I know both from Eleanor and from looking at the completeness of my datasets before/after detwinning that Detwin simply omits incomplete twin-pairs. I wonder how to assess the gains from implementation of what you suggest?
The question which keeps nagging at me is simply why detwinning works so badly--maybe your suggestion would improve it? I have some possibilities for other reasons why detwinning does not work well, but they don't seem powerful enough to overcome the apparent much-worseness of twinned data.
Again, it seems to me that the way to beat twinning is to model it from start to finish.
JPK
-----Original Message-----
From: CCP4 bulletin board [mailto:[log in to unmask]] On Behalf Of Kay Diederichs
Sent: Wednesday, May 11, 2016 5:30 PM
To: [log in to unmask]
Subject: Re: [ccp4bb] Dano--Sign Convention
Jacob,
my intuition was also wrong - the math is such that the "calculating the
anomalous difference" and "detwin" operations can be exchanged. In other
words, what I suggested to be possibly different options a) and b) are
actually mathematically equivalent. Thus, none of them has an advantage
over the other.
It remains the option to decide what to do with DANOtw(h1) if DANOtw(h2)
is unknown. I think that the decision should be done in a probabilistic
way: if the error from omitting DANOtrue(h1) (or equivalently, setting
it to zero) is likely higher than the error from just taking DANOtw(h1)
as DANOtrue(h1), then one should do the latter.
This should be the case when the magnitude of DANOtw(h1) is high, and it
should also be the case when the twinning fraction tf is low (so a test
combining both criteria would be to compare DANOtw(h1)/tf against
<DANO>/0.5).
I don't know if CCP4 detwin does anything like this, but it could be
implemented - it might improve the anom Patterson and the maps from the
detwinned data.
best,
Kay
Am 11.05.16 um 21:18 schrieb Keller, Jacob:
>> in your other posting, you gave the example (using the notation of the
> detwin documentation):
>
> DANOTw(h1) = 52
> DANOTw(h2) = 28
>
> and if you use the formulas that you cite below, you get the true values
> back:
>
> DANOtrue(h1) = (0.6*52 - 0.4*28)/(1-0.8) = 100
> DANOtrue(h2) = (0.6*28 - 0.4*52)/(1-0.8) = -20
>
> So then, what's problematic?
>
> ======================
>
> Nothing--I think you're right, and my intuition was wrong. Thanks for correcting my mis-statement.
>
> What would come of detwinning Dano rather than I+/-? I wonder, like you, whether anything would be gained or lost?
>
> JPK
>
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