Z Score Approximation

The objective is to find an expansion of equation 1 for large values of $Z$. In this case a Taylor expansion will not be useful. The solution is found using an integral recurrence relation.

Consider the integral:

$$I_n = \int_x^{\infty} w^{-n} \exp(-w^2) \, dw.$$ (8)

Differentiating $(-w^{-n-1}\exp(-w^2)/2)$ and then integrating gives:

$$I_n + \frac{n+1}{2} I_{n+2} = \left. -\frac{1}{2} w^{-n-1}\exp(-w^2) \right\vert^{\infty}_{x}$$ (9)

$$= \frac{1}{2} x^{-n-1} \exp(-x^2).$$ (10)

Now if $x > L$ then $\vert I_n\vert < L^{-n} \vert I_0 \vert$. Therefore, for large Z, this enables the error function (related to $I_0$) to be expanded in terms of $I_n$ where the absolute values of $I_n$ continue to decrease as $n$ increases. Specifically:

$$I_0 = \frac{1}{2} x^{-1} \exp(-x^2) - \frac{1}{2} I_2$$ (11)

$$I_2 = \frac{1}{2} x^{-3} \exp(-x^2) - \frac{3}{2} I_4$$ (12)

$$I_4 = \frac{1}{2} x^{-5} \exp(-x^2) - \frac{5}{2} I_6$$ (13)

and so on. Rearranging these terms then gives:

$$I_0 = \frac{1}{2} x^{-1} \left( 1 - \frac{1}{2} x^{-2} + \frac{3}{4} x^{-4} \right) \exp(-x^2) - \frac{15}{8} I_6$$ (14)

where $\vert I_6 \vert < L^{-6} \vert I_0 \vert$ for $x > L$.

Equations 2 and 1, and the fact that $\ensuremath{\mathrm{erf}}(\infty) = 1$ can be used to relate $p$ and $Z$, giving:

$$p = \frac{1}{\sqrt{\pi}} I_0.$$ (15)

where $x = \frac{Z}{\sqrt{2}}$.

Therefore by taking the logarithm, and neglecting the $I_6$ term, the asymptotic expansion for the Z statistic is:

$$\log(p) \approx -\frac{1}{2}\log(2\pi) -\frac{1}{2} Z^2 - \log(Z) + \log(1 - Z^{-2} +3 Z^{-4}).$$ (16)