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, > > Shem Cristobal > Project Manager, Stratpoint Global Outsourcing, Inc. > Unit 601 Tower 1 Globe Telecom Plaza > Pioneer St corner Madisson St > Mandaluyong City, Philippines 1552 > Email: [email protected] > Website: http://www.stratpoint.com/ > > Microsoft Certified Application Developer for .NET > Microsoft Certified Solution Developer for .NET > Microsoft Certified Technology Specialist - SQL Server 2005 > Microsoft Certified Technology Specialist - MS Project 2007 >
