James, I've been watching the thread with fascination, but also the
confusion of wild ignorance. I've finally realised why.
What I've missed is: what exactly makes the question so important?
I've understood what brought it up, if course, but not the consequence
of getting it wrong.
Frank
Sent from tiny silly touch screen
------------------------------------------------------------------------
*From:* James Holton <jmhol...@lbl.gov>
*Sent:* Saturday, 16 October 2021 20:01
*To:* CCP4BB@JISCMAIL.AC.UK
*Subject:* Re: [ccp4bb] am I doing this right?
Thank you everyone for your thoughtful and thought-provoking responses!
But, I am starting to think I was not as clear as I could have been
about my question. I am actually concerning myself with background, not
necessarily Bragg peaks. With Bragg photons you want the sum, but for
background you want the average.
What I'm getting at is: how does one properly weight a zero-photon
observation when it comes time to combine it with others? Hopefully
they are not all zero. If they are, check your shutter.
So, ignoring Bragg photons for the moment (let us suppose it is a
systematic absence) what I am asking is: what is the variance, or,
better yet,what is the WEIGHT one should assign to the observation of
zero photons in a patch of 10x10 pixels?
In the absence of any prior knowledge this is a difficult question, but
a question we kind of need to answer if we want to properly measure data
from weak images. So, what do we do?
Well, with the "I have no idea" uniform prior, it would seem that
expectation (Epix) and variance (Vpix) would be k+1 = 1 for each pixel,
and therefore the sum of Epix and Vpix over the 100 independent pixels is:
Epatch=Vpatch=100 photons
I know that seems weird to assume 100 photons should have hit when we
actually saw none, but consider what that zero-photon count, all by
itself, is really telling you:
a) Epix > 20 ? No way. That is "right out". Given we know its Poisson
distributed, and that background is flat, it is VERY unlikely you have E
that big when you saw zero. Cross all those E values off your list.
b) Epix=0 ? Well, that CAN be true, but other things are possible and
all of them are E>0. So, most likely E is not 0, but at least a little
bit higher.
c) Epix=1e-6 ? Yeah, sure, why not?
d) Epix= -1e-6 ? No. Don't be silly.
e) If I had to guess? Meh. 1 photon per pixel? That would be k+1
I suppose my objection to E=V=0 is because V=0 implies infinite
confidence in the value of E, and that we don't have. Yes, it is true
that we are quite confident in the fact that we did not see any photons
this time, but the remember that E and V are the mean and variance that
you would see if you did a million experiments under the same
conditions. We are trying to guess those from what we've got. Just
because you've seen zero a hundred times doesn't mean the 101st
experiment won't give you a count. If it does, then maybe Epatch=0.01
and Epix=0.0001? But what do you do before you see your first photon?
All you can really do is bracket it.
But what if you come up with a better prior than "I have no idea" ?
Well, we do have other pixels on the detector, and presuming the
background is flat, or at least smooth, maybe the average counts/pixel
is a better prior?
So, let us consider an ideal detector with 1e6 independent pixels. Let
us further say that 1e5 background photons have hit that detector. I
want to still ignore Bragg photons because those have a very different
prior distribution to the background. Let us say we have masked off all
the Bragg areas.
The average overall background is then 0.1 photons/pixel. Let us assign
that to the prior probability Ppix = 0.1. Now let us look again at that
patch of 10x10 pixels with zero counts on it. We expected to see 10,
but got 0. What are the odds of that? Pretty remote. Less than 1 in a
million.
I suspect in this situation where such an unlikely event has occurred it
should perhaps be given a variance larger than 100. Perhaps quite a bit
larger? Subsequent "sigma-weighted" summation would then squash its
contribution down to effectively 0. So, relative to any other
observation with even a shred of merit it would have no impact. Giving
it V=0, however? That can't be right.
But what if Ppix=0.01 ? Then we expect to see zero counts on our
100-pixel patch about 1/3 of the time. Same for 1-photon observations.
Giving these two kinds of observations the same weight seems more
sensible, given the prior.
Another prior might be to take the flux and sample thickness into
account. Given the cross section of light elements the expected
photons/pixel on most any detector would be:
Ppix = 1.2e-5*flux*exposure*thickness*omega/Npixels
where:
Ppix = expected photons/pixel
Npixels = number of pixels on the detector
omega = fraction of scattered photons that hit it (about 0.5)
thickness = thickness of sample and loop in microns
exposure = exposure time in seconds
flux = incident beam flux in photons/s
1.2e-5 = 1e-4 cm/um * 1.2 g/cm^3 * 0.2 cm^2/g (cross section of oxygen)
If you don't know anything else about the sample, you can at least know
that.
Or am I missing something?
-James Holton
MAD Scientist
On 10/16/2021 12:47 AM, Kay Diederichs wrote:
> Dear Gergely,
>
> with " 10 x 10 patch of pixels ", I believe James means that he
observes 100 neighbouring pixels each with 0 counts. Thus the
frequentist view can be taken, and results in 0 as the variance, right?
>
> best,
> Kay
>
>
> On Fri, 15 Oct 2021 21:07:26 +0000, Gergely Katona
<gergely.kat...@gu.se> wrote:
>
>> Dear James,
>>
>> Uniform distribution sounds like “I have no idea”, but a uniform
distribution does not go from -inf to +inf. If I believe that every
count from 0 to 65535 has the same probability, then I also expect
counts with an average of 32768 on the image. It is not an objective
belief in the end and probably not a very good idea for an X-ray
experiment if the number of observations are small. Concerning which
variance is the right one, the frequentist view requires frequencies
to be observed. In the absence of frequencies, there is no error
estimate. Bayesians at least can determine a single distribution as an
answer without observations and that will be their prior belief of the
variance. Again, I would avoid a uniform a priori distribution for the
variance. For a Poisson distribution the convenient conjugate prior is
the gamma distribution. It can control the magnitude of k and strength
of belief with its location and scale parameter, respectively.
>>
>> Best wishes,
>>
>> Gergely
>>
>> Gergely Katona, Professor, Chairman of the Chemistry Program Council
>> Department of Chemistry and Molecular Biology, University of Gothenburg
>> Box 462, 40530 Göteborg, Sweden
>> Tel: +46-31-786-3959 / M: +46-70-912-3309 / Fax: +46-31-786-3910
>> Web: http://katonalab.eu <http://katonalab.eu>, Email:
gergely.kat...@gu.se
>>
>> From: CCP4 bulletin board <CCP4BB@JISCMAIL.AC.UK> On Behalf Of
James Holton
>> Sent: 15 October, 2021 18:06
>> To: CCP4BB@JISCMAIL.AC.UK
>> Subject: Re: [ccp4bb] am I doing this right?
>>
>> Well I'll be...
>>
>> Kay Diederichs pointed out to me off-list that the k+1 expectation
and variance from observing k photons is in "Bayesian Reasoning in
Data Analysis: A Critical Introduction" by Giulio D. Agostini.
Granted, that is with a uniform prior, which I take as the Bayesean
equivalent of "I have no idea".
>>
>> So, if I'm looking to integrate a 10 x 10 patch of pixels on a weak
detector image, and I find that area has zero counts, what variance
shall I put on that observation? Is it:
>>
>> a) zero
>> b) 1.0
>> c) 100
>>
>> Wish I could say there are no wrong answers, but I think at least
two of those are incorrect,
>>
>> -James Holton
>> MAD Scientist
>> On 10/13/2021 2:34 PM, Filipe Maia wrote:
>> I forgot to add probably the most important. James is correct, the
expected value of u, the true mean, given a single observation k is
indeed k+1 and k+1 is also the mean square error of using k+1 as the
estimator of the true mean.
>>
>> Cheers,
>> Filipe
>>
>> On Wed, 13 Oct 2021 at 23:17, Filipe Maia
<fil...@xray.bmc.uu.se<mailto:fil...@xray.bmc.uu.se>> wrote:
>> Hi,
>>
>> The maximum likelihood estimator for a Poisson distributed variable
is equal to the mean of the observations. In the case of a single
observation, it will be equal to that observation. As Graeme
suggested, you can calculate the probability mass function for a given
observation with different Poisson parameters (i.e. true means) and
see that function peaks when the parameter matches the observation.
>>
>> The root mean squared error of the estimation of the true mean from
a single observation k seems to be sqrt(k+2). Or to put it in another
way, mean squared error, that is the expected value of (k-u)**2, for
an observation k and a true mean u, is equal to k+2.
>>
>> You can see some example calculations at
https://colab.research.google.com/drive/1eoaNrDqaPnP-4FTGiNZxMllP7SFHkQuS?usp=sharing
<https://colab.research.google.com/drive/1eoaNrDqaPnP-4FTGiNZxMllP7SFHkQuS?usp=sharing>
>>
>> Cheers,
>> Filipe
>>
>> On Wed, 13 Oct 2021 at 17:14, Winter, Graeme (DLSLtd,RAL,LSCI)
<00006a19cead4548-dmarc-requ...@jiscmail.ac.uk<mailto:00006a19cead4548-dmarc-requ...@jiscmail.ac.uk>>
wrote:
>> This rang a bell to me last night, and I think you can derive this
from first principles
>>
>> If you assume an observation of N counts, you can calculate the
probability of such an observation for a given Poisson rate constant
X. If you then integrate over all possible value of X to work out the
central value of the rate constant which is most likely to result in
an observation of N I think you get X = N+1
>>
>> I think it is the kind of calculation you can perform on a napkin,
if memory serves
>>
>> All the best Graeme
>>
>>
>> On 13 Oct 2021, at 16:10, Andrew Leslie - MRC LMB
<and...@mrc-lmb.cam.ac.uk<mailto:and...@mrc-lmb.cam.ac.uk>> wrote:
>>
>> Hi Ian, James,
>>
>> I have a strong feeling that I have seen this
result before, and it was due to Andy Hammersley at ESRF. I’ve done a
literature search and there is a paper relating to errors in analysis
of counting statistics (se below), but I had a quick look at this and
could not find the (N+1) correction, so it must have been somewhere
else. I Have cc’d Andy on this Email (hoping that this Email address
from 2016 still works) and maybe he can throw more light on this. What
I remember at the time I saw this was the simplicity of the correction.
>>
>> Cheers,
>>
>> Andrew
>>
>> Reducing bias in the analysis of counting statistics data
>> Hammersley,
AP<https://www.webofscience.com/wos/author/record/2665675
<https://www.webofscience.com/wos/author/record/2665675>> (Hammersley,
AP) Antoniadis,
A<https://www.webofscience.com/wos/author/record/13070551
<https://www.webofscience.com/wos/author/record/13070551>> (Antoniadis, A)
>> NUCLEAR INSTRUMENTS & METHODS IN PHYSICS RESEARCH SECTION
A-ACCELERATORS SPECTROMETERS DETECTORS AND ASSOCIATED EQUIPMENT
>> Volume
>> 394
>> Issue
>> 1-2
>> Page
>> 219-224
>> DOI
>> 10.1016/S0168-9002(97)00668-2
>> Published
>> JUL 11 1997
>>
>>
>> On 12 Oct 2021, at 18:55, Ian Tickle
<ianj...@gmail.com<mailto:ianj...@gmail.com>> wrote:
>>
>>
>> Hi James
>>
>> What the Poisson distribution tells you is that if the true count
is N then the expectation and variance are also N. That's not the
same thing as saying that for an observed count N the expectation and
variance are N. Consider all those cases where the observed count is
exactly zero. That can arise from any number of true counts, though
as you noted larger values become increasingly unlikely. However those
true counts are all >= 0 which means that the mean and variance of
those true counts must be positive and non-zero. From your results
they are both 1 though I haven't been through the algebra to prove it.
>>
>> So what you are saying seems correct: for N observed counts we
should be taking the best estimate of the true value and variance as
N+1. For reasonably large N the difference is small but if you are
concerned with weak images it might start to become significant.
>>
>> Cheers
>>
>> -- Ian
>>
>>
>> On Tue, 12 Oct 2021 at 17:56, James Holton
<jmhol...@lbl.gov<mailto:jmhol...@lbl.gov>> wrote:
>> All my life I have believed that if you're counting photons then the
>> error of observing N counts is sqrt(N). However, a calculation I just
>> performed suggests its actually sqrt(N+1).
>>
>> My purpose here is to understand the weak-image limit of data
>> processing. Question is: for a given pixel, if one photon is all you
>> got, what do you "know"?
>>
>> I simulated millions of 1-second experiments. For each I used a "true"
>> beam intensity (Itrue) between 0.001 and 20 photons/s. That is, for
>> Itrue= 0.001 the average over a very long exposure would be 1 photon
>> every 1000 seconds or so. For a 1-second exposure the observed
count (N)
>> is almost always zero. About 1 in 1000 of them will see one photon, and
>> roughly 1 in a million will get N=2. I do 10,000 such experiments and
>> put the results into a pile. I then repeat with Itrue=0.002,
>> Itrue=0.003, etc. All the way up to Itrue = 20. At Itrue > 20 I never
>> see N=1, not even in 1e7 experiments. With Itrue=0, I also see no N=1
>> events.
>> Now I go through my pile of results and extract those with N=1, and
>> count up the number of times a given Itrue produced such an event. The
>> histogram of Itrue values in this subset is itself Poisson, but with
>> mean = 2 ! If I similarly count up events where 2 and only 2 photons
>> were seen, the mean Itrue is 3. And if I look at only zero-count events
>> the mean and standard deviation is unity.
>>
>> Does that mean the error of observing N counts is really sqrt(N+1) ?
>>
>> I admit that this little exercise assumes that the distribution of
Itrue
>> is uniform between 0.001 and 20, but given that one photon has been
>> observed Itrue values outside this range are highly unlikely. The
>> Itrue=0.001 and N=1 events are only a tiny fraction of the whole.
So, I
>> wold say that even if the prior distribution is not uniform, it is
>> certainly bracketed. Now, Itrue=0 is possible if the shutter didn't
>> open, but if the rest of the detector pixels have N=~1, doesn't this
>> affect the prior distribution of Itrue on our pixel of interest?
>>
>> Of course, two or more photons are better than one, but these days with
>> small crystals and big detectors N=1 is no longer a trivial situation.
>> I look forward to hearing your take on this. And no, this is not a
trick.
>>
>> -James Holton
>> MAD Scientist
>>
>>
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