Re: statistics

*To*: mathgroup at smc.vnet.net*Subject*: [mg5876] Re: [mg5835] statistics*From*: daiyanh at mindspring.com (萩原大太郎)*Date*: Sun, 2 Feb 1997 01:31:38 -0500*Sender*: owner-wri-mathgroup at wolfram.com

<much interesting part snip> >experiment seems to show that: > >Expected Mean -> m +/- 1.00 s/Sqrt[n] >Expected Sqrt(Variance) -> s +/- 0.38 s/Sqrt[n] >Expected Skewness -> 0 +/- 2.32 1/Sqrt[n] >Expected Kurtosis -> 0 +/- 4.31 1/Sqrt[n] > >If this assumption is correct (in form), then what are the 0.38, 2.32 and >4.31 analytically? Comment on the first two. The first one is well known in sampling stat. The coefficient is 1 since the population is the whole real line which makes population size infinite. The second one is a bit tricky. Essentially this is chi-square dist, but you demand the stat for the Sqrt of chi-square. If you can live with the stat for Variance alone, then Expected Variance -> s^2 +/- Sqrt(2/(n-1)) s^2 Now, I don't know if I can do this. But the above can be transformed as below, using the well known formula Sqrt(1+x)=1+(1/2)x+... assuming small x: Expected Sqrt(Variance) -> s +/- 1/Sqrt(2(n-1)) s In this case, I get the numerical coefficient 1/Sqrt(2)=0.7. The apparent discrepancy comes from my expanding the Sqrt(1+x) above. To be a real masochist, however, and to carry out the computation of the third and fourth, you need to determine the Jacobian and integrate it to find out the stat's dist function. Then further integrate it with the stat as the integrand (you get the mean) and then with (x-u)^2 (you get the variance for the stat x). I don't know how efficiently you can do this with Mathematica. My feeling is that you will be stomped already at the second case. Try the expected Variance (not its Sqrt) to see if it agrees with the analytic sol'n above. And make sure you use unbiased def for s. Reference: Statistics for Physicists, BR Martin, Acedemic Press, 1971.