SuppressionStrategy
About
Description
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 $$R$$. For these purposes we'll assume that $$U()$$ is time-independent, though this isn't entirely realistic as seasonal factors, for example, will influence transmission rates and so how painful certain suppression measures are. 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 (which isn't a very onerous restriction): $$\mathbb P(\Lambda=\lambda_i)=p_i$$ for $$i=1,\ldots, n$$, where $$p_i>0$$ and $$0<r_1<\ldots<r_n$$. If $$n\ge3$$ then we can consider the effect of adjusting the probabilities $$(p_1, p_2, p_3)$$ by a multiple of $$(\Delta p_1, \Delta p_2, \Delta p_3)=(\lambda_2-\lambda_3, \lambda_3-\lambda_1, \lambda_1-\lambda_2)$$ since that preserves $$\sum_i p_i$$ and $$\sum_i p_i\lambda_i$$. If $$\Delta V = V(\lambda_1)\Delta p_1+V(\lambda_2)\Delta p_2+V(\lambda_3)\Delta p_3\ge0$$ then we can improve the overall value by adding on a positive multiple of $$(\Delta p_1, \Delta p_2, \Delta p_3)$$, while if $$\Delta V\le0$$ we can use a negative multiple. Either way we can change $$p_1, p_2, p_3$$ until one of them becomes $$0$$. This demonstrates that we only need to consider $$n=1$$ or $$n=2$$.
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 $$R$$?
I think this is the most interesting question here. Do you (as the government) go for $$n=1$$ which means a steady state of continuous medium-level suppression trying to fix $$R$$ at $$1$$, or do you go for $$n=2$$ which means deliberately bouncing between strong and weak suppression? Of course we can't properly answer this question here because it would require a large amount of real-world data that we don't have, but we can at least see what we can say about this objectively to see if that gives any useful feeling for the question. In the next section we'll look at some real-world data and I'll state my hunch (which is that $$n=2$$).
In the real world the two strategies could blur together somewhat since tracking $$R=1$$ ($$\Lambda=0$$) would involve some trial and error as you nudge $$R$$ up or down a bit according to whether infections are seen to be under control or getting too high, but there would still be a distinction. I think this $$n=1$$ strategy is the one that's most obvious to people, though it isn't necessarily the best. $$n=2$$ could seem counterintuitive because if the disease prevalence is relatively low, and if the current value of $$R$$ is $$1$$ so things are nicely under control, you are asking people to increase distancing measures to get $$R$$ down even further, and they will surely ask you what the point is of doing that. On the face of it, it will seem as though there is no benefit in enduring the pain of $$R=0.8$$ when infections have already stabilised at a manageable level, but the point is that further reducing disease prevalence is a kind of investment, and can be redeemed later when you get to indulge in a period of (for example) $$R=1.5$$.
(Somewhat speculative, but I have got the impression from the way we've opened up since April, and from the way that the progressive relaxation of measures was couched in terms of how much we can afford to do while still keeping the virus under control, that we might have fallen into the trap of thinking that it's necessarily optimal to open up as much as we can while keeping $$R\le1$$. That is, we might be following the $$n=1$$ strategy by default.)
Let's say that $$\Lambda$$ takes the two values $$a$$ and $$b$$, where $$-\infty