# RE: [R] Simplify formula for heterogeneity

From: Ted Harding <Ted.Harding_at_nessie.mcc.ac.uk>
Date: Fri 27 May 2005 - 02:05:17 EST

On 26-May-05 Stefaan Lhermitte wrote:
> 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

If I have understood your formula correctly (and you are applying it to a vector X of length n) then it seems that your H reduces to

sqrt[(Si(n*(Xi - Xbar)^2) + Sj(n*(Xj - Xbar)^2))/n]

• sqrt[2*(n-1)var(X)] = sd(X)*sqrt(2*(n-1))

(where Xbar is the mean of the values in X).

So I don't see what the special point of H is anyway. But at least this simplifies it1

Best wishes,
Ted.

> 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) 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 + X²) - (A + X²)² ) / n² ]
>
> This is failry easy in an iterative process, since before we continue
> with the next step we set:
> A = (A + X²)
> B = (B + X)
>
> Can anybody help me to do something comparable for H? Any other help to
> calculate H easily in an iterative process is also welcome!
>
>
> Kind regards,
> Stef
>
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E-Mail: (Ted Harding) <Ted.Harding@nessie.mcc.ac.uk> Fax-to-email: +44 (0)870 094 0861
```Date: 26-May-05                                       Time: 17:05:13
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