lecture10.tex

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\title{Mathematical Biostatistics Boot Camp 2: 
Lecture 10, Case Control Data}
\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}

%\section{Table of contents}
\frame{
  \frametitle{Table of contents}
  \tableofcontents
}


\section{Case-control methods}
\begin{frame}\frametitle{Case-control methods}
\begin{center}
\ttfamily
  \begin{tabular}{llll}
& \multicolumn{2}{c}{Lung cancer} & \\ \cline{2-3}
Smoker & Cases & Controls & Total \\ \hline
Yes & 688 & 650 & 1338 \\
No  & 21  & 59  &   80 \\ \hline
    & 709 & 709 & 1418 
  \end{tabular}
\normalfont
\end{center}
\begin{itemize}
\item Case status obtained from records 
\item Cannot estimate $P(\mbox{Case} ~|~ \mbox{Smoker})$
\item Can estimate $P(\mbox{Smoker} ~|~ \mbox{Case})$
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Continued}
\begin{itemize}
\item Can estimate odds ratio b/c 
\begin{eqnarray*}
& &\frac{Odds(\mbox{case} ~|~ \mbox{smoker})}{Odds(\mbox{case} ~|~ \mbox{smoker}^c)} \\ \\
& = & \frac{Odds(\mbox{smoker} ~|~ \mbox{case})}{Odds(\mbox{smoker} ~|~ \mbox{case}^c)}\\
\end{eqnarray*}
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Proof}
$C$ - case, $S$ - smoker

\begin{eqnarray*}
& &\frac{Odds(\mbox{case} ~|~ \mbox{smoker})}{Odds(\mbox{case} ~|~ \mbox{smoker}^c)}\\ \\
& = & \frac{P(C ~ | ~ S) / P(\bar{C} ~|~ S)}{P(C ~|~ \bar{S})/P(\bar{C} ~|~ \bar{S})} \\ \\
& = & \frac{P(C, S) / P(\bar{C}, S)}{P(C, \bar{S})/P(\bar{C}, \bar{S})} \\ \\
& = & \frac{P(C, S) P(\bar{C}, \bar{S})}{P(C,\bar{S}) P(\bar{C}, S)}
\end{eqnarray*}
Exchange $C$ and $S$ and the result is obtained
\end{frame}

\begin{frame}\frametitle{Notes}
\begin{itemize}
\item Sample $OR$ is $\frac{n_{11}n_{22}}{n_{12}n_{21}}$
\item Sample $OR$ is unchanged if a row or column is multiplied by a constant
\item Invariant to transposing
\item Is related to $RR$ 
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Notes continued}
 \begin{eqnarray*}
OR & = & \frac{P(S ~|~ C)/P(\bar{S} ~|~ C)}{P(S ~|~ \bar{C})/P(\bar{S} ~|~ \bar{C})} \\ \\
   & = & \frac{P(C ~|~ S)/P(\bar{C} ~|~ S)}{P(C ~|~ \bar{S})/P(\bar{C} ~|~ \bar{S})} \\ \\
   & = & \frac{P(C ~|~ S)}{P(C ~|~ \bar{S})}
         \frac{P(\bar{C} ~|~ \bar{S})}{P(\bar{C} ~|~ S)} \\ \\
   & = & RR \times \frac{1 - P(C ~|~ \bar{S})}{1 - P(C ~|~ S)}
  \end{eqnarray*}
\begin{itemize}
\item $OR$ approximate $RR$ if $P(C ~|~ \bar{S})$ and $P(C ~|~ S)$ are small
(or if they are nearly equal)
\end{itemize}
\end{frame}

\section{Rare disease assumption}
\begin{frame}\frametitle{Rare disease assumption}
\begin{center}
\ttfamily
  \begin{tabular}{llll}
&\multicolumn{2}{c}{Disease} & \\ \cline{2-3}
Exposure & Yes  & No & Total \\ \hline
Yes      &   9  &    1 & 10   \\
No       &   1  &  999 & 1000 \\ \hline
         &  10  & 1000 & 1010 
  \end{tabular}
\normalfont
\end{center}
\begin{itemize}
\item Cross-sectional data 
\item $\hat{P(D)} = 10 / 1010 \approx .01$
\item $\hat{OR} = (9 \times 999) / (1 \times 1) = 8991$
\item $\hat{RR} = (9 / 10) / (1 / 1000) = 900$
\item $D$ is rare in the sample
\item $D$ is not rare among the exposed
\end{itemize}
\end{frame}


\begin{frame}\frametitle{Notes}
\begin{itemize}
\item $OR = 1$ implies no association
\item $OR > 1$ positive association
\item $OR < 1$ negative association
\item For retrospective CC studies, $OR$ 
  can be interpreted prospectively
\item For diseases that are rare among \\
  the cases and controls, the $OR$ \\ approximates
  the $RR$
\item Delta method SE for log $OR$ is
$$
\sqrt{\frac{1}{n_{11}} + \frac{1}{n_{12}} + \frac{1}{n_{21}} + \frac{1}{n_{22}}}
$$
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Example}
\begin{center}
\ttfamily
  \begin{tabular}{llll}
& \multicolumn{2}{c}{Lung cancer} & \\ \cline{2-3}
Smoker & Cases & Controls & Total \\ \hline
Yes & 688 & 650 & 1338 \\
No  & 21  & 59  &   80 \\ \hline
    & 709 & 709 & 1418 
  \end{tabular}
\normalfont
\end{center}
~\footnote{Data from Agresti, Categorical Data Analysis, second edition}
\begin{itemize}
\item $\hat{OR} = \frac{688 \times 59}{21 \times 650} = 3.0$
\item $\hat{SE}_{\log \hat{OR}} = \sqrt{\frac{1}{688} + \frac{1}{650} + \frac{1}{21} + \frac{1}{59}} = .26$
\item $CI = \log(3.0) \pm 1.96 \times .26 = [.59, 1.61]$
\item The estimated odds of lung cancer for smokers are $3$ times
  that of the odds for non-smokers with an interval of 
  $[\exp(.59), \exp(1.61)] = [1.80, 5.00]$
\end{itemize}
\end{frame}

\section{Exact inference for the odds ratio}
\begin{frame}\frametitle{Exact inference for the OR}
\begin{center}
\ttfamily
  \begin{tabular}{llll}
& \multicolumn{2}{c}{Lung cancer} & \\ \cline{2-3}
Smoker & Cases & Controls & Total \\ \hline
Yes & 688 & 650 & 1338 \\
No  & 21  & 59  &   80 \\ \hline
    & 709 & 709 & 1418 
  \end{tabular}
\normalfont
\end{center}\ \\
\begin{itemize}
\item $X$ the number of smokers for the cases 
\item $Y$ the number of smokers for the controls
\item Calculate an exact CI for the odds ratio
\item Have to eliminate a nuisance parameter
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Notation}
\begin{itemize}
\item $\logit(p) = \log\{p / (1 - p)\}$ is the {\bf log-odds}
\item Differences in $\logit s$ are log-odds {\em ratios}
\item $\logit\{P(\mbox{Smoker} ~|~ \mbox{Case})\} = \delta$
  \begin{itemize}
  \item   $P(\mbox{Smoker} ~|~ \mbox{Case}) = e^\delta /(1 + e^\delta)$ 
  \end{itemize}
\item $\logit\{P(\mbox{Smoker} ~|~ \mbox{Control})\} = \delta + \theta$
  \begin{itemize}
  \item  $P(\mbox{Smoker} ~|~ \mbox{Control}) = e^{\delta + \theta} / (1 + e^{\delta+\theta})$   \end{itemize}
\item $\theta$ is the log-odds ratio
\item $\delta$ is the nuisance parameter
\end{itemize}
\end{frame}

\begin{frame}\frametitle{Notation}
\begin{itemize}
\item $X$ is binomial with $n_1$ trials and success probability $e^\delta /(1 + e^\delta)$
\item $Y$ is binomial with $n_2$ trials and success probability $e^{\delta + \theta} / (1 + e^{\delta+\theta})$
\end{itemize}
\begin{eqnarray*}
P(X = x) & = &
\left(\begin{array}{c} n_1 \\ x \end{array}\right)\left\{\frac{e^\delta}{1 + e^\delta}\right\}^x
\left\{\frac{1}{1 + e^\delta}\right\}^{n_1-x} \\
& = & \left(\begin{array}{c} n_1 \\ x \end{array}\right)e^{x\delta}
\left\{\frac{1}{1 + e^\delta}\right\}^{n_1}
\end{eqnarray*}
\end{frame}

\begin{frame}
$$
P(X = x) = \left(\begin{array}{c} n_1 \\ x \end{array}\right)e^{x\delta}
\left\{\frac{1}{1 + e^\delta}\right\}^{n_1}
$$ \ \\
$$
P(Y = z-x) =  \left(\begin{array}{c} n_2 \\ z-x \end{array}\right)e^{(z-x)\delta + (z-x)\theta}
\left\{\frac{1}{1 + e^{\delta+\theta}}\right\}^{n_2}
$$\ \\ 
$$
P(X+Y = z) = \sum_u P(X = u)P(Y = z - u)
$$ \ \\
$$
P(X = x ~|~ X+Y = z) = \frac{P(X = x)P(Y = z - x)}{\sum_u P(X = u) P(Y = z-u)}
$$
\end{frame}

\begin{frame}\frametitle{Non-central hypergeometric distribution}
$$
P(X = x ~|~ X+Y = z; \theta)
= \frac{
\left(\begin{array}{c} n_1 \\ x \end{array}\right)
\left(\begin{array}{c} n_2 \\ z-x \end{array}\right) e^{x\theta}
}{
\sum_u 
\left(\begin{array}{c} n_1 \\ u \end{array}\right)
\left(\begin{array}{c} n_2 \\ z-u \end{array}\right) e^{u\theta}
}
$$
\begin{itemize}
\item $\theta$ is the log odds ratio 
\item This distribution is used to calculate exact hypothesis tests for
  $H_0:\theta = \theta_0$
\item Inverting exact tests yields exact confidence intervals for the odds
  ratio
\item Simplifies to the hypergeometric distribution for $\theta = 0$
\end{itemize} 
\end{frame}

\end{document}