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I was reading the section 3.2 of this paper. Above equation (3.2), the authors say

"The posterior on the background estimate is then conservatively taken to be the Poisson posterior using a uniform prior for a sideband ... "

Then they show the equation \begin{equation} P(\mu_{BG}) = \Gamma(7N_{pass}/100+1.1/7) \end{equation} How could one possibly connect the Poisson posterior to a gamma function? The authors said that they used a uniform prior, and the posterior function is a Poisson distribution. Then is there a reason why the posterior equals to a Gamma function? I tried to explain this for an hour but couldn't.

In the equation $\mu_{BG}$ is the number of expected background events. ${N}_{pass}$ is defined in the paragraph above the equation 3.2. But I think the meaning of $N_{pass}$ is not important, since the point of my question is about the functional form.

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  • $\begingroup$ Please paste in whatever context is necessary to understand & answer your question. We want this thread to remain valuable even if the link goes dead. (Few people will try to answer if they need to read a paper first ...) $\endgroup$ – kjetil b halvorsen Nov 26 '20 at 15:51
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You misread the equation, it actually states $$P(\mu_{BG}) = \Gamma(7N_{pass}/100 + 1, 1/7),$$ with a "," instead of ".".

Indeed, it seems to the Gamma distribution with its two parameters, which is the conjugate prior distribution of a Poisson likelihood.

I do not understand the connection to the rest of the text in the paper, maybe someone with a better understanding of the subject may verify that the math display actually refers to a Gamma prior.

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