On Wed, 16 Jun 2004, Prof Brian Ripley wrote:
Aargh don't press keys this late in the day or I'll get the wrong ones.
According to Abramowitz and Stegun section 7.1,
erf z = \frac{2}{\sqrt{pi}} \int^z_0 e^_{-t^2} dt
Now pnorm(x) = \frac{1}{sqrt{2\pi}} \int^x_{-\infty} e^_{-u^2/2} du
so pnorm(x) - 1/2 = \frac{1}{sqrt{2\pi}} \int^x_0 e^_{-u^2/2} du
Now substitute t = u/sqrt{2}
pnorm(x) - 1/2 = \frac{1}{sqrt{2\pi}} \int^{x/\sqrt{2}}_0 e^_{-t^2} dt\sqrt{2}
so pnorm(x) - 1/2 = 1/2 erf(x/\sqrt{2}) or erf(z) = 2 pnorm(x\sqrt{2}) - 1
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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