Assuming that around half of those cases are infectious and another half carry the new variant, there could be around 25 cases with the new variant in Switzerland. What is the probability that they establish a sustained transmission chain in Switzerland? 4/n
Using branching process models, we can calculate this probability as P = 1 - 1/Re^n where Re is the effective reproduction number and n is the number of initial cases. 5/n
The effective reproduction number in Switzerland is arguably slightly above 1 at the moment. For Re = 1.05, we obtain a probability of 70% that the new variant will spread in Switzerland. For Re = 1.1, the probability would be 91%. 6/n https://sciencetaskforce.ch/epidemiologische-lagebeurteilung-21-dezember-2020/
But there is more to it. The equation above assumes that the number of secondary cases are geometrically distributed (k = 1). However, #SARSCoV2 exhibits superspreading and the number of secondary cases is highly overdispersed (k ~ 0.5). 7/n https://www.eurosurveillance.org/content/10.2807/1560-7917.ES.2020.25.4.2000058
That means that many infected cases do not transmit at all, and only a few infected individuals have the potential to initiate a sustained transmission chain. 8/n
Assuming k = 0.5, we obtain a probability of 58% (for Re = 1.05) and 79% (for Re = 1.1) that the 25 cases will initiate a sustained transmission chain in Switzerland. 9/n
Nevertheless, we have to assume that the new variant might start to spread in Switzerland during the coming weeks, particularly because it might have increased transmissibility and therefore a higher Re. End. 11/n
You can follow @C_Althaus.
Tip: mention @twtextapp on a Twitter thread with the keyword “unroll” to get a link to it.

Latest Threads Unrolled:

By continuing to use the site, you are consenting to the use of cookies as explained in our Cookie Policy to improve your experience.