T Score Approximation

An expansion of the basic definition 3 is sought for small values of $p$, which correspond to large values of $T$. Once again this is achieved by integrating by parts and deriving a recurrence relation. However, before doing this a change of variable in equation 3 is required to get it into a more useful form. Setting $y=\frac{D}{D+w^2}$ and combining equations 3 and  5 gives:

$$p = \frac{1}{\sqrt{D} \beta(\frac{D}{2},\frac{1}{2})} \int_T... ...fty} \left( 1 + \frac{w^2}{D} \right)^{-\frac{D+1}{2}} \, dw$$ (20)

Now, consider the integral:

$$I_{m,n} = \int_T^{\infty} (1 + a w^2)^{-m} w^{-n} \, dw$$ (21)

Using the fact that $(1 + a w^2)^{-m+1} = (1 + a w^2) (1 + a w^2)^{-m}$ gives:

$$I_{m-1,n} = I_{m,n} + a I_{m,n-2}$$ (22)

and differentiating, then integrating the function $(1 + a w^2)^{-m} w^{-n}$ gives:

$$-2am I_{m+1,n-1} -n I_{m,n+1} = Q_{m,n}$$ (23)

$$= \left. ( 1 + a w^2)^{-m} w^{-n} \right\vert _T^{\infty}$$ (24)

Combining equations 22 and 23 gives:

$$I_{m,n} = -a^{-1} ( 1 + 2 \frac{m-1}{n+1} )^{-1} I_{m,n+2} - a^{-1} ( 1 + 2 \frac{m-1}{n+1} )^{-1} \frac{1}{n+1} Q_{m-1,n+1}$$ (25)

Substituting $a=D^{-1}$, $m = (D+1)/2$ gives:

$$I_{m,n} = - D \frac{n+1}{n+D} I_{m,n+2} - \frac{D}{n+D} Q_{m-1,1}$$ (26)

so that

$$I_{m,0} = -I_{m,2} - Q_{m-1,1}$$ (27)

$$I_{m,2} = - \frac{3D}{D+2} I_{m,4} - \frac{D}{D+2} Q_{m-1,3}$$ (28)

$$I_{m,4} = - \frac{5D}{D+4} I_{m,6} - \frac{D}{D+4} Q_{m-1,5}$$ (29)

By combining equations 20 and 21 to give

$$p = \frac{1}{\sqrt{D} \beta(\frac{D}{2},\frac{1}{2})} I_{m,0}$$ (30)

and then substituting equations 27 to 29 and taking the logarithm, the approximation for $\log(p)$ is:

$$\log(p) \approx -\frac{1}{2} \log(D) - \log(\beta(\frac{D}{2},\frac{1}{2})) - \log(T) - \frac{D-1}{2} \log\left(1 + \frac{T^2}{D}\right)$$

$$+ \log\left(1 - \frac{D}{D+2} T^{-2} + \frac{3 D^2}{(D+2)(D+4)} T^{-4} \right).$$ (31)