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Dear Colin,
I think you are right: the question is quite open to interpretation.
Here is a Bayesian model, which shows that finding 100 pixels with 0 has almost
exactly 42% chance (which should come to no surprise to the fans of
Hitchhiker’s guide to Galaxy).
https://colab.research.google.com/
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 distrib
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I am sorry, this was a dead-end idea multinomial distribution with 0 trials is
not defined.
Gergely
From: CCP4 bulletin board On Behalf Of Gergely Katona
Sent: 21 October, 2021 15:29
To: CCP4BB@JISCMAIL.AC.UK
Subject: Re: [ccp4bb] am I doing this right?
Dear Randy and Kay,
My solution would i
Dear Randy and Kay,
My solution would involve a multinomial distribution for assigning the counts
to pixels.
Something like this:
rate ≈ Gamma(1,1)
total_counts ≈ Poisson(rate)
probs ≈ Dirichlet (alpha=1, for all pixels)
pixel_counts ≈ Multinomial (total_counts, probs of the different pixels
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 insignifi
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.
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 lengt
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.
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 th
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