SuppressionStrategy - Revision history

Alex: /* Conclusion */

2021-01-15T09:24:22Z

Conclusion

← Older revision		Revision as of 03:24, 15 January 2021
Line 102:		Line 102:
	= Conclusion =		= Conclusion =

	* If further or stronger restrictions are imposed when $$R$$ is already less than $$1$$, then this may be a useful thing even though it seems masochistic, because it "builds up credit" in a way that ~~may~~ be ~~more efficient~~ overall.		* If further or stronger restrictions are imposed when $$R$$ is already less than $$1$$ and disease levels are relatively low, then this may well be a useful thing even though it seems masochistic, because it "builds up credit" in a way that could be better overall.
	* If the policy alternates between strong and weak restrictions then this may be a good thing even if it seems as though the policy-makers can't make up their minds.		* If the policy alternates between strong and weak restrictions then this may be a good thing even if it seems as though the policy-makers can't make up their minds.
	* In the idealisation considered here a policy with three or more levels of restrictions is not useful (though it is still possible that such a multi-level policy makes sense due to this idealisation being too idealistic).		* In the idealisation considered here a policy with three or more levels of restrictions is not useful (though it is still possible that such a multi-level policy makes sense due to this idealisation being too idealistic).

Alex: /* Real world data */

2021-01-15T09:22:58Z

Real world data

← Older revision		Revision as of 03:22, 15 January 2021
Line 89:		Line 89:

	(*) It's common to use the [https://en.wikipedia.org/wiki/Gamma_distribution Gamma distribution] here because it's a very convenient two-parameter family of non-negative distributions, not because there is a particular reason to expect the distribution to have this precise form. (And there isn't usually enough data to fit more than two parameters.) If $$X\sim\Gamma(\alpha,\beta)$$ then $$V(\lambda)=U(\theta^\alpha)$$, where $$\theta=1+(\alpha/\beta)\lambda$$, and $$V''(\lambda)=(\alpha/\beta)^2\theta^{2\alpha-2}U''(\theta^\alpha)+(\alpha(\alpha-1)/\beta^2)\theta^{\alpha-2}U'(\theta^\alpha)$$, so provided $$\alpha>1$$, which is what everyone thinks, then $$V()$$ is "more convex" than $$U()$$ because $$V''$$ has an extra term that is always positive.		(*) It's common to use the [https://en.wikipedia.org/wiki/Gamma_distribution Gamma distribution] here because it's a very convenient two-parameter family of non-negative distributions, not because there is a particular reason to expect the distribution to have this precise form. (And there isn't usually enough data to fit more than two parameters.) If $$X\sim\Gamma(\alpha,\beta)$$ then $$V(\lambda)=U(\theta^\alpha)$$, where $$\theta=1+(\alpha/\beta)\lambda$$, and $$V''(\lambda)=(\alpha/\beta)^2\theta^{2\alpha-2}U''(\theta^\alpha)+(\alpha(\alpha-1)/\beta^2)\theta^{\alpha-2}U'(\theta^\alpha)$$, so provided $$\alpha>1$$, which is what everyone thinks, then $$V()$$ is "more convex" than $$U()$$ because $$V''$$ has an extra term that is always positive.

			= An alternative (and simpler) argument =

			Given a period of $$R=R_1$$ for time $$T_1$$ followed by a period of $$R=R_2$$ for time $$T_2$$, then swapping these two periods around (doing $$R=R_2$$ for time $$T_2$$ first) doesn't change anything before or after the two periods. At least, this is true in the idealisation we've been considering here where the policy in each period multiplies the disease prevalence by a factor independent of the incoming prevalence: granted this, it doesn't matter which order we multiply these two factors. But during the two periods, it's better to take the smaller $$R$$ first since you'd then enjoy lower disease prevalence for longer.

			This means given an alternating sequence of $$R=R_1$$ and $$R=R_2$$ strategies, as prescribed by the "Only two values of $$R$$ are needed" argument, you would keeping swap adjacent periods until all of the $$R_1$$s come before all of the $$R_2$$s (assuming $$R_1<R_2$$ for the sake of argument). So you would end up with at most one change of strategy during the whole epidemic as you find yourself engaging in a single large period of $$R_1$$ followed by a large period of $$R_2$$. (Actually carrying this off would require being able to predict the timing of the end of the epidemic, which isn't easy in practice.)

			The implied strategy is quite stark then: do all your lockdown to start with. In such a situation when your idealisation leads you to an extreme conclusion, it makes sense to check that the idealisation won't have broken down somewhere along the way. The kinds of things that could limit it are: (i) not being able to sustain the required lockdown for long enough: it's not completely true that lockdown+letup+lockdown is the same as lockdown+lockdown+letup because not everything can be put off indefinitely (otherwise this argument would imply that you could do lockdown for 10,000 years and then switch - i.e., it would obviously prove too much), (ii) the maths changes when the number of cases gets very low: it's not true that you can reduce cases until only $$10^{-100}$$ people are infected and then expect to be able to free everything up on the grounds that it would take a very long time for the virus to double its way back from such a low level. In reality you can extinguish the virus to zero incidence, as some countries have done, though then you have to consider the cost of protecting yourself from reinfection from the rest of the world. Or you could maintain the virus at a very low level that is consistent with the rate of reinfection from the rest of the world (which you would then obviously try to keep to a minimum).

			At any rate, this argument would suggest that, at least before you hit limits like (i) and (ii) above (and possibly even then), it's better to go in hard first and never let the prevalence get high.

	= Conclusion =		= Conclusion =

Alex: /* Addendum to the previous section */

2020-12-12T19:28:37Z

Addendum to the previous section

← Older revision		Revision as of 13:28, 12 December 2020
Line 42:		Line 42:

	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 $$R$$? =

Alex: /* Only two values of $$R$$ are needed */

2020-12-12T19:28:25Z

Only two values of $$R$$ are needed

← Older revision		Revision as of 13:28, 12 December 2020
Line 39:		Line 39:
	= Only two values of $$R$$ are needed =		= 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 $$0<r_1<\ldots<r_n$$. 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 $$n=2$$.		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.		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.

Alex: /* Description */

2020-12-12T19:27:17Z

Description

← Older revision		Revision as of 13:27, 12 December 2020
Line 21:		Line 21:
	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.)		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 $$U()$$.		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:		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:

Alex: /* Description */

2020-12-12T19:14:49Z

Description

← Older revision		Revision as of 13:14, 12 December 2020
Line 15:		Line 15:
	\[R=e^{\lambda T}.\]		\[R=e^{\lambda T}.\]

	But because the time taken to infect someone is not a constant, a better formula is		But because the time taken to infect someone is not a constant, [https://plus.maths.org/content/epidemic-growth-rate a better formula] is

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

Alex: /* Description */

2020-12-12T19:08:10Z

Description

← Older revision		Revision as of 13:08, 12 December 2020
Line 35:		Line 35:
	$$V()$$ 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.		$$V()$$ 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 $$-\infty<E[\Lambda]<0$$ we may as well increase $$\Lambda$$ a bit to improve $$\E[V(\Lambda)]$$ 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 $$\mathbb P(\Lambda=-\infty)=\mathbb P(R=0)=0$$.		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 =		= Only two values of $$R$$ are needed =

Alex: /* Real world data */

2020-12-11T00:11:53Z

Real world data

← Older revision		Revision as of 18:11, 10 December 2020
Line 98:		Line 98:

	For an example of a strong/weak policy, one could combine the $$R=1.4$$ state in mid-September at the height of the loosening with $$R=0.8$$ of May (or possibly November during "Lockdown 2"). This would have to be taken in the ratio of something like 1:2, for example one month of $$R=1.4$$ followed by two of $$R=0.8$$. The [https://web.archive.org/web/20200520063031/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do restrictions in place on 20 May 2020 can be found here] while [https://web.archive.org/web/20200909063459/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do those in place on 13 September 2020 can be found here].		For an example of a strong/weak policy, one could combine the $$R=1.4$$ state in mid-September at the height of the loosening with $$R=0.8$$ of May (or possibly November during "Lockdown 2"). This would have to be taken in the ratio of something like 1:2, for example one month of $$R=1.4$$ followed by two of $$R=0.8$$. The [https://web.archive.org/web/20200520063031/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do restrictions in place on 20 May 2020 can be found here] while [https://web.archive.org/web/20200909063459/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do those in place on 13 September 2020 can be found here].

			However, if the strongest lockdown we can manage only has $$R$$ as low as $$0.7-0.8$$, which is what we've managed hitherto, then intuition suggests that the benefits of a two-level strategy are probably not that great. Roughly speaking, if you do something like mix $$R=0.8$$ with $$R=1.25$$ in equal proportions then the average $$R$$ value (a proxy for the overall benefit) is only $$1.025$$. You'd get a more useful gain if you could mix something like $$R=0.5$$ with $$R=2$$ for an average $$R$$ of $$1.25$$ (we should really be working with growth not $$R$$, but this is just a rough thought), though I don't know if an $$R=0.5$$ regime is practically possible.

	(*) It's common to use the [https://en.wikipedia.org/wiki/Gamma_distribution Gamma distribution] here because it's a very convenient two-parameter family of non-negative distributions, not because there is a particular reason to expect the distribution to have this precise form. (And there isn't usually enough data to fit more than two parameters.) If $$X\sim\Gamma(\alpha,\beta)$$ then $$V(\lambda)=U(\theta^\alpha)$$, where $$\theta=1+(\alpha/\beta)\lambda$$, and $$V''(\lambda)=(\alpha/\beta)^2\theta^{2\alpha-2}U''(\theta^\alpha)+(\alpha(\alpha-1)/\beta^2)\theta^{\alpha-2}U'(\theta^\alpha)$$, so provided $$\alpha>1$$, which is what everyone thinks, then $$V()$$ is "more convex" than $$U()$$ because $$V''$$ has an extra term that is always positive.		(*) It's common to use the [https://en.wikipedia.org/wiki/Gamma_distribution Gamma distribution] here because it's a very convenient two-parameter family of non-negative distributions, not because there is a particular reason to expect the distribution to have this precise form. (And there isn't usually enough data to fit more than two parameters.) If $$X\sim\Gamma(\alpha,\beta)$$ then $$V(\lambda)=U(\theta^\alpha)$$, where $$\theta=1+(\alpha/\beta)\lambda$$, and $$V''(\lambda)=(\alpha/\beta)^2\theta^{2\alpha-2}U''(\theta^\alpha)+(\alpha(\alpha-1)/\beta^2)\theta^{\alpha-2}U'(\theta^\alpha)$$, so provided $$\alpha>1$$, which is what everyone thinks, then $$V()$$ is "more convex" than $$U()$$ because $$V''$$ has an extra term that is always positive.

Alex: /* Real world data */

2020-12-10T14:26:40Z

Real world data

← Older revision		Revision as of 08:26, 10 December 2020
Line 99:		Line 99:
	For an example of a strong/weak policy, one could combine the $$R=1.4$$ state in mid-September at the height of the loosening with $$R=0.8$$ of May (or possibly November during "Lockdown 2"). This would have to be taken in the ratio of something like 1:2, for example one month of $$R=1.4$$ followed by two of $$R=0.8$$. The [https://web.archive.org/web/20200520063031/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do restrictions in place on 20 May 2020 can be found here] while [https://web.archive.org/web/20200909063459/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do those in place on 13 September 2020 can be found here].		For an example of a strong/weak policy, one could combine the $$R=1.4$$ state in mid-September at the height of the loosening with $$R=0.8$$ of May (or possibly November during "Lockdown 2"). This would have to be taken in the ratio of something like 1:2, for example one month of $$R=1.4$$ followed by two of $$R=0.8$$. The [https://web.archive.org/web/20200520063031/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do restrictions in place on 20 May 2020 can be found here] while [https://web.archive.org/web/20200909063459/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do those in place on 13 September 2020 can be found here].

	(*) It's common to use the [https://en.wikipedia.org/wiki/Gamma_distribution Gamma distribution] here because it's a very convenient two-parameter family of non-negative distributions, not because there is a particular reason to expect the distribution to have this precise form. (And there isn't usually enough data to fit more than two parameters.) If $$X\sim\Gamma(\alpha,\beta)$$ then $$V(\lambda)=U(\theta^\alpha)$$, where $$\theta=1+(\alpha/\beta)\lambda$$, and $$V''(\lambda)=(\alpha/\beta)^2\theta^{2\alpha-2}U''(\theta^\alpha)+(\alpha(\alpha-1)/\beta^2)\theta^{\alpha-2}U'(\theta^\alpha)$$, so provided $$\alpha>1$$ which is what everyone thinks, then $$V()$$ is more convex than $$U()$$ because $$V''$$ has an extra term that is always positive.		(*) It's common to use the [https://en.wikipedia.org/wiki/Gamma_distribution Gamma distribution] here because it's a very convenient two-parameter family of non-negative distributions, not because there is a particular reason to expect the distribution to have this precise form. (And there isn't usually enough data to fit more than two parameters.) If $$X\sim\Gamma(\alpha,\beta)$$ then $$V(\lambda)=U(\theta^\alpha)$$, where $$\theta=1+(\alpha/\beta)\lambda$$, and $$V''(\lambda)=(\alpha/\beta)^2\theta^{2\alpha-2}U''(\theta^\alpha)+(\alpha(\alpha-1)/\beta^2)\theta^{\alpha-2}U'(\theta^\alpha)$$, so provided $$\alpha>1$$, which is what everyone thinks, then $$V()$$ is "more convex" than $$U()$$ because $$V''$$ has an extra term that is always positive.

	= Conclusion =		= Conclusion =

Alex: /* Real world data */

2020-12-10T14:18:43Z

Real world data

← Older revision		Revision as of 08:18, 10 December 2020
Line 87:		Line 87:
	In this graph, the green line is the [https://www.gov.uk/guidance/the-r-number-in-the-uk Government Office for Science and SAGE estimate of $$R$$], with weekly updates starting 15 May 2020. The documentation suggests it might lag by 21 days, but it looks a bit like 21 may not be enough and you have to bring the graph forward by around 30 days to match the blue line from August 2020. (Maybe the 21 days is only the lag from antigen test to publication and doesn't take into account the time between infection and test?)		In this graph, the green line is the [https://www.gov.uk/guidance/the-r-number-in-the-uk Government Office for Science and SAGE estimate of $$R$$], with weekly updates starting 15 May 2020. The documentation suggests it might lag by 21 days, but it looks a bit like 21 may not be enough and you have to bring the graph forward by around 30 days to match the blue line from August 2020. (Maybe the 21 days is only the lag from antigen test to publication and doesn't take into account the time between infection and test?)

	The blue line is the ratio of smoothed case counts (from the [https://coronavirus.data.gov.uk/ UK Covid-19 dashboard]) at day $$d$$ and $$d+14$$, where $$d$$ is the x-coordinate on this graph. This effectively brings it forward by 7 days which should roughly account for incubation+testing+reporting delays. This is naturally in units of growth ($$\lambda$$) rather than $$R$$, but I'm making the y-axis $$R$$ rather than growth rate because I think people are more used to $$R$$. The conversion formula used here is $$R=(1+\lambda/\beta)^\alpha$$, where $$\alpha=1.99$$ and $$\beta=0.336$$. This is what $$r(\lambda)$$ (see above) works out as if the mean generation time, $$X$$, is distributed as ~~[https://en.wikipedia.org/wiki/Gamma_distribution~~ $$\Gamma(\alpha,\beta)$$], using the (shape,rate) notation. These values of $$\alpha$$, $$\beta$$ were fitted to the SAGE estimates (green line), which is possible because they publish both $$R$$ and growth rate. So even if this conversion formula isn't quite right, it should therefore make the blue and green lines comparable. In any case when $$R$$ is close to $$1$$, $$R-1$$ is going to be close to proportional to the growth rate.		The blue line is the ratio of smoothed case counts (from the [https://coronavirus.data.gov.uk/ UK Covid-19 dashboard]) at day $$d$$ and $$d+14$$, where $$d$$ is the x-coordinate on this graph. This effectively brings it forward by 7 days which should roughly account for incubation+testing+reporting delays. This is naturally in units of growth ($$\lambda$$) rather than $$R$$, but I'm making the y-axis $$R$$ rather than growth rate because I think people are more used to $$R$$. The conversion formula used here is $$R=(1+\lambda/\beta)^\alpha$$, where $$\alpha=1.99$$ and $$\beta=0.336$$. This is what $$r(\lambda)$$ (see above) works out as if the mean generation time, $$X$$, is distributed as $$\Gamma(\alpha,\beta)$$ (*), using the (shape,rate) notation. These values of $$\alpha$$, $$\beta$$ were fitted to the SAGE estimates (green line), which is possible because they publish both $$R$$ and growth rate. So even if this conversion formula isn't quite right, it should therefore make the blue and green lines comparable. In any case when $$R$$ is close to $$1$$, $$R-1$$ is going to be close to proportional to the growth rate.

	The blue line probably overestimates $$R$$ because testing has been steadily increasing, so we're likely to be finding a greater proportion of cases as time goes on. In particular in April there was a sudden explosion in available tests, which may account for the blue line being at $$R~=1$$ then when the true value was probably more like $$0.8$$. At a guess, later blue line entries are closer to their true values.		The blue line probably overestimates $$R$$ because testing has been steadily increasing, so we're likely to be finding a greater proportion of cases as time goes on. In particular in April there was a sudden explosion in available tests, which may account for the blue line being at $$R~=1$$ then when the true value was probably more like $$0.8$$. At a guess, later blue line entries are closer to their true values.
Line 98:		Line 98:

	For an example of a strong/weak policy, one could combine the $$R=1.4$$ state in mid-September at the height of the loosening with $$R=0.8$$ of May (or possibly November during "Lockdown 2"). This would have to be taken in the ratio of something like 1:2, for example one month of $$R=1.4$$ followed by two of $$R=0.8$$. The [https://web.archive.org/web/20200520063031/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do restrictions in place on 20 May 2020 can be found here] while [https://web.archive.org/web/20200909063459/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do those in place on 13 September 2020 can be found here].		For an example of a strong/weak policy, one could combine the $$R=1.4$$ state in mid-September at the height of the loosening with $$R=0.8$$ of May (or possibly November during "Lockdown 2"). This would have to be taken in the ratio of something like 1:2, for example one month of $$R=1.4$$ followed by two of $$R=0.8$$. The [https://web.archive.org/web/20200520063031/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do restrictions in place on 20 May 2020 can be found here] while [https://web.archive.org/web/20200909063459/https://www.gov.uk/government/publications/coronavirus-outbreak-faqs-what-you-can-and-cant-do/coronavirus-outbreak-faqs-what-you-can-and-cant-do those in place on 13 September 2020 can be found here].

			(*) It's common to use the [https://en.wikipedia.org/wiki/Gamma_distribution Gamma distribution] here because it's a very convenient two-parameter family of non-negative distributions, not because there is a particular reason to expect the distribution to have this precise form. (And there isn't usually enough data to fit more than two parameters.) If $$X\sim\Gamma(\alpha,\beta)$$ then $$V(\lambda)=U(\theta^\alpha)$$, where $$\theta=1+(\alpha/\beta)\lambda$$, and $$V''(\lambda)=(\alpha/\beta)^2\theta^{2\alpha-2}U''(\theta^\alpha)+(\alpha(\alpha-1)/\beta^2)\theta^{\alpha-2}U'(\theta^\alpha)$$, so provided $$\alpha>1$$ which is what everyone thinks, then $$V()$$ is more convex than $$U()$$ because $$V''$$ has an extra term that is always positive.

	= Conclusion =		= Conclusion =