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?

  • $\begingroup$ Hi, could you provide some details about what is your doubt? If we assume no interactions/weak coupling between subpopulations, the formula seems rather straightforward. $\endgroup$
    – Anatoly
    Jul 21, 2014 at 14:25
  • $\begingroup$ Maybe it is straightforward, but I don't understand why we should average the time to extinction. If the time to extinction for a population of size $N/n$ is $k \exp({\epsilon N/n})$, then why do we divide this time by $n$ when the $n$ subpopulations are evolving towards extinction independently? I also don't understand why the explanation is talking about "next extinction" since I'd have assumed that all subpopulation would be extinct in $k \exp({\epsilon N/n})$. Maybe, the formula assumes that the time to extinction for $n$ subpopulations is uniformly distributed in $k \exp({\epsilon N/n})$. $\endgroup$ Jul 21, 2014 at 15:43
  • $\begingroup$ Hi Robert, thank you for your comments. I have just posted an answer, hoping that it could be useful to you. I also agree with you that some points in the book calculations are not clearly explained and would deserve some clarification/verification. $\endgroup$
    – Anatoly
    Jul 22, 2014 at 10:48

1 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.

  • $\begingroup$ +1. Thank you. Very nice answer. A couple of questions: when you describe the "time to first extinction", is there a reason why to choose the distribution $T/n, T/(n−1), T/(n−2) ..., T$ instead of something else? I guess this is a choice, since it is possible that the time to extinction follows another distributions. Furthermore, what is calculated as "total time to extinction" seems to me as a cumulative time because the successive extinctions are considering the time of the previous extinctions. Is this correct? $\endgroup$ Jul 25, 2014 at 15:05
  • $\begingroup$ Yes, that distribution was assumed arbitrarily. This is one of the questionable issues that the Authors might have done better to elucidate in their book. Regarding the "total time to extinction", if we assume that all subpopulations begin to evolve towards extinction at the same time, it necessarily represents a cumulative time (whose meaning and utility are questionable as well, in my opinion!). $\endgroup$
    – Anatoly
    Jul 25, 2014 at 18:43
  • $\begingroup$ Great. Thank you for your explanation. I will accept this answer as correct, although I want to wait a bit in case anyone wants to add something to the conversation. $\endgroup$ Jul 25, 2014 at 20:08

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