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\noindent
13 September 2007 \\
 {\bf Estimating an Odds Ratio}  \\


 
\noindent
Eric Rasmusen, erasmuse@Indiana.edu, 
 http://www.rasmusen.org/x/2007/oddsratio.pdf 

  

 Let    $n(radon)$  denote the probability someone was exposed to radon and $n(unexposed) = 1- n(radon)$ be the probability he was not. Let $P(cancer|radon) $ be the probability someone gets cancer after having been exposed to radon and   $P(healthy|radon) $  be the probability he does not get cancer, and let     $P(cancer|unexposed) $  and  $P(healthy|unexposed) $ be the equivalents for the unexposed.
Finally, let $\theta$  and $\gamma$ be the probability that  cancer and helathy observations are sampled. Then the 
  expected proportions in the sample  are as shown in this table: 

 \bigskip

\begin{tabular} {lcc}
  & Radon & Unexposed  \\
  &  &      \\
\cline{2-3}
  &  \multicolumn{1}{|c|}{ }  & \multicolumn{1}{|c|}{ } \\
 Cancer  &\multicolumn{1}{|c|}{$ \theta P(cancer|radon) \cdot n(radon)$}  &\multicolumn{1}{|c|}{$\theta P(cancer|unexposed)  \cdot
n(unexposed) $  }    \\
  &     \multicolumn{1}{|c|}{ }  & \multicolumn{1}{|c|}{ } \\
\cline{2-3}
  &      \multicolumn{1}{|c|}{ }  & \multicolumn{1}{|c|}{ } \\
Healthy &\multicolumn{1}{|c|}{$\gamma P(healthy|radon)\cdot n(radon) $   } &\multicolumn{1}{|c|}{$\gamma   P(healthy|unexposed)      
\cdot n(unexposed)     $ }  \\
  &       \multicolumn{1}{|c|}{ }  & \multicolumn{1}{|c|}{ } \\
\cline{2-3}
\end{tabular}
 
\bigskip

The odds ratio is  the ratio of the odds of getting cancer when exposed tor adon to the odds of getting cancer when unexposed: 
\begin{equation} \label{odds1}
 {\rm Odds \;Ratio} = \frac{ \theta P(cancer|radon) \cdot n(radon)/[\gamma P(healthy|radon)\cdot n(radon) ]    }{\theta P(cancer|unexposed)  \cdot
n(unexposed)/[ \gamma   P(healthy|unexposed)  \cdot    
\cdot n(unexposed) ]     },
\end{equation}
which can be simplified thus: 
\begin{equation} \label{odds2}
 {\rm Odds \;Ratio} = \frac{   P(cancer|radon)  /   P(healthy|radon)     }{  P(cancer|unexposed)   
 /      P(healthy|unexposed)       }
\end{equation}

The sampling probabilities $\gamma$ and $\theta$ have cancelled out. Thus, even if we sample cancer observations more heavily than healthy  observations, we still will get the same expected value of the odds ratio. 

  This is true regardless of the functional form of the distribution of the disturbances. They can be normal, or logistic, or uniform. But it is especially noteworthy for the logistic distribution, because for that distribution, the most important parameter is $\beta$, which is exactly the odds ratio. Thus, the value computed using equation (\ref{odds1}) is an unbiased estimate of $\beta$. If the disturbances had a normal distribution, on the other hand, we could still use equation (\ref{odds1}) as an unbiased estimate of the odds ratio, but it would be a complicated function of the $\beta$ and $\sigma$ parameters and we could not recover parameter estimates from our observed values.  

   
 

The odds ratio is not useful in itself. It is hard to understand. But it is an approximation of something more interesting: the relative risk. The relative risk is the ratio of the probability of getting cancer if radon-exposed  to the probability of getting cancer if unexposed, and if cancer is rare whether a person is exposed or not ($P(healthy|radon) \approx 1 \approx P(healthy|unexposed)$) then the odds ratio  is approximately equal to the relative risk:
\begin{equation} \label{odds2}
 {\rm Relative\;Risk}  =\frac{   P(cancer|radon)       }{  P(cancer|unexposed)   
        }  \approx  \frac{   P(cancer|radon)  /   P(healthy|radon)     }{  P(cancer|unexposed)   
 /      P(healthy|unexposed)       } = {\rm Odds \;Ratio}  
\end{equation}

 


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