WIP section 2.1

1 files changed, 16 insertions, 9 deletions

diff --git a/supplements/main.tex b/supplements/main.tex
index af9bded..d8c87d3 100644
--- a/supplements/main.tex
+++ b/supplements/main.tex
@@ -168,19 +168,26 @@ be expanded to:
 
 \section{Model Inference}
 
-\subsection{Background rate}
+\subsection{Exogenous Intensity}
 \label{sec:background}
 
-Because the seasonal variations in gunshot rates (Figure SX) remain consistent throughout the study period, we assume these are inherent to the infection process and not purely driven by noise or social contagion. Instead of having a constant background rate, we capture seasonal variations as a periodic sinusoidal function. We first compute the aggregate background rate of all the nodes, based on the number of infections each day.
-
-\begin{equation}
-M(t) = \mu_0 + A \sin(\omega t + \phi)
-\end{equation}
-
-Because we are modeling annual fluctuations, we know that the period is one year, i.e. $\omega=2\pi/365.24$. We learn the other three parameters using non-linear least squares estimates with the Gauss-Newton algorithm. This yields 
+Because the seasonal variations in gunshot rates (Figure~\ref{fig:background})
+remain consistent throughout the study period, we assume these are inherent to
+the infection process and not purely driven by noise or social contagion. We
+model these seasonal variations by a periodic sinusoidal function.
 
+Let $M(t)$ denote the number of infections occurring on day $t$ across all
+individuals. We assume the following form:
+\begin{displaymath}
+    M(t) = A\big[1 + \rho \sin(\omega t + \phi)\big]
+\end{displaymath}
+Because we are modeling annual fluctuations, we know that the period is one
+year, \emph{i.e.} $\omega=\frac{2\pi}{365.24}$. The remaining three parameters
+($A$, $\rho$ and $\phi$) are learnt using non-linear least squares estimates
+with the Gauss-Newton algorithm. This yields:
 \begin{equation}
-M(t) = 3.78 + 1.63 \sin(\frac{2\pi}{365.24} t + 4.36)
+    M(t) = 3.78\left[1 + 0.43 \sin\left(\frac{2\pi}{365.24}
+    t + 4.36\right)\right]
 \end{equation}
 
 \begin{figure}