[Aavso-photometry] CCD 'fainter-than' question

[Michael Newberry]

We define a 3-sigma detection as having SNR = 3.0. Then the issue becomes 
how the SNR is calulated at a given magnitude so one can determine what 
magnitude corresponds to SNR=3. Computing the SNR simply by dividing (Star 
Signal - Sky Signal) by the Noise from the sky annulus is *not* the correct 
way to compute the SNR because it hide certain values which must be added to 
make the statistics correct. First, there are 2 general strategies for 
estimating the noise value (I'll call it "sigma"). One is to use the scatter 
of values in the sky annulus and the other is to go back to theory and use 
the readout noise, gain, and exposure time to estimate it. Both methods have 
their supporting arguments. But in either case, you would not just divide 
the measured or calulated sigma into the net signal to get the SNR of a 
photometric measurement. I introduced the correct formulae in a 1991 paper, 
here:

http://adsabs.harvard.edu/abs/1991PASP..103..122N
About half way down the page, click [Send PDF].

There's lots of information in that paper. I hope it helps clarify the 
issues.

Michael

----- Original Message ----- 
From: "Wolfgang Renz" <w_renz at onlinehome.de>
To: "Jim Roe" <jroe at jamesroe.com>
Cc: "AAVSO-PHOTOMETRY" <aavso-photometry at mira.aavso.org>
Sent: Wednesday, May 28, 2008 11:44 AM
Subject: Re: [Aavso-photometry] CCD 'fainter-than' question


> Hi Jim
>
> That part was just the basic general math/stat part and I adapted my
> language away from pure math/stat to make it easier to understand.
> But you are right that it lost accuracy by doing so not well enough.
>
> To continue your example:
>
> How is the SNR of a star calculated in aperture photometry ?
>
> Most programs use (and output) the net ADU counts (which is the
> sum of the actual ADU counts in the star aperture minus the number
> of pixel in the star aperture times the mean/median/mode of the
> background level in the sky annulus).
>
> I never assumed that the SNR isn't integrated on a pixel by pixel
> base over all pixel in the star aperture.
> But long way back I incorrectly thought that the SNR is calculated
> in a formula using the MEAN net ADU count per pixel  divided by
> the ADU stddev in the sky annulus. But this is not the case.
> The SNR is actually calculated in a formula using the total net ADU
> count in the star aperture divided by the ADU stddev in the sky
> annulus.
>
> Clear skies
> Wolfgang
>
> -- 
> Wolfgang Renz, Karlsruhe, Germany
> Rz.BAV = WRe.vsnet = RWG.AAVSO
>
>
>
> ----- Original Message ----- 
> From: "Jim Roe"
> Cc: "AAVSO-PHOTOMETRY" <aavso-photometry at mira.aavso.org>
> Sent: Wednesday, May 28, 2008 2:23 PM
> Subject: Re: [Aavso-photometry] CCD 'fainter-than' question
>
>> Wolfgang Renz wrote:
>>> Hi Bob
>>>
>>> If one assumes that the noise in the sky annulus is normal distributed,
>>> then the following applies for the probability that a star measurement
>>> is still just background noise:
>>> <http://en.wikipedia.org/wiki/68-95-99.7_rule>
>>> <http://en.wikipedia.org/wiki/Standard_deviation#Rules_for_normally_distributed_data>
>>> <http://upload.wikimedia.org/wikipedia/commons/8/8c/Standard_deviation_diagram.svg>
>>> <http://upload.wikimedia.org/wikipedia/commons/thumb/8/8c/Standard_deviation_diagram.svg/400px-Standard_deviation_diagram.svg.png>
>>>
>>> So the probability for a measurement to lie outside the following
>>> ranges and to be still just a background detection are as follows:
>>> +/- 1 sigma   100% - 68.27% = 31.73%
>>> +/- 2 sigma   100% - 95.45% = 4.55%
>>> +/- 3 sigma   100% - 99.73% = 0.27%
>>> +/- 4 sigma   100% - 99.9934% = 0.0066%
>>> +/- 5 sigma   100% - 99.99994% = 0.00006%
>>> +/- 6 sigma   100% - 99.9999998% = 0,0000002%
>>> +/- 7 sigma   100% - 99.99999999974 % = 0.00000000026%
>>> But as there are no 'darker than the background' stars, one must
>>> half the above probabilities for just 'brighter than the background'
>>> stars.
>>
>> A precise definition, Wolfgang, but I would try to amplify it with a
>> numerical example.  First off, these results must be applied pixel by 
>> pixel.
>>
>> Thus, consider a background average of 100 and a sigma of 10.  The
>> probability that a pixel in the measuring aperture that measures 130 (+
>> 3 sigma) is just a background outlier is, as stated above, 0.0027/2.
>> But how many pixels are in your measuring aperture?  The math is pi r
>> squared where r is the radius of the aperture in pixels.  Suppose r = 6
>> then the number of pixels is 113.  Thus, the probability that at least
>> one of the background pixels in the measurement aperture is as bright as
>> 130 is only 15%.
>>
>> But one should never claim a "detection" based upon one pixel.  There
>> should be several contiguous pixels at or above the threshold before
>> claiming such.  Even then, a bit of caution is warranted as cosmic ray
>> strikes can produce some very star-like images.
>>
>> Jim Roe [ROE]
>
>
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