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Re: Death of Rmerge

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Fri, 1 Jun 2012 13:07:25 +0100

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 ```On 1 June 2012 03:22, Edward A. Berry <[log in to unmask]> wrote: > Leo will probably answer better than I can, but I would say I/SigI counts > only > the present reflection, so eliminating noise by anisotropic truncation > should > improve it, raising the average I/SigI in the last shell. We always include unmeasured reflections with I/sigma(I) = 0 in the calculation of the mean I/sigma(I) (i.e. we divide the sum of I/sigma(I) for measureds by the predicted total no of reflections incl unmeasureds), since for unmeasureds I is (almost) completely unknown and therefore sigma(I) is effectively infinite (or at least finite but large since you do have some idea of what range I must fall in). A shell with = 2 and 50% completeness clearly doesn't carry the same information content as one with the same and 100% complete; therefore IMO it's very misleading to quote including only the measured reflections. This also means we can use a single cut-off criterion (we use mean I/sigma(I) > 1), and we don't need another arbitrary cut-off criterion for completeness. As many others seem to be doing now, we don't use Rmerge, Rpim etc as criteria to estimate resolution, they're just too unreliable - Rmerge is indeed dead and buried! Actually a mean value of I/sigma(I) of 2 is highly statistically significant, i.e. very unlikely to have arisen by chance variations, and the significance threshold for the mean must be much closer to 1 than to 2. Taking an average always increases the statistical significance, therefore it's not valid to compare an _average_ value of I/sigma(I) = 2 with a _single_ value of I/sigma(I) = 3 (taking 3 sigma as the threshold of statistical significance of an individual measurement): that's a case of "comparing apples with pears". In other words in the outer shell you would need a lot of highly significant individual values >> 3 to attain an overall average of 2 since the majority of individual values will be < 1. > F/sigF is expected to be better than I/sigI because dx^2 = 2Xdx, > dx^2/x^2 = 2dx/x, dI/I = 2* dF/F  (or approaches that in the limit . . .) That depends on what you mean by 'better': every metric must be compared with a criterion appropriate to that metric. So if we are comparing I/sigma(I) with a criterion value = 3, then we must compare F/sigma(F) with criterion value = 6 ('in the limit' of zero I), in which case the comparison is no 'better' (in terms of information content) with I than with F: they are entirely equivalent. It's meaningless to compare F/sigma(F) with the criterion value appropriate to I/sigma(I): again that's "comparing apples and pears"! Cheers -- Ian```