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:I'm talking rubbish it's fine as x[n]-mean[n] = (n-1)/n*(x[n]-mean[n-1]). Sorry! --[[User:Cfp|cfp]] 22:20, 15 August 2006 (UTC)
--[[User:Snoopy67|Gabriel]] ([[User talk:Snoopy67|talk]]) 21:35, 17 July 2011 (UTC)
I think, the advantages of Welford's algorithm should be toned down. It is no more numerically stable than the direct two-pass method.
See http://www.johndcook.com/blog/2008/09/26/comparing-three-methods-of-computing-standard-deviation/
and http://www.jstor.org/stable/1267176 .
== Online algorithm ==
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