Today I thought a lot about very simple population models, and there are still a few things that bug me.
Consider a simple discrete exponential growth function:
$$ n(t+1) = n(t) + n(t) b - n(t) d = n (1+r) $$
And recursively expanding the formula for an time interval $T$ yields
$$ n(T) = n_0 (1+r)^{T} \tag{Eq.1}$$
where:
- $b$ and $d$ are rates (birth and death) expressed as a proportion of the population (thus $d$ ranges from $0$ to $1$)
- $r=b-d$ is the overall growth rate including both births and death
(These things might not be relevant but I'm not sure so I'll be very explicit with the notation)
This equation comes intuitively considering that at each time the population is the same as the previous time frame plus some births minus some deaths.
Now let's consider the continuous time exponential growth.
First of all, I have not found a simple way to pass from the discrete formula to the continuous one (maybe it's not possible?).
Anyway to obtain it one could do the same type of reasoning as above, we can write
$$ \frac{dn}{dt} = bn - dn = rn$$
Because the variation in the population size should be proportional to the births and inversely proportional to the deaths.
Integrating this formula for $t$ between $t=0$ and $t$, and for $n$ between $n_0$ and $N(t)$ we obtain:
$$ N(t) = n_0 \cdot e^{rt} \tag{Eq.2}$$
Now I tried to plug some numbers in to simulate the results and I found out that they are in fact different... but shouldn't they be the representation of the same phenomenon?
Here is the plot for the 2 equations
Maybe the error is thinking that r must be the same? But if so what is the relationship between the discrete and continuous $r$?
In the continuous model the rate assumes instantaneous compounding. If $1+r$ is the rate (as a factor) for the discrete model and you use that rate for continuous compounding then the equation is, as you say $$ n = n_0e^{\rho t} . $$ For small values of $r$, $e^r \approx 1+r$ .
In this model, births and deaths occur at any time, not just at discrete intervals. Think about what happens when you open an account earning $6\%$ annual interest. If the bank adds the interest to your account every millisecond then at the end of a year your initial investment will be $e^{0.06} = 1.061836\ldots$ as large, not just $1.06$.
So there is no contradiction in the two curves. The continuous one should be higher, and it is.
If you want to match the curves you have to reverse engineer the rate $s$ for the continuous model. That means you want $$ e^s = 1+r $$ so $s = \ln(1+r)$.
That gives $s=0.058\ldots$ for $r = 0.06$.