<div dir="ltr"><div>Dear All,</div><div></div><div>In the context of this discussion, I suggest the readers / discussion participants to have a look at new ideas on: Information Content of images, SNR, DQEs, A priori Probabilities, Shannon's Channel Information capacity,etc., expressed in: Van Heel & Schatz, arXiv 2020. (There is also an easy-to-find Youtube lecture on the issue by "marin van heel").</div><div>Cheers,</div><div>Marin<br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Tue, Oct 26, 2021 at 3:58 AM Jacob Keller <<a href="mailto:jacobpkeller@gmail.com">jacobpkeller@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="ltr"><div>Hi All,</div><div><br></div><div>haven't been following CCP4BB for a while, then I come back to this juicy Holtonian thread!</div><div><br></div><div>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?</div><div><br></div><div>Jacob</div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Fri, Oct 22, 2021 at 3:25 AM Gergely Katona <<a href="mailto:gergely.katona@gu.se" target="_blank">gergely.katona@gu.se</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div lang="EN-US">
<div>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif">Hi,<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif">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.<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif">Details of the calculation below:<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif"><a href="https://urldefense.proofpoint.com/v2/url?u=https-3A__colab.research.google.com_drive_1NK43-5F3r1rH5lBTDS2rzIFDFNWqFfekrZ-3Fusp-3Dsharing&d=DwMFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=SH8-k6_ktONw_SFkmg4v6mJm95Tfznxi63jj8FQjahI&e=" target="_blank">https://colab.research.google.com/drive/1NK43_3r1rH5lBTDS2rzIFDFNWqFfekrZ?usp=sharing</a><u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif">Best wishes,<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif">Gergely<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif"><u></u> <u></u></span></p>
<div>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif">Gergely Katona, Professor, Chairman of the Chemistry Program Council<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif">Department of Chemistry and Molecular Biology, University of Gothenburg<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif" lang="SV">Box 462, 40530 Göteborg, Sweden<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif" lang="SV">Tel: +46-31-786-3959 / M: +46-70-912-3309 / Fax: +46-31-786-3910<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif">Web: <a href="https://urldefense.proofpoint.com/v2/url?u=http-3A__katonalab.eu&d=DwMFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=WtAab9SqD68vXGNd_F4ilEWXlIeUv4CmaJeYQuDfROQ&e=" target="_blank">http://katonalab.eu</a>, Email: <a href="mailto:gergely.katona@gu.se" target="_blank">gergely.katona@gu.se</a><u></u><u></u></span></p>
</div>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif"><u></u> <u></u></span></p>
<div>
<div style="border-color:rgb(225,225,225) currentcolor currentcolor;border-style:solid none none;border-width:1pt medium medium;padding:3pt 0in 0in">
<p class="MsoNormal"><b><span style="font-size:11pt;font-family:Calibri,sans-serif">From:</span></b><span style="font-size:11pt;font-family:Calibri,sans-serif"> CCP4 bulletin board <<a href="mailto:CCP4BB@JISCMAIL.AC.UK" target="_blank">CCP4BB@JISCMAIL.AC.UK</a>>
<b>On Behalf Of </b>Nave, Colin (DLSLtd,RAL,LSCI)<br>
<b>Sent:</b> 21 October, 2021 19:21<br>
<b>To:</b> <a href="mailto:CCP4BB@JISCMAIL.AC.UK" target="_blank">CCP4BB@JISCMAIL.AC.UK</a><br>
<b>Subject:</b> Re: [ccp4bb] am I doing this right?<u></u><u></u></span></p>
</div>
</div>
<p class="MsoNormal"><u></u> <u></u></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB">Congratulations to James for starting this interesting discussion.
<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96);background:white none repeat scroll 0% 0%" lang="EN-GB">For those who are like me, nowhere near a black belt in statistics, t</span><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB">he
thread has included a number of distributions. I have had to look up where these apply and investigate their properties.
<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96);background:white none repeat scroll 0% 0%" lang="EN-GB">As an example,
</span><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB"><u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB">“The P<span style="background:white none repeat scroll 0% 0%">oisson 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.”<u></u><u></u></span></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96);background:white none repeat scroll 0% 0%" lang="EN-GB">A useful calculator for distributions can be found at<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB"><a href="https://urldefense.proofpoint.com/v2/url?u=https-3A__keisan.casio.com_menu_system_000000000540&d=DwMFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=vSFYZr4wbWSz1xVDpjU0ou57-qAxsXr5zxmefXAblf8&e=" target="_blank"><span style="color:rgb(0,32,96)">https://keisan.casio.com/menu/system/000000000540</span></a><u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB">a specific example is at<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB"><a href="https://urldefense.proofpoint.com/v2/url?u=https-3A__keisan.casio.com_exec_system_1180573179&d=DwMFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=e_uXyWeSVfcQs_ypUwBLU_b120FEMZZRNsd4OMw7MP0&e=" target="_blank"><span style="color:rgb(0,32,96)">https://keisan.casio.com/exec/system/1180573179</span></a><u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB">where cumulative probabilities for a Poisson distribution can be found given values for x and lambda.<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB">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.
<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB">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! <u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB">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? <u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB">Thanks to all for the interesting points.<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(0,32,96)" lang="EN-GB">Colin</span><span style="font-size:11pt;font-family:Calibri,sans-serif;color:rgb(31,73,125)" lang="EN-GB"><u></u><u></u></span></p>
<div>
<div style="border-color:rgb(225,225,225) currentcolor currentcolor;border-style:solid none none;border-width:1pt medium medium;padding:3pt 0in 0in">
<p class="MsoNormal"><b><span style="font-size:11pt;font-family:Calibri,sans-serif">From:</span></b><span style="font-size:11pt;font-family:Calibri,sans-serif"> CCP4 bulletin board <<a href="mailto:CCP4BB@JISCMAIL.AC.UK" target="_blank">CCP4BB@JISCMAIL.AC.UK</a>>
<b>On Behalf Of </b>Randy John Read<br>
<b>Sent:</b> 21 October 2021 13:23<br>
<b>To:</b> <a href="mailto:CCP4BB@JISCMAIL.AC.UK" target="_blank">CCP4BB@JISCMAIL.AC.UK</a><br>
<b>Subject:</b> Re: [ccp4bb] am I doing this right?<u></u><u></u></span></p>
</div>
</div>
<p class="MsoNormal"><span lang="EN-GB"><u></u> <u></u></span></p>
<div>
<p class="MsoNormal"><span lang="EN-GB">Hi Kay,<u></u><u></u></span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-GB"><u></u> <u></u></span></p>
</div>
<p class="MsoNormal"><span lang="EN-GB">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. <u></u><u></u></span></p>
<div>
<p class="MsoNormal"><span lang="EN-GB"><u></u> <u></u></span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-GB">Best wishes,<u></u><u></u></span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-GB">Randy<u></u><u></u></span></p>
<div>
<p class="MsoNormal" style="margin-bottom:12pt"><span lang="EN-GB"><u></u> <u></u></span></p>
<blockquote style="margin-top:5pt;margin-bottom:5pt">
<div>
<p class="MsoNormal"><span lang="EN-GB">On 21 Oct 2021, at 13:03, Kay Diederichs <<a href="mailto:kay.diederichs@uni-konstanz.de" target="_blank">kay.diederichs@uni-konstanz.de</a>> wrote:<u></u><u></u></span></p>
</div>
<p class="MsoNormal"><span lang="EN-GB"><u></u> <u></u></span></p>
<div>
<div>
<p class="MsoNormal" style="margin-bottom:12pt"><span lang="EN-GB">Randy,<br>
<br>
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 <a href="https://urldefense.proofpoint.com/v2/url?u=https-3A__www.jiscmail.ac.uk_cgi-2Dbin_wa-2Djisc.exe-3FA2-3DCCP4BB-3Bbdd31b04.2110&d=DwMFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=mhUwmSlF2o-ffQtdYI3zFn1XhBTscxvjXTfk5O47CjM&e=" target="_blank">
https://www.jiscmail.ac.uk/cgi-bin/wa-jisc.exe?A2=CCP4BB;bdd31b04.2110</a> 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.<br>
<br>
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.<br>
<br>
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.<br>
<br>
Best wishes,<br>
Kay<br>
<br>
On Thu, 21 Oct 2021 09:00:23 +0000, Randy John Read <<a href="mailto:rjr27@CAM.AC.UK" target="_blank">rjr27@CAM.AC.UK</a>> wrote:<br>
<br>
<u></u><u></u></span></p>
<blockquote style="margin-top:5pt;margin-bottom:5pt">
<p class="MsoNormal"><span lang="EN-GB">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.<br>
<br>
On 21 Oct 2021, at 09:54, Randy John Read <<a href="mailto:rjr27@cam.ac.uk" target="_blank">rjr27@cam.ac.uk</a><<a href="mailto:rjr27@cam.ac.uk" target="_blank">mailto:rjr27@cam.ac.uk</a>>> wrote:<br>
<br>
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.<br>
<br>
I think you're trying to say the same thing here, Kay. Is that right?<br>
<br>
Best wishes,<br>
<br>
Randy<br>
<br>
On 21 Oct 2021, at 08:51, Kay Diederichs <<a href="mailto:kay.diederichs@uni-konstanz.de" target="_blank">kay.diederichs@uni-konstanz.de</a><<a href="mailto:kay.diederichs@uni-konstanz.de" target="_blank">mailto:kay.diederichs@uni-konstanz.de</a>>> wrote:<br>
<br>
Hi Ian,<br>
<br>
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).<br>
<br>
best wishes,<br>
Kay<br>
<br>
On Wed, 20 Oct 2021 13:08:33 +0100, Ian Tickle <<a href="mailto:ianjt05@GMAIL.COM" target="_blank">ianjt05@GMAIL.COM</a><<a href="mailto:ianjt05@GMAIL.COM" target="_blank">mailto:ianjt05@GMAIL.COM</a>>> wrote:<br>
<br>
Hi Kay<br>
<br>
Can I just confirm that your result Iobs=0.01 sigIobs=0.01 is the estimate<br>
of the true average intensity *per pixel* for a patch of 100 pixels? So<br>
then the total count for all 100 pixels is 1 with variance also 1, or in<br>
general for k observed counts in the patch, expectation = variance = k+1<br>
for the total count, irrespective of the number of pixels? If so then that<br>
agrees with my own conclusion. It makes sense because Iobs=0.01<br>
sigIobs=0.01 cannot come from a Poisson process (which obviously requires<br>
expectation = variance = an integer), whereas the total count does come<br>
from a Poisson process.<br>
<br>
The difference from my approach is that you seem to have come at it via the<br>
individual pixel counts whereas I came straight from the Agostini result<br>
applied to the whole patch. The number of pixels seems to me to be<br>
irrelevant for the whole patch since the design of the detector, assuming<br>
it's an ideal detector with DQE = 1 surely cannot change the photon flux<br>
coming from the source: all ideal detectors whatever their pixel layout<br>
must give the same result. The number of pixels is then only relevant if<br>
one needs to know the average intensity per pixel, i.e. the total and s.d.<br>
divided by the number of pixels. Note the pixels here need not even<br>
correspond to the hardware pixels, they can be any arbitrary subdivision of<br>
the detector surface.<br>
<br>
Best wishes<br>
<br>
-- Ian<br>
<br>
<br>
On Tue, 19 Oct 2021 at 12:39, Kay Diederichs <<a href="mailto:kay.diederichs@uni-konstanz.de" target="_blank">kay.diederichs@uni-konstanz.de</a><<a href="mailto:kay.diederichs@uni-konstanz.de" target="_blank">mailto:kay.diederichs@uni-konstanz.de</a>>><br>
wrote:<br>
<br>
James,<br>
<br>
I am saying that my answer to "what is the expectation and variance if I<br>
observe a 10x10 patch of pixels with zero<br>
counts?" is Iobs=0.01 sigIobs=0.01 (and Iobs=sigIobs=1 if there is only<br>
one pixel) IF the uniform prior applies. I agree with Gergely and others<br>
that this prior (with its high expectation value and variance) appears<br>
unrealistic.<br>
<br>
In your posting of Sat, 16 Oct 2021 12:00:30 -0700 you make a calculation<br>
of Ppix that appears like a more suitable expectation value of a prior to<br>
me. A suitable prior might then be 1/Ppix * e^(-l/Ppix) (Agostini §7.7.1).<br>
The Bayesian argument is IIUC that the prior plays a minor role if you do<br>
repeated measurements of the same value, because you use the posterior of<br>
the first measurement as the prior for the second, and so on. What this<br>
means is that your Ppix must play the role of a scale factor if you<br>
consider the 100-pixel experiment.<br>
However, for the 1-pixel experiment, having a more suitable prior should<br>
be more important.<br>
<br>
best,<br>
Kay<br>
<br>
<br>
<br>
<br>
On Mon, 18 Oct 2021 12:40:45 -0700, James Holton <<a href="mailto:jmholton@LBL.GOV" target="_blank">jmholton@LBL.GOV</a><<a href="mailto:jmholton@LBL.GOV" target="_blank">mailto:jmholton@LBL.GOV</a>>> wrote:<br>
<br>
Thank you very much for this Kay!<br>
<br>
So, to summarize, you are saying the answer to my question "what is the<br>
expectation and variance if I observe a 10x10 patch of pixels with zero<br>
counts?" is:<br>
Iobs = 0.01<br>
sigIobs = 0.01 (defining sigIobs = sqrt(variance(Iobs)))<br>
<br>
And for the one-pixel case:<br>
Iobs = 1<br>
sigIobs = 1<br>
<br>
but in both cases the distribution is NOT Gaussian, but rather<br>
exponential. And that means adding variances may not be the way to<br>
propagate error.<br>
<br>
Is that right?<br>
<br>
-James Holton<br>
MAD Scientist<br>
<br>
<br>
<br>
On 10/18/2021 7:00 AM, Kay Diederichs wrote:<br>
Hi James,<br>
<br>
I'm a bit behind ...<br>
<br>
My answer about the basic question ("a patch of 100 pixels each with<br>
zero counts - what is the variance?") you ask is the following:<br>
<br>
1) we all know the Poisson PDF (Probability Distribution Function)<br>
P(k|l) = l^k*e^(-l)/k! (where k stands for for an integer >=0 and l is<br>
lambda) which tells us the probability of observing k counts if we know l.<br>
The PDF is normalized: SUM_over_k (P(k|l)) is 1 when k=0...infinity is 1.<br>
2) you don't know before the experiment what l is, and you assume it is<br>
some number x with 0<=x<=xmax (the xmax limit can be calculated by looking<br>
at the physics of the experiment; it is finite and less than the overload<br>
value of the pixel, otherwise you should do a different experiment). Since<br>
you don't know that number, all the x values are equally likely - you use a<br>
uniform prior.<br>
3) what is the PDF P(l|k) of l if we observe k counts? That can be<br>
found with Bayes theorem, and it turns out that (due to the uniform prior)<br>
the right hand side of the formula looks the same as in 1) : P(l|k) =<br>
l^k*e^(-l)/k! (again, the ! stands for the factorial, it is not a semantic<br>
exclamation mark). This is eqs. 7.42 and 7.43 in Agostini "Bayesian<br>
Reasoning in Data Analysis".<br>
3a) side note: if we calculate the expectation value for l, by<br>
multiplying with l and integrating over l from 0 to infinity, we obtain<br>
E(P(l|k))=k+1, and similarly for the variance (Agostini eqs 7.45 and 7.46)<br>
4) for k=0 (zero counts observed in a single pixel), this reduces to<br>
P(l|0)=e^(-l) for a single observation (pixel). (this is basic math; see<br>
also §7.4.1 of Agostini.<br>
5) since we have 100 independent pixels, we must multiply the<br>
individual PDFs to get the overall PDF f, and also normalize to make the<br>
integral over that PDF to be 1: the result is f(l|all 100 pixels are<br>
0)=n*e^(-n*l). (basic math). A more Bayesian procedure would be to realize<br>
that the posterior PDF P(l|0)=e^(-l) of the first pixel should be used as<br>
the prior for the second pixel, and so forth until the 100th pixel. This<br>
has the same result f(l|all 100 pixels are 0)=n*e^(-n*l) (Agostini § 7.7.2)!<br>
6) the expectation value INTEGRAL_0_to_infinity over l*n*e^(-n*l) dl is<br>
1/n . This is 1 if n=1 as we know from 3a), and 1/100 for 100 pixels with<br>
0 counts.<br>
7) the variance is then INTEGRAL_0_to_infinity over<br>
(l-1/n)^2*n*e^(-n*l) dl . This is 1/n^2<br>
<br>
I find these results quite satisfactory. Please note that they deviate<br>
from the MLE result: expectation value=0, variance=0 . The problem appears<br>
to be that a Maximum Likelihood Estimator may give wrong results for small<br>
n; something that I've read a couple of times but which appears not to be<br>
universally known/taught. Clearly, the result in 6) and 7) for large n<br>
converges towards 0, as it should be.<br>
What this also means is that one should really work out the PDF instead<br>
of just adding expectation values and variances (and arriving at 100 if all<br>
100 pixels have zero counts) because it is contradictory to use a uniform<br>
prior for all the pixels if OTOH these agree perfectly in being 0!<br>
<br>
What this means for zero-dose extrapolation I have not thought about.<br>
At least it prevents infinite weights!<br>
<br>
Best,<br>
Kay<br>
<br>
<br>
<br>
<br>
<br>
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Randy J. Read<br>
Department of Haematology, University of Cambridge<br>
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------<br>
Randy J. Read<br>
Department of Haematology, University of Cambridge<br>
Cambridge Institute for Medical Research Tel: + 44 1223 336500<br>
The Keith Peters Building Fax: + 44 1223 336827<br>
Hills Road E-mail: <a href="mailto:rjr27@cam.ac.uk%3cmailto:rjr27@cam.ac.uk" target="_blank">
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<p class="MsoNormal"><span style="font-size:9pt;font-family:Helvetica,sans-serif;color:black" lang="EN-GB">------<u></u><u></u></span></p>
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<p class="MsoNormal"><span style="font-size:9pt;font-family:Helvetica,sans-serif;color:black" lang="EN-GB">Randy J. Read<u></u><u></u></span></p>
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<p class="MsoNormal"><span style="font-size:9pt;font-family:Helvetica,sans-serif;color:black" lang="EN-GB">Department of Haematology, University of Cambridge<u></u><u></u></span></p>
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<p class="MsoNormal"><span style="font-size:9pt;font-family:Helvetica,sans-serif;color:black" lang="EN-GB">Cambridge Institute for Medical Research Tel: + 44 1223 336500<u></u><u></u></span></p>
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</blockquote></div><br clear="all"><div><br></div>-- <br><div dir="ltr"><div dir="ltr"><p class="MsoNormal" style="color:rgb(34,34,34)">+++++++++++++++++++++++++++++++++++++++++++++++++</p><p class="MsoNormal" style="color:rgb(34,34,34)">Jacob Pearson Keller</p><p class="MsoNormal" style="color:rgb(34,34,34)">Assistant Professor</p><p class="MsoNormal" style="color:rgb(34,34,34)">Department of Pharmacology and Molecular Therapeutics</p><p class="MsoNormal" style="color:rgb(34,34,34)">Uniformed Services University</p><p class="MsoNormal" style="color:rgb(34,34,34)">4301 Jones Bridge Road</p><p class="MsoNormal" style="color:rgb(34,34,34)">Bethesda MD 20814</p><p class="MsoNormal" style="color:rgb(34,34,34)"><a href="mailto:jacob.keller@usuhs.edu" target="_blank"><span style="color:blue">jacob.keller@usuhs.edu</span></a>; <a href="mailto:jacobpkeller@gmail.com" target="_blank"><span style="color:blue">jacobpkeller@gmail.com</span></a></p><p class="MsoNormal" style="color:rgb(34,34,34)"><span style="background-image:initial;background-position:initial;background-repeat:initial">Cell: (301)592-7004</span></p><p class="MsoNormal" style="color:rgb(34,34,34)"><span style="background-image:initial;background-position:initial;background-repeat:initial">+++++++++++++++++++++++++++++++++++++++++++++++++</span></p></div></div></div>
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