% \documentclass[serif]{beamer} % Serif for Computer Modern math font.
\documentclass[serif, handout]{beamer} % Handout to ignore pause statements

\hypersetup{colorlinks,linkcolor=,urlcolor=red}


\usefonttheme{serif} % Looks like Computer Modern for non-math text -- nice!
\setbeamertemplate{navigation symbols}{} % Suppress navigation symbols
% \usetheme{Berlin} % Displays sections on top
\usetheme{Frankfurt}  % Displays section titles on top: Fairly thin but still swallows some material at bottom of crowded slides
%\usetheme{Berkeley}

\usepackage[english]{babel}
\usepackage{amsmath} % for binom
% \usepackage{graphicx} % To include pdf files!
% \definecolor{links}{HTML}{2A1B81}
% \definecolor{links}{red}
\setbeamertemplate{footline}[frame number] 

\mode<presentation>

\title{Moment-generating functions\footnote{ This slide show is an open-source document. See last slide for copyright information.}}
\subtitle{STA 302: Fall 2015}
\date{} % To suppress date

\begin{document}

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


\begin{frame}
\frametitle{The change of variables formula}
\framesubtitle{Let $X$ be a random variable.} 
Let $Y=g(X)$. There are two ways to get $E(Y)$. \pause 
\vspace{3mm}
\begin{enumerate}
    \item Derive the distribution of $Y$ and compute
\begin{displaymath}
    E(Y) = \int_{-\infty}^\infty y \, f_{_Y}(y) \, dy
\end{displaymath}  \pause
    \item Use the distribution of $X$, and calculate 
\begin{displaymath}
    E(g(X)) = \int_{-\infty}^\infty g(x) \, f_{_X}(x) \, dx
\end{displaymath}  \pause
\end{enumerate}
Big theorem: These two expressions are equal.
\end{frame}


\begin{frame}
\frametitle{The change of variables formula is very general}
\framesubtitle{Including but not limited to }  \pause
\begin{columns}   % Use Beamer's columns to use more of the margins!
\column{1.2\textwidth}
\begin{itemize}
    \item[] $E(g(X)) = \int_{-\infty}^\infty g(x) \, f_{_X}(x) \, dx$ \pause
    \item[] 
    \item[] $E(g(\mathbf{X})) =  
\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty g(x_1, \ldots, x_p) \, 
f_{_\mathbf{X}}(x_1, \ldots, x_p) \, dx_1 \ldots dx_p $ \pause
    \item[] 
    \item[] $E\left(g(X)\right) = \sum_x g(x) p_{_X}(x) $
\end{itemize}
\end{columns}
\end{frame}


\begin{frame}
\frametitle{Moment-generating functions}
%\framesubtitle{} 
{\LARGE
\begin{displaymath}
    M_{_Y}(t) =  E(e^{Yt})  \pause =
\left\{ \begin{array}{l} % ll means left left
   \int_{-\infty}^\infty e^{yt} \, f_{_Y}(y) \, dy  \\ 
   \\
   \sum_y e^{yt} p_{_Y}(y) 
   \end{array}  \right.   % Need that crazy invisible right period!
\end{displaymath} 
} % End size

\end{frame}

\begin{frame}
\frametitle{Properties of moment-generating functions} \pause
%\framesubtitle{} 
\begin{itemize}
    \item Moment-generating functions can be used to generate moments.  \pause
To get $E(Y^k)$, differentiate $M_{_Y}(t)$ with respect to $t$. Differentiate $k$ times and set $t=0$. \pause
    \item[] 
    \item Moment-generating functions correspond uniquely to probability distributions.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{The function $M(t)$ is like a fingerprint of the probability distribution.} \pause
%\framesubtitle{} 
\begin{itemize}
    \item[] $Y \sim N(\mu,\sigma^2)$ if and only if $M_{_Y}(t) =   e^{\mu t + \frac{1}{2}\sigma^2t^2}$. \pause
    \item[] 
    \item[] $Y \sim \chi^2(\nu)$ if and only if $M_{_Y}(t) = (1-2t)^{-\nu/2}$ for $t < \frac{1}{2}$.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Normal: $M(t) = e^{\mu t + \frac{1}{2}\sigma^2t^2}$}
%\framesubtitle{} 
\begin{center}
\includegraphics[width=3in]{NormalMGF}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Chi-squared: $M(t) = (1-2t)^{-\nu/2}$}
%\framesubtitle{} 
\begin{center}
\includegraphics[width=3in]{ChisqMGF}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Example: Using moment-generating functions to prove distribution facts}
%\framesubtitle{} 
Let $X \sim N(\mu,\sigma^2)$. Show $Y = \frac{X-\mu}{\sigma} \sim N(0,1)$
\end{frame}



\begin{frame}
\frametitle{Facts about moment-generating functions}
\framesubtitle{Use these to find distributions of \emph{functions} of random variables} 
\begin{itemize}
    \item $M_{aY}(t) = M_Y(at)$ \pause
    \item $M_{_{Y+a}}(t) = e^{at}M_Y(t)$ \pause
    \item If $Y_1, \ldots, Y_n$ are independent, $M_{_{\sum_{i=1}^n Y_i}}(t) = \prod_{i=1}^n M_{Y_i}(t)$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{A standard example}
\framesubtitle{Using $M_{_{\sum_{i=1}^n X_i}}(t) = \prod_{i=1}^n M_{X_i}(t)$}  \pause

Let $X_1, \ldots, X_n \stackrel{i.i.d.}{\sim} N(\mu,\sigma^2)$, with $Y = \sum_{i=1}^n X_i$. Find the probability distribution of $Y$. \pause

\vspace{3mm}

How about $\overline{X}$?  \pause Recall $M_{aY}(t) = M_Y(at)$.

\end{frame}

\begin{frame}
\frametitle{Another standard example}
% \framesubtitle{Using $M_{_{\sum_{i=1}^n X_i}}(t) = \prod_{i=1}^n M_{X_i}(t)$}  \pause

Let $X_1, \ldots, X_n \stackrel{ind.}{\sim} \chi^2(\nu_i)$, and $Y = \sum_{i=1}^n X_i$. Find the probability distribution of $Y$.

\end{frame}


\begin{frame}
\frametitle{Less well known}
\framesubtitle{But very useful later} 
If  $W=W_1+W_2$ with $W_1$ and $W_2$ independent, $W\sim\chi^2(\nu_1+\nu_2)$ and $W_2\sim\chi^2(\nu_2)$ then $W_1\sim\chi^2(\nu_1)$. 
\end{frame}


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


\begin{frame}
\frametitle{Copyright Information}

This slide show was prepared by  \href{http://www.utstat.toronto.edu/~brunner}{Jerry Brunner},
Department of Statistical Sciences, 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:

\vspace{5mm}
\href{http://www.utstat.toronto.edu/~brunner/oldclass/302f15} {\small\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/302f15}}

\end{frame}


\end{document}

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

# R code for plots of normal MGFs
tt = seq(from=-1,to=1,by=0.05)
mu = 0; sigsq = 1
zero = exp(mu*tt + 0.5*sigsq*tt^2)
mu = 1; one = exp(mu*tt + 0.5*sigsq*tt^2)
mu = -1; minusone = exp(mu*tt + 0.5*sigsq*tt^2)
x = c(tt,tt,tt); y = c(zero,one,minusone)
plot(x,y,pch=' ',xlab='t',ylab = 'M(t)')
lines(tt,zero,lty=1)
lines(tt,one,lty=2)
lines(tt,minusone,lty=3)
title("Fingerprints of the normal distribution")
# Legend
x1 <- c(-0.4,0) ; y1 <- c(4,4) ; lines(x1,y1,lty=1)
text(0.25,4,expression(paste(mu," =  0, ",sigma^2," = 1")))
x2 <- c(-0.4,0) ; y2 <- c(3.75,3.75) ; lines(x2,y2,lty=2)
text(0.25,3.75,expression(paste(mu," =  1, ",sigma^2," = 1")))
x3 <- c(-0.4,0) ; y3 <- c(3.5,3.5) ; lines(x3,y3,lty=3)
text(0.25,3.5,expression(paste(mu," = -1, ",sigma^2," = 1")))


# R code for plots of chi-squared MGFs
tt = seq(from=-0.25,to=0.25,by=0.005)
nu = 1; one = (1-2*tt)^(-nu/2)
nu = 2; two = (1-2*tt)^(-nu/2)
nu = 3; three = (1-2*tt)^(-nu/2)
x = c(tt,tt,tt); y = c(one,two,three)
plot(x,y,pch=' ',xlab='t',ylab = 'M(t)')
lines(tt,one,lty=1)
lines(tt,two,lty=2)
lines(tt,three,lty=3)
title("Fingerprints of the chi-squared distribution")
# Legend
x1 <- c(-0.2,-0.1) ; y1 <- c(2.5,2.5) ; lines(x1,y1,lty=1)
text(-0.05,2.5,expression(paste(nu," =  1")))
x2 <- c(-0.2,-0.1) ; y2 <- c(2.3,2.3) ; lines(x2,y2,lty=2)
text(-0.05,2.3,expression(paste(nu," =  2")))
x3 <- c(-0.2,-0.1) ; y3 <- c(2.1,2.1) ; lines(x3,y3,lty=3)
text(-0.05,2.1,expression(paste(nu," =  3")))