Hi All, haven't been following CCP4BB for a while, then I come back to this juicy Holtonian thread!
Sorry for being more practical, but could you use a windowed approach: integrate values of the same pixel/relp combo (roxel?) over time (however that works with frame slicing) to estimate the error over many frames, then shrink the window incrementally, see whether the successive plotted shrinking-window values show a trend? Most likely flat for background? This could be used for the spots as well as background. Would this lose the precious temporal info? Jacob On Fri, Oct 22, 2021 at 3:25 AM Gergely Katona <gergely.kat...@gu.se> wrote: > Hi, > > > > I have more estimates to the same problem using a multinomial data > distribution. I should have realized that for prediction, I do not have to > deal with infinite likelihood of 0 trials when observing only 0s on an > image. Whenever 0 photons generated by the latent process, the image is > automatically empty. With this simplification, I still have to hide behind > mathematical convenience and use Gamma prior for the latent Poisson > process, but observing 0 counts just increments the beta parameter by 1 > compared to the prior belief. With equal photon capture probabilities, the > mean counts are about 0.01 and the std is about 0.1 with > rate≈Gamma(alpha=1, beta=0.1) prior . With a symmetric Dirichlet prior to > the capture probabilities, the means appear unchanged, but the predicted > stds starts high at very low concentration parameter and level off at high > concentration parameter. This becomes more apparent at high photon counts > (high alpha of Gamma distribution). The answer is different if we look at > the std across the detector plane or across time of a single pixel. > > Details of the calculation below: > > > > > https://colab.research.google.com/drive/1NK43_3r1rH5lBTDS2rzIFDFNWqFfekrZ?usp=sharing > > > > 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 *Nave, > Colin (DLSLtd,RAL,LSCI) > *Sent:* 21 October, 2021 19:21 > *To:* CCP4BB@JISCMAIL.AC.UK > *Subject:* Re: [ccp4bb] am I doing this right? > > > > Congratulations to James for starting this interesting discussion. > > > > For those who are like me, nowhere near a black belt in statistics, the > thread has included a number of distributions. I have had to look up where > these apply and investigate their properties. > > As an example, > > “The Poisson distribution is used to model the # of events in the future, > Exponential distribution is used to predict the wait time until the very > first event, and Gamma distribution is used to predict the wait time until > the k-th event.” > > A useful calculator for distributions can be found at > > https://keisan.casio.com/menu/system/000000000540 > > a specific example is at > > https://keisan.casio.com/exec/system/1180573179 > > where cumulative probabilities for a Poisson distribution can be found > given values for x and lambda. > > > > The most appropriate prior is another issue which has come up e.g. is a > flat prior appropriate? I can see that a different prior would be > appropriate for different areas of the detector (e.g. 1 pixel instead of > 100 pixels) but the most appropriate prior seems a bit arbitrary to me. One > of James’ examples was 10^5 background photons distributed among 10^6 > pixels – what is the most appropriate prior for this case? I presume it is > OK to update the prior after each observation but I understand that it can > create difficulties if not done properly. > > > > Being able to select the prior is sometimes seen as a strength of Bayesian > methods. However, as a strong advocate of Bayesian methods once put it, > this is a bit like Achilles boasting about his heel! > > > > I hope for some agreement among the black belts. It would be good to end > up with some clarity about the most appropriate probability distributions > and priors. Also, have we got clarity about the question being asked? > > > > Thanks to all for the interesting points. > > > > Colin > > *From:* CCP4 bulletin board <CCP4BB@JISCMAIL.AC.UK> *On Behalf Of *Randy > John Read > *Sent:* 21 October 2021 13:23 > *To:* CCP4BB@JISCMAIL.AC.UK > *Subject:* Re: [ccp4bb] am I doing this right? > > > > Hi Kay, > > > > No, I still think the answer should come out the same if you have good > reason to believe that all the 100 pixels are equally likely to receive a > photon (for instance because your knowledge of the geometry of the source > and the detector says the difference in their positions is insignificant, > i.e. part of your prior expectation). Unless the exact position of the spot > where you detect the photon is relevant, detecting 1 photon on a big pixel > and detecting the same photon on 1 of 100 smaller pixels covering the same > area are equivalent events. What should be different in the analysis, if > you're thinking about individual pixels, is that the expected value for a > photon landing on any of the pixels will be 100 times lower for each of the > smaller pixels than the single big pixel, so that the expected value of > their sum is the same. You won't get to that conclusion without having a > different prior probability for the two cases that reflects the 100-fold > lower flux through the smaller area, regardless of the total power of the > source. > > > > Best wishes, > > Randy > > > > On 21 Oct 2021, at 13:03, Kay Diederichs <kay.diederi...@uni-konstanz.de> > wrote: > > > > Randy, > > I must admit that I am not certain about my answer, but I lean toward > thinking that the result (of the two thought experiments that you describe) > is not the same. I do agree that it makes sense that the expectation value > is the same, and the math that I sketched in > https://www.jiscmail.ac.uk/cgi-bin/wa-jisc.exe?A2=CCP4BB;bdd31b04.2110 > actually shows this. But the variance? To me, a 100-pixel patch with all > zeroes is no different from sequentially observing 100 pixels, one after > the other. For the first of these pixels, I have no idea what the count is, > until I observe it. For the second, I am less surprised that it is 0 > because I observed 0 for the first. And so on, until the 100th. For the > last one, my belief that I will observe a zero before I read out the pixel > is much higher than for the first pixel. The variance is just the inverse > of the amount of error (squared) that we assign to our belief in the > expectation value. And that amount of belief is very different. I find it > satisfactory that the sigma goes down with the sqrt() of the number of > pixels. > > Also, I don't find an error in the math of my posting of Mon, 18 Oct 2021 > 15:00:42 +0100 . I do think that a uniform prior is not realistic, but this > does not seem to make much difference for the 100-pixel thought experiment. > > We could change the thought experiment in the following way - you observe > 99 pixels with zero counts, and 1 with 1 count. Would you still say that > both the big-pixel-single-observation and the 100-pixel experiment should > give expectation value of 2 and variance of 2? I wouldn't. > > Best wishes, > Kay > > On Thu, 21 Oct 2021 09:00:23 +0000, Randy John Read <rj...@cam.ac.uk> > wrote: > > Just to be a bit clearer, I mean that the calculation of the expected > value and its variance should give the same answer if you're comparing one > pixel for a particular length of exposure with the sum obtained from either > a larger number of smaller pixels covering the same area for the same > length of exposure, or the sum from the same pixel measured for smaller > time slices adding up to the same total exposure. > > On 21 Oct 2021, at 09:54, Randy John Read <rj...@cam.ac.uk< > mailto:rj...@cam.ac.uk <rj...@cam.ac.uk>>> wrote: > > I would think that if this problem is being approached correctly, with the > right prior, it shouldn't matter whether you collect the same signal > distributed over 100 smaller pixels or the same pixel measured for the same > length of exposure but with 100 time slices; you should get the same > answer. So I would want to formulate the problem in a way where this > invariance is satisfied. I thought it was, from some of the earlier > descriptions of the problem, but this sounds worrying. > > I think you're trying to say the same thing here, Kay. Is that right? > > Best wishes, > > Randy > > On 21 Oct 2021, at 08:51, Kay Diederichs <kay.diederi...@uni-konstanz.de< > mailto:kay.diederi...@uni-konstanz.de <kay.diederi...@uni-konstanz.de>>> > wrote: > > Hi Ian, > > it is Iobs=0.01 and sigIobs=0.01 for one pixel, but adding 100 pixels each > with variance=sigIobs^2=0.0001 gives 0.01 , yielding a 100-pixel-sigIobs > of 0.1 - different from the 1 you get. As if repeatedly observing the same > count of 0 lowers the estimated error by sqrt(n), where n is the number of > observations (100 in this case). > > best wishes, > Kay > > On Wed, 20 Oct 2021 13:08:33 +0100, Ian Tickle <ianj...@gmail.com< > mailto:ianj...@gmail.com <ianj...@gmail.com>>> wrote: > > Hi Kay > > Can I just confirm that your result Iobs=0.01 sigIobs=0.01 is the estimate > of the true average intensity *per pixel* for a patch of 100 pixels? So > then the total count for all 100 pixels is 1 with variance also 1, or in > general for k observed counts in the patch, expectation = variance = k+1 > for the total count, irrespective of the number of pixels? If so then that > agrees with my own conclusion. It makes sense because Iobs=0.01 > sigIobs=0.01 cannot come from a Poisson process (which obviously requires > expectation = variance = an integer), whereas the total count does come > from a Poisson process. > > The difference from my approach is that you seem to have come at it via the > individual pixel counts whereas I came straight from the Agostini result > applied to the whole patch. The number of pixels seems to me to be > irrelevant for the whole patch since the design of the detector, assuming > it's an ideal detector with DQE = 1 surely cannot change the photon flux > coming from the source: all ideal detectors whatever their pixel layout > must give the same result. The number of pixels is then only relevant if > one needs to know the average intensity per pixel, i.e. the total and s.d. > divided by the number of pixels. Note the pixels here need not even > correspond to the hardware pixels, they can be any arbitrary subdivision of > the detector surface. > > Best wishes > > -- Ian > > > On Tue, 19 Oct 2021 at 12:39, Kay Diederichs < > kay.diederi...@uni-konstanz.de<mailto:kay.diederi...@uni-konstanz.de > <kay.diederi...@uni-konstanz.de>>> > wrote: > > James, > > I am saying that my answer to "what is the expectation and variance if I > observe a 10x10 patch of pixels with zero > counts?" is Iobs=0.01 sigIobs=0.01 (and Iobs=sigIobs=1 if there is only > one pixel) IF the uniform prior applies. I agree with Gergely and others > that this prior (with its high expectation value and variance) appears > unrealistic. > > In your posting of Sat, 16 Oct 2021 12:00:30 -0700 you make a calculation > of Ppix that appears like a more suitable expectation value of a prior to > me. A suitable prior might then be 1/Ppix * e^(-l/Ppix) (Agostini §7.7.1). > The Bayesian argument is IIUC that the prior plays a minor role if you do > repeated measurements of the same value, because you use the posterior of > the first measurement as the prior for the second, and so on. What this > means is that your Ppix must play the role of a scale factor if you > consider the 100-pixel experiment. > However, for the 1-pixel experiment, having a more suitable prior should > be more important. > > best, > Kay > > > > > On Mon, 18 Oct 2021 12:40:45 -0700, James Holton <jmhol...@lbl.gov< > mailto:jmhol...@lbl.gov <jmhol...@lbl.gov>>> wrote: > > Thank you very much for this Kay! > > So, to summarize, you are saying the answer to my question "what is the > expectation and variance if I observe a 10x10 patch of pixels with zero > counts?" is: > Iobs = 0.01 > sigIobs = 0.01 (defining sigIobs = sqrt(variance(Iobs))) > > And for the one-pixel case: > Iobs = 1 > sigIobs = 1 > > but in both cases the distribution is NOT Gaussian, but rather > exponential. And that means adding variances may not be the way to > propagate error. > > Is that right? > > -James Holton > MAD Scientist > > > > On 10/18/2021 7:00 AM, Kay Diederichs wrote: > Hi James, > > I'm a bit behind ... > > My answer about the basic question ("a patch of 100 pixels each with > zero counts - what is the variance?") you ask is the following: > > 1) we all know the Poisson PDF (Probability Distribution Function) > P(k|l) = l^k*e^(-l)/k! (where k stands for for an integer >=0 and l is > lambda) which tells us the probability of observing k counts if we know l. > The PDF is normalized: SUM_over_k (P(k|l)) is 1 when k=0...infinity is 1. > 2) you don't know before the experiment what l is, and you assume it is > some number x with 0<=x<=xmax (the xmax limit can be calculated by looking > at the physics of the experiment; it is finite and less than the overload > value of the pixel, otherwise you should do a different experiment). Since > you don't know that number, all the x values are equally likely - you use a > uniform prior. > 3) what is the PDF P(l|k) of l if we observe k counts? That can be > found with Bayes theorem, and it turns out that (due to the uniform prior) > the right hand side of the formula looks the same as in 1) : P(l|k) = > l^k*e^(-l)/k! (again, the ! stands for the factorial, it is not a semantic > exclamation mark). This is eqs. 7.42 and 7.43 in Agostini "Bayesian > Reasoning in Data Analysis". > 3a) side note: if we calculate the expectation value for l, by > multiplying with l and integrating over l from 0 to infinity, we obtain > E(P(l|k))=k+1, and similarly for the variance (Agostini eqs 7.45 and 7.46) > 4) for k=0 (zero counts observed in a single pixel), this reduces to > P(l|0)=e^(-l) for a single observation (pixel). (this is basic math; see > also §7.4.1 of Agostini. > 5) since we have 100 independent pixels, we must multiply the > individual PDFs to get the overall PDF f, and also normalize to make the > integral over that PDF to be 1: the result is f(l|all 100 pixels are > 0)=n*e^(-n*l). (basic math). A more Bayesian procedure would be to realize > that the posterior PDF P(l|0)=e^(-l) of the first pixel should be used as > the prior for the second pixel, and so forth until the 100th pixel. This > has the same result f(l|all 100 pixels are 0)=n*e^(-n*l) (Agostini § > 7.7.2)! > 6) the expectation value INTEGRAL_0_to_infinity over l*n*e^(-n*l) dl is > 1/n . This is 1 if n=1 as we know from 3a), and 1/100 for 100 pixels with > 0 counts. > 7) the variance is then INTEGRAL_0_to_infinity over > (l-1/n)^2*n*e^(-n*l) dl . This is 1/n^2 > > I find these results quite satisfactory. Please note that they deviate > from the MLE result: expectation value=0, variance=0 . The problem appears > to be that a Maximum Likelihood Estimator may give wrong results for small > n; something that I've read a couple of times but which appears not to be > universally known/taught. Clearly, the result in 6) and 7) for large n > converges towards 0, as it should be. > What this also means is that one should really work out the PDF instead > of just adding expectation values and variances (and arriving at 100 if all > 100 pixels have zero counts) because it is contradictory to use a uniform > prior for all the pixels if OTOH these agree perfectly in being 0! > > What this means for zero-dose extrapolation I have not thought about. > At least it prevents infinite weights! > > Best, > Kay > > > > > > ######################################################################## > > 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 > <http://www.jiscmail.ac.uk/CCP4BB%3chttp:/www.jiscmail.ac.uk/CCP4BB>>, a > mailing list hosted by www.jiscmail.ac.uk<http://www.jiscmail.ac.uk/ > <http://www.jiscmail.ac.uk%3chttp:/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<http://www.jiscmail.ac.uk/CCP4BB > <http://www.jiscmail.ac.uk/CCP4BB%3chttp:/www.jiscmail.ac.uk/CCP4BB>>, a > mailing list hosted by www.jiscmail.ac.uk<http://www.jiscmail.ac.uk > <http://www.jiscmail.ac.uk%3chttp:/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<http://www.jiscmail.ac.uk/CCP4BB > <http://www.jiscmail.ac.uk/CCP4BB%3chttp:/www.jiscmail.ac.uk/CCP4BB>>, a > mailing list hosted by www.jiscmail.ac.uk<http://www.jiscmail.ac.uk > <http://www.jiscmail.ac.uk%3chttp:/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<http://www.jiscmail.ac.uk/CCP4BB > <http://www.jiscmail.ac.uk/CCP4BB%3chttp:/www.jiscmail.ac.uk/CCP4BB>>, a > mailing list hosted by www.jiscmail.ac.uk<http://www.jiscmail.ac.uk > <http://www.jiscmail.ac.uk%3chttp:/www.jiscmail.ac.uk>>, terms & > conditions are available at https://www.jiscmail.ac.uk/policyandsecurity/ > > ------ > Randy J. Read > Department of Haematology, University of Cambridge > Cambridge Institute for Medical Research Tel: + 44 1223 336500 > The Keith Peters Building Fax: + 44 1223 > 336827 > Hills Road E-mail: > rj...@cam.ac.uk<mailto:rj...@cam.ac.uk> > Cambridge CB2 0XY, U.K. > www-structmed.cimr.cam.ac.uk<http://www-structmed.cimr.cam.ac.uk/> > > > ________________________________ > > To unsubscribe from the CCP4BB list, click the following link: > https://www.jiscmail.ac.uk/cgi-bin/WA-JISC.exe?SUBED1=CCP4BB&A=1 > > ------ > Randy J. Read > Department of Haematology, University of Cambridge > Cambridge Institute for Medical Research Tel: + 44 1223 336500 > The Keith Peters Building Fax: + 44 1223 > 336827 > Hills Road E-mail: > rj...@cam.ac.uk<mailto:rj...@cam.ac.uk> > Cambridge CB2 0XY, U.K. > www-structmed.cimr.cam.ac.uk<http://www-structmed.cimr.cam.ac.uk> > > > ######################################################################## > > 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/ > > > > > > ------ > > Randy J. Read > > Department of Haematology, University of Cambridge > > Cambridge Institute for Medical Research Tel: + 44 1223 336500 > > The Keith Peters Building Fax: + 44 1223 > 336827 > > Hills Road E-mail: > rj...@cam.ac.uk <rj...@cam.ac.uk> > > Cambridge CB2 0XY, U.K. > www-structmed.cimr.cam.ac.uk > > > > > ------------------------------ > > 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. 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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 > -- +++++++++++++++++++++++++++++++++++++++++++++++++ Jacob Pearson Keller Assistant Professor Department of Pharmacology and Molecular Therapeutics Uniformed Services University 4301 Jones Bridge Road Bethesda MD 20814 jacob.kel...@usuhs.edu; jacobpkel...@gmail.com Cell: (301)592-7004 +++++++++++++++++++++++++++++++++++++++++++++++++ ######################################################################## 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/
