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I am trying to learn basic reproduction number and have a very basic question.

In a given time window, If number of infected people grows exponentially, for example

$N_{d+1}=1.15N_{d}$

Can we say R0 in this time window is $1.15$?


If yes, then for covid19, in almost all countries, at the beginning number of infected grows exponentially, and this coefficient is close to $1.15$, but why people estimate R0 to be $2.0 - 5.0$? ($2.0>1.15$, and I assume after a while, the spread will slow down, so, the coefficient should be even smaller than $1.15$)

kjetil b halvorsen
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Haitao Du
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2 Answers2

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Described Quantity

The $1.15$ quantity is similar to the initial growth rate, which we can calculate via the slope of a line comparing time and the natural-log of new cases. The initial growth rate is not $R_0$, but can be used to calculate $R_0$. Depending on the assumed model, the calculation will differ.

$R_0$ from initial growth rate

First, let's go through an SIR model. Let $\lambda$ indicate the initial growth rate, and $\frac{1}{\mu + \delta}$ indicate the infectious period. Where $\mu$ is the birth rate (i.e. new susceptibles) and $\delta$ is the recovery rate. Therefore $R_0$ can be calculated as $$R_0 = \lambda \times \frac{1}{\mu + \delta} +1$$

However, this approach assumes that the infection follows the SIR model, which may not be true. Instead, we can suppose an SEIR model, where there is a latent period. Let the latent period be $\frac{1}{\sigma}$, where $\sigma$ is transition rate between E and I compartments. Then $R_0$ can be calculated as $$R_0 = (\frac{\lambda}{\mu + \delta} +1)\times(\frac{\lambda}{\sigma} +1)$$

There are many approaches to estimate $R_0$, but this is one approach

$R_0$ vs $R_t$

There are also two quantities that differ. $R_0$ is the number of secondary cases from a single case in a population of all susceptibles. $R_t$ which is the effective reproductive number, which is the reproductive number at a singular instance of time.

Here is why the distinction is important: as more of the population moves to the Removed compartment, the $R_t$ will move closer to 1 (and eventually drop below 1). $R_0$ is the same for this population, since it is a special case of $R_t$ when the entire population is in the Susceptible compartment

pzivich
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Say, initially, each person passes the virus on to two new people (this is an extreme simplification, not everybody passes the virus on to the same amount of people, but it is how the simple SIR model works works). That is $R_0 = 2$

Then the number of cases grows for each new generation like 1, 2, 4, 8, 16, 32, etc.

However the number 1.15 that you refer to is the growth in time. Those increase in cases per generation 1, 2, 4, 8, 16, 32 may happen slowly or fast. They can be the same $R_0$ but different grow rates.

The initial grow rate relates to $\beta - \gamma$ and the reproduction number relates to $\frac{\beta}{\gamma}$. (Note that if the grow rate is positive then the reproduction rate is above 1.)


For example,

  • the reproduction rate for HIV/aids may be between 2 or 5. However, it is not like the reproduction occurs at a daily frequency and instead it may take months or years before an infected person get's to infect those 2 to 5 other people.
  • For a disease like the common cold, flu or sars, the reproduction may be between 2 or 5 as well, but now it will take about the order of a week that a person infects those 2 to 5 other people.

In the (average) time period that somebody is infectious $\frac{1}{\gamma}$ there will (initally) be $R_0$ newly people infected and $1$ person recovered. So the growth rate (the exponent) will be:

$$K = \frac{R_0-1}{1/\gamma} = \beta - \gamma$$

Then the (initial) growth is like $e^{Kt}$ and in one day you get an increase of $e^{\beta-\gamma}$

Sextus Empiricus
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