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phi_notes.tex
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phi_notes.tex
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\documentclass[12pt]{article}
\usepackage{./commands}
\begin{document}
\section*{Strength and speed of intervention}
\subsection*{Renewal equation}
$$i_s(t) = i(s) k(t-s) S(t)/N$$
$$i(t) = S(t)/N \int{i(s) k(t-s)} ds \approx \int{i(s) k(t-s)}$$
$$ = \int i(t-\tau) k(\tau) d\tau $$
\subsection*{Strength}
$\Rx = \int{k(\tau) d\tau}$ is the reproductive number, and
$g(\tau) = k(\tau)/\Rx$ is the ``intrinsic'' generation interval.
If we have an intervention that reduces $k(\tau)$ to $\hat k(\tau) \equiv k(\tau)/L(\tau)$ then the reproductive number after intervention is $\hatR = \int{\hat k(\tau) d\tau}$.
We define the \emph{strength} of the intervention to be $\strength = \Rx/\hatR$ -- the proportional amount by which the intervention reduces transmission.
We can then show that \strength\ is the harmonic mean of $L$, weighted by the generation distribution $g$.
The outbreak can be controlled ($\hatR<1$) iff $\strength > \Rx$.
\subsubsection*{Math}
$$\theta = \frac{R}{\int{k(\tau)/L(\tau)d\tau}}$$
$$ = \frac{1}{\int{g(\tau)/L(\tau)d\tau}}$$
$$ = 1/\left\langle 1/L(\tau) \right\rangle_{g(\tau)}$$
\subsection*{Speed}
The solution $r$ of $\int{k(\tau) \exp(-r\tau) d\tau} = 1$ is the
exponential rate of spread, and $b(\tau) = k(\tau) \exp(-r\tau)$ is the initial ``backward'' generation distribution.
Under the intervention $L(\tau)$, the exponential rate of spread $\hat r$ solves $\hat r = \int{\hat k(\tau) \exp(-\hat r\tau) d\tau}$.
We define the \emph{speed} of the intervention to be $\speed = r-\hat r$ -- the amount by which the intervention slows down spread.
We can then show that $\exp(\speed\tau)/L(\tau)$ has a weighted mean of 1 over the initial backward generation distribution. Thus \speed\ is an approximation of the rate of exponential growth of $L(\tau)$
The outbreak can be controlled ($\hat r<0$) iff $\speed > r$.
\subsubsection*{Math}
$$2 = \int{\hat k(\tau)\exp(-\hat r(\tau)) d\tau}$$
$$ = \int{\frac{k(\tau)}{L(\tau)}\exp(-(r-\speed)\tau) d\tau} $$
$$ = \int{k(\tau)\exp(-r\tau)\frac{\exp(\speed\tau)}{L(\tau)} d\tau} $$
$$ = \left\langle \frac{\exp(\speed\tau)}{L(\tau)} \right\rangle_{b(\tau)}$$
\end{document}