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\begin{center}   
{\Large \textbf{STA 302 Formulas}}\\   
\vspace{1 mm}
\end{center}


\noindent
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$M_y(t) = E(e^{yt})$ & ~~~~~ & $M_{ay}(t) = M_y(at)$ \\
$M_{y+a}(t) = e^{at}M_y(t)$ & ~~~~~ & 
$M_{\sum_{i=1}^n y_i}(t) = \prod_{i=1}^n M_{y_i}(t)$
\\
$y \sim N(\mu,\sigma^2)$ means $M_y(t) = e^{\mu t + \frac{1}{2}\sigma^2t^2}$
& ~~~~~ & 
$W \sim \chi^2(\nu)$ means $M_W(t) = (1-2t)^{-\nu/2}$
\\
If $W_1, \ldots, W_n \stackrel{ind}{\sim} \chi^2(\nu_i)$, then 
$\sum_{i=1}^n W_i \sim \chi^2(\sum_{i=1}^n \nu_i)$
& ~~~~~ &
If $Z \sim N(0,1)$ then $Z^2 \sim \chi^2(1)$
\\
\multicolumn{3}{l}{If  $W=W_1+W_2$ with $W_1$ and $W_2$ independent, $W\sim\chi^2(\nu_1+\nu_2)$, $W_2\sim\chi^2(\nu_2)$ then $W_1\sim\chi^2(\nu_1)$} \\ 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\parbox{7 cm}{Columns of  $\mathbf{A}$ \emph{linearly dependent} means there is a vector $\mathbf{v} \neq \mathbf{0}$ with $\mathbf{Av} = \mathbf{0}$.} & ~~~~~ &
\parbox{7 cm}{Columns of  $\mathbf{A}$ \emph{linearly independent} means that $\mathbf{Av} = \mathbf{0}$ implies $\mathbf{v} = \mathbf{0}$.}
\\
\multicolumn{3}{l}{$\mathbf{A}$ \emph{positive definite} means  $\mathbf{v}^\prime \mathbf{Av} > 0$ for all vectors $\mathbf{v} \neq \mathbf{0}$.} \\
$\boldsymbol{\Sigma} = \mathbf{CD} \mathbf{C}^\prime$
& ~~~~~ & 
$\boldsymbol{\Sigma}^{-1} = \mathbf{C} \mathbf{D}^{-1} \mathbf{C}^\prime$
\\
$\boldsymbol{\Sigma}^{1/2} = \mathbf{CD}^{1/2} \mathbf{C}^\prime$
& ~~~~~ &
$\boldsymbol{\Sigma}^{-1/2} = \mathbf{CD}^{-1/2} \mathbf{C}^\prime$
\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$cov(\mathbf{y}) = 
E\left\{(\mathbf{y}-\boldsymbol{\mu}_y)(\mathbf{y}-\boldsymbol{\mu}_y)^\prime\right\}$ 
& ~~~~~ & 
$C(\mathbf{y,t}) = E\left\{ (\mathbf{y}-\boldsymbol{\mu}_y)
                             (\mathbf{t}-\boldsymbol{\mu}_t)^\prime\right\}$
\\
$cov(\mathbf{y}) = E\{\mathbf{yy}^\prime\} - \boldsymbol{\mu}_y\boldsymbol{\mu}_y^\prime$
& ~~~~~ &
$cov(\mathbf{Ay}) = \mathbf{A}cov(\mathbf{y}) \mathbf{A}^\prime$
\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$M_{\mathbf{y}}(\mathbf{t}) = E(e^{\mathbf{t}^\prime\mathbf{y}})$ 
& ~~~~~ & 
$M_{\mathbf{Ay}}(\mathbf{t}) = M_{\mathbf{y}}(\mathbf{A}^\prime\mathbf{t})$
\\
$M_{\mathbf{y}+\mathbf{c}}(\mathbf{t}) = e^{\mathbf{t}^\prime\mathbf{c}} M_{\mathbf{y}}(\mathbf{t})$
 & ~~~~~ & 
$\mathbf{y} \sim N_p(\boldsymbol{\mu}, \boldsymbol{\Sigma})$ means $M_{\mathbf{y}}(\mathbf{t}) = e^{\mathbf{t}^\prime\boldsymbol{\mu} + \frac{1}{2} \mathbf{t}^\prime \boldsymbol{\Sigma} \mathbf{t}}$
\\
\multicolumn{3}{l}{$\mathbf{y}_1$ and $\mathbf{y}_2$ are independent if and only if
$M_{(\mathbf{y}_1,\mathbf{y}_2)}\left(\mathbf{t}_1,\mathbf{t}_2\right)
= M_{\mathbf{y}_1}(\mathbf{t}_1) M_{\mathbf{y}_2}(\mathbf{t}_2)$} \\
If $\mathbf{y} \sim N_p(\boldsymbol{\mu}, \boldsymbol{\Sigma})$, then $\mathbf{Ay} \sim N_q(\mathbf{A}\boldsymbol{\mu}, \mathbf{A}\boldsymbol{\Sigma}\mathbf{A}^\prime)$,
 & ~~~~~ &
and $W = (\mathbf{y}-\boldsymbol{\mu})^\prime
           \boldsymbol{\Sigma}^{-1}(\mathbf{y}-\boldsymbol{\mu}) \sim \chi^2(p)$
\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$y_i = \beta_0 + \beta_1 x_{i1} + \cdots + \beta_k x_{ik} + \epsilon_i$
& ~~~~~ & 
$\epsilon_1, \ldots, \epsilon_n$ independent $N(0,\sigma^2)$
\\
$\mathbf{y} = \mathbf{X} \boldsymbol{\beta} + \boldsymbol{\epsilon}$ with $\boldsymbol{\epsilon} \sim N_n(\mathbf{0},\sigma^2\mathbf{I}_n)$
& ~~~~~ &
$\widehat{\boldsymbol{\beta}} = (\mathbf{X}^\prime \mathbf{X})^{-1} 
                   \mathbf{X}^\prime \mathbf{y} \sim N_{k+1}(\boldsymbol{\beta}, \sigma^2(\mathbf{X}^\prime \mathbf{X})^{-1})$
\\
$\widehat{\mathbf{y}} = \mathbf{X}\widehat{\boldsymbol{\beta}} = \mathbf{Hy}$, where 
$\mathbf{H} = \mathbf{X}(\mathbf{X}^\prime \mathbf{X})^{-1} 
                   \mathbf{X}^\prime $
& ~~~~~ &
$\widehat{\boldsymbol{\epsilon}} = \mathbf{y} - \widehat{\mathbf{y}} = (\mathbf{I}-\mathbf{H})\mathbf{y}$ 
\\
$\widehat{\boldsymbol{\beta}}$ and $\widehat{\boldsymbol{\epsilon}}$ are independent under normality.
& ~~~~~ &
$\frac{SSE}{\sigma^2} = \frac{\hat{\boldsymbol{\epsilon}}^\prime \hat{\boldsymbol{\epsilon}}}{\sigma^2}  \sim \chi^2(n-k-1)$ 
\\
$\sum_{i=1}^n(y_i-\overline{y})^2 = \sum_{i=1}^n(y_i-\widehat{y}_i)^2 + \sum_{i=1}^n(\widehat{y}_i-\overline{y})^2$
& ~~~~~ &
$SST=SSE+SSR$ and $R^2 = \frac{SSR}{SST}$
\\
$T = \frac{Z}{\sqrt{W/\nu}} \sim t(\nu)$
& ~~~~~ &
$F = \frac{W_1/\nu_1}{W_2/\nu_2} \sim F(\nu_1,\nu_2)$
\\
$T = \frac{\mathbf{a}^\prime \widehat{\boldsymbol{\beta}}-\mathbf{a}^\prime \boldsymbol{\beta}}
             {\sqrt{MSE \, \mathbf{a}^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{a}}} \sim t(n-k-1)$
& ~~~~~ &
$T = \frac{y_0-\mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}}
             {\sqrt{MSE \, (1+\mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}} \sim t(n-k-1)$
\\
\multicolumn{3}{l}{$F = \frac{(\mathbf{C}\widehat{\boldsymbol{\beta}}-\mathbf{t})^\prime
            (\mathbf{C}(\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{C}^\prime)^{-1}
            (\mathbf{C}\widehat{\boldsymbol{\beta}}-\mathbf{t})}
           {q \, MSE} 
           = \frac{SSR(full)-SSR(reduced)}{q \, MSE} \sim F(q,n-k-1)$, 
           where $MSE = \frac{SSE}{n-k-1}$} 
\\
\parbox{7 cm}{$ F = \left( \frac{p}{1-p} \right) \left( \frac{n-k-1}{q}  \right) ~\Leftrightarrow~ p = \frac{qF}{qF+n-k-1}$, where $p = \frac{R^2(full)-R^2(reduced)}{1-R^2(reduced)}$}
& ~~~~~ &
$r_{xy} = \frac{\sum_{i=1}^n (x_i-\overline{x})(y_i-\overline{y})}
               {\sqrt{\sum_{i=1}^n (x_i-\overline{x})^2} \sqrt{\sum_{i=1}^n (y_i-\overline{y})^2}}$
\\
\end{tabular}
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\noindent
{\small
This formula sheet 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:
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\begin{center}
\href{http://www.utstat.toronto.edu/~brunner/oldclass/302f15} {\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/302f15}}
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\multicolumn{3}{l}{Columns of  $\mathbf{A}$ \emph{linearly independent} means that $\mathbf{Av} = \mathbf{0}$ implies $\mathbf{v} = \mathbf{0}$.} \\ 

\multicolumn{3}{l}{Columns of  $\mathbf{A}$ \emph{linearly dependent} means there is a vector $\mathbf{v} \neq \mathbf{0}$ with $\mathbf{Av} = \mathbf{0}$.} \\


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\multicolumn{3}{l}{Columns of  $\mathbf{A}$ \emph{linearly dependent} means there is a vector $\mathbf{v} \neq \mathbf{0}$ with $\mathbf{Av} = \mathbf{0}$.} \\
\multicolumn{3}{l}{Columns of  $\mathbf{A}$ \emph{linearly independent} means that $\mathbf{Av} = \mathbf{0}$ implies $\mathbf{v} = \mathbf{0}$.} \\ 
$\boldsymbol{\Sigma} = \mathbf{CD} \mathbf{C}^\prime$
& ~~~~~ & 
$\boldsymbol{\Sigma}^{-1} = \mathbf{C} \mathbf{D}^{-1} \mathbf{C}^\prime$
\\
$\boldsymbol{\Sigma}^{1/2} = \mathbf{CD}^{1/2} \mathbf{C}^\prime$
& ~~~~~ &
$\boldsymbol{\Sigma}^{-1/2} = \mathbf{CD}^{-1/2} \mathbf{C}^\prime$
\\

\noindent
The vectors $\mathbf{x}_1, \ldots, \mathbf{x}_p$ are \emph{linearly dependent} if there is a set of scalars $a_1, \ldots, a_p$, not all zero, with $a_1 \mathbf{x}_1 + a_2 \mathbf{x}_2 + \cdots + a_p \mathbf{x}_p = \mathbf{0}$.

\vspace{2mm}

\noindent
The vectors $\mathbf{x}_1, \ldots, \mathbf{x}_p$ are \emph{linearly independent} if  $a_1 \mathbf{x}_1 + a_2 \mathbf{x}_2 + \cdots + a_p \mathbf{x}_p = \mathbf{0}$ implies $a_1 = \cdots = a_p = 0$.


\\
$ \log\left(\frac{\pi_i}{1-\pi_i} \right) = 
    \beta_0 + \beta_1 x_{i,1} + \ldots + \beta_k x_{i,k}$
& ~~~~~ &
   $\pi_i  =  \frac{e^{\beta_0 + \beta_1 x_{i,1} + \ldots + \beta_k x_{i,k}}}
                    {1+e^{\beta_0 + \beta_1 x_{i,1} + \ldots + \beta_k x_{i,k}}}$
