Polishing 2.1

1 files changed, 26 insertions, 31 deletions

diff --git a/supplements/main.tex b/supplements/main.tex
index 50de5a3..b6cc505 100644
--- a/supplements/main.tex
+++ b/supplements/main.tex
@@ -185,48 +185,43 @@ 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}
+\begin{displaymath}
     M(t) = 3.78\left[1 + 0.43 \sin\left(\frac{2\pi}{365.24}
     t + 4.36\right)\right]
-\end{equation}
+\end{displaymath}
 
 \begin{figure}
 \centering
 \includegraphics[width=\textwidth]{background.pdf}
-\caption{The background rate $M(t)$ learned to describe the data. Each dot represents the number of infections (fatal and nonfatal) that occurred on a given day.}
+\caption{Aggregated number of infections. Each blue dot represents the number
+    of infections (fatal and nonfatal) that occurred on a given day. The green
+curve is the sinusoidal function fit to the data.}
 \label{fig:background}
 \end{figure}
 
-Because we do not know \emph{a priori} which infections are due to the background rate versus peer contagion,
-can use the rate of fluctuations but not the scale
-we cannot use the base rate and amplitude found. 
-[set variable for the value, assume linear relationship between base and fluctuations]
-$A = a\mu_0$
-
-\begin{align}
-M(t) &= \mu_0' + 0.43 \mu_0 \sin\left(\frac{2\pi}{365.24} t + 4.36\right) \\
-     &= \mu_0' \left[1 + 0.43 \sin\left(\frac{2\pi}{365.24} t + 4.36\right) \right]
-\end{align}
-
-
-Finally, we convert the aggregate background rate $M(t)$ to an individual background rate $\mu(t)$ for each person. To do this, note that the aggregate number of shootings in a given day is the sum of each person's instantaneous rate over the course of that day, i.e.
-\begin{align}
-M(t) &= \sum_{v} \int_{t-1}^{t} \mu(t') dt' \\
-     &= |V| \int_{t-1}^{t} \mu(t') dt'
-\end{align}
+Because we do not know \emph{a priori} the relative importance of the exogenous
+intensity and peer contagion, we only keep $\rho$, $\omega$ and $\phi$ from the
+fitted parameters. In other words, we only keep the parameters characterizing
+the seasonal fluctuations; the base amplitude $A$ of the exogenous intensity
+will be inferred together with the exciting functions parameters.
 
-To simplify this calculation, we assume that the individual background rate for each person is constant over the course of a day, i.e. $\int_{t-1}^{t} \mu(t') dt' = \mu(t)$.
-[explain why this doesn't really affect results]
-This yields the result
-
-\begin{align}
-\mu(t) &= \frac{M(t)}{|V|} \\
-       &= \frac{\mu_0'}{|V|} \left[1 + 0.43 \sin\left(\frac{2\pi}{365.24} t + 4.36\right) \right] \\
-       &= \mu_0 \left[1 + 0.43 \sin\left(\frac{2\pi}{365.24} t + 4.36\right) \right]
-\end{align}
-where $\mu_0=\mu_0'/|V|$.
+Finally, we relate the aggregate number of infections to the node-level
+exogenous intensity. By definition:
+\begin{displaymath}
+    M(t) = \sum_{v\in V}\int_{t-1}^t \mu(s)ds = |V|\int_{t-1}^t\mu(s)ds
+\end{displaymath}
+where we used that the exogenous intensity is shared across the nodes. Assuming
+that that $\mu$ is approximately constant over the course of one
+day\footnote{The time resolution in our dataset is the day, so we only need to
+approximate $\mu$ at the day level.}, we get $M(t) = |V|\mu(t)$. Hence we
+obtain the following form for the exogenous intensity:
+\begin{equation}
+    \mu(t) = \mu_0\left[1 + 0.43 \sin\left(\frac{2\pi}{365.24}
+    t + 4.36\right)\right]
+\end{equation}
+where $\mu_0 = \frac{A}{|V|}$.
 
-\subsection{Kernel function parameters}
+\subsection{Kernel Function Parameters}
 
 We learn parameters using