Hi Eleanor I reran my simulation including your method. Let's just look at the two extreme cases in my original table: sd(F+) = 10, sd(F-) = 1 (5th row) and sd(F+) = 1000, sd(F-) = 100 (last row: the case where my method performs worst but still better than the others!). I used the same 'true' values as before: F+ = 578, F- = 1369, Fmean = 973.
For the first case, the values of su(Fmean) for the methods 'unweighted', 'variance-weighted', 'RMS-error-weighted' (my method) and 'sd-weighted' (your method) are 5 1 5 1 respectively, so as expected the two weighted methods give the lowest variances. Similarly the mean biases are: 0 388 0 323, so again as expected the weighted methods are the most biased. The RMS errors are: 5 388 5 323 so in this case the 'variance-weighted' and 'sd-weighted' methods both perform poorly with the highest overall errors, though your method is somewhat better than the 'variance-weighted' one. For each statistic I'm including the first 3 values for comparison but they are of course exactly the same as before: only the last value for the 'sd-weighted' method is new. For the second case with sd(F+) = 1000, sd(F-) = 100 and the same values for the true Fs, the results from the 4 methods in the same order as before are: su(Fmean): 386 99 171 115. bias(Fmean): 87 389 274 339. RMSE(Fmean): 396 402 323 358. So again the results for su(Fmean) and bias(Fmean) are as expected (i.e. the 'sd-weighted' method performs intermediate between 'unweighted' and 'variance-weighted'). However using the RMSE measure the 'RMS-error-weighted' method performs best. Using the RMSE measure as the definitive metric (which IMO is the only sensible procedure), in no case does the 'RMS-error-weighted' method perform worse than any of the others. Cheers -- Ian On 13 March 2017 at 08:22, Ian Tickle <ianj...@gmail.com> wrote: > > Eleanor > > I notice that you are calculating a weighted mean using 1/sd(I) as the > weight whereas in least squares one would of course normally use > variance-weighting, i.e. 1/sd(I)^2. I assume this is intentional and is a > way to reduce the bias effect of weighting, so that the resulting bias of > the mean would be intermediate between that of an unweighted and a > variance-weighted mean, though the variance of the mean would no longer be > a minimum. > > Assuming that it was indeed intentional, I will see if I can find the > program I used to simulate the various ways of calculating the mean using > the MSE as the measure of accuracy and see how this method compares with > the others. > > Cheers > > -- Ian > > > > On 12 March 2017 at 17:58, Eleanor Dodson <eleanor.dod...@york.ac.uk> > wrote: > >> You read: >> h k l IPLUS SIGIPLUS INEG SIGINEG >> Then program calculates this: >> >> SIGIMEAN = SIGIPLUS*SIGINEG/(SIGIPLUS+SIGINEG) >> >> IMEAN = (IPLUS/SIGIPLUS + INEG/SIGINEG)*SIGIMEAN >> >> ie: IMEAN = ( IPLUS*SIGINEG + INEG * SIGIPLUS ) / /(SIGIPLUS+SIGINEG) >> >> Is that the right thing to do? Not sure! >> >> Eleanor Dodson >> >> >> >> On 11 March 2017 at 15:18, Karthikeyan Subramanian <skarthi...@gmail.com> >> wrote: >> >>> Dear CCP4bb, >>> >>> How IMEAN and SIGIMEAN is calculated in scalepack2mtz if the input is >>> with anomalous intensity (obtained from HKL2000). Any guide/reference is >>> highly appreciated. >>> >>> Thanks in advance, >>> >>> With regards >>> >>> Karthikeyan S. >>> >>> Principal Scientist >>> >>> IMTECH, Chandigarh >>> >> >> >
