From cd9a037c2c4c9a9e65575c39928e88fb27385eef Mon Sep 17 00:00:00 2001
From: tlatorre <tlatorre@uchicago.edu>
Date: Mon, 21 Oct 2019 14:21:16 -0500
Subject: add appendix about the combination of poisson and binomial random
 variables

---
 doc/sddm.tex | 19 +++++++++++++++++++
 1 file changed, 19 insertions(+)

(limited to 'doc/sddm.tex')

diff --git a/doc/sddm.tex b/doc/sddm.tex
index 7da66e9..f4919e2 100644
--- a/doc/sddm.tex
+++ b/doc/sddm.tex
@@ -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.
-- 
cgit 

