4
$\begingroup$

I've been studing population growth models, but there's something I can not find that's fustrating. That's a formula for the variance in population growth. I know that other models can be aplied, but I want to start with the simplest case.

Let's assume a population is growing exponentially as defined by: $N(t) = N(0)W^{t}$

The parameters $W$(absolute fitness in my definition) tell us that the time for division (if the species reproduces binarally) is given by an exponential distribution with $\lambda = \dfrac{\ln W}{\ln 2}$

So I see a model for $N(t)$ and a random variable associated with this model. What is the pdf, expected value and variance of $N(t)$? The expected value I expect it to be given by the first equation (or the value to be the same) but what about the variance?

$\endgroup$
  • $\begingroup$ The exponential growth without noise is the deterministic approximation. Now knowing that this means that the time for division is given by an exponential distribution, what is the variance in the population size? $\endgroup$ – Diogo Santos Aug 14 '16 at 10:53
  • $\begingroup$ My bad, I didn't read the question right. Removing my embarrassing comment. $\endgroup$ – Memming Aug 14 '16 at 12:05
  • $\begingroup$ Basically, you have a Poisson process where the rate doubles whenever there's a (birth) event. The mean grows exponentially, and the variance should be overdispersed since it's a mixture of Poisson... $\endgroup$ – Memming Aug 14 '16 at 12:32
  • $\begingroup$ Not exactly... Because whenever there's a birth the time for the birth from the individuals that did not divide is the same as before minus the time of birth... But there is also the two new individuals... $\endgroup$ – Diogo Santos Aug 14 '16 at 13:05
  • $\begingroup$ By the way, as my final interest is the frequency of this in a larger population, I think diffusion theory, and Kolgomorov forward and backward equations might help... I just don't know enough to use it $\endgroup$ – Diogo Santos Aug 14 '16 at 13:09
4
$\begingroup$

Check out Kendall (1949), section 2. He shows that the pdf is a geometric distribution. In your notation, it's $P_n(t) = N(0)W^{-t}\left(1-W^{-t}\right)^{n-1}$, which implies that the mean is indeed $E[N(t)] = N(0)W^t$ and the variance is $\text{Var}[N(t)] = N(0) W^t(W^t-1)$. (Be careful -- his definition of $\lambda$ differs from yours by a factor of $\ln 2$.)

$\endgroup$
  • $\begingroup$ so, my answer isn't incorrect (I'll go ahead and leave it up), but this is the full solution. $\endgroup$ – Ben Bolker Aug 15 '16 at 19:39
  • $\begingroup$ yes... I accepted this one but yours already point to the right direction $\endgroup$ – Diogo Santos Aug 15 '16 at 20:01
4
$\begingroup$

If $W$ is a parameter, I'd expect (in the simplest case) that $W$ itself is a constant (i.e. it doesn't have a distribution). (By the way, I would normally use $T_{1/2}$ to denote the doubling time, $\ln 2/\ln W$, and would use $\lambda = \ln W$ for the exponential rate, i.e. $W^t= \exp(\ln W)^t = \exp(t \ln W) = \exp(\lambda t)$. I don't normally see people express rates scaled by $\ln 2$.)

In the case of a continuous-time birth-death process (the technical jargon for your model), I believe the analytical answer will be that the standard deviation of the population size at time $t$ will grow at the same exponential rate as the mean population size, and that the variance will grow twice as fast (because it has units of $\textrm{population size}^2$), i.e. $\propto \exp(2 \lambda t)$)

I don't have a derivation/reference handy, will dig around for one. This is fairly basic stochastic-process math, but stochastic-process math also gets pretty hairy-looking (to a biologist) quickly. Bartlett 1960 (Stochastic population models in ecology and epidemiology) is one place to look for basic results, although I'm sure there are many others. This recent PhD thesis gives a thorough introduction to the math of birth-death processes ... although, not to my disappointment, a canned derivation of the variance result I'm too lazy to derive for you.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.