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I am simulating a simple detector, imaging a very faint astronomical source (so that some dozens of (uncorrelated) photons are detected in every frame)

Currently I am first sampling from Poisson-distribution to determine how many photons are to arrive in a given frame.

Then for every photon I have to determine the pixel it will hit: I have for every pixel a number, that gives the probability, that if a photon comes, it will hit that pixel. (so that the sum over all pixels is 1.) I have tools to sample N hits from this PSF (the pixelwise hit probabilities are proportional to the Point Spread Function of the telescope)

It would however be more convenient, if I could simply multiply the PSF with the expected number of photons per frame, and sample for each pixel from this Poisson distribution.

Are the two methods equivalent? If not, which of them is more correct?

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    $\begingroup$ From what I read you have all the tools to decide this numerically. $\endgroup$ – my2cts Jan 7 at 17:11
  • $\begingroup$ The coefficient of linear expansion $\alpha$ is defined as $\ell + \Delta\ell = \ell(1 + \alpha)\Delta T$ $\endgroup$ – Yuvraj Jan 7 at 17:23
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If $X$ is a random variable, which is distributed according to a Poissonian distribution with rate parameter $\lambda_1$, we write $X\sim \textrm{Pois}(\lambda_1)$. Now, if $Y\sim \textrm{Pois}(\lambda_2)$ then $$Z=X+Y\sim \textrm{Pois}(\lambda_1) + \textrm{Pois}(\lambda_2) = \textrm{Pois}(\lambda_1 + \lambda_2)$$ In your case you have the same rate parameter $\lambda$ for each pixel, $X_{\textrm{pixel}} = \textrm{Pois}(\lambda)$. Thus, if you have $N$ pixels per frame the photon number per frame is distributed as $X_{\textrm{frame}} = \textrm{Pois}(N\lambda)$. Hence, you are definitely allowed to sample from a Poissonian distribution for each pixel, if you assume that the counts on each pixel are statistically independent.

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  • $\begingroup$ But I do not have the same rate for every pixel. Rather I have a distribution which specifies the probability that an incoming photon will hit that pixel (and not other). And a global rate. $\endgroup$ – b.Lorenz Jan 7 at 20:27
  • $\begingroup$ That was not stated in the question. It would be great, if you could edit your question, such that this becomes clear. For me, the setup is currently unclear. Two further remarks: (1) You could look into joint density distribution functions. However, this probably doesn't simplify your calculation. (2) I recommend that such kind of questions are ask on a statistic side, e.g. stats.stackexchange.com However, be aware that they are rather mathematical and that you should provide your mathematical model in greater detail. $\endgroup$ – Semoi Jan 8 at 6:57

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