The EM and Cryo-EM world might have something to say about that perhaps. Isn’t RIP phasing coming from what you are describing as well?
Just curious, Jürgen __________________________________________ Jürgen Bosch, PhD, MBA Center for Global Health & Diseases Case Western Reserve University https://www.linkedin.com/in/jubosch/ CEO & Co-Founder at InterRayBio, LLC Johns Hopkins University Bloomberg School of Public Health Department of Biochemistry & Molecular Biology On Oct 17, 2021, at 13:24, James Holton <jmhol...@lbl.gov> wrote: 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 to be a gamma distribution? Not that that really narrows things down since there are an infinite number of them, but is that really the "i have no idea" prior? Or just a convenient closed-form choice? I've only just recently 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. 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