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F Score Approximation

An expansion of the basic definition 4 is sought for small values of $p$, which correspond to large values of $F$. This is acheived in a very similar way to the T score approximation in the previous section. Once again integration by parts is used to derive a recurrence relation. Firstly, a change of variable in equation 4 is required to get it into a more useful form. Setting $y=\frac{D_2}{D_2+D_1w}$ and combining equations 4 and  5 gives:

\begin{displaymath}
p = \frac{(\frac{D_1}{D_2})^{\frac{D_1}{2}}}{\beta(\frac{D_2...
...{D_2} \right)^{-\frac{D_1+D_2}{2}}w^{-(1-\frac{D_1}{2})} \, dw
\end{displaymath} (41)

Now, consider the integral:

\begin{displaymath}
I_{m,n} = \int_F^{\infty} (1 + a w)^{-m} w^{-n} \, dw
\end{displaymath} (42)

Using the fact that $(1 + a w)^{-m+1} = (1 + a w) (1 + a w)^{-m}$ gives:
\begin{displaymath}
I_{m-1,n} = I_{m,n} + a I_{m,n-1}
\end{displaymath} (43)

and differentiating, then integrating the function $(1 + a w)^{-m} w^{-n}$ gives:
$\displaystyle -am I_{m+1,n} -n I_{m,n+1}$ $\textstyle =$ $\displaystyle Q_{m,n}$ (44)
  $\textstyle =$ $\displaystyle \left. ( 1 + a w)^{-m} w^{-n} \right\vert _F^{\infty}$ (45)

Combining equations 43 and 44 gives:

\begin{displaymath}
I_{m,n} = \frac{-1}{a(n+m-1)} ( nI_{m,n+1} + Q_{m-1,n} )
\end{displaymath} (46)

Combining equations 41 and 42 to gives:

\begin{displaymath}
p = \frac{(\frac{D_1}{D_2})^{\frac{D_1}{2}}}{\beta(\frac{D_2}{2},\frac{D_1}{2})} I_{m,n}
\end{displaymath} (47)

with $a=\frac{D_1}{D_2}$, $n=1-\frac{D_1}{2}$ and $m=\frac{D_1+D_2}{2}$. Taking the logarithm and repeatedly substituting in equation 46, the approximation for $\log(p)$ is:
$\displaystyle \log(p)$ $\textstyle \approx$ $\displaystyle \frac{D_1}{2}\log(\frac{D_1}{D_2}) - \log(\beta(\frac{D_2}{2},\frac{D_1}{2})) - (m-1)\log(1 + aF) - n\log(F)$  
    $\displaystyle + \log\left(\frac{2}{D_1}\left(1-\frac{D_2}{D_1}\frac{(2-D_1)}{(2...
..._2}{D_1})^2\frac{(2-D_1)(4-D_1)}{(2+D_2)(4+D_2)}F^{-2} + \cdots \right. \right.$  
    $\displaystyle + \left. \left. (-1)^q(\frac{D_2}{D_1})^q\left(\prod_{n=1}^{q}\frac{(2n-D_1)}{(2n+D_2)}\right)F^{-q} + \cdots \right)\right)$ (48)



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Mark Jenkinson 2004-01-21