Diff for "FAQ/eorder" - CBU statistics Wiki
location: Diff for "FAQ/eorder"
Differences between revisions 2 and 6 (spanning 4 versions)
Revision 2 as of 2009-01-12 16:36:43
Size: 386
Editor: PeterWatson
Comment:
Revision 6 as of 2009-01-12 16:54:42
Size: 583
Editor: PeterWatson
Comment:
Deletions are marked like this. Additions are marked like this.
Line 3: Line 3:
The expected value of a Normal Order Statistic may be computed as the absolute value of The expected value of  the i-th Normal Order Statistic from a sample of size, n, may be computed as the absolute value of
Line 5: Line 5:
$$\Phi^{-1}(U(i))$$ $$\Phi^{-1}(U(i,n))$$
Line 7: Line 7:
where $$\Phi^{-1}$$ is the inverse Normal cumulative density function which converts a probability to a z-score. where $$\Phi^{-1}$$ is the inverse Normal cumulative density function which converts a probability, such as U(i,n), to a z-score.
Line 9: Line 9:
where U(i) is given [ http://en.wikipedia.org/wiki/Normal_probability_plot: here.] where the U(i,n) are given as U(i) [http://en.wikipedia.org/wiki/Normal_probability_plot here.]


U(i,n) =
$$ 1- 0.5^{1/n} & i=1 \\
\frac{i-0.3175}{n+0.365} & i=2, \ldots, n-1 \\
0.5^{1/n} & i=n
\end{tabular}
\right.

Quick formula for the expected value of a Normal order statistic

The expected value of the i-th Normal Order Statistic from a sample of size, n, may be computed as the absolute value of

$$\Phi^{-1}(U(i,n))$$

where $$\Phi^{-1}$$ is the inverse Normal cumulative density function which converts a probability, such as U(i,n), to a z-score.

where the U(i,n) are given as U(i) [http://en.wikipedia.org/wiki/Normal_probability_plot here.]

U(i,n) = $$ 1- 0.5^{1/n} & i=1 \\ \frac{i-0.3175}{n+0.365} & i=2, \ldots, n-1 \\ 0.5^{1/n} & i=n \end{tabular} \right.

None: FAQ/eorder (last edited 2013-03-08 10:17:56 by localhost)