Time to extinction

Question

Suppose that a population has a rate of extinction equal to $k \exp({-\epsilon N})$ where $\epsilon$ and $k$ are constant and $N$ is the population size. This also means that the time to extinction is given by $k \exp({\epsilon N})$. By the way, when I say "time to extinction", it usually means the time needed for a population to reach a disease-free state, so the extinction part actually refers to an infectious agent such as measles.

Now, if we divide this population in $n$ subpopulations, I suppose the time to extinction should be $k \exp({\epsilon N/n})$. However, what I don't understand is that the total time to extinction (in all subpopulations) is calculated as follows:

$$\text{total time to extinction} = \frac{k}{n \exp({-\epsilon N/n})} + \frac{k}{(n-1) \exp({-\epsilon N/n})} + ... + \frac{k}{\exp({-\epsilon N/n})} < k(1+log(n))\exp({\epsilon N/n})$$

And the explication (quoted from this book) is this:

The above formula comes from calculating the average time to the first extinction when $n$ subpopulations are infected, followed by the average time to the next extinction given that now only $n − 1$ subpopulations are infected, proceeding iteratively until all populations are disease free.

But I don't understand why we are calculating average times for the next extinction. As I understand it, the subpopulations are independent and I think, concurrently undergoing changes until extinction.

Does anyone know why the equation above is correct?

Answer 1

The time to extinction for a population of total size $N$ divided in $n$ subpopulations is $T=k \exp({\epsilon N/n})$. If we assume no interactions between subpopulations, the within-subpopulation time to extinction is distributed around a mean value equal to $T$.

Since we expect the extinction of $n$ subpopulations to be completed over the time $T$ (a key concept in this calculations is that the subpulations evolve towards extinction independently, each with a different extinction time), we then can estimate the "time to first extinction" by assuming that, on average, for a group of $n$ subpopulations, this will occur after a time equal to $T/n$. Under this assumption, the average time to second, third, and successive extinctions can be estimated as $T/(n-1)$, $T/(n-2)$, and so on. We then get that the total time to extinction is given by $$ \sum\limits_{j=1}^n T/j=k (\gamma + log(n))\exp({\epsilon N/n})$$

A confounding element is that in the book this formula is reported as a disequality after substituting $\gamma$ with $1$ (I believe that this has been probably done to "simplify the message" avoiding the use of $\gamma$, and to highlight the concept that the overall result is larger than $T=k \exp({\epsilon N/n})$). The Authors also highlight that, for increasing $n$, the overall result rapidly decreases with respect to the extinction time expected for a non-divided, randomly-mixed population of size $N$ (given by $k \exp({\epsilon N})$).

It is also to be pointed out that the Authors do not provide any analysis of the validity of the above mentioned assumption.

I hope that this answer could be useful to you. I also found two interesting papers on this topic here and here. I hope that these links can be useful to you as well. Interestingly, the same confounding simplification ("unexplained" substitution of $\gamma$ with $1$) is present in the first of these two papers.