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Simo · #8 November 24th, 2008, 09:06 AM

actually...solving the integral equation...

$\frac{1}{\sigma \sqrt{2\pi }}\int_{x=\bar{X}}^{+\infty }\exp \left( -\frac{\left( x-\mu \right) ^{2}}{2\sigma ^{2}}\right) dx=\frac{K}{\sigma \sqrt{2\pi }}\int_{-\infty }^{x=\bar{X}}\exp \left( -\frac{\left( x-\mu \right)^{2}}{2\sigma ^{2}}\right) dx$

with the new variable
$y=\frac{x-\mu }{\sigma \sqrt{2}}$

the former become
$\int_{y=-\infty }^{\frac{\bar{X}-\mu }{\sigma \sqrt{2}}}\exp(-y^{2})dy=K\int_{y=\frac{\bar{X}-\mu }{\sigma \sqrt{2}}}^{+\infty }\exp(-y^{2})dy$
(note that in the previous the constant terms that appear on both sides are omitted)
solving the integrals become

${erf}(\frac{\bar{X}-\mu }{\sigma \sqrt{2}})-(-1)=K\left( (+1)-{erf}(\frac{\bar{X}-\mu }{\sigma \sqrt{2}})\right)$

${erf}(\frac{\bar{X}-\mu }{\sigma \sqrt{2}})=\frac{K-1}{K+1}$

$\mu =\bar{X}-{erf}^{-1}\left( \frac{K-1}{K+1}\right) \sigma \sqrt{2}$

at the end, it was quite straightforward.
i've tested it numerically...works.

regards
simo