poisson posterior = gamma function x uniform prior? 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.
 A: 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.
A: The reason that using uniform prior distribution (note that uniform distribution is a special case of Gamma distribution for $\alpha = 1$ and $\beta = 1$) will result in Gamma posterior when assuming the data comes from Poisson distribution, is that Gamma distribution is conjugate prior to the Poisson distribution.
Using conjugate prior is convenient because you can be sure that no matter what the results of the experiment will be like, posterior distribution of the unknown parameter will be from the same distribution family as the prior distribution.
