Dear James, I am sorry to nitpick, but this is the answer to "what is my belief of expectation and variance if I observe a 10x10 patch of pixels with zero counts?" This will heavily depend on my model. When I make predictions like this, my intention is not to replace the data with a "new and improved" data that is closer to the Truth and deposit in some database from the position of authority.
I would simply use it to validate my model. Well, my model expects the Iobs to be 0.01, but in fact it is 0. This may make me slightly worried, but then I look at the posterior distribution and I see 0 with highest posterior probability so I relax a bit that I do not have to throw out my model outright. Still, a better model may be out there. For a Bayesian the data is fixed and holy, the model may change. And the question rarely manifests like that one does not have to spend a lot of time pondering about if a uniform distribution of the rate is compatible with my belief in some quantum process. Bayesian folks are pragmatic. Your question about "what is my belief about the slope and intercept of a line that is the basis of some time-dependent random process given my observations" is more relevant. It is straightforward to implement as a Bayesian network to answer this question and it will give you predictions that looks deceptively like the data. Here, you only care about your prior belief about the magnitude of slope and intercept, the belief about what the rate may be independent of time is quite irrelevant and so are the predictions they may make. And I guess you would not intend to deposit images that were generated by the predictions of these posterior models and the "new and improved data". 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 -----Original Message----- From: CCP4 bulletin board <CCP4BB@JISCMAIL.AC.UK> On Behalf Of James Holton Sent: 18 October, 2021 21:41 To: CCP4BB@JISCMAIL.AC.UK Subject: Re: [ccp4bb] am I doing this right? 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, 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/
