SuppressionStrategy

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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 $$0<R\le R_0$$. Our goal (imagining ourselves to be the government) is to choose the best distribution for $$R$$. Here $$R_0$$ is the usual reproduction number in the absence of any distancing measures, which serves as an upper bound for $$R$$ under distancing. For example, $$\mathbb P(1<R<2)=1/4$$ would mean that we want be spending $$1/4$$ of the time with $$R$$ between $$1$$ and $$2$$. Let us suppose there is some utility function $$U(R)$$, increasing in $$R$$, that represents the benefit to society of maintaining a constant reproduction of $$R$$. Additive constants don't matter to $$U()$$, so it may be helpful to think of $$U(R)$$ negative and increasing up to $$U(R_0)=0$$. $$R=R_0$$ represents normal life in the absence of a pandemic, while negative $$U(R)$$ for $$R<R_0$$ represents the level of pain at implementing various levels of distancing/lockdown. Note that $$U(R)$$ is the utility when you fix the reproduction number at $$R$$; the possibility of varying this is what we're going to look at here and is represented by a mixture of different values of $$U()$$. For example, the strategy described in the preceding paragraph would have value $$\frac13(U(4)+2U(\frac12))$$. It seems reasonable that there is some version of the function $$U()$$ that represents reality to some extent, though of course it will be hard to agree upon any particular version of $$U()$$. We shall therefore try to see what we can deduce that doesn't depend on a precise $$U()$$, starting with observations that are more-or-less completely $$U()$$-independent, and ending with speculation that depends on a broad form of $$U()$$.

For these purposes it's more convenient to work with the growth rate $$\lambda$$, which we can consider as a random variable $$\Lambda$$ related to $$R$$ by $$R=r(\Lambda)$$. However, we shall also keep $$R$$ in the picture because it's probably more intuitive to people now as we've had experience of reproduction rates in the news. Using growth rate as the random variable, we can define $$V(\lambda)=U(r(\lambda))$$ and formulate the problem like this:

We wish to maximise

\[\mathbb E[V(\Lambda)]\]

subject to

\[\mathbb E[\Lambda]=0.\]

$$V()$$ is some increasing function and $$\Lambda\le\Lambda_0$$, where $$\Lambda_0$$ is defined by $$r(\Lambda_0)=R_0$$. This is meant to capture the time-independent aspect of the problem of medium/long term control of the virus: decide how long you want to spend at different lockdown levels, and factor out "implementation details" of how/when to switch from one level to another. Obviously in reality such "details" are very important and may in practice be a sticking point but still I think it's interesting to consider what kinds of distributions are likely to make sense.

Note that we don't consider the possibility of $$\mathbb E[\Lambda]<0$$ even though this would also result in disease suppression, because $$V()$$ is increasing so if $$-\infty<\mathbb E[\Lambda]<0$$ we may as well increase $$\Lambda$$ a bit to improve $$\mathbb E[V(\Lambda)]$$ and maintain suppression. And $$\mathbb E[\Lambda]=-\infty$$ is a zero-Covid strategy (leading to extinction of the disease) which we are not considering here, so in particular we must have $$\mathbb P(\Lambda=-\infty)=\mathbb P(R=0)=0$$.

Only two values of $$R$$ are needed

Equivalently, only two values of $$\Lambda$$ are needed. Suppose that $$\Lambda$$ only takes finitely many values: $$\mathbb P(\Lambda=\lambda_i)=p_i$$ for $$i=1,\ldots, n$$, where $$p_i>0$$ and $$0