Re: Software to perform inhomogeneous ensemble photometry

[Stupendous Man <richmond@stupendous.cis.rit.edu>]

  I:

> > Yes, it would be even better to figure out some sort of
> > uncertainty for each individual measurement, based on a combination
> > of the SNR for that measurement and the overall nightly scatter,
> > but I don't know how to do that properly.

  Michael K.

> ...the way I do it is just adding the 1/SNR and the comp star 1-sigma 
> in quadrature:
>  
>    err = sqrt( (1/SNR)^2 + comp_sigma^2)
> 
> Why I find this useful is you can see when clouds rolled through or bad 
> tracking occurred and such because in those cases the low SNR causes 
> that term to dominate. Not all data points are necessarily created 
> equal.

  Yes, your method will clearly show where the clouds rolled in.
That is a Good Thing.

  On the other hand, if I understand this correctly, if conditions
were perfect, it is possible that the 1/SNR uncertainty could
account for the entire frame-to-frame scatter ... right?  In that
case, your method of combining the two uncertainties in quadrature
is taking the actual uncertainty and multiplying it by sqrt(2).
  
  So your technique will always give a useful relative measure
of the uncertainty, but might sometimes overestimate the true
uncertainty.  Still, it's more useful than mine :-/

> This and the zeropoint question are related. For each frame Mira makes 
> a solution based on the inputted magnitudes of the comp stars. It does 
> a least squares on the differences between the instrumental magnitudes 
> and the given magnitudes. The effectively spreads the error out among 
> all the comp stars. For each frame, each star will end up being the 
                                      ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> same distance away from it's "true" magnitude. This creates the 
  ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> situation where all the comp stars end up with the same 1-sigma. The 
> zeropoint error stamped in the header is the uncertainty in the least 
> squares fit.

  Hmmm.  Giving equal weights to all stars in the solution, 
even though some are bright and some are faint?  Sounds fishy to
me -- in theory.  In practice, if the only comparison stars you
choose are all roughly the same brightness, and the same brightness
as your target, this may turn out fine.

                                           Michael Richmond