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For each SNP the Cochrane-Armitage trend-test statistic is computed from a 3x2 contingency table

  Cases Controls Row totals
AA $N_{11}$ $N_{12}$ $N_{1*}$
AB $N_{21}$ $N_{22}$ $N_{2*}$
BB $N_{31}$ $N_{32}$ $N_{3*}$
Column totals $N_{*1}$ $N_{*2}$ $N$

$N_{11}$ ($N_{12}$) and $N_{21}$ ($N_{22}$) and $N_{31}$ ($N_{32}$) are the total number of occurrences of genotype AA, AB and BB, respectively, for the cases (controls), and

\begin{eqnarray*}
N_{*1} &=& N_{11} + N_{21} + N_{31},\\
N_{*2} &=& N_{12} + N_...
... \\
N &=& N_{11} + N_{12} + N_{21} + N_{22} + N_{31} + N_{32}.
\end{eqnarray*}

The Cochrane-Armitage trend-test statistic with weights $(x_1, x_2, x_3)$ is defined as

\begin{eqnarray*}
\frac{ (\frac{1}{N} \cdot
(x_1\cdot( N_{*2}\cdot N_{11} - N_...
...*})
-( x_1\cdot N_{1*} +x_2\cdot N_{2*} + x_3\cdot N_{3*})^2)},
\end{eqnarray*}

The MAX statistics is equal to the maximum of the three Cochrane-Armitage trend-test statistic obtained when assuming a recessive ( $(x_1, x_2, x_3)=(0,0,1)$) model, a dominant model ( $(x_1, x_2, x_3)=(0,1,1)$) and an additive model ( $(x_1, x_2, x_3)=(0,1,2)$). When assuming an additive model( $(x_1, x_2, x_3)=(0,1,2)$), the Cochrane-Armitage trend-test statistic is equal to

\begin{eqnarray*}
\frac{N \cdot ( N\cdot E_{1} - N_{*1}\cdot E_{2} )^2}{N_{*1} \cdot ( N - N_{*1} ) \cdot (N\cdot E_{3}-(E_{2})^2)},
\end{eqnarray*}

where $E_{1} &=& N_{21} + 2\cdot N_{31}$, $E_{2} &=& N_{2*} + 2\cdot N_{3*}$ and $E_{3} &=& N_{2*} + 4\cdot N_{3*}$.



Marit Holden 2008-08-07