oh i see... my bad... thanks for clearing things up.
On Mon, Apr 25, 2011 at 12:06 AM, Sean Owen <[email protected]> wrote: > sumX is definitely not (necessarily) 0, no. > > Let mx and my be the means of X and Y. To compute the centered sum of XY, > we > are computing: > sum( (X-mx)(Y-my) ) > > This is > > sum( XY - mx*Y - my*X + my*mx ) > or > sum(XY) - sum(mx*Y) - sum(my*X) + sum(my*mx) > or > sum(XY) - mx*sum(Y) - my*sum(X) + n*my*mx > > The last observation is that n*my = sum(Y) so you can cancel terms 2 and 4. > (Or because n*mx = sum(X) you could also as well cancel 3 and 4) > > This leaves > > sum(XY) - my*sum(X) > > On Sun, Apr 24, 2011 at 4:54 PM, Shem Cristobal <[email protected] > >wrote: > > > It was assumed that sumX is 0 so > > > > double centeredSumXY = sumXY - meanY * sumX; > > > > reduces to > > > > double centeredSumXY = sumXY; > > > > right? > > > > > > > > > > On Sun, Apr 24, 2011 at 11:36 PM, Sean Owen <[email protected]> wrote: > > > > > It's correct. :) Want to show your working or me to post the > derivation? > > > > > > On Sun, Apr 24, 2011 at 4:07 PM, Shem Cristobal < > > [email protected] > > > >wrote: > > > > > > > Has anyone seen this line 235 (and line 313) of > > AbstractSimilarity.class? > > > > > > > > double centeredSumXY = sumXY - meanY * sumX; > > > > > > > > This is so wrong and should be replaced by > > > > > > > > double centeredSumXY = sumXY - meanY * meanX; > > > > > > > > > > > > > > > > Best regards, > > > > > > > > @shemcristobal > > > > > > > > > > > > > > Best regards, @shemcristobal
