%                       Large sample tools for grad SEM
% Notes and comments are after the end of the document

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% Comment this out for handout
\title{Large sample tools\footnote{See last slide for copyright information.}}
\subtitle{STA2053 Fall 2022}
\date{} % To suppress date

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Overview}
\tableofcontents
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Foundations}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Sample Space $\Omega$, $\omega \in \Omega$}
% \framesubtitle{Usually infinite in theory} 
  \begin{itemize}
    \item $\Omega$ is a set, the underlying sample space. % \pause
    % \item It could literally be the universe of websites from which we intend to sample. \pause
    \item $\EuScript{F}$ is a class of subsets of $\Omega$.  % \pause
    % \item If could be the class of all subsets (if $\Omega$ is countable). \pause
    \item There is a probability measure $\EuScript{P}$ defined on the elements of $\EuScript{F}$. % \pause
    % \item Maybe each website is equally likely to be chosen (with replacement). 
  \end{itemize} \pause
  
\vspace{5mm}

\begin{center}
{\Large Probability space $(\Omega,\EuScript{F},\EuScript{P})$}
\end{center}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Random variables are functions from $\Omega$ into the set of real numbers} \pause

{\LARGE
\begin{displaymath}
    Pr\{X \in B\} = Pr(\{\omega \in \Omega: X(\omega) \in B \})
\end{displaymath}
}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Random Sample $X_1(\omega), \ldots, X_n(\omega)$} \pause
%\framesubtitle{} 
  \begin{itemize}
    \item $T = T(X_1, \ldots, X_n)$ \pause
    \item $T = T_n(\omega)$ \pause
    \item Let $n \rightarrow \infty$ to see what happens for large samples.
  \end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[fragile]
\frametitle{{\LARGE Modes of Convergence}} \pause
%\framesubtitle{} 
{\LARGE 
  \begin{itemize}
    \item Almost Sure Convergence
    \item Convergence in Probability
    \item Convergence in Distribution
  \end{itemize} }
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{frame}
\frametitle{Almost Sure Convergence}
We say that $T_n$ converges \emph{almost surely} to $T$, and write 
$T_n \stackrel{a.s.}{\rightarrow} T$ if \pause

\begin{displaymath}
    Pr\{\omega:\, \lim_{n \rightarrow \infty} T_n(\omega) = T(\omega)\}=1.
\end{displaymath} \pause


  \begin{itemize}
    \item Acts like an ordinary limit, except possibly on a set of probability zero. % \pause
    \item All the usual rules apply. \pause
    \item Called convergence with probability one or sometimes strong convergence. \pause
    \item In this course, convergence will usually be to a constant.% \pause
  \end{itemize}
\begin{displaymath}
    Pr\{\omega:\, \lim_{n \rightarrow \infty} T_n(\omega) = c \}=1.
\end{displaymath}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{LLN}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Strong Law of Large Numbers}
%\framesubtitle{} 

Let $X_1, \ldots, X_n$ be independent with common expected value $\mu$. \pause

\vspace{10mm}

{\huge
\begin{displaymath}
    \overline{X}_n \stackrel{a.s.}{\rightarrow} E(X_i) = \mu 
\end{displaymath}  } \pause

The only condition required for this to hold is the existence of the expected value.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Probability is long run relative frequency} \pause
  \begin{itemize}
    \item  Statistical experiment: Probability of ``success" is $\theta$. % \pause
    \item Carry out the experiment many times independently.%  \pause
    \item Code the results $X_i=1$ if success, $X_i=0$ for failure, $i = 1, 2, \ldots$ 
  \end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Sample proportion of successes converges to the probability of success} 
\framesubtitle{Recall $X_i=0$ or $1$.} 
 \pause
{\Large
\begin{eqnarray*}
    E(X_i) &=& \sum_{x=0}^1 x \, Pr\{X_i = x\} \\ % \pause
%           &&\\
           &=& 0\cdot (1-\theta) + 1\cdot \theta \\ % \pause
%           &&\\
           &=& \theta
\end{eqnarray*} }

\pause

Relative frequency is 

{\Large
\begin{displaymath}
    \frac{1}{n}\sum_{i=1}^n X_i  \pause = \overline{X}_n  % \pause \stackrel{a.s.}{\rightarrow} \theta 
\end{displaymath} }
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Simulation}
\framesubtitle{Using pseudo-random number generation by computer}  \pause
  \begin{itemize}
    \item Estimate almost any probability that's hard to figure out. \pause 
    \item Statistical power \pause 
    \item Weather model \pause 
    \item Performance of statistical methods 
    \item[] \pause 
    \item Tests or confidence intervals for estimated probabilities. 
  \end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{}
\begin{center}
{\Large Back to the Law of Large Numbers}
\end{center}
%\framesubtitle{} 
 
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{frame}
\frametitle{Recall the Change of Variables formula: Let $Y = g(X)$}  \pause
%\framesubtitle{} 

{\LARGE
\begin{displaymath}
    E(Y) = \int_{-\infty}^\infty y \, f_{_Y}(y) \, dy \pause = 
           \int_{-\infty}^\infty g(x) \, f_{_X}(x) \, dx
\end{displaymath} }  \pause

Or, for discrete random variables  % \pause

{\LARGE
\begin{displaymath}
    E(Y) = \sum_y y \, p_{_Y}(y) = 
           \sum_x g(x) \, p_{_X}(x)
\end{displaymath} }  \pause

This is actually a big theorem, not a definition.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Applying the change of variables formula}
\framesubtitle{To approximate $E[g(X)]$}   \pause
Simulate $X_1, \ldots, X_n$ from the distribution of $X$.  \pause Calculate 
{\LARGE
\begin{eqnarray*}
\frac{1}{n}\sum_{i=1}^n g(X_i)  \pause &=& \frac{1}{n}\sum_{i=1}^n Y_i  \pause
      \stackrel{a.s.}{\rightarrow} E(Y) \\ \\  \pause
      &=& E(g(X))
\end{eqnarray*} }
\end{frame}

\begin{frame}
\frametitle{So for example}
%\framesubtitle{} 
{\LARGE
\begin{eqnarray*}
    \frac{1}{n}\sum_{i=1}^n X_i^k &\stackrel{a.s.}{\rightarrow}& E(X^k) \\
    &&\\ \pause 
   \frac{1}{n}\sum_{i=1}^n U_i^2 V_i W_i^3 &\stackrel{a.s.}{\rightarrow}& E(U^2VW^3)
\end{eqnarray*} }

\pause

\vspace{5mm}

That is, sample moments converge almost surely to population moments.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Approximate an integral: $\int_{-\infty}^{\infty} h(x) \, dx$}
\framesubtitle{Where $h(x)$ is a nasty function.}

\pause

Let $f(x)$ be a density with $f(x)>0$ wherever $h(x)\neq 0$.

\pause

\begin{eqnarray*}
    \int_{-\infty}^{\infty} h(x) \, dx \pause
        & = & \int_{-\infty}^{\infty} \frac{h(x)}{f(x)} f(x) \, dx \\ \pause 
        & = & E\left[ \frac{h(X)}{f(X)}\right] \\ \pause 
        & = & E[g(X)],  
\end{eqnarray*} 

 \pause

So 
  \begin{itemize}
    \item Sample $X_1, \ldots, X_n$ from the distribution with density $f(x)$ \pause 
    \item Calculate $Y_i = g(X_i) = \frac{h(X_i)}{f(X_i)}$ for $i=1, \ldots, n$ \pause 
    \item Calculate $\overline{Y}_n \stackrel{a.s.}{\rightarrow} E[Y]= E[g(X)]$ \pause 
    \item Confidence interval for $\mu = E[Y]$ is routine.
  \end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Convergence in Probability}
We say that $T_n$ converges \emph{in probability} to $T$, % \pause 
and write 
$T_n \stackrel{P}{\rightarrow} T$ % \pause 
if for all $\epsilon>0$, \pause
{\LARGE
\begin{displaymath}
    \lim_{n \rightarrow \infty} P\{\omega: |T_n(\omega)-T(\omega)| < \epsilon  \}=1
\end{displaymath} } \pause

For us, convergence will usually be to a constant:
{\LARGE
\begin{displaymath}
    \lim_{n \rightarrow \infty} P\{|T_n-c|<\epsilon  \}=1
\end{displaymath} } \pause

Convergence in probability (say to $c$) means no matter how small the interval around $c$, for large enough $n$ (that is, for all $n>N_1$) the probability of getting that close to $c$ is as close to one as you like. \pause

We will seldom use the definition in this class.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Weak Law of Large Numbers}


{\huge
\begin{displaymath}
    \overline{X}_n \stackrel{p}{\rightarrow}  \mu 
\end{displaymath} }

\pause

  \begin{itemize}
    \item Almost Sure Convergence implies Convergence in Probability
    \item Strong Law of Large Numbers implies Weak Law of Large Numbers
  \end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Consistency}

\begin{frame}
\frametitle{Consistency}
\framesubtitle{$T = T(X_1, \ldots, X_n)$ is a statistic estimating a parameter $\theta$} \pause

The statistic $T_n$ is said to be \emph{consistent} for $\theta$ if $T_n \stackrel{P}{\rightarrow} \theta$  \pause for all $\theta$ in the parameter space.

\pause

{\LARGE
\begin{displaymath}
    \lim_{n \rightarrow \infty}
            P\{|T_n-\theta|<\epsilon  \}=1
\end{displaymath} }

\pause

\vspace{5mm}

The statistic $T_n$ is said to be \emph{strongly consistent} for $\theta$ if $T_n \stackrel{a.s.}{\rightarrow} \theta$.  \pause

\vspace{5mm}

Strong consistency implies ordinary consistency. 
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Consistency is great but it's not enough.} \pause

{\LARGE
\begin{displaymath}
    T_n \stackrel{a.s.}{\rightarrow} \theta \pause \Rightarrow 
U_n = T_n + \frac{100,000,000}{n} \pause \stackrel{a.s.}{\rightarrow} \theta
\end{displaymath} }
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Consistency of the Sample Variance }

\pause

%{\LARGE
\begin{eqnarray*}
\widehat{\sigma}^2_n &=& \frac{1}{n}\sum_{i=1}^n (X_i-\overline{X})^2 \\ \\ \pause 
                     &=& \frac{1}{n}\sum_{i=1}^n X_i^2 - \overline{X}^2
\end{eqnarray*} 
%}

\pause

\vspace{5mm}

By SLLN, $\overline{X}_n \stackrel{a.s.}{\rightarrow}\mu$ and
$\frac{1}{n}\sum_{i=1}^n X_i^2 \stackrel{a.s.}{\rightarrow} E(X^2) = \sigma^2+\mu^2$.

\pause

\vspace{5mm}

Because the function $g(x,y)=x-y^2$ is continuous,

\pause

\vspace{5mm}

\begin{displaymath}
    \widehat{\sigma}^2_n = 
g\left(\frac{1}{n}\sum_{i=1}^n X_i^2,\overline{X}_n\right)  \pause
\stackrel{a.s.}{\rightarrow}
g(\sigma^2+\mu^2,\mu) \pause = \sigma^2+\mu^2 - \mu^2 = \pause \sigma^2
\end{displaymath}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{CLT}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Convergence in Distribution}
\framesubtitle{Sometimes called \emph{Weak Convergence}, or \emph{Convergence in Law}} 

\pause

Denote the cumulative distribution functions of $T_1, T_2, \ldots$ by $F_1(t), F_2(t), \ldots$ respectively, and denote the cumulative distribution function of $T$ by $F(t)$.

\pause

\vspace{5mm}

We say that $T_n$ converges \emph{in distribution} to $T$, and write 
$T_n \stackrel{d}{\rightarrow} T$ if \pause for every point $t$ at which $F$ is continuous, \pause
{\LARGE
\begin{displaymath}
    \lim_{n \rightarrow \infty} F_n(t) = F(t)
\end{displaymath} } \pause
Again, we will seldom use this definition directly.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Univariate Central Limit Theorem}

Let $X_1, \ldots, X_n$ be a random sample from a distribution with expected value $\mu$ and variance $\sigma^2$. Then \pause

{\LARGE
\begin{displaymath}
    Z_n = \frac{\sqrt{n}(\overline{X}_n-\mu)}{\sigma}
\stackrel{d}{\rightarrow} Z \sim N(0,1)
\end{displaymath} }

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Connections among the Modes of Convergence} \pause

{\LARGE
   \begin{itemize}
   \item $ T_n \stackrel{a.s.}{\rightarrow} T  \Rightarrow 
           T_n \stackrel{p}{\rightarrow} T \Rightarrow 
           T_n \stackrel{d}{\rightarrow} T $.
           
\pause

\vspace{5mm}
           
   \item If $a$ is a constant, %\pause 
$ T_n \stackrel{d}{\rightarrow} a 
                     \Rightarrow T_n \stackrel{p}{\rightarrow} a$.
   \end{itemize} }
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Sometimes we say the distribution of the sample mean is approximately normal, or asymptotically normal.} \pause
%\framesubtitle{} 
  \begin{itemize}
    \item This is justified by the Central Limit Theorem. \pause
    \item But it does \emph{not} mean that $\overline{X}_n$ converges in distribution to a normal random variable. \pause
    \item The Law of Large Numbers says that $\overline{X}_n$ converges almost surely (and in probability) to a constant, $\mu$. \pause
    \item So $\overline{X}_n$ converges to $\mu$ in distribution as well.
  \end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Why would we say that for large $n$, the sample mean is approximately $N(\mu,\frac{\sigma^2}{n})$?} 

\pause

\vspace{5mm}

Have 
$Z_n = \frac{\sqrt{n}(\overline{X}_n-\mu)}{\sigma} \pause
\stackrel{d}{\rightarrow} Z \sim N(0,1)$. 

\pause

{\footnotesize
\begin{eqnarray*}
    Pr\{\overline{X}_n \leq x\}  \pause
    & = & 
    Pr\left\{ \frac{\sqrt{n}(\overline{X}_n-\mu)}{\sigma} \leq
    \frac{\sqrt{n}(x-\mu)}{\sigma}\right\}    \\ \pause 
    & = & 
    Pr\left\{ Z_n \leq \frac{\sqrt{n}(x-\mu)}{\sigma}\right\}  \pause 
    \approx   \Phi\left( \frac{\sqrt{n}(x-\mu)}{\sigma} \right)
\end{eqnarray*} }

\pause

Suppose $Y$ is \emph{exactly} $N(\mu,\frac{\sigma^2}{n})$: 

\pause

{\footnotesize
\begin{eqnarray*}
    Pr\{Y \leq x\}  \pause
    & = & 
    Pr\left\{ \frac{\sqrt{n}(Y-\mu)}{\sigma} \leq
    \frac{\sqrt{n}(x-\mu)}{\sigma}\right\}    \\ \pause 
    & = & 
    Pr\left\{ Z_n \leq \frac{\sqrt{n}(x-\mu)}{\sigma}\right\}  \pause 
    =  \Phi\left( \frac{\sqrt{n}(x-\mu)}{\sigma} \right)
\end{eqnarray*} 
} % End size
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Convergence of random vectors}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[allowframebreaks] % Continue frame onto several slides. Pause does not seem to work.
\frametitle{Convergence of random vectors} 

{\footnotesize
\begin{enumerate}
   \item Definitions (All quantities in boldface are vectors in $\mathbb{R}^m$ unless otherwise stated )   
   \begin{enumerate}
      \item[$\star$] $ \mathbf{T}_n \stackrel{a.s.}{\rightarrow} \mathbf{T}$ means
            $P\{\omega:\, \lim_{n \rightarrow \infty} \mathbf{T}_n(\omega) = \mathbf{T}(\omega)\}=1$.  
      \item[$\star$] $ \mathbf{T}_n \stackrel{P}{\rightarrow} \mathbf{T}$ means
            $\forall \epsilon>0,\,\lim_{n \rightarrow \infty}
            P\{||\mathbf{T}_n-\mathbf{T}||<\epsilon  \}=1$. 
      \item[$\star$] $ \mathbf{T}_n \stackrel{d}{\rightarrow} \mathbf{T}$ means
            for every continuity point $\mathbf{t}$ of $F_\mathbf{T}$, 
            $\lim_{n \rightarrow \infty}F_{\mathbf{T}_n}(\mathbf{t}) = F_\mathbf{T}(\mathbf{t})$.  
   \end{enumerate} 
   \item $ \mathbf{T}_n \stackrel{a.s.}{\rightarrow} \mathbf{T} \Rightarrow
           \mathbf{T}_n \stackrel{P}{\rightarrow} \mathbf{T} \Rightarrow
           \mathbf{T}_n \stackrel{d}{\rightarrow} \mathbf{T} $.   
   \item \label{dsop} If $\mathbf{a}$ is a vector of constants, $ \mathbf{T}_n \stackrel{d}{\rightarrow} \mathbf{a} 
                     \Rightarrow \mathbf{T}_n \stackrel{P}{\rightarrow} \mathbf{a}$.  
   \item Strong Law of Large Numbers (SLLN): Let $\mathbf{X}_1, \ldots \mathbf{X}_n$ be independent and identically distributed 
         random vectors with finite first moment, and let $\mathbf{X}$ be a general random vector from the same distribution. Then 
         $ \overline{\mathbf{X}}_n \stackrel{a.s.}{\rightarrow} E(\mathbf{X})$.   

  \item Central Limit Theorem: Let $\mathbf{X}_1, \ldots, \mathbf{X}_n$ be i.i.d. random vectors with expected value vector $\boldsymbol{\mu}$ and covariance matrix $\boldsymbol{\Sigma}$. Then $\sqrt{n}(\overline{\mathbf{X}}_n-\boldsymbol{\mu})$ converges in distribution to a multivariate normal with mean \textbf{0} and covariance matrix $\boldsymbol{\Sigma}$.

\framebreak

   \item \label{slutd} Slutsky Theorems for Convergence in Distribution: 
        \begin{enumerate}
            \item \label{slutcond}
                  If $\mathbf{T}_n \in \mathbb{R}^m$, 
                     $\mathbf{T}_n \stackrel{d}{\rightarrow} \mathbf{T}$ and if
                     $f:\,\mathbb{R}^m \rightarrow \mathbb{R}^q$ 
                     (where $q \leq m$) is
                      continuous except possibly on a set $C$ with 
                      $P(\mathbf{T} \in C)=0$, then
                      $f(\mathbf{T}_n) \stackrel{d}{\rightarrow} f(\mathbf{T})$. 
            \item  \label{slutdiffd}
                  If $\mathbf{T}_n \stackrel{d}{\rightarrow} \mathbf{T}$ and
                     $(\mathbf{T}_n - \mathbf{Y}_n) \stackrel{P}{\rightarrow} 0$,
                     then $\mathbf{Y}_n \stackrel{d}{\rightarrow} \mathbf{T}$. 
            \item  \label{slutstackd}
                  If $\mathbf{T}_n \in \mathbb{R}^d$,
                     $\mathbf{Y}_n \in \mathbb{R}^k$, 
                     $\mathbf{T}_n \stackrel{d}{\rightarrow} \mathbf{T}$ and
                     $\mathbf{Y}_n \stackrel{P}{\rightarrow} \mathbf{c}$, then 
                     \begin{displaymath}
                        \left( \begin{array}{cc}  
                        \mathbf{T}_n \\ \mathbf{Y}_n
                        \end{array} \right)
                          \stackrel{d}{\rightarrow}
                        \left( \begin{array}{cc}  
                        \mathbf{T} \\ \mathbf{c}
                        \end{array} \right)
                     \end{displaymath}
        \end{enumerate}

\framebreak

   \item \label{slutp} Slutsky Theorems for Convergence in Probability:
        \begin{enumerate}
            \item \label{slutconp}
                  If $\mathbf{T}_n \in \mathbb{R}^m$, 
                     $\mathbf{T}_n \stackrel{P}{\rightarrow} \mathbf{T}$ and if
                      $f:\,\mathbb{R}^m \rightarrow \mathbb{R}^q$ 
                     (where $q \leq m$) is
                      continuous except possibly on a set $C$ with 
                      $P(\mathbf{T} \in C)=0$, then
                      $f(\mathbf{T}_n) \stackrel{P}{\rightarrow} f(\mathbf{T})$.
            \item \label{slutdiffp}
                  If $\mathbf{T}_n \stackrel{P}{\rightarrow} \mathbf{T}$ and
                     $(\mathbf{T}_n - \mathbf{Y}_n) \stackrel{P}{\rightarrow} 0$,
                     then $\mathbf{Y}_n \stackrel{P}{\rightarrow} \mathbf{T}$.
            \item \label{slutstackp}
                  If $\mathbf{T}_n \in \mathbb{R}^d$,
                     $\mathbf{Y}_n \in \mathbb{R}^k$, 
                     $\mathbf{T}_n \stackrel{P}{\rightarrow} \mathbf{T}$ and
                     $\mathbf{Y}_n \stackrel{P}{\rightarrow} \mathbf{Y}$, then
                     \begin{displaymath}
                        \left( \begin{array}{cc}  
                        \mathbf{T}_n \\ \mathbf{Y}_n
                        \end{array} \right)
                          \stackrel{P}{\rightarrow}
                        \left( \begin{array}{cc}  
                        \mathbf{T} \\ \mathbf{Y}
                        \end{array} \right)
                     \end{displaymath}
        \end{enumerate}

\framebreak

   \item \label{delta} Delta Method (Theorem of Cram\'{e}r, Ferguson p. 45):  Let $g: \mathbb{R}^d \rightarrow \mathbb{R}^k$  be such that the elements of \.{g}$(\mathbf{x}) = 
\left[ \frac{\partial g_i}{\partial x_j} \right]_{k \times d}$ are continuous in a neighborhood of $\boldsymbol{\theta} \in \mathbb{R}^d$.  If $\mathbf{T}_n$ is a sequence of $d$-dimensional random vectors such that $\sqrt{n}(\mathbf{T}_n-\boldsymbol{\theta}) \stackrel{d}{\rightarrow} \mathbf{T}$,  then 
$\sqrt{n}(g(\mathbf{T}_n)-g(\boldsymbol{\theta})) \stackrel{d}{\rightarrow} 
\mbox{\.{g}} (\boldsymbol{\theta}) \mathbf{T}$.  In particular, if
$\sqrt{n}(\mathbf{T}_n-\boldsymbol{\theta}) \stackrel{d}{\rightarrow} \mathbf{T} 
\sim N(\mathbf{0},\mathbf{\Sigma})$, then $\sqrt{n}(g(\mathbf{T}_n)-g(\boldsymbol{\theta})) \stackrel{d}{\rightarrow} 
\mathbf{Y} \sim N(\mathbf{0},
\mbox{\.{g}}(\boldsymbol{\theta})\mathbf{\Sigma}\mbox{\.{g}}(\boldsymbol{\theta}) ^\prime)$.
\end{enumerate}

}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{An application of the Slutsky Theorems}

  \begin{itemize}
    \item Let $X_1, \ldots, X_n \stackrel{i.i.d.}{\sim}\,?(\mu,\sigma^2)$ \pause 
    \item By CLT, $Y_n = \sqrt{n}(\overline{X}_n-\mu) \stackrel{d}{\rightarrow} Y
\sim N(0,\sigma^2)$ \pause 
    \item Let $\widehat{\sigma}_n$ be \emph{any} consistent estimator of $\sigma$. \pause 
    \item Then by \ref{slutd}.\ref{slutstackd}, 
                        $\mathbf{T}_n = \left( \begin{array}{cc}  
                        Y_n \\ \widehat{\sigma}_n
                        \end{array} \right)
                          \stackrel{d}{\rightarrow}
                        \left( \begin{array}{cc}  
                        Y \\ \sigma
                        \end{array} \right) = \mathbf{T}  $ \pause 
    \item The function $f(x,y)=x/y$ is continuous except if $y=0$ \\
so by \ref{slutd}.\ref{slutcond}, \pause 
  \end{itemize}

\begin{displaymath}
f(\mathbf{T}_n) = \frac{\sqrt{n}(\overline{X}_n-\mu)}{\widehat{\sigma}_n} \pause
\stackrel{d}{\rightarrow} f(\mathbf{T}) \pause = \frac{Y}{\sigma} \pause \sim N(0,1) 
\end{displaymath}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Another application: Asymptotic normality of the sample variance}

  \begin{itemize}
    \item Let $X_1, \ldots, X_n \stackrel{i.i.d.}{\sim}\,?(\mu,\sigma^2)$, and 
    $\widehat{\sigma}^2_n = \frac{1}{n}\sum_{i=1}^n(X_i-\overline{X}_n)^2$. \pause
    \item Want to show $\sqrt{n}\left( \widehat{\sigma}^2_n -\sigma^2\right)$ converges to a normal. \pause
    \item Substitute $\mu$ for $\overline{X}_n$? Look at $\frac{1}{n}\sum_{i=1}^n(X_i-\mu)^2$? \pause
    \item If so, it's easy.
        \begin{itemize}
            \item Let $Y_i = (X_i-\mu)^2$ \pause
            \item $E(Y_i) = \sigma^2$ \pause
            \item $Var(Y_i) = E(Y_i^2) - \left(E(Y_i)\right)^2 \pause = 
            E(X_i-\mu)^4-\sigma^4 \pause = \sigma^2_y$. \pause
            \item $\overline{Y}_n = \frac{1}{n}\sum_{i=1}^n(X_i-\mu)^2$ \pause
            \item By CLT, $\sqrt{n}\left( \overline{Y}_n -\sigma^2 \right)
            \stackrel{d}{\rightarrow} Y \sim N(0,\sigma^2_y)$.
        \end{itemize}
  \end{itemize}
  \end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Show 
$\sqrt{n}\left( \widehat{\sigma}^2_n - \sigma^2 \right) - \sqrt{n}\left( \overline{Y}_n -\sigma^2 \right) \stackrel{p}{\rightarrow} 0$}
\framesubtitle{See \ref{slutd}.\ref{slutdiffd}} 
\pause
\begin{eqnarray*}
    \widehat{\sigma}^2_n 
    & = & \frac{1}{n}\sum_{i=1}^n(X_i-\overline{X}_n)^2 \\
    & = & \frac{1}{n}\sum_{i=1}^n(X_i {\color{red} -\mu+\mu }
                                -\overline{X}_n)^2 \\ \pause
    & = & \frac{1}{n}\sum_{i=1}^n \left[(X_i-\mu)^2 + 2(X_i-\mu)(\mu-\overline{X}_n) 
    + (\mu-\overline{X}_n)^2 \right] \\ \pause
    &&\\
    & = & \ldots  \\ \pause
    & = & \frac{1}{n}\sum_{i=1}^n(X_i-\mu)^2 - (\overline{X}_n-\mu)^2 \\ \pause
    & = &  \overline{Y}_n - (\overline{X}_n-\mu)^2
\end{eqnarray*}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Using $\widehat{\sigma}^2_n = \overline{Y}_n - (\overline{X}_n-\mu)^2$}
%\framesubtitle{} 
{\footnotesize
\begin{eqnarray*}
    \sqrt{n}\left( \overline{Y}_n - \sigma^2 \right) - \sqrt{n}\left( \widehat{\sigma}^2_n -\sigma^2 \right) 
    & = & \pause  \sqrt{n}\left( \overline{Y}_n - {\color{red}\widehat{\sigma}^2_n} \right) \\ \pause 
    & = & \sqrt{n}\left( \overline{Y}_n - {\color{red}(\overline{Y}_n - (\overline{X}_n-\mu)^2)} \right) \\ \pause 
    & = & \sqrt{n}\left( \overline{X}_n-\mu\right)^2  \\ \pause
    & = & \sqrt{n}(\overline{X}_n-\mu) \,\cdot\, (\overline{X}_n-\mu)\\ \pause
\end{eqnarray*}
  \begin{itemize}
    \item First term goes in distribution to $X \sim N(0,\sigma^2)$ by CLT. \pause
    \item Second term goes to zero in probability by LLN. \pause
    \item $\left(\begin{array}{c} \sqrt{n}(\overline{X}_n-\mu) \\ \overline{X}_n-\mu
    \end{array}\right) \stackrel{d}{\rightarrow}
    \left(\begin{array}{c} X \\ 0 \end{array}\right)$ by \ref{slutd}.\ref{slutstackd}. \pause
    \item By continuous mapping \ref{slutd}.\ref{slutcond}, 
    $\sqrt{n}(\overline{X}_n-\mu) \,\cdot\, (\overline{X}_n-\mu) \stackrel{d}{\rightarrow} X\cdot 0 = 0$ \pause
    \item Convergence in distribution to a constant implies convergence in probability (Rule~\ref{dsop})\pause, so the difference converges in probability to zero, and the result follows by \ref{slutd}.\ref{slutdiffd} \hspace{4mm} $\blacksquare$
  \end{itemize}
} % End size
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{The Result}
%\framesubtitle{} 
  \begin{itemize}
    \item Because the difference between $\sqrt{n}\left( \widehat{\sigma}^2_n -\sigma^2 \right)$ and $\sqrt{n}\left( \frac{1}{n}\sum_{i=1}^n(X_i-\mu)^2 -\sigma^2 \right)$ goes to zero in probability, they converge in distribution to the same target. \pause
    \item Target is $N(0,\sigma^2_y)$ \pause
    \item $\sigma^2_y = E(X_i-\mu)^4-\sigma^4$.
  \end{itemize}
\end{frame}




\begin{comment}
\end{comment}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Delta Method}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Univariate delta method}

In the multivariate Delta Method~\ref{delta}, the matrix $\mbox{\.{g}}(\boldsymbol{\theta})$ is a Jacobian. The univariate version % 
of the delta method says that \pause \vspace{3mm}

If $\sqrt{n}\left( T_n- \theta \right) \stackrel{d}{\rightarrow} T$ and $g^{\prime\prime}(x)$ is continuous in a neighbourhood of  $\theta$, then 
\begin{displaymath}
    \sqrt{n}\left( g(T_n)- g(\theta) \right) \stackrel{d}{\rightarrow} 
     g^\prime(\theta) \, T.
\end{displaymath}

\vspace{5mm}
\pause
When using the Central Limit Theorem, \emph{especially} if there is a $\theta \neq \mu$ in the model, it's safer to write 
\begin{displaymath}
    \sqrt{n}\left( g(\overline{X}_n)- g(\mu) \right) \stackrel{d}{\rightarrow} 
     g^\prime(\mu) \, T.
\end{displaymath} 
and then substitute for $\mu$ in terms of $\theta$.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Delta Method Example}
%\framesubtitle{} 
  \begin{itemize}
    \item[] $X_1, \ldots, X_n \stackrel{i.i.d.}{\sim} \mbox{Poisson}(\lambda)$ \pause
    \item[] $E(X_i)=Var(X_i) = \lambda$ \pause
    \item[] $\frac{\sqrt{n}(\overline{X}_n-\lambda)}{\sqrt{\overline{X}_n}} 
            \stackrel{d}{\rightarrow} Z_1 \sim N(0,1)$ \pause
    \item[] Confidence interval 
    $\left(\overline{X}_n - z_{\alpha/2} \sqrt{\frac{\overline{X}_n}{n}}~,~
           \overline{X}_n + z_{\alpha/2} \sqrt{\frac{\overline{X}_n}{n}}\right)$ \pause
    \item[] 
    \item[] Maybe we can do better.
  \end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Delta Method says $ \sqrt{n}\left( g(T_n)- g(\theta) \right) \stackrel{d}{\rightarrow} g^\prime(\theta) \, T$}
%\framesubtitle{} 
  \begin{itemize}
    \item[] $\sqrt{n}(\overline{X}_n-\lambda) \stackrel{d}{\rightarrow} X \sim N(0,\lambda)$. \pause
    \item[] $ \sqrt{n}\left( g(\overline{X}_n)- g(\lambda) \right) \pause \stackrel{d}{\rightarrow} g^\prime(\lambda) \, X \sim N(0,g^\prime(\lambda)^2\lambda)$ \pause
    \item Choose $g$ to make the variance not depend on $\lambda$. \pause
    \item How about $g(\lambda) = 2\sqrt{\lambda}$ \pause
    \item[] $g^\prime(\lambda) = 2 \, \frac{1}{2}\lambda^{-1/2} = \pause \frac{1}{\sqrt{\lambda}}$. \pause
    \item Variance of the target is $g^\prime(\lambda)^2\lambda = 1$. \pause

    \item[] So,  
  \end{itemize}

\begin{equation*}
\sqrt{n}\left(2\sqrt{\overline{X}_n}-2\sqrt{\lambda}\right) \stackrel{d}{\rightarrow} Z_2 \sim N(0,1). 
\end{equation*}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{$\sqrt{n}\left(2\sqrt{\overline{X}_n}-2\sqrt{\lambda}\right) \stackrel{d}{\rightarrow} Z_2 \sim N(0,1)$}
%\framesubtitle{} 
%{\LARGE
\begin{eqnarray*}
    0.95 & \approx & P\left\{ -z_{\alpha/2} < \sqrt{n}\left(2\sqrt{\overline{X}_n}-2\sqrt{\lambda}\right) < z_{\alpha/2} \right\} \\ \pause
    & = &  P\left\{ -\frac{z_{\alpha/2}}{2\sqrt{n}} <  \sqrt{\overline{X}_n}-\sqrt{\lambda}
                     < \frac{z_{\alpha/2}}{2\sqrt{n}} \right\} \\ \pause
    & = &  P\left\{ \sqrt{\overline{X}_n}-\frac{z_{\alpha/2}}{2\sqrt{n}} < \sqrt{\lambda}
                     < \sqrt{\overline{X}_n}+\frac{z_{\alpha/2}}{2\sqrt{n}} \right\} \\ \pause
    & = &  P\left\{ \left(\sqrt{\overline{X}_n}-\frac{z_{\alpha/2}}{2\sqrt{n}}\right)^2 < \lambda
                     <  \left(\sqrt{\overline{X}_n}+\frac{z_{\alpha/2}}{2\sqrt{n}}\right)^2 \right\}.
\end{eqnarray*}  \pause
%} % End size
Compare 
$P\left\{\overline{X}_n - z_{\alpha/2} \sqrt{\frac{\overline{X}_n}{n}} < \lambda <
           \overline{X}_n + z_{\alpha/2} \sqrt{\frac{\overline{X}_n}{n}}\right\}$

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{The delta method comes from Taylor's Theorem} \pause
\textbf{Taylor's Theorem}: Let the $n$th derivative $f^{(n)}$ be continuous
in $[a,b]$ and differentiable in $(a,b)$, with $x$ and $x_0$ in $(a,b)$. Then
there exists a point $\xi$ between $x$ and $x_0$ such that \pause

\begin{eqnarray*}
    f(x) & = & f(x_0) \;+\; f^\prime(x_0)\,(x-x_0) 
                \;+\; \frac{f^{\prime\prime}(x_0)(x-x_0)^2}{2!}
                \;+\; \ldots  \\
         & + &  \frac{f^{(n)}(x_0)(x-x_0)^n}{n!} \;+\;
               \frac{f^{(n+1)}(\xi)(x-x_0)^{n+1}}{(n+1)!}
\end{eqnarray*} \pause
where $R_n = \frac{f^{(n+1)}(\xi)(x-x_0)^{n+1}}{(n+1)!}$ is called the
\emph{remainder term}. \pause 

If $R_n \rightarrow 0$ as $n \rightarrow \infty$, \pause the
resulting infinite series is called the \emph{Taylor Series} for $f(x)$.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Taylor's Theorem with two terms plus remainder}
\framesubtitle{Very common in applications}  \pause
Let $g(x)$ be a function for which $g^{\prime\prime}(x)$ is continuous in an open interval containing $x=\theta$. \pause Then
{\Large
\begin{displaymath}
    g(x) = \pause g(\theta) + g^\prime(\theta)(x-\theta) + 
           \frac{g^{\prime\prime}(\theta^*)(x-\theta)^2}{2!}
\end{displaymath} \pause
} % End size
where $\theta^*$ is between $x$ and $\theta$. 
\end{frame}

\begin{frame}
\frametitle{Delta method}
\framesubtitle{Using $g(x) = g(\theta) + g^\prime(\theta)(x-\theta) + 
           \frac{1}{2}g^{\prime\prime}(\theta^*)(x-\theta)^2$} \pause
Let  $\sqrt{n}(T_n-\theta) \stackrel{d}{\rightarrow} T$ \pause
so that $T_n \stackrel{p}{\rightarrow} \theta$. \pause


{\footnotesize
\begin{eqnarray*}
    \sqrt{n}\left({\color{red}g(T_n)}-g(\theta)\right) \pause & = & 
     \sqrt{n}\left(
     {\color{red} g(\theta) + g^\prime(\theta)(T_n-\theta) + 
           \frac{1}{2}g^{\prime\prime}(\theta_n^*)(T_n-\theta)^2}
     -g(\theta)\right) \\ \pause
& = &  \sqrt{n}\left(g^\prime(\theta)(T_n-\theta) + 
           \frac{1}{2}g^{\prime\prime}(\theta_n^*)(T_n-\theta)^2 \right) \\ \pause
& = & g^\prime(\theta) \, \sqrt{n}(T_n-\theta) \\
&&    ~~~+ \pause \frac{1}{2}g^{\prime\prime}(\theta_n^*) \cdot 
      \sqrt{n}(T_n-\theta) \cdot (T_n-\theta) \\  \pause
&\stackrel{d}{\rightarrow}& \pause g^\prime(\theta) T + 0
\end{eqnarray*} 
} % End size
\end{frame}

% Need a homely example

\begin{frame}
\frametitle{That was fun, but it was all univariate.} % \pause



  \begin{itemize}
    \item The multivariate CLT establishes convergence to a multivariate normal. % \pause
    \item Vectors of MLEs are approximately multivariate normal for large samples. % \pause 
    \item The multivariate delta method can yield the asymptotic distribution of useful functions of the MLE vector.
  \end{itemize}

\vspace{15mm}

% \pause

We need to look at random vectors and the multivariate normal distribution.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{frame}
\frametitle{Copyright Information}

This slide show was prepared by  \href{http://www.utstat.toronto.edu/brunner}{Jerry Brunner},
Department of Statistics, University of Toronto. It is licensed under a 
\href{http://creativecommons.org/licenses/by-sa/3.0/deed.en_US}
     {Creative Commons Attribution - ShareAlike 3.0 Unported License}. Use any part of it as you like and share the result freely. The \LaTeX~source code is available from the course website:
\href{http://www.utstat.toronto.edu/brunner/oldclass/2053f22} {\small\texttt{http://www.utstat.toronto.edu/brunner/oldclass/2053f22}}
\end{frame}


\end{document}

\sin^{-1}\left(\sqrt{\overline{X}_m}\right)
\sin^{-1}\left(\sqrt{\theta}\right)



\begin{displaymath}
    \sqrt{m}\left( g(\overline{X}_m)- g(\theta) \right) \stackrel{d}{\rightarrow} 
    Y \sim N\left(0,(g^\prime(\theta))^2\theta(1-\theta)\right).
\end{displaymath} }


CLT says $\sqrt{n}(\overline{X}_n-\lambda) \stackrel{d}{\rightarrow} T \sim N(0,\lambda)$.



$\sqrt{n}\left( g(\overline{X}_n)- g(\lambda) \right)
  = \sqrt{n}\left( 2\sqrt{\overline{X}_n}- 2\sqrt{\lambda} \right) 
 \stackrel{d}{\rightarrow} Z \sim N(0,1)$

\Pr\{-z < 2\sqrt{n}\left(\sqrt{\overline{X}_n}-\sqrt{\lambda}\right) < z \}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{}
%\framesubtitle{} 
  \begin{itemize}
    \item 
    \item 
    \item 
  \end{itemize}
\end{frame}

{\LARGE
\begin{displaymath}
    
\end{displaymath} }