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This is slightly interesting. Based only on fairly mild assumptions, it means you only want to be aiming for at most two states: relatively locked down or relatively relaxed (or with $$n=1$$, aim for just one state), and not using anything in between. Of course this is only absolutely true within the idealisation we have used and doesn't automatically invalidate the multi-tiered approach which many countries have adopted. In the real world you don't know exactly what $$R$$ you will get from a particular set of measures, so it may be useful to have a responsive set of measures that are able to nudge $$R$$ down or up a bit.
This is slightly interesting. Based only on fairly mild assumptions, it means you only want to be aiming for at most two states: relatively locked down or relatively relaxed (or with $$n=1$$, aim for just one state), and not using anything in between. Of course this is only absolutely true within the idealisation we have used and doesn't automatically invalidate the multi-tiered approach which many countries have adopted. In the real world you don't know exactly what $$R$$ you will get from a particular set of measures, so it may be useful to have a responsive set of measures that are able to nudge $$R$$ down or up a bit.
= Addendum to the previous section =
This is a technical section showing that the assumption about $$R$$ taking finitely many values isn't necessary. This section can be skipped without losing the sense of the discussion.
Given $$\epsilon>0$$ consider the intervals $$I_k=(\Lambda_0-(k+1)\epsilon,\Lambda_0-k\epsilon]$$ and $$A_k=(-\infty,\Lambda_0-k\epsilon]$$.
Because $$\mathbb E[\Lambda]=0$$ and $$\Lambda$$ is bounded above, we know that $$\mathbb E[|\Lambda|]<\infty]$$ and $$\mathbb E[\Lambda;A_k]\to0$$ as $$k\to\infty$$, so
there is an $$N$$ such that $$\Lambda_0-N\epsilon<0$$ and $$\mathbb E[\Lambda;A_N]>-\epsilon$$. (The notation $$\mathbb E[X;S]$$ refers to the expectation over the subset $$S$$, not a conditional expectation.)
Using such an $$N$$, we can partition $$(-\infty,\Lambda_0]$$ into the $$N+1$$ intervals $$A_N, I_0, I_1, \ldots I_{N-1}$$ and create a new random variable $$\Lambda'$$ with
$$\mathbb P(\Lambda'=\Lambda_0-k\epsilon)=\mathbb P(\Lambda\in I_k)$$ for $$k=0,\ldots,N-1$$ and $$\mathbb P(\Lambda'=\Lambda_0-N\epsilon)=\mathbb P(\Lambda\in A_N)$$. We may consider $$\Lambda$$ and $$\Lambda'$$ coupled, with $$\Lambda'\ge\Lambda$$ and $$\Lambda'\le\Lambda+\epsilon$$ over $$I_0\cup\ldots\cup I_{N-1}$$. Then $$\mathbb E[\Lambda';A_N]\le0\le\mathbb E[\Lambda;A_N]+\epsilon\le\mathbb E[\Lambda+\epsilon]+\epsilon$$ and $$\mathbb E[\Lambda';I_0\cup\ldots\cup I_{N-1}]\le\mathbb E[\Lambda+\epsilon;I_0\cup\ldots\cup I_{N-1}]$$, so $$0=\mathbb E[\Lambda]\le\mathbb E[\Lambda']\le2\epsilon$$ and since $$V()$$ is increasing, $$\mathbb E[V(\Lambda')]\ge\mathbb E[V(\Lambda)]$$.
Finally, define another random variable $$\Lambda''$$ by $$\Lambda''=\Lambda'$$ with probability $$1-\delta$$ and $$\Lambda''=-1$$ with probability $$\delta$$. Setting $$\delta$$ to $$\mathbb E[\Lambda']/(\mathbb E[\Lambda']+1)$$, we see that $$\delta\le2\epsilon$$ and $$\mathbb E[\Lambda'']=0$$. We also have $$\mathbb E[V(\Lambda'')]\ge \delta V(-1)+(1-\delta)\mathbb E[V(\Lambda)]$$, so we can approach the value $$\mathbb E[V(\Lambda)]$$ as closely as we like with this finitely-valued $$\Lambda''$$ (which, by the previous argument, means we can approach $$\mathbb E[V(\Lambda)]$$ arbitrarily closely using a two-valued version of $$\Lambda$$).


= Should we use 1 or 2 values of $$R$$? =
= Should we use 1 or 2 values of $$R$$? =

Revision as of 13:28, 12 December 2020

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The purpose of this is to take a broad look from a certain zoomed-out point of view at what kind of suppression strategies might be best. Here "suppression" is meant to mean that you intend to maintain infections in a low-level holding pattern until there is some favourable change in circumstances such as widespread immunity from a vaccine. I'm not trying to argue here against either infection-induced herd immunity or against zero-Covid/elimination strategies, just looking at what kinds of steady-state suppression strategies might work better than others. As of November 2020, most countries in the world are following a suppression strategy of some sort.

The reason I'm writing this is that when Covid-19 came on the scene in early 2020 I had a slight hunch that the best strategy of this form would be to alternate between strong and weak suppression. The discussion here is meant to explore the plausibility of this from a simplified point of view - it does not claim to prove it.

The abstraction here is that you just look at what values of the basic reproduction number, $$R$$, are best to aim at. This is an idealisation of the process of controlling a disease and ignores all sorts of things like how $$R$$ differs between regions and subcommunities, how to actually achieve a particular value of $$R$$, what the generation time is, how efficient it is to swap between different suppression states, what the effect of suppression on the distribution of infectivity time, to what extent the concept of $$R$$ actually makes sense, and so on, yet I would like to pursue it to see if there is any useful intuition to be obtained at this level. Is it better to keep $$R$$ hovering around 1, or is it better to mix strong suppression phases with relaxed phases? Should there be more than two sorts of such phases? Recently (as of November 2020) this has become topical again in the UK as we are trading off a relaxation over the Christmas period with some extra suppression before and after.

A suppression strategy means keeping the average growth rate of the disease equal to zero (or less). This is similar to keeping the multiplicative average $$R$$ at 1, though not quite the same (see below). For example if you alternate $$R=4$$ for 1 week and $$R=\frac12$$ for 2 weeks then the net effect over three weeks will be similar to an $$R$$ of $$4\times\frac12\times\frac12 = 1$$. This is because, to first approximation, $$R$$ is related to the growth rate $$\lambda$$ and the average time for an infected person to infect another (the mean generation time), $$T$$ by

\[R=e^{\lambda T}.\]

But because the time taken to infect someone is not a constant, a better formula is

\[R=r(\lambda)=\mathbb E[e^{-\lambda X}]^{-1},\]

where $$X$$ is the time to infection for each infectee, considered as a random variable. (Note, the above expectation uses a different probability space from that considered next.)

Consider the reproduction rate $$R$$ to be a random variable satisfying $$0

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