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\mode<presentation>

\title{Expected Value, Variance and Covariance\footnote{ This slide show is an open-source document. See last slide for copyright information.}}
\subtitle{STA 256: Fall 2018}
\date{} % To suppress date

\begin{document}

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

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

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Expected Value}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Definition}
%\framesubtitle{Discrete or Continuous} 
The expected value of a discrete random variable is  \pause
{\LARGE
\begin{displaymath}
    E(X) = \sum_x x \, p_x(x)
\end{displaymath} \pause
} % End size

\begin{itemize}
     \item Provided $\sum_x |x| \, p_x(x) < \infty$. \pause If the sum diverges, the expected value does not exist. \pause
     \item Existence is only an issue for infinite sums (and integrals over infinite intervals).
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Expected value is an average} \pause
%\framesubtitle{} 
\begin{itemize}
    \item Imagine a very large jar full of balls. \pause This is the population. \pause
    \item The balls are numbered $x_1, \ldots, x_N$. \pause These are measurements carried out on members of the population. \pause
    \item Suppose for now that all the numbers are different. \pause
    \item A ball is selected at random; all balls are equally likely to be chosen. \pause
    \item Let $X$ be the number on the ball selected. \pause 
    \item $P(X = x_i) = \frac{1}{N}$. \pause
\end{itemize}
\begin{eqnarray*}
    E(X) & = & \sum_x x \, p_x(x)  \\ \pause
         & = & \sum_{i=1}^N x_i  \frac{1}{N} \\ \pause
         & = & \frac{\sum_{i=1}^N x_i}{N} \\ 
\end{eqnarray*}
% The common average, or arithmetic mean.
\end{frame}

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

\begin{frame}
\frametitle{For the jar full of numbered balls, $E(X) = \frac{\sum_{i=1}^N x_i}{N}$}
%\framesubtitle{} 
{\footnotesize
\begin{itemize}
    \item This is the common average, or arithmetic mean. \pause
    \item Suppose there are ties. \pause
    \item Unique values are $v_i$, for $i=1, \ldots, n$. \pause
    \item Say $n_1$ balls have value $v_1$, \pause and $n_2$ balls have value $v_2$, and \ldots \pause 
    $n_n$ balls have value $v_n$. \pause
    \item Note $n_1 + \cdots + n_n = N$\pause, and $P(X=v_j) = \frac{n_j}{N}$. \pause 
\end{itemize}
\begin{eqnarray*}
    E(X) & = & \frac{\sum_{i=1}^N x_i}{N}  \\ \pause
         & = & \sum_{j=1}^n n_j v_j  \frac{1}{N} \\ \pause
         & = & \sum_{j=1}^n  v_j  \frac{n_j}{N} \\ \pause
         & = & \sum_{j=1}^n v_j  P(X=v_j) \\ 
\end{eqnarray*}
} % End size
\end{frame}

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

\begin{frame}
\frametitle{Compare $E(X)=\sum_{j=1}^n v_j  P(X=v_j) \mbox{ and } \sum_x x \, p_x(x)$} \pause
%\framesubtitle{} 
\begin{itemize}
    \item Expected value is a generalization of the idea of an average, or mean. \pause
    \item Specifically a \emph{population} mean. \pause
    \item It is often just called the ``mean."
\end{itemize}
\end{frame}


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

\begin{frame}
\frametitle{Definition}
%\framesubtitle{Discrete or Continuous} 
The expected value of a continuous random variable is  \pause
{\LARGE
\begin{displaymath}
    E(X) = \int_{-\infty}^\infty x \, f_x(x) \, dx
\end{displaymath} \pause
} % End size

\begin{itemize}
     \item Provided $ \int_{-\infty}^\infty |x| \, f_x(x) \, dx < \infty$. \pause If the integral diverges, the expected value does not exist. 
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{}
%\framesubtitle{} 
{\LARGE
The expected value is the physical balance point. 
} % End size
\end{frame}

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

\begin{frame}
\frametitle{Sometimes the expected value does not exist}
\framesubtitle{Need $ \int_{-\infty}^\infty |x| \, f_x(x) \, dx < \infty$}  \pause
For the Cauchy distribution, $f(x) = \frac{1}{\pi(1+x^2)}$.  \pause
\begin{eqnarray*}
    E(|X|) & = & \int_{-\infty}^\infty |x| \, \frac{1}{\pi(1+x^2)} \, dx \\ \pause
     & = & 2 \int_0^\infty \frac{x}{\pi(1+x^2)} \, dx \\ \pause
     && u = 1+x^2, ~du = 2x \, dx \\  \pause
     & = & \frac{1}{\pi} \int_1^\infty \frac{1}{u} \, du \\ \pause
     & = & \ln u |_1^\infty \\ \pause
     & = & \infty - 0 \pause = \infty \pause
\end{eqnarray*} 
So to speak. \pause When we say an integral ``equals" infinity, we just mean it is unbounded above.
\end{frame}

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

\begin{frame}
\frametitle{Existence of expected values} \pause
%\framesubtitle{} 
\begin{itemize}
    \item If it is not mentioned in a general problem, existence of expected values is assumed. \pause
    \item Sometimes, the answer to a specific problem is ``Oops! The expected value dies not exist." \pause
    \item You never need to show existence unless you are explicitly asked to do so. \pause
    \item If you do need to deal with existence, Fubini's Theorem can help with multiple sums or integrals. \pause
    \begin{itemize}
        \item Part One says that if the integrand is positive, the answer is the same when you switch order of integration, even when the answer is ```$\infty$." \pause
        \item Part Two says that if the integral converges absolutely, you can switch order of integration. \pause For us, absolute convergence just means that the expected value exists.
    \end{itemize}
\end{itemize}
\end{frame}

% Now a break for some examples, called sample problems.

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

% Markov's inequalty is here in the text. 

\begin{frame}
\frametitle{The change of variables formula for expected value}
\framesubtitle{Theorems $A$ and $B$ in Chapter 4}  \pause 
Let $X$ be a random variable and $Y=g(X)$. \pause There are two ways to get $E(Y)$. \pause 
\vspace{3mm}
\begin{enumerate}
    \item Derive the distribution of $Y$ \pause and compute
\begin{displaymath}
    E(Y) = \int_{-\infty}^\infty y \, f_{_Y}(y) \, dy
\end{displaymath}  \pause
    \item Use the distribution of $X$ \pause 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[] $E\left(g(X)\right) = \sum_x g(x) p_{_X}(x)$ \pause
    \item[] 
    \item[] $E(g(\mathbf{X})) =  
\sum_{x_1} \cdots \sum_{x_p} g(x_1, \ldots, x_p) \, 
p_{_\mathbf{X}}(x_1, \ldots, x_p) $
\end{itemize}
\end{columns}
\end{frame}

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

\begin{frame}
\frametitle{Example: Let $Y=aX$. Find $E(Y)$.} \pause
%\framesubtitle{} 

\begin{eqnarray*}
    E(aX) & = & \sum_x ax \, p_{_X}(x) \\ \pause
          & = & a \, \sum_x x \, p_{_X}(x) \\ \pause
          & = & a \, E(X)  \pause
\end{eqnarray*} 
So $E(aX) = aE(X)$.
\end{frame}

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

\begin{frame}
\frametitle{Show that the expected value of a constant is the constant.} \pause
%\framesubtitle{} 

\begin{eqnarray*}
    E(a)  & = & \sum_x a  \, p_{_X}(x) \\ \pause
          & = & a \, \sum_x \, p_{_X}(x) \\ \pause
          & = & a  \cdot 1  \\ \pause
          & = & a  \pause
\end{eqnarray*} 
So $E(a) = a$.
\end{frame}

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

\begin{frame}
\frametitle{$E(X+Y) = E(X)+E(Y)$} \pause
%\framesubtitle{} 
\begin{columns}  
\column{1.2\textwidth} % Single wide column to use more of the margin
\begin{eqnarray*}
    E(X+Y) & = & \int_{-\infty}^\infty \int_{-\infty}^\infty (x+y) \, f_{xy}(x,y) \, dx \, dy \\ \pause
           & = & \int_{-\infty}^\infty \int_{-\infty}^\infty x \, f_{xy}(x,y) \, dx \, dy +
                 \int_{-\infty}^\infty \int_{-\infty}^\infty y \, f_{xy}(x,y) \, dx \, dy \\ \pause
           & = & E(X) + E(Y) 
\end{eqnarray*} 
\end{columns}
\end{frame}

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

\begin{frame}
\frametitle{Putting it together}
%\framesubtitle{} 
{\LARGE
\begin{displaymath}
    E(a + bX + cY) = a + b \, E(X) + c \, E(Y)
\end{displaymath} \pause
} % End size
And in fact, \pause
{\LARGE
\begin{displaymath}
    E\left(\sum_{i=1}^n a_iX_i \right) = \sum_{i=1}^n a_iE(X_i)
\end{displaymath} \pause
} % End size

You can move the expected value sign through summation signs and constants. \pause Expected value is a linear transformation. % E(X+Y) = E(X)+E(Y) and E(aX) = aE(X)
\end{frame}

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

\begin{frame}
\frametitle{$ E\left(\sum_{i=1}^n X_i \right) = \sum_{i=1}^n E(X_i)$, but in general} \pause
%\framesubtitle{} 
{\LARGE
\begin{displaymath}
    E(g(X)) \neq g(E(X))
\end{displaymath} \pause
} % End size
\vspace{6mm}

Unless $g(x)$ is a linear function. \pause So for example, \pause
\begin{itemize}
    \item[]  $E(\ln(X)) \neq \ln(E(X))$ \pause
    \item[]  $E(\frac{1}{X}) \neq \frac{1}{E(X)}$ \pause
    \item[]  $E(X^k) \neq \left( E(X)\right)^k$  \pause
\end{itemize} \vspace{3mm}
That is, the statements are not true in general. They might be true for some distributions.
\end{frame}

% Should I talk about expected values of indicators? I'm running short of time in 2018.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Variance}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Variance of a random variable $X$} \pause
% \framesubtitle{Definition}
Let $E(X) = \mu$ (The Greek letter ``mu"). \pause
{\LARGE
\begin{displaymath}
    Var(X) = E\left( (X-\mu)^2 \right)
\end{displaymath}  \pause
} % End size
\begin{itemize}
    \item The average (squared) difference from the average. \pause
    \item It's a measure of how spread out the distribution is. \pause
    \item Another measure of spread is the standard deviation, the square root of the variance. 
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Variance rules} \pause
%\framesubtitle{} 
\begin{itemize}
    \item[] $Var(a+bX) = b^2Var(X)$  \pause
    \item[]
    \item[] $Var(X) = E(X^2)-[E(X)]^2$ 
\end{itemize}
\end{frame}

% More sample questions here.

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

\begin{frame}
\frametitle{Famous Russian Inequalities} \pause
\framesubtitle{Very useful later} 
Because the variance is a measure of spread or dispersion, it places limits on how much probability can be out in the tails of a probability density of probability mass function. \pause To see this, we will \pause
\begin{itemize}
    \item Prove Markov's inequalty. \pause
    \item Use Markov's inequality to prove Chebyshev's inequality. \pause
    \item Look at some examples.
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Markov's inequality}\pause
%\framesubtitle{} 
Let $Y$ be a random variable with $P(Y \geq 0)=1$ \pause and $E(Y)=\mu$. \pause Then for any $t>0$, \pause
$P(Y \geq t) \leq E(Y)/t$. \pause \hspace{5mm} Proof: \pause
\begin{eqnarray*}
    E(Y) & = & \int_{-\infty}^\infty y f(y) \, dy \\ \pause
         & = &  \int_{-\infty}^t y f(y) \, dy + \int_t^\infty y f(y) \, dy \\ \pause
         & \geq & \int_t^\infty {\color{red}y} f(y) \, dy \\ \pause
         & \geq & \int_t^\infty {\color{red}t} f(y) \, dy \\ \pause
         & = & {\color{red}t} \int_t^\infty f(y) \, dy \\ \pause
         & = &  t \, P(Y \geq t) \pause
\end{eqnarray*} 
So $P(Y \geq t) \leq E(Y)/t$.
\end{frame}

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

\begin{frame}
\frametitle{Chebyshev's inequality} \pause
%\framesubtitle{} 
Let $X$ be a random variable with $E(X)=\mu$ and $Var(X)=\sigma^2$. \pause Then for any $k>0$, \pause

{\LARGE
\begin{displaymath}
    P(|X-\mu| \geq k\sigma) \leq \frac{1}{k^2}
\end{displaymath} \pause
} % End size

\begin{itemize}
    \item We are measuring distance from the mean \emph{in units of the standard deviation}. \pause
    \item The probability of observing a value of $X$ more than 2 standard deviations away from the mean cannot be more than one fourth. \pause
    \item This is true for \emph{any} random variable that has a standard deviation. \pause
    \item For the normal distribution, $ P(|X-\mu| \geq 2\sigma) \approx .0455 \pause < \frac{1}{4}$. 
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Proof of Chebyshev's inequality}
\framesubtitle{Markov says $P(Y \geq t) \leq E(Y)/t$}  \pause
In Markov's inequality, let $Y=(X-\mu)^2$ and $t = k^2\sigma^2$. \pause Then \pause

\begin{eqnarray*}
    P\left((X-\mu)^2 \geq k^2\sigma^2\right) & \leq & \frac{E\left((X-\mu)^2\right)}{k^2\sigma^2} \\ \pause
                                  & = & \frac{\sigma^2}{k^2\sigma^2} \pause = \frac{1}{k^2}\\ \pause 
\end{eqnarray*} 
So $P(|X-\mu| \geq k\sigma) \leq \frac{1}{k^2}$. 
\end{frame}

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

\begin{frame}
\frametitle{Example}
\framesubtitle{Chebyshev says $P(|X-\mu| \geq k\sigma) \leq \frac{1}{k^2}$}  \pause
Let $X$ have density $f(x)=e^{-x}$ for $x \geq 0$ (standard exponential). \pause We know $E(X) = Var(X) = 1$. \pause Find $P(|X-\mu| \geq 3\sigma)$ and compare with Chebyshev's inequality. \pause \vspace{3mm}

$F(x) = 1-e^{-x}$ for $x \geq 0$, so \pause
\begin{eqnarray*}
    P(|X-\mu| \geq 3\sigma) & = & P(X<-2) + P(X > 4) \\ \pause
                              & = & 1 - F(4) \\ \pause
                              & = & e^{-4} \\ \pause
                              & \approx & 0.01831564 \\ \pause
\end{eqnarray*}

Compared to $\frac{1}{3^2} = \frac{1}{9} = 0.11$.
\end{frame}

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

\begin{frame}
\frametitle{Conditional Expectation}
\framesubtitle{The idea}  \pause
Consider jointly distributed random variables $X$ and $Y$. \pause
\begin{itemize}
    \item For each possible value of $X$, there is a conditional distribution of $Y$.  \pause
    \item Each conditional distribution has an expected value (population mean). \pause
    \item If you could estimate $E(Y|X=x)$, it would be a good way to predict $Y$ from $X$. \pause
    \item Estimation comes later (in STA260).
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Definition of Conditional Expectation} \pause
%framesubtitle{} 
If $X$ and $Y$ are discrete, the conditional expected value of $Y$ given $X$ is \pause

{\LARGE
\begin{displaymath}
    E(Y|X=x) = \sum_y y \, p_{y|x}(y|x)
\end{displaymath} \pause
} % End size
If $X$ and $Y$ are continuous, \pause
{\LARGE
\begin{displaymath}
    E(Y|X=x) = \int_{-\infty}^\infty y \, f_{y|x}(y|x) \, dy
\end{displaymath} 
} % End size
\end{frame}

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

\begin{frame}
\frametitle{Double Expectation: $ E(Y) = E[E(Y|X)]$}
\framesubtitle{Theorem A on page 149}  \pause
To make sense of this, note  \pause
\begin{itemize}
    \item While $E(Y|X=x) = \int_{-\infty}^\infty y \, f_{y|x}(y|x) \, dy$ is a real-valued function of $x$,  \pause
    \item $E(Y|X)$ is a random variable, a function of the random variable $X$. \pause
    \item $E(Y|X) = g(X) = \int_{-\infty}^\infty y \, f_{y|x}(y|X) \, dy$. \pause
    \item So that in $E[E(Y|X)] = E[g(X)]$, the outer expected value is with respect to the probability distribution of $X$. \pause
\end{itemize}
\begin{eqnarray*}
    E[E(Y|X)] & = &  E[g(X)] \\ \pause
              & = & \int_{-\infty}^\infty g(x) \, f_x(x) \, dx \\ \pause
              & = & \int_{-\infty}^\infty \left( 
                    \int_{-\infty}^\infty y \, f_{y|x}(y|x) \, dy
                    \right) \, f_x(x) \, dx
\end{eqnarray*}
\end{frame}

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

\begin{frame}
\frametitle{Proof of the double expectation formula}
\framesubtitle{Book calls it the ``Law of Total Expectation"}  \pause
\begin{eqnarray*}
    E[E(Y|X)] \pause
    & = & \int_{-\infty}^\infty \left( 
          \int_{-\infty}^\infty y \, f_{y|x}(y|x) \, dy
          \right) \, f_x(x) \, dx \\ \pause
    & = &  \int_{-\infty}^\infty 
          \int_{-\infty}^\infty y \, \frac{f_{xy}(x,y)}{f_x(x)} \, dy
           \, f_x(x) \, dx \\ \pause 
    & = &  \int_{-\infty}^\infty 
          \int_{-\infty}^\infty y \, f_{xy}(x,y) \, dy \, dx  \\ \pause
    & = &   E(Y)
\end{eqnarray*} 
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Covariance} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Definition of Covariance}  \pause
% \framesubtitle{}
Let $X$ and $Y$ be jointly distributed random variables with $E(X)=\mu_x$ and $E(Y)=\mu_y$. The \emph{covariance} between $X$ and $Y$ is \pause

{\LARGE
\begin{displaymath}
    Cov(X,Y) = E[(X-\mu_x)(Y-\mu_y)]
\end{displaymath} \pause
} % End size
\begin{itemize}
    \item If values of $X$ that are above average tend to go woth values of $Y$ that are above average, the covariance will be positive. \pause
    \item If values of $X$ that are above average tend to go woth values of $Y$ that are \emph{below} average, the covariance will be negative. \pause
    \item Covariance means they vary together. \pause
    \item You could think of $Var(X)=E[(X-\mu_x)^2]$ as $Cov(X,X)$.
\end{itemize}
\end{frame}

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

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

\begin{frame}
\frametitle{Properties of Covariance} \pause
%\framesubtitle{} 
\begin{itemize}
    \item[] $Cov(X,Y) = E(XY)-E(X)E(Y)$ \pause
    \item[] If $X$ and $Y$ are independent, $Cov(X,Y)= 0$ \pause
    \item[] If $Cov(X,Y)= 0$, it does \emph{not} follow that $X$ and $Y$ are independent. \pause
    \item[] $Cov(a+X,b+Y) \pause = Cov(X,Y)$ \pause
    \item[] $Cov(aX,bY) \pause = abCov(X,Y)$ \pause % Prove as HW
    \item[] $Cov(X,Y+Z) \pause = Cov(X,Y) + Cov(X,Z)$ \pause
    % \item[] $Cov(aW+bX,cY+dZ) =$ \pause \\ ~~~ $acCov(W,Y) + adCov(W,Z) + bcCov(X,Y) + bdCov(X,Z)$ \pause % Maybe just put this in HW.
    \item[] $Var(aX+bY) = a^2Var(X)+b^2Var(Y) + 2abCov(X,Y)$  \pause
%    \item[] 
     \item[] If $X_1, \ldots, X_n$ are ind. \pause 
     $Var(\sum_{i=1}^n X_i) = \sum_{i=1}^n Var(X_i)$ \pause
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Correlation}
%\framesubtitle{Population correlation, Greek letter ``rho"} \pause
{\LARGE
\begin{displaymath}
    Corr(X,Y) = \rho = \frac{Cov(X,Y)}{\sqrt{Var(X) \, Var(Y)}}
\end{displaymath} \pause
} % End size
\begin{itemize}
    \item $-1 \leq \rho \leq 1$ \pause
    \item  Scale free: $Corr(aX,bY) = Corr(X,Y)$
\end{itemize}
\end{frame}

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\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/256f18} {\small\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/256f18}}
\end{frame}
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\end{document}


and $Var(Y) = \sigma^2$

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

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

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
\includegraphics[width=2in]{BivariateNormal}
\end{center}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{}
%\framesubtitle{} 
{\LARGE
\begin{eqnarray*}
    m_1 & = & a + b \\
    m_2 & = & c + d 
\end{eqnarray*} 
} % End size
\end{frame}

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

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