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Michael
06/29/12 9:06pm

Your function does not appear to be a PDF -- the integral (over all x, or x, n) is greater than 1. In fact it is very close to

In[86]:= PDF[NormalDistribution[0, \[Pi]^(1/4) \[Sigma]], x]

Out[86]= E^(-(x^2/(2 Sqrt[\[Pi]] \[Sigma]^2)))/(Sqrt[2] \[Pi]^(
3/4) \[Sigma])

which is different by a factor of

\[Pi]^(1/4)/(Sqrt[2] Gamma[n/2])

I hope that helps indicate where the problem is.

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Author Date Posted
DistributionFitTest question mani 06/29/12 07:32am
Re: DistributionFitTest question Michael 06/29/12 9:06pm
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