On Thu, Sep 15, 2011 at 12:18 PM, Viktor Cerovski <[email protected]> wrote: > You(r subtracted) mean (is) this: > > stddev _1e12+1e12+1e6?@$0 > 0.288461
If I am allowed to know the mean of 1e12+1e6?@$0 -- if I am allowed to treat the data as non-random -- then, yes. But that was not really my point. The point, I thought, was to come up with an implementation of standard deviation that is stable, numerically, when the mean is much larger than the deviation. And, of course, Welford's approach does achieve that. But Welford's algorithm assumes a scalar computing architecture. Meanwhile, it seems to me that subtracting the computed mean achieves approximately the same thing that Welford's approach achieves, but modularized to eliminate the "scalar architecture" aspect. That said, if there are generic cases where Welford's approach behaves better than subtracting the computed mean from the data set, I would be interested in hearing about them. (But I doubt they exist, except in hand crafted special cases with no general utility, because Welford's algorithm is doing essentially the same thing, with a running average.) -- Raul ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
