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\title[Stat 609 Lecture 20]{Stat 609: Mathematical Statistics \\ Lecture 20}
\author[UW-Madison]{Jun Shao}
\institute[Statistics]{%
%  Department of Statistics\\
%  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{Lecture 20: Multivariate convergence and\\ the Central Limit Theorem}\vspace{-1mm}
	\begin{alertblock}{Convergence in distribution for random vectors}
		Let $Z,Z_1,Z_2,...$ be random vectors on ${\cal R}^k$.
		
		If the cdf of $Z$ is continuous,
		then we can define convergence in distribution of $Z_n$ to $Z$ by 
		$ \lim_{n\rightarrow \infty}F_{Z_n}(x) = F_Z(x)$,
for every $x \in {\cal R}^k$.

		But this is not good enough if $F_Z$ is not always continuous.
		
		We can adopt the following definition.
	\end{alertblock}\vspace{-2mm}
	\begin{block}{Definition.}
		Let $Z,Z_1,Z_2,...$ be random vectors on ${\cal R}^k$.
		If, for every $c \in {\cal R}^k$,
		$c' Z_n$ converges in distribution to $c' Z$, then we say that
		$Z_n$ converges in distribution to $Z$.
	\end{block}\vspace{-2mm}
	\begin{block}{}
		\begin{itemize}
			\item
			For any constant vector $c \in {\cal R}^k$,
			$c' Z_n$ is a linear combination of components of $Z_n$.
			\item
			Note that in this definition, the convergence of $c'Z_n$ has to be true for
			every $c$ (not just some $c$).
		\end{itemize}
	\end{block}}
	\frame{\vspace{-2mm}
		\begin{block}{}
			\begin{itemize}
				\item
				If $Z_n$ converges in distribution to $Z$, then every component of $Z_n$
				converges in distribution to the corresponding component of $Z$.
				\item
				The converse is not true: if every component of $Z_n$
				converges in distribution to the corresponding component of $Z$,
				$Z_n$ does  not necessarily converge in distribution to $Z$,
				because each component corresponds to a particular $c$ only.
				(A counter-example is given below.)
				\vspace{1mm}
				
				This is different from convergence in probability and convergence
				almost surely.
				\item
				Theorems 5.5.12 and 5.5.13 can be
				extended to the case of random vectors.
			\end{itemize}
		\end{block}\vspace{-2mm}
		\begin{alertblock}{A counter-example}
			Let $Z = (X,Y) $ have the joint pdf \vspace{-2mm}
			\vspace{-2mm}
			\[ f(x,y) = \frac{1}{2\pi} e^{-(x^2+y^2)/2} [1+ g(x,y)] , \quad \quad (x,y) \in {\cal R}^2 \vspace{-1mm} \]
			\[ g(x,y) = \left\{ \br{ll}
			xy & \quad -1<x<1 , \ -1<y<1 \vspace{1mm} \\
			0 & \quad \mbox{otherwise} \er \right.  \]
		\end{alertblock}}
		\frame{\vspace{-2mm}
			\begin{alertblock}{}
				In Chapter 4 we showed that $X \sim N(0,1)$ and $Y \sim N(0,1)$, but the joint distribution of $Z$ is not normal.
				\vspace{2mm}
				
				Let $Z_n = (X_n,Y_n)$, where for each $n$, the random variables $X_n$ and $Y_n$ are
				independent, $X_n \sim N(0,1)$,  and
				$Y_n \sim N(0,1)$.
				\vspace{2mm}
				
				Since $X_n$'s have the same $N(0,1)$ distribution for all $n$, obviously that $X_n$ ``converges" in distribution to
				$X \sim N(0,1)$; similarly, $Y_n $ converges in distribution to $Y \sim N(0,1)$.
				\vspace{2mm}
				
				But the joint distribution of $Z_n=(X_n,Y_n)$ for every $n$  is \vspace{-2mm}
				\[ f_{Z_n}(x,y) = \frac{1}{2\pi} e^{-(x^2+y^2)/2}  , \quad \quad (x,y) \in {\cal R}^2 \vspace{-1mm} \]
				since $X_n$ and $Y_n$ are independent. \vspace{2mm}
				
				Since $Z_n$'s have the same distribution for every $n$, $Z_n$ converges in distribution to $Z_1$, which has a different distribution
				from $Z$. \vspace{2mm}
				
				Thus, $Z_n$ does not converge in distribution to $Z$ although each component of $Z_n$ converges in distribution to the corresponding component of $Z$.
			\end{alertblock}}
			\frame{
				\begin{block}{}
					The convergence in distribution is equivalent to the convergence in chf (Theorem C7 or M3(ii)) so that the convergence in chf is a tool to study convergence in distribution.
					\vspace{1mm}
					
					We can also use Theorem 2.3.12 or M3(i) to establish convergence in distribution by showing the convergence in mgf, but we have to know the existence of mgf's.  
				\end{block}\vspace{-2mm}
				\begin{alertblock}{Example.}
					Let $X_1,..., X_n  $ be iid random variables. 
					
					We want to show that there does not exist a sequence
					of real numbers $\{ c_n \}$ such that
					$\lim_{n\rightarrow \infty} \sum_{i=1}^n (X_i -c_i) $ exists almost surely, unless $P(X_1 =c)=1$ for a constant $c$. 
					\vspace{1mm} 
					
					Suppose that
					$Y = \lim_{n\rightarrow \infty} \sum_{i=1}^n (X_i -c_i) $ exists almost surely. 
					
					For any $n$, the chf of
					$Y_n = \sum_{i=1}^n (X_i -c_i) $ is \vspace{-2mm} 
					\[ \phi_{Y_n} (t) = \prod_{i=1}^n \phi_{X_1}(t) e^{-\dot{\imath} tc_i}
					= [\phi_{X_1}(t)]^n e^{-\dot{\imath} t(c_1+\cdots + c_n)}\vspace{-2mm}  \]
					If $Y$ exists almost surely, then $Y_n$ converges in distribution to $Y$ and 
			\end{alertblock}}
			\frame{\vspace{-2mm}
				\begin{alertblock}{}
							\vspace{-2mm} 
						\[ \lim_{n\rightarrow \infty}  \bigg|[\phi_{X_1}(t)]^n e^{-\dot{\imath}t(c_1+\cdots + c_n)} \bigg|
						= \lim_{n\rightarrow \infty}  |\phi_{X_1}(t)|^n = |\phi_Y (t)| .\vspace{-2mm}  \]
						However, $\lim_{n\rightarrow \infty}  |\phi_{X_1}(t)|^n$ is either 0 or 1, depending on whether 
						$ |\phi_{X_1}(t)| < 1$ or $=1$, which means that $|\phi_Y(t)| $ must be either 0 or 1. 
						\vspace{1mm} 
						
						Since $|\phi_Y (t)|$ is continuous and $\phi_Y (0)=1$,
						$|\phi_Y(t)| =1$ for all $t$ and hence  $|\phi_{X_1} (t)|=1$ for all $t$. 
						\vspace{1mm}
						
						This proves that 
						$P(X_1 =c)=1$ for a constant $c$.
					\end{alertblock}\vspace{-2mm}
\begin{block}{}
The Central Limit Theorem (CLT) is one of the most important theorems in probability and statistics.

It derives the limiting distribution of a sequence of normalized random variables/vectors.
\end{block}\vspace{-2mm}
\begin{block}{Theorem 5.5.15 (Central Limit Theorem)}
Let $X_1,X_2,...$ be iid random variables with $E(X_1)=\mu$ and Var$(X_i)=\sigma^2 < \infty$.
Then, for any $x \in {\cal R}$, \vspace{-3mm}
\[\lim_{n\rightarrow \infty} P(\sqrt{n}(\bar{X}-\mu)/\sigma
\leq x ) = \Phic (x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-t^2/2}dt \vspace{-2mm} \]
That is, $\sqrt{n}(\bar{X}-\mu)/\sigma $ converges in distribution to
$Z \sim N(0,1)$.
\end{block}}
\frame{\vspace{-2mm}
\begin{block}{}
\begin{itemize}
\item
Normality comes from sums of iid random variables without distributional assumption except the finiteness of the variance.
\vspace{-1mm}
\item
The assumption of finite variance is essentially necessary for convergence of a sume of iid random variables to normality.
\end{itemize}
\end{block}\vspace{-2mm}
\begin{block}{Proof.}
We apply Theorem C7, i.e., we try to show that the chf of \vspace{-2mm} 
\[ Z_n = \frac{\sqrt{n}(\bar{X}-\mu)}{\sigma} = \frac{1}{\sqrt{n}}\sum_{i=1}^n \frac{X_i-\mu}{\sigma}  \vspace{-3mm} \]
converges to the chf of $N(0,1)$, which is $e^{-t^2/2}$. 
\vspace{1mm}

Let $\phi$ be the chf of $(X_i - \mu)/\sigma$ (not depending on $i$ since $X_i$'s are iid).  \vspace{1mm} 

From the properties of chf, the chf of $Z_n$ is \vspace{-2mm}
\[ \phi_{Z_n}(t) = \left[ \phi \left(\frac{t}{\sqrt{n}}\right)\right]^n . \vspace{-2mm} \]
It remains to show that $\lim_{n\rightarrow \infty} \phi_{Z_n}(t) = e^{- t^2/2}$ for $t \in {\cal R}$.
\vspace{1mm} 

Since $E((X_i - \mu)/\sigma) =0$ and Var$((X_i - \mu)/\sigma) = 1$, by Theorem C1 and Taylor's expansion,  
\end{block}}
\frame{
\begin{block}{}
\vspace{-2mm}
\[ \phi (s  ) = 1 - \frac{s^2}{2} +R(s) \quad \quad \mbox{as $|s| \rightarrow 0$} \vspace{-2mm} \]
where\vspace{-2mm} 
\[ \lim_{|s| \rightarrow 0} R(s)/s^2 =0 \vspace{-2mm} \] 
Then, for any $t \in {\cal R}$, \vspace{-2mm}
\[ \lim_{n \rightarrow \infty } n R \left( \frac{t}{\sqrt{n}}\right) =0 \vspace{-2mm} \]
and hence \vspace{-2mm} 
\beqrs
\lim_{n\rightarrow \infty} \phi_{Z_n}\left( t \right) &=&
\lim_{n\rightarrow \infty} \left[ \phi \left(\frac{t}{\sqrt{n}}\right)\right]^n \\
 &= & \lim_{n\rightarrow \infty} \left[1 - \frac{t^2}{2n} + R\left(\frac{t}{\sqrt{n}}\right)\right]^n \\
& = & \lim_{n\rightarrow \infty}
\left[1-  \frac{1}{n} \left\{\frac{t^2}{2} +
n R \left(\frac{t}{\sqrt{n}}\right) \right\} \right]^n \\
&= & \lim_{n\rightarrow \infty} \left[1 - \frac{1}{n} \left(\frac{t^2}{2}\right)\right]^n \\
& =&  e^{- t^2/2}
\eeqrs
\end{block}}
\frame{
\begin{block}{The multivariate CLT}
Let $X_1,X_2,...$ be iid random vectors on ${\cal R}^k$
 with $E(X_1)=\mu$ and finite covariance matrix $\Sigma$.
Then $\sqrt{n}(\bar{X}-\mu) $ converges in distribution to
a random vector $X \sim N(0,\Sigma )$, the $k$-dimensional normal distribution
with mean 0 and covariance matrix $\Sigma$.
\end{block}\vspace{-2mm}
\begin{block}{Proof.}
By the definition of convergence in distribution for random vectors, we need to show that for any constant vector $c \in {\cal R}^k$,
$\sqrt{n} c'(\bar{X} - \mu) $ converges in distribution to $c'X $.

From the result discussed earlier, we know that $c'X \sim N(0, c'\Sigma c)$.

By the CLT, \vspace{-2mm}
\[ \frac{\sqrt{n}(c'\bar{X} - E(c'\bar{X}))}{\sqrt{{\rm Var}(c'X)}}
= \frac{\sqrt{n} (c'\bar{X} - c'\mu )}{\sqrt{c' \Sigma c}} \vspace{-2mm} \]
converges in distribution to $Z \sim N(0,1)$.

Note that $\sqrt{c'\Sigma c} Z \sim N(0, c'\Sigma c)$.

The proof is then completed because we can show from the definition that, if $Y_n $ converges in distribution to $Y$, then
$aY_n $ converges in distribution to $aY$.
\end{block}
}
\frame{
    \begin{alertblock}{Example 5.5.16.}
    Let $X_1,...,X_n$ be a random sample from a $negative$-$binomial(r,p)$ distribution.

    Recall that \vspace{-2mm}
    \[ \mu = E(X_1)= \frac{r(1-p)}{p} , \quad \quad \sigma^2 = {\rm Var}(X_1) = \frac{r(1-p)}{p^2} \vspace{-2mm} \]
    Then, the CLT tells us that the sequence of normalized random variables \vspace{-2mm}
    \[ \frac{\sqrt{n}(\bar{X} - \mu)}{\sigma} = \frac{\sqrt{n} ( \bar{X} - r(1-p)/p)}{\sqrt{r(1-p)/p^2}} , \quad \quad n=1,2,... \vspace{-2mm}\]
    converges in distribution to $Z \sim N(0,1)$.
    \vspace{2mm}

    This provides a way to approximately calculate probabilities that are very difficult to compute exactly.
    \vspace{2mm}

For example, if $r=10$, $p=1/2$ and $n =30$, using the fact that $\sum_{i=1}^{30} X_i \sim negative$-$binomial(nr, p)$,
we can exactly compute
     \end{alertblock}}
\frame{\vspace{-2mm}
\begin{alertblock}{}
\vspace{-4mm}
\[ P(\bar{X}\leq 11) = P\left( \sum_{i=1}^{30} X_i \leq 330 \right)
= \sum_{x=0}^{330} {300 + x -1 \choose x } (0.5)^{300+x} = 0.8916 \vspace{-2mm} \]
The CLT gives us the approximation \vspace{-2mm}
\beqrs
 P(\bar{X}\leq 11) &=& P\left( \frac{\sqrt{30}(\bar{X}- 10)}{\sqrt{20}} \leq \frac{\sqrt{30}(11-20)}{\sqrt{20}} \right)\\
& \approx & \Phic \left(\frac{\sqrt{30}(11-20)}{\sqrt{20}} \right)
= \Phic (1.2247 ) = 0.8888 \vspace{-2mm}
\eeqrs
where $\Phic (x)$ is the cdf of $N(0,1)$.
\end{alertblock}
\vspace{-2mm}
\begin{block}{Normal approximation to binomial}
Let $Z_n \sim binomial(n, p)$, $n=1,2,...$

To apply the CLT, note that each $Z_n$ is a sum of $n$ independent Bernoulli
random variables $X_1,...,X_n$, i.e., \vspace{-2mm}
\[ Z_n = X_1 + \cdots + X_n , \quad \quad n=1,2,... \vspace{-2mm} \]
\[ X_1,X_2,... \quad \mbox{are iid with} \ \ E(X_i) = p, \ \ {\rm Var}(X_i) = p(1-p) \]
 \end{block}
 }

\frame {
    \begin{block}{}
Then, $\bar{X}= Z_n/n$ and\vspace{-2mm}
\[ \frac{Z_n - np}{\sqrt{np(1-p)}} = \frac{\sqrt{n}(\bar{X}-p)}{\sqrt{p(1-p)}}
\mbox{ converges in distribution to $Z \sim N(0,1)$} \vspace{-2mm} \]
Consequently,
for any $m$, \vspace{-2mm}
\[ P(Z_n \leq m) = P\left(  \frac{Z_n - np}{\sqrt{np(1-p)}} \leq  \frac{m - np}{\sqrt{np(1-p)}}\right)
\approx \Phic \left(  \frac{m - np}{\sqrt{np(1-p)}} \right) \]
     \end{block}\vspace{-2mm}
     \begin{alertblock}{}
From the previous two examples,
to approximate a probability of the form $P(Z_n \leq x)$, we always first normalize $Z_n $ to $[Z_n-E(Z_n)]/\sqrt{{\rm Var}(Z_n)}$ and then
approximate the probability by $\Phic (  [x-E(Z_n)]/\sqrt{{\rm Var}(Z_n)})$.
\vspace{1mm}

In fact, a little bit more can be done using the following result.
     \end{alertblock}\vspace{-2mm}
\begin{block}{P\'{o}lya's theorem}
If a sequence of random vectors $Z_n$ converges in distribution to $Z$ and $Z$ has
a continuous cdf $F_Z$ on  ${\cal R}^k$, then\vspace{-2mm}
\[ \lim_{n \rightarrow \infty} \sup_{x \in {\cal R}^k}
| F_{Z_n} (x) - F_Z(x) | = 0.  \]
     \end{block}}
     \frame{\vspace{-2mm}
\begin{block}{}
\begin{itemize}
\item
If the limiting cdf $F_Z$ is continuous, then
 the convergence of $F_{Z_n}$ to $F_Z$ is not only for every $x$, but also uniformly for all $x \in {\cal R}^k$.
\item
This result implies the following useful result:

If $Z_n$ converges in distribution to $Z$ with a continuous $F_Z$ and
$c_n \in {\cal R}^k$ with $c_n \rightarrow c$, then $ F_{Z_n} (c_n ) \rightarrow F_Z(c)$.
\end{itemize}
\end{block}\vspace{-2mm}
\begin{alertblock}{Asymptotic normality}
If $Y_n$, $n=1,2,...$, is a sequence of random variables and $\mu_n$ and $\sigma_n> 0$, $n=1,2,...$, are constants
(typically $\mu_n = E(Y_n)$ and $\sigma_n^2 = {\rm Var}(Y_n)$ if they are finite) such that
$(Y_n-\mu_n)/\sigma_n$ converges in distribution to $N(0,1)$, then for any sequence $\{ c_n \}$ of constants, \vspace{-2mm}
\[  \lim_{n\rightarrow \infty} \left| P ( Y_n \leq c_n ) -
\Phic \left(\frac{c_n-\mu_n}{\sigma_n}\right) \right| =0,\]
i.e., $P( Y_n \leq c_n ) $ can be approximated by
$ \Phic \big(\frac{c_n-\mu_n}{\sigma_n}\big)$, regardless of whether
$c_n $, $\mu_n$, or $\sigma_n$ has a limit.
\vspace{2mm}

Since $ \Phic \big(\frac{t-\mu_n}{\sigma_n}\big)$ is
the cdf of $N(\mu_n,\sigma_n^2)$,
$Y_n$ is said to be {\bf asymptotically distributed}
as $N(\mu_n,\sigma_n^2)$
or simply {\bf asymptotically normal}.
     \end{alertblock}
}
\frame{
\begin{block}{Other forms of CLT}
There are other forms of CLT for non-iid sequences of random variables/vectors.

We introduce without proof the following three.
\end{block}\vspace{-2mm}
\begin{block}{Lindeberg's CLT}
 Let $X_{n}$, $n=1,2,...$, be independent
random variables with
\vspace{-2mm}
\[ 0 <\sigma_n^2 = {\rm Var} \left(\sum_{j=1}^{n} X_{j} \right)
< \infty,  \quad n=1,2,... \vspace{-3mm}\]
If \vspace{-1mm}
\[ \lim_{n \rightarrow \infty} \frac{1}{\sigma_n^2} \sum_{j=1}^{n} E \left[ (X_{j} - E(X_{j}))^2
I_{\{ |X_{j} - E(X_{j}) | > \epsilon \sigma_n \}} \right] =0
 \ \ \ \hbox{for any $\epsilon >0$},\]
then \vspace{-2mm}
\[  \frac{1}{\sigma_n } \sum_{j=1}^{n} (X_{j} - E(X_{j})) \mbox{ converges in distribution to } N(0,1). \]
\end{block}}
\frame{ \vspace{-2mm}
\begin{alertblock}{Liapounov's CLT}
 Let $X_{n}$, $n=1,2,...$, be independent
random variables with
\vspace{-2mm}
\[ \lim_{n\rightarrow \infty} \frac{1}{\sigma_n^{2+\delta}}
\sum_{j=1}^{n } E|X_{j} - EX_{j}|^{2+\delta} = 0
\ \ \ \hbox{for some $\delta >0$.} \vspace{-2mm}\]
Then the same conclusion as in Lindeberg's CLT holds.

The condition in Liapounov's CLT is stronger than the condition in Lindeberg's CLT,
Liapounov's condition is easier to verify.
\end{alertblock}\vspace{-2mm}
\begin{block}{The CLT for a linear combination of iid random variables}
Let $X_{n}$, $n=1,2,...$, be iid
random variables with $E(X_1)=\mu$ and ${\rm Var}(X_1)=\sigma^2 < \infty$.
Let $ c_{n} \in {\cal R}$, $n=1,2,...$, be constants such that\vspace{-3mm}
\[  \lim_{n\rightarrow \infty} \left(
\max_{1 \leq i \leq n}
c_{i}^2 \bigg/ \sum_{i=1}^{n} c_{i}^2\right) =0 .\vspace{-2mm}\]
Then \vspace{-2mm}
\[ \sum_{i=1}^{n} \frac{ c_{i}(X_{i}-\mu )}{\sigma} \bigg/
 \sqrt{  \sum_{i=1}^{n} c_{i}^2}
\mbox{ converges in distribution to }  N(0,1)  .\]
\end{block}
}
\frame{
\begin{alertblock}{Example.} 
We apply 
 Liapounov's CLT to independent random variables $X_1,X_2,...$ satisfying $P(X_j=
\pm j^a) = P(X_j=0)= 1/3$, where $a>0$, $j=1,2,$....
\vspace{1mm} 

Note that $E(X_j)=0$ and for all $j$ \vspace{-2mm} 
\[ \sigma_n^2 = {\rm Var} \left(\sum_{j=1}^n X_{j}\right)
= \sum_{j=1}^n \mbox{Var}(X_{j})
= \frac{2}{3} \sum_{j=1}^n j^{2a} . \vspace{-2mm} \]
We need to show that, for some $\delta >0$, \vspace{-2mm} 
\[ \lim_{n\rightarrow \infty}  \frac{1}{\sigma_n^{2+\delta}}
\sum_{j=1}^n E|X_j - EX_j|^{2+\delta} = 0 \vspace{-2mm} \]
For any $\delta >0 $, \vspace{-2mm} 
\[ \sum_{j=1}^n E| X_j - E(X_j)| ^{ 2+\delta } =
\frac{2}{3}\sum_{j=1}^n  j ^{ (2 + \delta )a } . \vspace{-2mm}  \]
Since\vspace{-1mm} 
\[ \sum_{j=1}^{n-1}j^{t } \leq \sum_{j=1}^{n-1} \int_j^{j+1} x^{t }dx
\leq \sum_{j=2}^n j^{t}  \]
\end{alertblock}}
\frame{
\begin{alertblock}{}
and
\[ \sum_{j=1}^{n-1} \int_j^{j+1} x^{t }dx
= \int_1^n x^{t }dx = \frac{n^{t +1}-1}{t +1} ,\vspace{-2mm}  \]
we conclude that \vspace{-2mm} 
\[ \limn \frac{1}{n^{t+1}} \sum_{j=1}^n j^t = \frac{1}{t+1} \vspace{-2mm} \]
for any $t>0$. 
\vspace{1mm} 

Then \vspace{-2mm} 
\beqrs
\limn \frac{1}{\sigma_n^{2+\delta}}
\sum_{j=1}^n E|X_j - EX_j|^{2+\delta} \back & = & \back
\limn \frac{\frac{2}{3} \sum_{j=1}^n j^{(2+\delta )a}}{
\left( \frac{2}{3} \sum_{j=1}^n j^{2a}\right)^{1+\delta/2} } \\
& = & \back \limn \left( \frac{3}{2}\right)^{\delta /2}
\frac{(2a+1)^{1+\delta /2}}{(2+\delta )a+1} \frac{n^{(2+\delta )a+1}}{
n^{(2a+1)(1+\delta /2 )}} \\
& = & \back  \left( \frac{3}{2}\right)^{\delta /2}
\frac{(2a+1)^{1+\delta /2}}{(2+\delta )a+1} \limn \frac{1}{n^{\delta /2} }\\
& = & \back 0. 
\eeqrs

\end{alertblock}}
\end{document}

