and is denoted by \vspace{-2mm}
\[ \bar{X} = \frac{X_1+\cdots + X_n}{n} = \frac{1}{n} \sum_{i=1}^n X_i \vspace{-2mm} \]
\item
When $n \geq 2$ and $k=1$, 
 the {\bf sample variance} is defined as \vspace{-3mm}
\[ S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i- \bar{X})^2 \vspace{-4mm} \]
The {\bf sample standard deviation} is defined as $S = \sqrt{S^2}$.

When $n \geq 2$ and 
$k \geq 2$,
the {\bf sample covariance matrix} is  \vspace{-3mm}
\[ S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i- \bar{X})(X_i-\bar{X})'  \vspace{-3mm} \]
The diagonal elements of $S^2$ are sample variances and the off-diagonal elements are called
{\bf sample covariances}.
\end{itemize}
\end{block}}
\frame{\vspace{-2mm} 
\begin{block}{}
\begin{itemize}
	\item
		When  $k=1$, the $j$th {\bf sample moment} is defined as \vspace{-3.5mm} 
		\[ M_j = \frac{1}{n} \sum_{i=1}^n X_i^j , \quad\quad k=1,2,... \vspace{-3.5mm} \]
		and the $j$th {\bf sample central moment} is defined as   \vspace{-3.5mm}
		\[ \tilde{M}_j = \frac{1}{n} \sum_{i=1}^n (X_i- \bar{X})^j , \quad\quad k=2,3,... \vspace{-3mm} \]
	\item
When $k=1$, the {\bf empirical  cdf} is defined as \vspace{-3mm}
\[ F_n(x) = \frac{1}{n} \sum_{i=1}^n I(X_i \leq x) \quad\quad x \in {\cal R} , \vspace{-3mm} \]
where $I(X_i \leq x) =1$ if $X_i \leq x$ and $=0 $ if $X_i>x$, the indicator function of the set $\{ X_i \leq x \}$.

The empirical cdf is a discrete cdf, i.e., a step function with a jump of size $n^{-1}$ at each $X_i$.

It can be used to estimate the unknown cdf $F$.

For a fixed $x \in {\cal R}$, since each $I(X_i \leq x)$ is a Bernoulli random variable and $I(X_i \leq x)$, $i=1,...,n$, are independent and
have the sample probability $P( I(X_i\leq x)=1 ) = P(X_i \leq x) = F(x)$, the distribution of $n F_n(x)$ is $binomial (n,F(x))$.
\end{itemize}
\end{block}}
\frame{\vspace{-2mm}
\begin{alertblock}{}
	Sums formed from a random sample are useful statistics.
	
	We now study their properties.
\end{alertblock}\vspace{-2mm}
\begin{block}{Lemma 5.2.5.}
	Let $X_1,...,X_n$ be a random sample from a population and let $g(x)$ be a function such that
	$E[g(X_1)$ and Var$(g(X_1))$ exist. Then, \vspace{-2mm}
	\[ E \left[ \sum_{i=1}^n g(X_i)\right] = n E[g(X_1)] \quad \quad \mbox{and} \quad \quad  {\rm Var} \left( \sum_{i=1}^n g(X_i)\right) = n{\rm Var}(g(X_1)) \]
\end{block}\vspace{-2mm}
\begin{block}{Proof.}
The proof is simple and omitted. 
\end{block}
\vspace{-2mm}
	\begin{alertblock}{Theorem 5.2.6.}
		Let $X_1,...,X_n$ be a random sample from a population $F$ on ${\cal R}$ with mean $\mu$ and variance $\sigma^2$. Then\\
		a. $E(\bar{X}) = \mu$;\\
		b. Var$(\bar{X})=\sigma^2/n$;\\
		c. $E(S^2) = \sigma^2$.
	\end{alertblock}
}
\frame{
	\begin{alertblock}{Proof.}
		Letting $g(X_i) = X_i/n$, we can apply Lemma 5.2.5 to obtain \vspace{-2mm}
		\[ E(\bar{X}) = E \left[ \sum_{i=1}^n g(X_i)\right] = nE[g(X_1)] = n E(X_1/n) = E(X_1) = \mu \vspace{-2mm} \]
		\[ {\rm Var}(\bar{X}) = {\rm Var} \left[ \sum_{i=1}^n g(X_i)\right] = n {\rm Var}(g(X_1))
		= n{\rm Var}(X_1/n) = \frac{1}{n} {\rm Var}(X_1) = \frac{\sigma^2}{n} \vspace{-2mm} \]
		To show c, we use the formula (derived in the last lecture)
		\vspace{-2mm}
		\[ (n-1) S^2 = \sum_{i=1}^n [(X_i-\mu) - (\bar{X}-\mu)]^2
		= \sum_{i=1}^n (X_i-\mu)^2 - n(\bar{X}-\mu)^2 \vspace{-2mm} \]
		Applying Lemma 5.2.5 with $g(X_i) = (X_i-\mu)^2$, we obtain \vspace{-2mm}
		\beqrs
		(n-1)E(S^2) &=& E[(n-1)S^2]
		= E \left[  \sum_{i=1}^n (X_i-\mu)^2 \right] - E[n(\bar{X}-\mu)^2] \\
		&=& n E[(X_1-\mu)^2] - n E[(\bar{X}-\mu)^2]
		= n {\rm Var}(X_1)- n{\rm Var}(\bar{X}) \\
		& = &  (n-1) \sigma^2
		\eeqrs
	\end{alertblock}}
	\frame{\vspace{-2mm}
		\begin{block}{}
			The results in Theorem 5.2.6 are about the moments of $\bar{X}$ and $S^2$.
			\vspace{1mm}
			
			Since the sample mean $\bar{X}$ is a sum of independent random variable/vectors divided by a constant $n$,
			the results about a sum in Chapter 4 is useful to obtain the distribution of $\bar{X}$.
		\end{block}\vspace{-2mm}
\begin{alertblock}{Example 5.2.8.}
			\begin{itemize}
				\item
				As a direct consequence of Theorem 4.2.14 (additivity of normal distributions), we know that, if $X_1,...,X_n$ is a random sample from
				$N(\mu, \sigma^2)$, then $\bar{X} \sim N( \mu, \sigma^2/n)$.
				\item
				From the additivity of gamma distributions, we know that, if $X_1,...,X_n$ is a random sample from
				$gamma(\alpha ,\beta )$, then $\bar{X} \sim gamma( n\alpha , \beta / n)$.
			\item
			If $X_1,...,X_n$ is a random sample from $Poisson(\lambda )$, then $n \bar{X} \sim Poisson(n \lambda )$.
					\item
					If $X_1,...,X_n$ is a random sample from $binomial(m,p)$, then $n \bar{X} \sim binomial(nm, p)$.
						\item
						If $X_1,...,X_n$ is a random sample from $Cauchy(\mu,\sigma)$, then the sample mean
						$\bar{X} \sim Cauchy(\mu , \sigma )$!
						\end{itemize}
		\end{alertblock}}
			\frame{
\vspace{-2mm}
				\begin{block}{}
					The last result can be proved as follows.  
					\vspace{1mm}
				
					Let $X_1,...,X_n$ be a random sample from a population with chf $\phi (t)$. Then the chf of the sample mean is\vspace{-2mm}
\[					\phi_{\bar{X}}(t) = [ \phi (t/n)]^n.\vspace{-2mm} \]
					$Cauchy(\mu, \sigma)$ has chf $ \phi (t) =e^{\dot{\imath} \mu t - \sigma |t|}$ and hence\vspace{-2mm}
					\[ \phi_{\bar{X}}(t) = [\phi (t/n)]^n = (e^{\dot{\imath} \mu t/n - \sigma |t/n|})^n
					= e^{\dot{\imath} \mu t - \sigma |t|} = \phi (t), \quad \quad t \in {\cal R}\]
				\end{block}
		\vspace{-2mm}\begin{alertblock}{}
					The additivity of $Cauchy (\mu, \sigma )$ and $N(\mu, \sigma^2)$ distributions are in fact the special case of the following result. 
				\end{alertblock}
				\vspace{-2mm}
				\begin{block}{Theorem.}
					Let $\alpha \in [1,2]$ be a fixed constant. 
					The class of distributions corresponding to the class of chf's  
					$ e^{\dot{\imath} \mu t - \sigma |t|^\alpha }$, $t \in {\cal R}$, which is indexed by 
					$\mu \in {\cal R}$ and $\sigma >0$, is  additive. 
				\end{block}\vspace{-2mm}
				\begin{block}{}
					Note that the normal distribution family  is the special case of $\alpha =2$ and the Cauchy distribution family is the special case of $\alpha =1$.  
				\end{block}
			}
			\frame{
				\begin{alertblock}{Location-scale families}
					Suppose that $X_1,...,X_n$ is a random sample from a population in
					a location-scale family, i.e., the pdf of $X_i$ is of the form
					$\sigma^{-1} f((x-\mu )/\sigma)$ with a known pdf $f$ and parameters $\mu \in {\cal R}$ and $\sigma >0$.
					\vspace{1mm}
					
					From the discussion in Chapter 3, there exist random variables $Z_1,...,Z_n$ such that
					$X_i = \sigma Z_i + \mu$ and the pdf of each $Z_i$ is $f(x)$.
					\vspace{1mm}
					
					Furthermore $Z_1,...,Z_n$ are independent and, hence, $Z_1,...,Z_n$ is a random sample from the population with pdf $f(x)$.
					\vspace{1mm}
					
					The sample mean $\bar{X}$ and $\bar{Z}$ are related by \vspace{-2mm}
					\[ \bar{X} = \frac{1}{n} \sum_{i=1}^n X_i = \frac{1}{n} \sum_{i=1}^n (\sigma Z_i + \mu) =
					\frac{\sigma}{n} \sum_{i=1}^n Z_i + \mu = \sigma \bar{Z} + \mu \vspace{-2mm}\]
					Therefore, if we find that $g(x)$ is the pdf of $\bar{Z}$, then
					$\sigma^{-1} g((x-\mu)/\sigma )$ is the pdf of $\bar{X}$.
					\vspace{1mm}
					
					This argument has been used in the discussion of a random sample from the Cauchy distribution family.
					\vspace{1mm}
					
					The pdf $g$ may or may not be of a familiar pdf.
				\end{alertblock}}
				\frame{ \vspace{-2mm}
					\begin{block}{Exponential families}
						When the population of a random sample is in an exponential family, the joint distribution of some sums of functions of the sample  can be derived.
					\end{block}\vspace{-2mm}
					\begin{alertblock}{Theorem 5.2.11.}
						Suppose that $X_1,...,X_n$ is a random sample from a pdf or pmf
						\vspace{-2mm}
						\[ f_\theta (x ) = h(x) c(\theta ) \exp \left( \sum_{j=1}^k w_j(\theta ) t_j(x) \right) \vspace{-2mm} \]
						in an exponential family with parameter $\theta \in \Theta$. Define statistics \vspace{-2mm}
						\[ T_j = \sum_{i=1}^n t_j(X_i) , \quad \quad j=1,...,k \vspace{-2mm} \]
						If the set $\{ (w_1(\theta ),...,w_k(\theta )): \theta \in \Theta \}$ contains an open subset of ${\cal R}^k$, then
						the distribution of $T=(T_1,...,T_k)$ is in an exponential family with pdf \vspace{-2mm}
						\[ g_{\theta } ( t_1,...,t_k) = H(t_1,...,t_k) [c(\theta )]^n \exp \left( \sum_{j=1}^k w_j(\theta ) t_j \right) \]
					\end{alertblock}
				}
				\frame{
					\begin{alertblock}{Proof for the discrete case}
						The joint pmf of $X_1,...,X_n$ is \vspace{-2mm}
						\beqrs
						\prod_{i=1}^n f_\theta (x_i)
						&=& \prod_{i=1}^n \left[ h(x_i) c(\theta ) \exp \left( \sum_{j=1}^k w_j(\theta ) t_j(x_i) \right)\right]  \\
						& = &  \prod_{i=1}^n h(x_i) [c(\theta )]^n \exp \left( \sum_{j=1}^k w_j(\theta ) \sum_{i=1}^n t_j(x_i) \right)
						\eeqrs
						Then, the pmf of $T=(T_1,...,T_k) $ is \vspace{-2mm}
						\beqrs
						g_\theta (t_1,...,t_k) & = & P(T_1=t_1,...,T_k=t_k)
						= \sum_{t_j = \sum_i t_j(x_i), \ j=1,...,k}  \prod_{i=1}^n f_\theta (x_i)\\
						&=& \sum_{t_j = \sum_i t_j(x_i), \ j=1,...,k} \  \prod_{i=1}^n h(x_i) [c(\theta )]^n \exp \left( \sum_{j=1}^k w_j(\theta ) \sum_{i=1}^n t_j(x_i) \right) \\
						&=& \left[ \sum_{t_j = \sum_i t_j(x_i), \ j=1,...,k} \  \prod_{i=1}^n h(x_i) \right]
						[c(\theta )]^n \exp \left( \sum_{j=1}^k w_j(\theta ) t_j \right)
						\eeqrs
						
					\end{alertblock}
				}
				\frame{\vspace{-2mm}
					\begin{block}{Example 5.2.12}
						If $X_1,...,X_n$ is a random sample from Bernoulli trials, the joint pmf is \vspace{-2mm}
						\beqrs
						&& \prod_{i=1}^n I( x_i = 1 \mbox{ or } 0) p^{x_i} (1-p)^{1-x_i} \\
						&=&   \prod_{i=1}^n \left[ I( x_i = 1 \mbox{ or } 0) (1-p) \exp \left( x_i \log \frac{p}{1-p} \right) \right] \\
						&=&   \left[ \prod_{i=1}^n  I( x_i = 1 \mbox{ or } 0)\right] (1-p)^n \exp\left( \log\frac{p}{1-p} \sum_{i=1}^n x_i \right)
						\eeqrs
						\[ h(x_1,...,x_n) = \prod_{i=1}^n I( x_i = 1 \mbox{ or } 0), \quad \quad c(\theta ) = (1-p), \quad \quad
						w(\theta ) = \log\frac{p}{1-p} \]
						The sum $T= X_1+\cdots +X_n$ has pmf\vspace{-2mm}
						\[ P(T=t) = \sum_{x_1+\cdots + x_n = t} \prod_{i=1}^n I( x_i = 1 \mbox{ or } 0) (1-p)^n \exp \left( t \log\frac{p}{1-p} \right) \]
						We know that $T \sim binomial (n,p)$.
					\end{block}}

\end{document}


