Suppose the simplest population model where we track the size $y$ of a population:

$$\frac{dy}{dt} = ry$$

for a positive constant $r$ and some $y$ such that $y(0) > 0$.

For this population model there is a corresponding "replicator dynamic" model defined by

$$x_0 := \frac{1}{1+y}$$

$$x_1 := \frac{y}{1+y}$$

These are supposed to model "frequencies" of different variants in a population,


except its not clear to me what the different variants here. There is only a single species here so of course it is 100% of its own population or 100% of all populations being modeled here, so I feel that I must be misunderstanding what the "frequencies" of the replicator dynamic are. Can someone clarify what $x_0$ and $x_1$ correspond to here? Thank you!


1 Answer 1


The transformation carried here is a particular type of the more general one. As the linked text notes:

The continuous replicator equation on n types is equivalent to the Generalized Lotka–Volterra equation in n − 1 dimensions.

In the case described in the OP the number of types is $n=2$, although "Lotka-Volterra" is a bit too generous a term for the simple model in case.

So indeed an extra species is being added here with a zero growth rate. The advantage is that one can now talk about the relative fitness and otherwise use the rich theory developed for the fixed-size populations, since $x_0+x_1=1$ whereas $y$ may grow without limit. E.g., a lot of such intuition comes from the familiarity with Wright-Fisher and Moran models.

  • $\begingroup$ Thank you! Why is that needed for the replicator dynamic? I'm interested in using the replicator dynamic to prove ESS (obviously not in this model, instead suppose there are two populations $y_1$ and $y_1$ with growth rates $r_1$ and $r_2$ respectively, suppose $r_1 > r_2 > 0$ and I want to prove $y_1$ is an ESS), does the dummy population help for that somehow? $\endgroup$ Jul 26, 2021 at 15:00
  • 1
    $\begingroup$ @John_Krampf one wants the total population to normalize to $1$. If one has two populations, $y_1$ and $y_2$, one could construct $x_1=y_1/(y_1+y_2)$ and $x_2=y_2/(y_1+y_2)$, but this would behave bad when both populations vanish. So $1$ is a kind of pseudocount here. But I must warn, that I am not a game theorist. $\endgroup$
    – Roger V.
    Jul 26, 2021 at 15:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .