Re: [ccp4bb] am I doing this right?

James Holton Sun, 17 Oct 2021 10:25:16 -0700

Thank you Gergely.  That is interesting!

I don't mind at all making this Bayesian, as long as it works!

Something I'm not quite sure about: does the prior distribution HAVE tobe a gamma distribution? Not that that really narrows things down sincethere are an infinite number of them, but is that really the "i have noidea" prior? Or just a convenient closed-form choice? I've only justrecently heard of conjugate priors.


Much appreciate any thoughts you may have on this,

-James


On 10/16/2021 3:48 PM, Gergely Katona wrote:

Dear James,

If I understand correctly you are looking for a single rate parameter to 
describe the pixels in a block. It would also be possible to estimate the rates 
for individual pixels or estimate the thickness of the sample from the counts 
if you have a good model, that is where Bayesian methods really shine. I tested 
the simplest first Bayesian network with 10 and 100 zero count pixels, 
respectively:

https://colab.research.google.com/drive/1TGJx2YT9I-qyOT1D9_HCC7G7as1KXg2e?usp=sharing


The two posterior distributions are markedly different even if they start from 
the same prior distribution, which I find more intuitive than the frequentist 
treatment of uncertainty. You can test different parameters for the gamma prior 
or change to another prior distribution. It is possible to reduce the posterior 
distributions to their mean or posterior maximum, if needed. If you are looking 
for an alternative to the Bayesian perspective then this will not help, 
unfortunately.

Best wishes,

Gergely

-----Original Message-----
From: CCP4 bulletin board <CCP4BB@JISCMAIL.AC.UK> On Behalf Of James Holton
Sent: den 16 oktober 2021 21:01
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, 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-4FTGiNZxMllP7SFH
kQuS?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>
(Hammersley, AP) Antoniadis,
A<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

#####################################################################
###

To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1

This message was issued to members of
www.jiscmail.ac.uk/CCP4BB<http://www.jiscmail.ac.uk/CCP4BB>, a
mailing list hosted by
www.jiscmail.ac.uk<http://www.jiscmail.ac.uk/>, terms & conditions
are available at https://www.jiscmail.ac.uk/policyandsecurity/

________________________________

To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1


________________________________

To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1




--

This e-mail and any attachments may contain confidential, copyright and or 
privileged material, and are for the use of the intended addressee only. If you 
are not the intended addressee or an authorised recipient of the addressee 
please notify us of receipt by returning the e-mail and do not use, copy, 
retain, distribute or disclose the information in or attached to the e-mail.
Any opinions expressed within this e-mail are those of the individual and not 
necessarily of Diamond Light Source Ltd.
Diamond Light Source Ltd. cannot guarantee that this e-mail or any attachments 
are free from viruses and we cannot accept liability for any damage which you 
may sustain as a result of software viruses which may be transmitted in or with 
the message.
Diamond Light Source Limited (company no. 4375679). Registered in
England and Wales with its registered office at Diamond House,
Harwell Science and Innovation Campus, Didcot, Oxfordshire, OX11 0DE,
United Kingdom


________________________________

To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1

________________________________

To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1


________________________________

To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1

#####################################################################
###

To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1

This message was issued to members of www.jiscmail.ac.uk/CCP4BB, a
mailing list hosted by www.jiscmail.ac.uk, terms & conditions are
available at https://www.jiscmail.ac.uk/policyandsecurity/

######################################################################
##

To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1

This message was issued to members of www.jiscmail.ac.uk/CCP4BB, a
mailing list hosted by www.jiscmail.ac.uk, terms & conditions are
available at https://www.jiscmail.ac.uk/policyandsecurity/

########################################################################

To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1

This message was issued to members of www.jiscmail.ac.uk/CCP4BB, a mailing list 
hosted by www.jiscmail.ac.uk, terms & conditions are available at 
https://www.jiscmail.ac.uk/policyandsecurity/


########################################################################

To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1

This message was issued to members of www.jiscmail.ac.uk/CCP4BB, a mailing list 
hosted by www.jiscmail.ac.uk, terms & conditions are available at 
https://www.jiscmail.ac.uk/policyandsecurity/

Re: [ccp4bb] am I doing this right?

Reply via email to