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Consider the following population model: We start with a population of a single cell at time $t=0$. Each cell divides into $k$ new cells at random times $T$ distributed according to a probability density function $f(T)$. We now wait for a long time and pick a random cell in the current population. Assuming we are in the $t\to \infty$ regime, what will be the average remaining time before division?

For the case $k=1$, this is a renewal process and I believe that the average remaining time will be $\frac{\mathbb{E}\left[T^2\right]}{2\mathbb{E}\left[T\right]}$. However I am unsure how to extend it for $k>1$. Ideally, I would like to express it in terms of a large deviation principle but my textbooks (e.g. Karlin and Taylor, Athreya, Harris) do not cover this case (I think!). An exact reference or just a brief derivation would be enough.

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This is a Bellman-Harris process. See Chapter VI of The Theory of Branching Processes by Harris, and in particular Section 24. It analyzes ages, but I imagine you can get remaining lifetimes from that.

The rough idea is as follows. In the renewal case, the age distribution has density proportional to $\mathbf{P}[T > s]$. Here, it's instead proportional to $e^{-\alpha s} \mathbf{P}[T > s]$ for some parameter $\alpha > 0$ having to do with both $k$ and the distribution of $T$. Specifically, it satisfies $$ \mathbf{E}[e^{-\alpha T}] = \frac{1}{k}. $$

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I don't agree. I think the exponential growth biases it towards younger individuals, that is, lots of newborns. My back of the envelope calculation is the following. Suppose the distribution of birth times is 1 with prob 1/2 and 2 otherwise, so everything happens at times 1,2 etc. Then I think that over time the population will be 2/3 either newborns or those who have just given birth, and 1/3 those who are a year away from birth. This is because calling $a_n$ the proportion of the former, and $b_n$ the latter, $a_{n+1} = b_n + \frac 12 a_n$. If proportions are going to converge, so $b_n \approx \alpha n$ then $\alpha = \frac 13$. Then you can can calculate that a randomly chosen individual will have expected time $\frac 13 + \frac 23 \times \frac 32 = \frac 43$ whereas $\frac {ET}{2ET^2} = \frac 3 {10}$.

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    $\begingroup$ What do you not agree with? when $k=1$ the population is constant: it is only a single individual at all times. Each cell divides into $k$ offsprings only once. The case $k=1$ is simply described by a renewal process and the expected remaining time really is $ET/(2ET^2)$. For $k>1$ there is indeed an exponential growth and the result from renewal theory does no apply. In a nutshell, I posted here exactly because the formula I give does not hold... $\endgroup$
    – papad
    Commented Mar 25 at 4:21

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