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\begin{document}

\paragraph{Notations}
\begin{itemize}
    \item $n$ criminals, $V = \{1,\dots,n\}$
    \item $\alpha_{i,j}$ influence of criminal $i$ on criminal $j$.
    \item the infections form a set of pairs $C = \{(t_i, v_i),\; 1\leq i\leq m\}$
        where $t_i$ is the time and $v_i\in\{1,\dots, n\}$ is the identifier of
        the criminal.
\end{itemize}

\paragraph{Model}
The infections are modeled as a multivariate Hawkes process. The instantaneous rate
of infection of user $j$ at time $t$ is given by:
\begin{displaymath}
    \lambda_j(t) = \lambda(t) + \sum_{t_i<t}\alpha_{v_i, j}
    \phi(t-t_i)
\end{displaymath}

\emph{Technical concern:} stability of the process, there is a condition on the
$\alpha_{i,j}$ parameters so that the process is stable, i.e does not keep
accelerating forever. We probably do not need to worry about it too much:
empirically the process is stable.

\paragraph{Time kernel}
we choose an exponential kernel for $\phi$ (will allow the recursive trick
later on) $ \phi(t) = \mu e^{-\mu t} $

\paragraph{Base rate} $\lambda(t)$ captures the seasonal effect. TBD, worst
case is to choose a constant approximation, $\lambda(t) = \lambda_0$. We
define $\Lambda(T) \eqdef \int_{0}^T \lambda(t)dt$.

\paragraph{Edge model} TBD

\paragraph{Likelihood} The log-likelihood for observation horizon $T$ is given
by:
\begin{displaymath}
    \mathcal{L}(C) = \sum_{i=1}^m \log \lambda_{v_i}(t_i)
    - \sum_{j=1}^n\int_{0}^T \lambda_j(t)dt
\end{displaymath}
Note that we have:
\begin{displaymath}
    \int_{0}^T \lambda_j(t)dt = \int_{0}^T\lambda(t)dt + \sum_{i=1}^m
        \alpha_{v_i, j}\big[1-e^{-\mu(T-t_i)}\big]
\end{displaymath}

Hence the log-likelihood can be rewritten (after reordering a little bit):
\begin{displaymath}
    \mathcal{L}(C) = \sum_{i=1}^m\log \lambda_{v_i}(t_i)
    - \sum_{i=1}^m A_{v_i} \big[1-e^{-\mu(T-t_i)}\big]
    - n\Lambda(T)
\end{displaymath}
where $A_{v_i} = \sum_{j=1}^n \alpha_{v_i, j}$ (out-weight of criminal $v_i$),
or in complete form:
\begin{displaymath}
    \mathcal{L}(C) = \sum_{i=1}^m\log\left[ \lambda(t_i)
    + \sum_{j=1}^{i-1}\alpha_{v_j, v_i}\mu e^{-\mu(t_i-t_j)}\right]
    - \sum_{i=1}^m A_{v_i} \big[1-e^{-\mu(T-t_i)}\big]
    - n\Lambda(T)
\end{displaymath}

\paragraph{Computational trick} As written above, the log-likelihood can be
computed in time $O(m^2 + n^2)$. This is not linear in the input size, which is
$O(m + n^2)$. The $O(m^2)$ term comes from the first sum in the definition of
the log-likelihood. However, in the case of an exponential time kernel, using
a recursive trick, the complexity of this term can be reduced to $O(mn)$ for an
overall running time of $O(mn + n^2)$.

\emph{Question:} Is it worth it in our case? Yes if the number of victims
is significantly smaller than the number of infections. Is it the case?


\end{document}
