Added lecture 3

Author: bcaffo

expectation.pdf

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+\documentclass[aspectratio=169]{beamer}
+\mode<presentation>
+\usetheme{Hannover}
+\useoutertheme{sidebar}
+\usecolortheme{dolphin}
+
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{enumerate}
+
+%some bold math symbosl
+\newcommand{\Cov}{\mathrm{Cov}}
+\newcommand{\Var}{\mathrm{Var}}
+\newcommand{\brho}{\boldsymbol{\rho}}
+\newcommand{\bSigma}{\boldsymbol{\Sigma}}
+\newcommand{\btheta}{\boldsymbol{\theta}}
+\newcommand{\bbeta}{\boldsymbol{\beta}}
+\newcommand{\bmu}{\boldsymbol{\mu}}
+\newcommand{\bW}{\mathbf{W}}
+\newcommand{\one}{\mathbf{1}}
+\newcommand{\bH}{\mathbf{H}}
+\newcommand{\by}{\mathbf{y}}
+\newcommand{\bolde}{\mathbf{e}}
+\newcommand{\bx}{\mathbf{x}}
+
+\newcommand{\cpp}[1]{\texttt{#1}}
+
+
+
+\title{Mathematical Biostatistics Boot Camp: Lecture 3, Expectations}
+
+\author{Brian Caffo}
+\date{\today}
+\institute[Department of Biostatistics]{
+  Department of Biostatistics \\
+  Johns Hopkins Bloomberg School of Public Health\\
+  Johns Hopkins University
+}
+
+
+\begin{document}
+
+\frame{\titlepage}
+
+\frame{
+  \frametitle{Table of contents}
+  \tableofcontents
+}
+
+
+\section{Outline}
+\frame{
+  \frametitle{Outline}
+  \begin{enumerate}
+  \item Define expected values
+  \item Properties of expected values
+  \item Unbiasedness of the sample mean
+  \item Define variances 
+  \item Define the standard deviation
+  \item Calculate Bernoulli variance
+  \end{enumerate}
+}
+
+\section{Expected values}
+\subsection{Discrete random variables}
+\begin{frame}
+  \frametitle{Expected values}
+  \begin{itemize}
+  \item The {\bf expected value} or {\bf mean} of a random variable is the center of its
+    distribution
+  \item For discrete random variable $X$ with PMF $p(x)$, it is defined as follows
+    $$
+    E[X] = \sum_x xp(x).
+    $$
+    where the sum is taken over the possible values of $x$
+  \item $E[X]$ represents the center of mass of a collection of
+    locations and weights, $\{x, p(x)\}$
+  \end{itemize}
+\end{frame}
+
+
+\begin{frame}
+  \frametitle{Example}
+  \includegraphics[width=3.5in]{expectation.pdf}
+\end{frame}
+
+
+
+\begin{frame}
+  \frametitle{Example} 
+  \begin{itemize}
+  \item Suppose a coin is flipped and $X$ is declared $0$ or $1$ corresponding
+    to a head or a tail, respectively
+  \item What is the expected value of $X$? 
+    $$
+    E[X] = .5 \times 0 + .5 \times 1 = .5
+    $$
+  \item Note, if thought about geometrically, this answer is obvious; if two equal
+    weights are spaced at 0 and 1, the center of mass will be $.5$
+  \end{itemize}
+\end{frame}
+
+\begin{frame}\frametitle{Example} 
+  \begin{itemize}
+  \item Suppose that a die is tossed and $X$ is the number face up
+  \item What is the expected value of $X$?
+    $$
+    E[X] = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} +
+    3 \times \frac{1}{6} + 4 \times \frac{1}{6} +
+    5 \times \frac{1}{6} + 6 \times \frac{1}{6} = 3.5
+    $$
+  \item Again, the geometric argument makes this answer obvious without calculation.
+  \end{itemize}
+\end{frame}
+
+\subsection{Continuous random variables}
+\begin{frame}\frametitle{Continuous random variables}
+  \begin{itemize}
+  \item For a continuous random variable, $X$, with density, $f$, the expected
+    value is defined as follows
+    $$
+    E[X] = \int_{-\infty}^\infty t f(t)dt
+    $$
+  \item This definition borrows from the definition of center of mass for
+    a continuous body
+  \end{itemize}
+\end{frame}
+
+\begin{frame}\frametitle{Example}
+  \begin{itemize}
+  \item   Consider a density where $f(x) = 1$ for $x$
+    between zero and one
+  \item (Is this a valid density?)
+  \item Suppose that $X$ follows this density; what is its expected value? 
+    $$
+    E[X] = \int_{0}^{1} x dx = \left. \frac{x^2}{2} ~\right|_{0}^{1} = 1/2
+    $$
+  \end{itemize}
+\end{frame}
+
+\section{Rules about expected values}
+\begin{frame}\frametitle{Rules about expected values}
+  \begin{itemize}
+  \item The expected value is a linear operator 
+  \item If $a$ and $b$ are not random and $X$ and $Y$ 
+    are two random variables then
+    \begin{itemize}
+    \item $E[aX + b] = a E[X] + b$
+    \item $E[X + Y] = E[X] + E[Y]$
+    \end{itemize}
+  \item {\em In general} if $g$ is a function that is not linear,
+    $$
+    E[g(X)] \neq g(E[X])
+    $$
+  \item For example, in general, $E[X^2] \neq E[X]^2$ 
+  \end{itemize}
+\end{frame}
+
+\begin{frame}\frametitle{Example}
+  \begin{itemize}
+  \item  You flip a coin, $X$ and simulate a uniform random number $Y$, what 
+    is the expected value of their sum? 
+    $$
+    E[X + Y] = E[X] + E[Y] = .5 + .5 = 1
+    $$ 
+  \item Another example, you roll a coin twice. What is the expected value of
+    the average? 
+  \item Let $X_1$ and $X_2$ be the results of the two rolls
+    $$
+    E[(X_1 + X_2) / 2] = \frac{1}{2}(E[X_1] + E[X_2])
+    = \frac{1}{2}(3.5 + 3.5) = 3.5
+    $$
+  \end{itemize}
+\end{frame}
+
+
+\begin{frame}\frametitle{Example}
+  \begin{enumerate}
+  \item Let $X_i$ for $i=1,\ldots,n$ be a collection of random
+    variables, each from a distribution with mean $\mu$
+  \item Calculate the expected value of the sample average of the $X_i$
+  \end{enumerate}
+  \begin{eqnarray*}
+    E\left[ \frac{1}{n}\sum_{i=1}^n X_i\right]
+    & = & \frac{1}{n} E\left[\sum_{i=1}^n X_i\right] \\
+    & = & \frac{1}{n} \sum_{i=1}^n E\left[X_i\right] \\
+    & = & \frac{1}{n} \sum_{i=1}^n \mu =  \mu.
+  \end{eqnarray*}
+\end{frame}
+
+\begin{frame}
+  \frametitle{Remark}
+  \begin{itemize}
+  \item Therefore, the expected value of the {\bf sample mean} is the {\bf
+      population mean} that it's trying to estimate
+  \item When the expected value of an estimator is what its trying to estimate,
+    we say that the estimator is {\bf unbiased}
+  \end{itemize}
+\end{frame}
+
+
+\section{Variances}
+\begin{frame}\frametitle{The variance}
+  \begin{itemize}
+  \item The variance of a random variable is a measure of {\em spread}
+  \item If $X$ is a random variable with mean $\mu$, the variance of 
+    $X$ is defined as
+    $$
+    \Var(X) = E[(X - \mu)^2]
+    $$
+    the expected (squared) distance from the mean
+  \item Densities with a higher variance are more spread out than
+    densities with a lower variance
+  \end{itemize}
+\end{frame}
+
+\begin{frame}
+  \begin{itemize}
+  \item Convenient computational form
+    $$
+    \Var(X) = E[X^2] - E[X]^2
+    $$
+  \item If $a$ is constant then $\Var(aX) = a^2 \Var(X)$
+  \item The square root of the variance is called the {\bf standard deviation}
+  \item The standard deviation has the same units as $X$
+  \end{itemize}
+\end{frame}
+
+
+\begin{frame}\frametitle{Example}
+  \begin{itemize}
+  \item   What's the sample variance from the result of a toss of a die? 
+    \begin{itemize}
+    \item $E[X] = 3.5$ 
+    \item $E[X^2] = 1 ^ 2 \times \frac{1}{6} + 2 ^ 2 \times \frac{1}{6} +
+  3 ^ 2 \times \frac{1}{6} + 4 ^ 2 \times \frac{1}{6} +
+  5 ^ 2 \times \frac{1}{6} + 6 ^ 2 \times \frac{1}{6} = 15.17$ 
+    \end{itemize}
+  \item $\Var(X) = E[X^2] - E[X]^2 \approx 2.92$
+ \end{itemize}
+\end{frame}
+
+\begin{frame}\frametitle{Example}
+  \begin{itemize}
+  \item  What's the sample variance from the result of the toss of a coin
+    with probability of heads (1) of $p$? 
+    \begin{itemize}
+    \item   $E[X] = 0 \times (1 - p) + 1 \times p = p$
+    \item $E[X^2] = E[X] = p$ 
+    \end{itemize}
+  \item $\Var(X) = E[X^2] - E[X]^2 = p - p^2 = p(1 - p)$
+  \end{itemize}
+\end{frame}
+
+
+\begin{frame}\frametitle{Example} 
+  \begin{itemize}
+  \item Suppose that a random variable is such that $0 \leq X \leq 1$ and $E[X] = p$
+  \item Note $X^2 \leq X$ so that $E[X^2] \leq E[X] = p$
+  \item $\Var(X) = E[X^2] - E[X]^2 \leq E[X] - E[X]^2 = p(1-p)$
+  \item Therefore the Bernoulli variance is the largest possible for random variables bounded between $0$ and $1$
+  \end{itemize}
+\end{frame}
+
+\section{Chebyshev's inequality}
+\begin{frame}
+\frametitle{Interpreting variances}
+\begin{itemize}
+  \item Chebyshev's inequality is useful for interpreting variances
+  \item This inequality states that
+    $$
+    P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}
+    $$
+  \item For example, the probability that a random variable lies beyond $k$
+    standard deviations from its mean is less than $1/k^2$
+    \begin{eqnarray*}
+      2\sigma & \rightarrow & 25\% \\
+      3\sigma & \rightarrow & 11\% \\
+      4\sigma & \rightarrow &  6\% 
+    \end{eqnarray*}
+  \item Note this is only a bound; the actual probability might be
+    quite a bit smaller
+\end{itemize}
+\end{frame}
+
+\begin{frame} \frametitle{Proof of Chebyshev's inequality}
+  \begin{eqnarray*}
+    P(|X - \mu| > k\sigma) & = & \int_{\{x: |x-\mu| > k\sigma\}} f(x) dx \\
+& \leq & \int_{\{x:|x -\mu| > k\sigma\}}\frac{(x - \mu)^2}{k^2\sigma^2} f(x) dx \\
+& \leq & \int_{-\infty}^{\infty} \frac{(x - \mu)^2}{k^2\sigma^2} f(x) dx \\
+& = & \frac{1}{k^2}
+  \end{eqnarray*}
+\end{frame}
+
+\begin{frame} \frametitle{Example}
+  \begin{itemize}
+  \item IQs are often said to be distributed with a mean of $100$ and a sd of $15$
+  \item What is the probability of a randomly drawn person having an IQ higher than
+    $160$ or below $40$?
+  \item Thus we want to know the probability of a person being more
+    than $4$ standard deviations from the mean
+  \item Thus Chebyshev's inequality suggests that this will be no larger than 6\%
+  \item IQs distributions are often cited as being bell shaped, in which case this
+    bound is very conservative
+  \item The probability of a random draw from a bell curve being $4$
+    standard deviations from the mean is on the order of $10^{-5}$ (one
+    thousandth of one percent)
+  \end{itemize}
+\end{frame}
+
+\begin{frame} \frametitle{Example}
+  \begin{itemize}
+  \item A popular buzz phrase in industrial quality control is
+    Motorola's``Six Sigma'' whereby businesses are suggested to
+    control extreme events or rare defective parts
+  \item Chebyshev's inequality states that the probability of a ``Six
+    Sigma'' event is less than $1/6^2 \approx 3\%$
+  \item If a bell curve is assumed, the probability of a ``six sigma''
+    event is on the oder of $10^{-9}$ (one ten millionth of a percent)
+  \end{itemize}
+\end{frame}
+\end{document}