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authortlatorre <tlatorre@uchicago.edu>2019-10-21 14:21:16 -0500
committertlatorre <tlatorre@uchicago.edu>2019-10-21 14:21:16 -0500
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add appendix about the combination of poisson and binomial random variables
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@@ -635,6 +635,25 @@ measurement and many of them end up reading outside of the linear TAC region.
Therefore we also tag any events in which less than 70\% of the PMT hits have a
TAC value above 400.
+\appendix
+\chapter{Poisson Binomial}
+
+Suppose we have a Poisson process whose output is then subject to a binomial
+process. For example, we expect $\mu$ background events on average and we can
+detect them with probability $p$. What is the probability of detecting $n$
+background events?
+
+\begin{align}
+p(n) &= \sum_{N=n}^{\infty} P(n|N) P(N) \\
+ &= \sum_{N=n}^{\infty} \frac{N!}{n!(N-n)!} p^n (1-p)^{N-n} e^{-\mu}\frac{\mu^N}{N!} \\
+ &= \sum_{N=n}^{\infty} \frac{1}{n!(N-n)!} p^n (1-p)^{N-n} e^{-\mu}\mu^N \\
+ &= e^{-\mu} \frac{p^n}{n!} \sum_{N=n}^{\infty} \frac{1}{(N-n)!} (1-p)^{N-n} \mu^N \\
+ &= e^{-\mu} \frac{(\mu p)^n}{n!} \sum_{N=n}^{\infty} \frac{\left(\mu (1-p)\right)^{N-n}}{(N-n)!} \\
+ &= e^{-\mu p} \frac{(\mu p)^n}{n!}
+\end{align}
+
+Therefore the end result is a Poisson distribution with mean $\mu p$.
+
\begin{thebibliography}{9}
\bibitem{grossman2017}
Grossman, et al. \textit{Self-Destructing Dark Matter}. \href{https://arxiv.org/abs/1712.00455}{{\tt arXiv:1712.00455}}. Dec 2017.