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Consider a number of $x_0$ reproducing individuals and ignore death and limiting environmental factors. I've heard that the growth of such a population (of bacteria, insects, humans, etc...) can be modeled by the differential equation

$$\dot{x} = \lambda x , \quad \quad \quad x(0)=x_0\in \mathbb{R}^{+}.$$

This seems to be taken for absolutely granted in nearly every book and every lecture I've seen. But I still don't understand what assumptions I have to make to the reproduction rate of a single individual within the population to RIGOROUSLY derive this equation.

In a book ("Evolutionary Dynamics" by Martin Nowak) I read that the basic assumption behind this differential equation is a stochastic process, namely that the time until one individual $A$ produces another individual $A'$ is exponentially distributed around an average time $t_0>0$. This would mean, that the probability of $A$ producing the offspring $A'$ within the time interval $[0,t]$ is given by $1-\exp(-t/t_0)$.

Then why is the differential equation fully deterministic? Is $x(t)$ the probabilistic expectation value of individuals after time $t$? Is there somebody who can strictly derive the differential equation by just using the abolve stochastic process? Or is there another way to theoretically justify this equation? And why does it also work for human populations, where there is always TWO individuals producing offspring? Thanks!

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    $\begingroup$ Yes, $x$ is the expected value of the birth process. It also can work for the "paired" case, you just reduce the value of $\lambda$ and require sufficiently large initial conditions. But if some kind of interaction process is required then this no longer works. (For instance this occurs in second order chemical kinetics.) $\endgroup$
    – Ian
    Nov 12 '15 at 0:19
  • $\begingroup$ The equation is saying: the rate of change of the population is proportional to how many individuals are already in the population. This is to be expected. For example, bacteria divide given enough food, (this is a simple model, so I would think this is assumed), so the more bacteria there are, then the more divisions! But we know that the process is not continuous (there can't be half a bacteria!), but this is just a simple model to give one an approximation. $\endgroup$
    – Almentoe
    Nov 12 '15 at 0:25
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The stochastic process model that you seem to have in mind can be rigorously formulated as follows. When you have a population of size $N$, each member of the population is independently waiting to make another member of the population. If you assume that the process is Markov, then these waiting times must be exponentially distributed. (Note that this Markov hypothesis is an oversimplification, for various reasons.) We assume they all have the same mean $\frac{1}{\lambda}$. (This is also an oversimplification.) Then the time to wait for the next birth is the minimum of $N$ independent exponential variables with mean $\frac{1}{\lambda}$. This is itself exponential with mean $\frac{1}{N \lambda}$.

Now if the population is currently $N$, the average time to the next birth is $\frac{1}{N \lambda}$, which means that the average growth rate of the population is currently $\lambda N$, as in the exponential growth ODE.

Note that the model goes through the same with slightly different hypotheses. For instance, maybe only some fraction $0<p<1$ of the population is actively trying to reproduce. In this case the model goes through exactly the same except that $\lambda$ is replaced by $\lambda'=p \lambda$.

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  • $\begingroup$ thank you! why does the exponential distribution follow from the assumption that the process is Markov? I understand that the average current growth rate of the population is $\lambda N$. But its still not clear for me how I can use this to rigorously get to the differential equation. $\endgroup$
    – Joker123
    Nov 12 '15 at 23:15
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    $\begingroup$ @Joker123 The Markov property tells you that if you've already been waiting for time $t$, the chance that you will wait for an additional time $s$ is the same as if you hadn't waited for time $t$ already. Thus $P(T>s+t|T>t)=P(T>s)$ for the waiting times. The only variables with this property are exponentially distributed. There is a neat proof of this using a simple functional equation and a monotonicity/squeeze theorem argument. $\endgroup$
    – Ian
    Nov 12 '15 at 23:21
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    $\begingroup$ @Joker123 As for the DE, you can use an argument based on a finite time interval to proceed carefully. Basically, the probability of at least one birth in a time interval of length $\Delta t$ is $\int_0^{\Delta t} \frac{1}{N \lambda} e^{-s/(N \lambda)} ds$. You can estimate this integral for small $\Delta t$ using the FTC; you get just $\frac{\Delta t}{N \lambda}$. You can then show that the probability of more than one birth in such an interval is $O(\Delta t^2)$. From here you can readily compute the mean growth rate. $\endgroup$
    – Ian
    Nov 12 '15 at 23:26
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To make it simpler, think of it first in case of discrete time. You have a population of $x_k$, and on the next step each from these population produce on average $r$ kids, so if you don't allow deaths $$ x_{k+1} = x_k + rx_k = (1+r)x_k = (1+r)^kx_0 $$ which is obviously a growth function, exponential w.r.t time variable $k$. If you say that a production rate is $\mu\cdot \Delta t$ and a death rate of $\nu\cdot\Delta t$, saying that a $\mu$ fraction of population is added in a negligibly small time period $\Delta t$, and a $\nu$ fraction dies, you get $$ x_{t + \Delta t} = x_t + (\mu - \nu))x_t\Delta t $$ so $$ \dot x_t \approx \frac{x_{t + \Delta t} - x_t}{\Delta t} = (\mu - \nu)x_t. $$

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  • $\begingroup$ thanks for the great answer. but why did you assume that the production rate is of the form $\mu \delta t$? And what does "negligibly small" mean in this context? $\endgroup$
    – Joker123
    Nov 12 '15 at 23:04
  • $\begingroup$ @Joker123 It's linear approximation; $f(x+h)=f(x)+hf'(x)+o(h)$. $\endgroup$
    – Ian
    Nov 13 '15 at 0:13

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