Re: [R] Simplify formula for heterogeneity
Thank you very much Ted! I have been looking at your simplification for more then an hour, but I don't see how you did it. Could you perhaps, if it is not to much work, explain me how you reduced H? It would help me to understand what I am realy doing. Looking at the result, it seems indeed that H does add more information than sd already did. Intuitively I thought the square of the sum of all possible differences would not be related to the standard deviation. Looking at your result it seems it is related by a factor sqrt(2*(n-1)) so there is no special point in calculating H and I know I cannot trust my intuition anymore. Thanks again! Kind regards, Stef __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Simplify formula for heterogeneity
Dear R-ians, I'm looking for a computational simplified formula to calculate a measure for heterogeneity (let's say H ): H = sqrt [ (Si (Sj (Xi - Xj)² ) ) /n ] where: sqrt = square root Si = summation over i (= 0 to n) Sj = summation over j (= 0 to n) Xi = element of X with index i Xj = element of X with index j I can simplify the formula to: H = sqrt [ ( 2 * n * Si (Xi) - 2 Si (Sj ( Xi * Xj)) ) / n] Unfortunately this formula stays difficult in iterative programming, because I have to keep every element of X to calculate H. I know a computional simplified formula exists for the standard deviation (sd) that is much easier in iterative programming. Therefore I wondered I anybody knew about analog simplifications to simplify H: sd = sqrt [ ( Si (Xi - mean(X) )² ) /n ] - simplified computation - sqrt [ (n * Si( X² ) - ( Si( X ) )² )/ n² ] This simplied formula is much easier in iterative programming, since I don't have to keep every element of X. E.g.: I have a vector X[1:10] and I already have caculated Si( X[1:10]² ) (I will call this A) and Si( X ) (I will call this B). When X gets extendend by 1 element (eg. X[11]) it easy fairly simple to calculate sd(X[1:11]) without having to reuse the elements of X[1:10]. I just have to calculate: sd = sqrt [ (n * (A + X[11]²) - (A + X[11]²)² ) / n² ] This is failry easy in an iterative process, since before we continue with the next step we set: A = (A + X[11]²) B = (B + X[11]) Can anybody help me to do something comparable for H? Any other help to calculate H easily in an iterative process is also welcome! Thanx in advance! Kind regards, Stef __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
